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BMET Geometry


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Cifarelli Fiori Gloag Greenberg Jordan Sconyers




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Printed: September 1, 2011




Authors
Victor Cifarelli, Nick Fiori, Andrew Gloag, Dan Greenberg, Lori Jordan, Jim Sconyers, Bill Zahner


Editor
Annamaria Farbizio


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Contents


1 Basics of Geometry 1
1.1 Points, Lines, and Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Segments and Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Angles and Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.4 Midpoints and Bisectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1.5 Angle Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
1.6 Classifying Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
1.7 Chapter 1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
1.8 Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57


2 Coordinate Geometry 61
2.1 Vocabulary Self-Rating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
2.2 Distance and Midpoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
2.3 Parallel and Perpendicular . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
2.4 Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
2.5 Translating and Reflecting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
2.6 Rotating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92


3 Triangles and Congruence 99
3.1 Triangle Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
3.2 Congruent Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
3.3 Triangle Congruence using SSS and SAS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
3.4 Triangle Congruence using ASA, AAS, and HL . . . . . . . . . . . . . . . . . . . . . . . . . 133
3.5 Isosceles and Equilateral Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
3.6 Chapter 4 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
3.7 Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159


4 Right Triangle Trigonometry 162
4.1 The Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162


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4.2 Converse of the Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
4.3 Using Similar Right Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
4.4 Special Right Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
4.5 Tangent, Sine and Cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
4.6 Inverse Trigonometric Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
4.7 Chapter 8 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
4.8 Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214


5 Circles 217
5.1 Parts of Circles & Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
5.2 Properties of Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
5.3 Properties of Chords . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
5.4 Inscribed Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248
5.5 Angles of Chords, Secants, and Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259
5.6 Segments of Chords, Secants, and Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
5.7 Extension: Writing and Graphing the Equations of Circles . . . . . . . . . . . . . . . . . . . 282
5.8 Chapter 9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
5.9 Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288


6 Perimeter and Area 293
6.1 Triangles and Parallelograms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293
6.2 Trapezoids, Rhombi, and Kites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302
6.3 Areas of Similar Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
6.4 Circumference and Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
6.5 Areas of Circles and Sectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
6.6 Chapter 10 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331
6.7 Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334


7 Surface Area and Volume 337
7.1 Exploring Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
7.2 Surface Area of Prisms and Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
7.3 Surface Area of Pyramids and Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355
7.4 Volume of Prisms and Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
7.5 Volume of Pyramids and Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
7.6 Surface Area and Volume of Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
7.7 Extension: Exploring Similar Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386
7.8 Chapter 11 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392
7.9 Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394


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8 Analyzing Conic Sections 398
8.1 Introduction to Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
8.2 Circles and Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406
8.3 Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425
8.4 Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441
8.5 General Algebraic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454


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Chapter 1


Basics of Geometry


In this chapter, students will learn about the building blocks of geometry. We will start with what the basic
terms: point, line and plane. From here, students will learn about segments, midpoints, angles, bisectors,
angle relationships, and how to classify polygons.


1.1 Points, Lines, and Planes
Learning Objectives


• Understand the terms point, line, and plane.
• Draw and label terms in a diagram.


Review Queue
1. List and draw pictures of five geometric figures you are familiar with.
2. What shape is a yield sign?


3. Solve the algebraic equations.


(a) 4x − 7 = 29
(b) −3x + 5 = 17


Know What? Geometry is everywhere. Remember these wooden blocks that you played with as a kid?
If you played with these blocks, then you have been “studying” geometry since you were a child.
How many sides does the octagon have? What is something in-real life that is an octagon?
Geometry: The study of shapes and their spatial properties.


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Building Blocks
Point: An exact location in space.
A point describes a location, but has no size. Examples:


Table 1.1:


Label It Say It
A point A


Line: Infinitely many points that extend forever in both directions.
A line has direction and location is always straight.


Table 1.2:


Label It Say It
line g line g
←→PQ line PQ
←→QP line QP


Plane: Infinitely many intersecting lines that extend forever in all directions.


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Think of a plane as a huge sheet of paper that goes on forever.


Table 1.3:


Label It Say It
Plane M Plane M
Plane ABC Plane ABC


Example 1: What best describes San Diego, California on a globe?
A. point
B. line
C. plane
Solution: A city is usually labeled with a dot, or point, on a globe.
Example 2: What best describes the surface of a movie screen?
A. point
B. line
C. plane
Solution: The surface of a movie screen is most like a plane.
Beyond the Basics Now we can use point, line, and plane to define new terms.
Space: The set of all points expanding in three dimensions.
Think back to the plane. It goes up and down, and side to side. If we add a third direction, we have space,
something three-dimensional.
Collinear: Points that lie on the same line.


P,Q,R, S , and T are collinear because they are all on line w. If a point U was above or below line w, it
would be non-collinear.
Coplanar: Points and/or lines within the same plane.


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Lines h and i and points A, B,C,D,G, and K are coplanar in Plane J .
Line ←→KF and point E are non-coplanar with Plane J .
Example 3: Use the picture above to answer these questions.
a) List another way to label Plane J .
b) List another way to label line h.
c) Are K and F collinear?
d) Are E, B and F coplanar?
Solution:
a) Plane BDG. Any combination of three coplanar points that are not collinear would be correct.
b) ←→AB. Any combination of two of the letters A,C or B would also work.
c) Yes
d) Yes
Endpoint: A point at the end of a line.
Line Segment: A line with two endpoints. Or, a line that stops at both ends.
Line segments are labeled by their endpoints. Order does not matter.


Table 1.4:


Label It Say It
AB Segment AB
BA Segment BA


Ray: A line with one endpoint and extends forever in the other direction.
A ray is labeled by its endpoint and one other point on the line. For rays, order matters. When labeling,
put endpoint under the side WITHOUT an arrow.


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Table 1.5:


Label It Say It
−−→
CD Ray CD
←−−
DC Ray CD


Intersection: A point or line where lines, planes, segments or rays cross.


Example 4: What best describes a straight road connecting two cities?
A. ray
B. line
C. segment
D. plane
Solution: The straight road connects two cities, which are like endpoints. The best term is segment, or
C.
Example 5: Answer the following questions about the picture to the right.


a) Is line l coplanar with Plane V or W?
b) Are R and Q collinear?
c) What point is non-coplanar with either plane?
d) List three coplanar points in Plane W.
Solution:
a) No.
b) Yes.
c) S
d) Any combination of P,O,T and Q would work.
Further Beyond This section introduces a few basic postulates.
Postulates: Basic rules of geometry. We can assume that all postulates are true.
Theorem: A statement that is proven true using postulates, definitions, and previously proven theorems.


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Postulate 1-1: There is exactly one (straight) line through any two points.


Investigation 1-1: Line Investigation


1. Draw two points anywhere on a piece of paper.
2. Use a ruler to connect these two points.
3. How many lines can you draw to go through these two points?


Postulate 1-2: One plane contains any three non-collinear points.


Postulate 1-3: A line with points in a plane is also in that plane.


Postulate 1-4: The intersection of two lines will be one point.


Lines l and m intersect at point A.
Postulate 1-5: The intersection of two planes is a line.


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When making geometric drawings, you need to be clear and label all points and lines.
Example 6a: Draw and label the intersection of line ←→AB and ray −−→CD at point C.
Solution: It does not matter where you put A or B on the line, nor the direction that −−→CD points.


Example 6b: Redraw Example 6a, so that it looks different but is still true.
Solution:


Example 7: Describe the picture below using the geometric terms you have learned.
Solution: ←→AB and D are coplanar in Plane P, while←→BC and←→AC intersect at point C which is non-coplanar.


Know What? Revisited The octagon has 8 sides. In Latin, “octo” or “octa” means 8, so octagon,
literally means “8-sided figure.” An octagon in real-life would be a stop sign.


Review Questions
• Questions 1-5 are similar to Examples 6a and 6b.
• Questions 6-8 are similar to Examples 3 and 5.
• Questions 9-12 are similar to Examples 1, 2, and 4.
• Questions 13-16 are similar to Example 7.
• Questions 17-25 use the definitions and postulates learned in this lesson.


For questions 1-5, draw and label an image to fit the descriptions.


1. −−→CD intersecting AB and Plane P containing AB but not −−→CD.
2. Three collinear points A, B, and C and B is also collinear with points D and E.
3. −→XY ,−→XZ, and −−→XW such that −→XY and −→XZ are coplanar, but −−→XW is not.
4. Two intersecting planes, P and Q, with GH where G is in plane P and H is in plane Q.
5. Four non-collinear points, I, J,K, and L, with line segments connecting all points to each other.
6. Name this line in five ways.


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7. Name the geometric figure in three different ways.


8. Name the geometric figure below in two different ways.


9. What is the best possible geometric model for a soccer field? Explain your answer.
10. List two examples of where you see rays in real life.
11. What type of geometric object is the intersection of a line and a plane? Draw your answer.
12. What is the difference between a postulate and a theorem?


For 13-16, use geometric notation to explain each picture in as much detail as possible.


13.


14.


15.


16.


For 17-25, determine if the following statements are true or false.


17. Any two points are collinear.


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18. Any three points determine a plane.
19. A line is to two rays with a common endpoint.
20. A line segment is infinitely many points between two endpoints.
21. A point takes up space.
22. A line is one-dimensional.
23. Any four points are coplanar.
24. −→AB could be read “ray AB” or “ray “BA.”
25. ←→AB could be read “line AB” or “line BA.”


Review Queue Answers
1. Examples could be triangles, squares, rectangles, lines, circles, points, pentagons, stop signs (oc-


tagons), boxes (prisms), or dice (cubes).
2. A yield sign is a triangle with equal sides.
3. (a) 4x − 7 = 29


4x = 36
x = 9


(b) −3x + 5 = 17
− 3x = 12


x = −4


1.2 Segments and Distance
Learning Objectives


• Use the ruler postulate.
• Use the segment addition postulate.
• Plot line segments on the x − y plane.


Review Queue
1. Draw a line segment with endpoints C and D.
2. How would you label the following figure? List 2 different ways.


3. Draw three collinear points and a fourth that is coplanar.
4. Plot the following points on the x − y plane.


(a) (3, -3)
(b) (-4, 2)
(c) (0, -7)
(d) (6, 0)


Know What? The average adult human body can be measured in “heads.” For example, the average
human is 7-8 heads tall. When doing this, each person uses their own head to measure their own body.
Other measurements are in the picture to the right.
See if you can find the following measurements:


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• The length from the wrist to the elbow
• The length from the top of the neck to the hip
• The width of each shoulder


Measuring Distances
Distance: The length between two points.
Measure: To determine how far apart two geometric objects are.
The most common way to measure distance is with a ruler. In this text we will use inches and centimeters.
Example 1: Determine how long the line segment is, in inches. Round to the nearest quarter-inch.


Solution: To measure this line segment, it is very important to line up the “0” with the one of the
endpoints. DO NOT USE THE EDGE OF THE RULER.


From this ruler, it looks like the segment is 4.75 inches (in) long.
Inch-rulers are usually divided up by eight-inch (or 0.125 in) segments. Centimeter rulers are divided up
by tenth-centimeter (or 0.1 cm) segments.


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The two rulers above are NOT DRAWN TO SCALE, which means that the measured length is not
the distance apart that it is labeled.
Example 2: Determine the measurement between the two points to the nearest tenth of a centimeter.


Solution: Even though there is no line segment between the two points, we can still measure the distance
using a ruler.


It looks like the two points are 6 centimeters (cm) apart.
NOTE: We label a line segment, AB and the distance between A and B is shown below. m means measure.
The two can be used interchangeably.


Table 1.6:


Label It Say It
AB The distance between A and B
mAB The measure of AB


Ruler Postulate
Ruler Postulate: The distance between two points is the absolute value of the difference between the
numbers shown on the ruler.
The ruler postulate implies that you do not need to start measuring at “0”, as long as you subtract the
first number from the second. “Absolute value” is used because distance is always positive.
Example 3: What is the distance marked on the ruler below? The ruler is in centimeters.


Solution: Subtract one endpoint from the other. The line segment spans from 3 cm to 8 cm. |8−3| = |5| = 5
The line segment is 5 cm long. Notice that you also could have done |3 − 8| = | − 5| = 5.
Example 4: Draw CD, such that CD = 3.825 in.


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Solution: To draw a line segment, start at “0” and draw a segment to 3.825 in.


Put points at each end and label.


Segment Addition Postulate
First, in the picture below, B is between A and C. As long as B is anywhere on the segment, it can be
considered to be between the endpoints.


Segment Addition Postulate: If A, B, and C are collinear and B is between A and C, then AB+BC = AC.
For example, if AB = 5 cm and BC = 12 cm, then AC must equal 5 + 12 or 17 cm (in the picture above).
Example 5: Make a sketch of OP, where Q is between O and P.
Solution: Draw OP first, then place Q on the segment.


Example 6: In the picture from Example 5, if OP = 17 and QP = 6, what is OQ?
Solution: Use the Segment Additional Postulate.


OQ + QP = OP
OQ + 6 = 17


OQ = 17 − 6
OQ = 11


Example 7: Make a sketch of: S is between T and V. R is between S and T . TR = 6,RV = 23, and
TR = SV.
Solution: Interpret the first sentence first: S is between T and V.


Then add in what we know about R: It is between S and T . Put markings for TR = SV.


Example 8: Find SV, TS ,RS and TV from Example 7.


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Solution:
For SV: It is equal to TR, so SV = 6 cm.


For RS : RV = RS + SV For TS : TS = TR + RS For TV : TV = TR + RS + SV
23 = RS + 6 TS = 6 + 17 TV = 6 + 17 + 6
RS = 17 cm TS = 23 cm TV = 29 cm


Example 9: Algebra Connection For HK, suppose that J is between H and K. If HJ = 2x + 4, JK =
3x + 3, and KH = 22, find x.
Solution: Use the Segment Addition Postulate.


HJ + JK = KH
(2x + 4) + (3x + 3) = 22


5x + 7 = 22
5x = 15
x = 3


Distances on a Grid
You can now find the distances between points in the x − y plane if the lines are horizontal or vertical.
If the line is vertical, find the change in the y−coordinates.
If the line is horizontal, find the change in the x−coordinates.
Example 10: What is the distance between the two points shown below?


Solution: Because this line is vertical, look at the change in the y−coordinates.


|9 − 3| = |6| = 6


The distance between the two points is 6 units.
Example 11: What is the distance between the two points shown below?


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Solution: Because this line is horizontal, look at the change in the x−coordinates.


|(−4) − 3| = | − 7| = 7


The distance between the two points is 7 units.
Know What? Revisited The length from the wrist to the elbow is one head, the length from the top of
the neck to the hip is two heads, and the width of each shoulder one head width.


Review Questions
• Questions 1-8 are similar to Examples 1 and 2.
• Questions 9-12 are similar to Example 3.
• Questions 13-17 are similar to Examples 5 and 6.
• Questions 18 and 19 are similar to Example 7 and 8.
• Questions 20 and 21 are similar to Example 9.
• Questions 22-26 are similar to Examples 10 and 11.


For 1-4, find the length of each line segment in inches. Round to the nearest 18 of an inch.


1.


2.
3.


4.


For 5-8, find the distance between each pair of points in centimeters. Round to the nearest tenth.


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5.


6.


7.


8.


For 9-12, use the ruler in each picture to determine the length of the line segment.


9.


10.


11.


12.


13. Make a sketch of BT , with A between B and T .
14. If O is in the middle of LT , where exactly is it located? If LT = 16 cm, what is LO and OT?
15. For three collinear points, A between T and Q.


(a) Draw a sketch.
(b) Write the Segment Addition Postulate.
(c) If AT = 10 in and AQ = 5 in, what is TQ?


16. For three collinear points, M between H and A.
(a) Draw a sketch.


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(b) Write the Segment Addition Postulate.
(c) If HM = 18 cm and HA = 29 cm, what is AM?


17. For three collinear points, I between M and T .
(a) Draw a sketch.
(b) Write the Segment Addition Postulate.
(c) If IT = 6 cm and MT = 25 cm, what is AM?


18. Make a sketch that matches the description: B is between A and D. C is between B and D. AB =
7 cm, AC = 15 cm, and AD = 32 cm. Find BC, BD, and CD.


19. Make a sketch that matches the description: E is between F and G. H is between F and E. FH =
4 in, EG = 9 in, and FH = HE. Find FE,HG, and FG.


For 20 and 21, Suppose J is between H and K. Use the Segment Addition Postulate to solve for x. Then
find the length of each segment.


20. HJ = 4x + 9, JK = 3x + 3, KH = 33
21. HJ = 5x − 3, JK = 8x − 9, KH = 131


For 23-26, determine the vertical or horizontal distance between the two points.


23.


24.


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25.


26.


Review Queue Answers


1.


2. line l, MN
3.


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4.


1.3 Angles and Measurement
Learning Objectives


• Classify angles.
• Apply the Protractor Postulate and the Angle Addition Postulate.


Review Queue
1. Label the following geometric figure. What is it called?


2. Find a, XY and YZ.


3. B is between A and C on AC. If AB = 4 and BC = 9, what is AC?


Know What? Back to the building blocks. Every block has its own dimensions, angles and measurements.
Using a protractor, find the measure of the three outlined angles in the “castle” to the right.


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Two Rays = One Angle
In #1 above, the figure was a ray. It is labeled −→AB, with the arrow over the point that is NOT the endpoint.
When two rays have the same endpoint, an angle is created.
Angle: When two rays have the same endpoint.
Vertex: The common endpoint of the two rays that form an angle.
Sides: The two rays that form an angle.


Table 1.7:


Label It Say It
∠ABC Angle ABC
∠CBA Angle CBA


The vertex is B and the sides are −→BA and −→BC. Always use three letters to name an angle, ∠
SIDE-VERTEX-SIDE.
Example 1: How many angles are in the picture below? Label each one.


Solution: There are three angles with vertex U. It might be easier to see them all if we separate them.


So, the three angles can be labeled, ∠XUY (or ∠YUX), ∠YUZ (or ∠ZUY), and ∠XUZ (or ∠ZUX).


Protractor Postulate
We measure a line segment’s length with a ruler. Angles are measured with something called a protractor.
A protractor is a measuring device that measures how “open” an angle is. Angles are measured in degrees,
and labeled with a ◦ symbol.


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There are two sets of measurements, one starting on the left and the other on the right side of the protractor.
Both go around from 0◦ to 180◦. When measuring angles, always line up one side with 0◦, and see where
the other side hits the protractor. The vertex lines up in the middle of the bottom line.


Example 2: Measure the three angles from Example 1, using a protractor.


Solution: Just like in Example 1, it might be easier to measure these three angles if we separate them.


With measurement, we put an m in front of the ∠ sign to indicate measure. So, m∠XUY = 84◦, m∠YUZ = 42◦
and m∠XUZ = 126◦.
Just like the Ruler Postulate for line segments, there is a Protractor Postulate for angles.


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Protractor Postulate: For every angle there is a number between 0◦ and 180◦ that is the measure of
the angle. The angle’s measure is the difference of the degrees where the sides of the angle intersect the
protractor. For now, angles are always positive.
In other words, you do not have to start measuring an angle at 0◦, as long as you subtract one measurement
from the other.
Example 3: What is the measure of the angle shown below?


Solution: This angle is lined up with 0◦, so where the second side intersects the protractor is the angle
measure, which is 50◦.
Example 4: What is the measure of the angle shown below?


Solution: This angle is not lined up with 0◦, so use subtraction to find its measure. It does not matter
which scale you use.
Inner scale: 140◦ − 25◦ = 125◦


Outer scale: 165◦ − 40◦ = 125◦


Example 5: Use a protractor to measure ∠RST below.


Solution: Lining up one side with 0◦ on the protractor, the other side hits 100◦.


Classifying Angles
Angles can be grouped into four different categories.
Straight Angle: An angle that measures exactly 180◦.


Right Angle: An angle that measures exactly 90◦.


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This half-square marks right, or 90◦, angles.
Acute Angles: Angles that measure between 0◦ and 90◦.


Obtuse Angles: Angles that measure between 90◦ and 180◦.


Perpendicular: When two lines intersect to form four right angles.


Even though all four angles are 90◦, only one needs to be marked with the half-square.
The symbol for perpendicular is ⊥.


Table 1.8:


Label It Say It
l ⊥ m Line l is perpendicular to line m.
←→
AC ⊥ ←→DE Line AC is perpendicular to line DE.


Example 6: Name the angle and determine what type of angle it is.


Solution: The vertex is U. So, the angle can be ∠TUV or ∠VUT . To determine what type of angle it is,
compare it to a right angle.
Because it opens wider than a right angle, and less than a straight angle it is obtuse.
Example 7: What type of angle is 84◦? What about 165◦?


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Solution: 84◦ is less than 90◦, so it is acute. 165◦ is greater than 90◦, but less than 180◦, so it is obtuse.


Drawing an Angle


Investigation 1-2: Drawing a 50◦ Angle with a Protractor
1. Start by drawing a horizontal line across the page, 2 in long.


2. Place an endpoint at the left side of your line.
3. Place the protractor on this point, such that the bottom line of the protractor is on the line and the
endpoint is at the center. Mark 50◦ on the appropriate scale.


4. Remove the protractor and connect the vertex and the 50◦ mark.


This process can be used to draw any angle between 0◦ and 180◦. See http://www.mathsisfun.com/
geometry/protractor-using.html for an animation of this investigation.
Example 8: Draw a 135◦ angle.
Solution: Following the steps from above, your angle should look like this:


Now that we know how to draw an angle, we can also copy that angle with a compass and a ruler. Anytime
we use a compass and ruler to draw geometric figures, it is called a construction.


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Compass: A tool used to draw circles and arcs.
Investigation 1-3: Copying an Angle with a Compass and Ruler
1. We are going to copy the 50◦ angle from Investigation 1-2. First, draw a straight line, 2 inches long,
and place an endpoint at one end.


2. With the point (non-pencil side) of the compass on the vertex, draw an arc that passes through both
sides of the angle. Repeat this arc with the line we drew in #1.


3. Move the point of the compass to the horizontal side of the angle we are copying. Place the point where
the arc intersects this side. Open (or close) the “mouth” of the compass so that you can draw an arc that
intersects the other side and the arc drawn in #2. Repeat this on the line we drew in #1.


4. Draw a line from the new vertex to the arc intersections.


To watch an animation of this construction, see http://www.mathsisfun.com/geometry/construct-anglesame.
html


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Marking Angles and Segments in a Diagram
With all these segments and angles, we need to have different ways to label equal angles and segments.
Angle Markings


Segment Markings


Example 9: Write all equal angle and segment statements.


Solution: AD ⊥ ←→FC


m∠ADB = m∠BDC = m∠FDE = 45◦


AD = DE
FD = DB = DC


m∠ADF = m∠ADC = 90◦


Angle Addition Postulate
Like the Segment Addition Postulate, there is an Angle Addition Postulate.
Angle Addition Postulate: If B is on the interior of ∠ADC, then


m∠ADC = m∠ADB + m∠BDC


.


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Example 10: What is m∠QRT in the diagram below?


Solution: Using the Angle Addition Postulate, m∠QRT = 15◦ + 30◦ = 45◦.
Example 11: What is m∠LMN if m∠LMO = 85◦ and m∠NMO = 53◦?


Solution: m∠LMO = m∠NMO + m∠LMN, so 85◦ = 53◦ + m∠LMN.


m∠LMN = 32◦.


Example 12: Algebra Connection If m∠ABD = 100◦, find x.


Solution: m∠ABD = m∠ABC + m∠CBD. Write an equation.


100◦ = (4x + 2)◦ + (3x − 7)◦


100◦ = 7x◦ − 5◦


105◦ = 7x◦


15◦ = x


Know What? Revisited Using a protractor, the measurement marked in the red triangle is 90◦, the
measurement in the blue triangle is 45◦ and the measurement in the orange square is 90◦.


Review Questions
• Questions 1-10 use the definitions, postulates and theorems from this section.
• Questions 11-16 are similar to Investigation 1-2 and Examples 7 and 8.
• Questions 17 and 18 are similar to Investigation 1-3.
• Questions 19-22 are similar to Examples 2-5.


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• Question 23 is similar to Example 9.
• Questions 24-28 are similar to Examples 10 and 11.
• Questions 29 and 30 are similar to Example 12.


For questions 1-10, determine if the statement is true or false.


1. Two angles always add up to be greater than 90◦.
2. 180◦ is an obtuse angle.
3. 180◦ is a straight angle.
4. Two perpendicular lines intersect to form four right angles.
5. A construction uses a protractor and a ruler.
6. For an angle ∠ABC,C is the vertex.
7. For an angle ∠ABC, AB and BC are the sides.
8. The m in front of m∠ABC means measure.
9. Angles are always measured in degrees.


10. The Angle Addition Postulate says that an angle is equal to the sum of the smaller angles around it.


For 11-16, draw the angle with the given degree, using a protractor and a ruler. Also, state what type of
angle it is.


11. 55◦
12. 92◦
13. 178◦
14. 5◦
15. 120◦
16. 73◦
17. Construction Copy the angle you made from #12, using a compass and a ruler.
18. Construction Copy the angle you made from #16, using a compass and a ruler.


For 19-22, use a protractor to determine the measure of each angle.


19.


20.


21.


22.


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23. Interpret the picture to the right. Write down all equal angles, segments and if any lines are perpen-
dicular.


In Exercises 24-29, use the following information: Q is in the interior of ∠ROS . S is in the interior of
∠QOP. P is in the interior of ∠SOT . S is in the interior of ∠ROT and m∠ROT = 160◦, m∠SOT = 100◦, and
m∠ROQ = m∠QOS = m∠POT .


24. Make a sketch.
25. Find m∠QOP
26. Find m∠QOT
27. Find m∠ROQ
28. Find m∠SOP


Algebra Connection Solve for x.


29. m∠ADC = 56◦


30. m∠ADC = 130◦


Review Queue Answers
1. −→AB, a ray
2. XY = 3, YZ = 38
a − 6 + 3a + 11 = 41


4a + 5 = 41


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4a = 36
a = 9


3. Use the Segment Addition Postulate, AC = 13.


1.4 Midpoints and Bisectors
Learning Objectives


• Identify the midpoint of line segments.
• Identify the bisector of a line segment.
• Understand and use the Angle Bisector Postulate.


Review Queue


1. m∠SOP = 38◦, find m∠POT and m∠ROT .


2. Find the slope between the two numbers.


(a) (-4, 1) and (-1, 7)
(b) (5, -6) and (-3, -4)


3. Find the average of these numbers: 23, 30, 18, 27, and 32.


Know What? The building to the right is the Transamerica Building in San Francisco. This building
was completed in 1972 and, at that time was one of the tallest buildings in the world. In order to make
this building as tall as it is and still abide by the building codes, the designer used this pyramid shape.
It is very important in designing buildings that the angles and parts of the building are equal. What
components of this building look equal? Analyze angles, windows, and the sides of the building.


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Congruence
You could argue that another word for equal is congruent. But, the two are a little different.
Congruent: When two geometric figures have the same shape and size.


Table 1.9:


Label It Say It
AB BA AB is congruent to BA


Table 1.10:


Equal Congruent
=
used with measurement used with figures
mAB = AB = 5cm AB BA
m∠ABC = 60◦ ∠ABC ∠CBA


If two segments or angles are congruent, then they are also equal.


Midpoints
Midpoint: A point on a line segment that divides it into two congruent segments.


Because AB = BC, B is the midpoint of AC.
Midpoint Postulate: Any line segment will have exactly one midpoint.


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This postulate is referring to the midpoint, not the lines that pass through the midpoint.


There are infinitely many lines that pass through the midpoint.
Example 1: Is M a midpoint of AB?


Solution: No, it is not MB = 16 and AM = 34 − 16 = 18. AM must equal MB in order for M to be the
midpoint of AB.


Midpoint Formula
When points are plotted in the coordinate plane, we can use a formula to find the midpoint between them.
Here are two points, (-5, 6) and (3, 4).


It follows that the midpoint should be halfway between the points on the line. Just by looking, it seems
like the midpoint is (-1, 4).
Midpoint Formula: For two points, (x1, y1) and x2, y2, the midpoint is


( x1+x2
2 ,


y1+y2
2


)


.
Let’s use the formula to make sure (-1, 4) is the midpoint between (-5, 6) and (3, 2).


(−5 + 3
2


,


6 + 2
2


)


=
(−2


2
,


8
2


)


= (−1, 4)


Always use this formula to determine the midpoint.
Example 2: Find the midpoint between (9, -2) and (-5, 14).
Solution: Plug the points into the formula.


(


9 + (−5)
2


,


−2 + 14
2


)


=
(4
2
,


12
2


)


= (2, 6)


Example 3: If M(3,−1) is the midpoint of AB and B(7,−6), find A.


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Solution: Plug in what you know into the midpoint formula.


(7 + xA
2


,


−6 + yA
2


)


= (3,−1)
7 + xA


2
= 3 and −6 + yA


2
= −1


7 + xA = 6 and − 6 + yA = −2
xA = −1 and yA = 4
So, A is (−1, 4).


Segment Bisectors
Segment Bisector: A bisector cuts a line segment into two congruent parts and passes through the
midpoint.
Example 4: Use a ruler to draw a bisector of the segment.


Solution: First, find the midpoint. Measure the line segment. It is 4 cm long. To find the midpoint,
divide 4 cm by 2 because we want 2 equal pieces. Measure 2 cm from one endpoint and draw the midpoint.


To finish, draw a line that passes through Z.


A specific type of segment bisector is called a perpendicular bisector.
Perpendicular Bisector: A line, ray or segment that passes through the midpoint of another segment
and intersects the segment at a right angle.


AB BC


AC ⊥ ←→DE


Perpendicular Bisector Postulate: For every line segment, there is one perpendicular bisector.
Example 5: Which line is the perpendicular bisector of MN?


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Solution: The perpendicular bisector must bisect MN and be perpendicular to it. Only ←→OQ fits this
description. ←→SR is a bisector, but is not perpendicular.
Example 6: Algebra Connection Find x and y.


Solution: The line shown is the perpendicular bisector.


So, 3x − 6 = 21 And, (4y − 2)◦ = 90◦
3x = 27 4y = 92◦


x = 9 y = 23◦


Investigation 1-4: Constructing a Perpendicular Bisector
1. Draw a line that is 6 cm long, halfway down your page.


2. Place the pointer of the compass at an endpoint. Open the compass to be greater than half of the
segment. Make arc marks above and below the segment. Repeat on the other endpoint. Make sure the
arc marks intersect.


3. Use your straightedge to draw a line connecting the arc intersections.


This constructed line bisects the line you drew in #1 and intersects it at 90◦. To see an animation of this
investigation, go to http://www.mathsisfun.com/geometry/construct-linebisect.html.


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Congruent Angles
Example 7: Algebra Connection What is the measure of each angle?


Solution: From the picture, we see that the angles are equal.
Set the angles equal to each other and solve.


(5x + 7)◦ = (3x + 23)◦


2x◦ = 16◦


x = 8◦


To find the measure of ∠ABC, plug in x = 8◦ to (5x + 7)◦ → (5(8) + 7)◦ = (40 + 7)◦ = 47◦. Because
m∠ABC = m∠XYZ, m∠XYZ = 47◦ too.
Angle Bisectors
Angle Bisector: A ray that divides an angle into two congruent angles, each having a measure exactly
half of the original angle.


BD is the angle bisector of ∠ABC


∠ABD ∠DBC


m∠ABD = 1
2
m∠ABC


Angle Bisector Postulate: Every angle has exactly one angle bisector.
Example 8: Let’s take a look at Review Queue #1 again. Is OP the angle bisector of ∠SOT?


Solution: Yes, OP is the angle bisector of ∠SOT from the markings in the picture.
Investigation 1-5: Constructing an Angle Bisector


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1. Draw an angle on your paper. Make sure one side is horizontal.


2. Place the pointer on the vertex. Draw an arc that intersects both sides.


3. Move the pointer to the arc intersection with the horizontal side. Make a second arc mark on the
interior of the angle. Repeat on the other side. Make sure they intersect.


4. Connect the arc intersections from #3 with the vertex of the angle.


To see an animation of this construction, view http://www.mathsisfun.com/geometry/construct-anglebisect.
html.
Know What? Revisited The image to the right is an outline of the Transamerica Building from earlier
in the lesson. From this outline, we can see the following parts are congruent:


TR TC ∠TCR ∠TRC


RS CM ∠CIE ∠RAN


CI RA and ∠TMS ∠TS M
AN IE ∠IEC ∠ANR


TS TM ∠TCI ∠TRA


All the four triangular sides of the building are congruent to each other as well.


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Review Questions
• Questions 1-18 are similar to Examples 1, 4, 5 and 8.
• Questions 19-22 are similar to Examples 6 and 7.
• Question 23 is similar to Investigation 1-5.
• Question 24 is similar to Investigation 1-4.
• Questions 25-28 are similar to Example 2.
• Question 29 and 30 are similar to Example 3.


1. Copy the figure below and label it with the following information:


∠A ∠C
∠B ∠D


AB CD


AD BC


H is the midpoint of AE and DG, B is the midpoint of AC, GD is the perpendicular bisector of FA and EC
AC FE and FA EC


Find:


2. AB


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3. GA
4. ED
5. HE
6. m∠HDC
7. FA
8. GD
9. m∠FED


10. How many copies of triangle AHB can fit inside rectangle FECA without overlapping?


For 11-18, use the following picture to answer the questions.


11. What is the angle bisector of ∠TPR?
12. P is the midpoint of what two segments?
13. What is m∠QPR?
14. What is m∠TPS ?
15. How does VS relate to QT?
16. How does QT relate to VS ?
17. What is m∠QPV?
18. Is PU a bisector? If so, of what?


Algebra Connection For 19-22, use algebra to determine the value of variable in each problem.


19.


20.


21.


22.


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23. Construction Using your protractor, draw an angle that is 110◦. Then, use your compass to
construct the angle bisector. What is the measure of each angle?


24. Construction Using your ruler, draw a line segment that is 7 cm long. Then use your compass to
construct the perpendicular bisector, What is the measure of each segment?


For questions 25-28, find the midpoint between each pair of points.


25. (-2, -3) and (8, -7)
26. (9, -1) and (-6, -11)
27. (-4, 10) and (14, 0)
28. (0, -5) and (-9, 9)


Given the midpoint (M) and either endpoint of AB, find the other endpoint.


29. A(−1, 2) and M(3, 6)
30. B(−10,−7) and M(−2, 1)


Review Queue Answers
1. m∠POT = 38◦, m∠ROT = 57◦ + 38◦ + 38◦ = 133◦
2. (a) 7−1−1+4 = 63 = 2


(b) −4+6−3−5 = 2−8 = −14
3. 23+30+18+27+325 = 1305 = 26


1.5 Angle Pairs
Learning Objectives


• Recognize complementary angles supplementary angles, linear pairs, and vertical angles.
• Apply the Linear Pair Postulate and the Vertical Angles Theorem.


Review Queue


1. Find x.
2. Find y.
3. Find z.


Know What? A compass (as seen to the right) is used to determine the direction a person is traveling.
The angles between each direction are very important because they enable someone to be more specific
with their direction. A direction of 45◦ NW, would be straight out along that northwest line.


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What headings have the same angle measure? What is the angle measure between each compass line?


Complementary Angles
Complementary: Two angles that add up to 90◦.
Complementary angles do not have to be:


• congruent
• next to each other


Example 1: The two angles below are complementary. m∠GHI = x. What is x?


Solution: Because the two angles are complementary, they add up to 90◦. Make an equation.


x + 34◦ = 90◦


x = 56◦


Example 2: The two angles below are complementary. Find the measure of each angle.


Solution: The two angles add up to 90◦. Make an equation.


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8r + 9◦ + 7r + 6◦ = 90◦


15r + 15◦ = 90◦


15r = 75◦


r = 5◦


However, you need to find each angle. Plug r back into each expression.
m∠GHI = 8(5◦) + 9◦ = 49◦


m∠JKL = 7(5◦) + 6◦ = 41◦


Supplementary Angles
Supplementary: Two angles that add up to 180◦.
Supplementary angles do not have to be:


• congruent
• next to each other


Example 3: The two angles below are supplementary. If m∠MNO = 78◦ what is m∠PQR?


Solution: Set up an equation. However, instead of equaling 90◦, now it is 180◦.
78◦ + m∠PQR = 180◦


m∠PQR = 102◦


Example 4: What is the measure of two congruent, supplementary angles?
Solution: Supplementary angles add up to 180◦. Congruent angles have the same measure. So, 180◦÷2 =
90◦, which means two congruent, supplementary angles are right angles, or 90◦.


Linear Pairs
Adjacent Angles: Two angles that have the same vertex, share a side, and do not overlap.
∠PSQ and ∠QSR are adjacent.
∠PQR and ∠PQS are NOT adjacent because they overlap.


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Linear Pair: Two angles that are adjacent and the non-common sides form a straight line.


∠PSQ and ∠QSR are a linear pair.
Linear Pair Postulate: If two angles are a linear pair, then they are supplementary.
Example 5: Algebra Connection What is the measure of each angle?


Solution: These two angles are a linear pair, so they add up to 180◦.


(7q − 46)◦ + (3q + 6)◦ = 180◦


10q − 40◦ = 180◦


10q = 220◦


q = 22◦


Plug in q to get the measure of each angle. m∠ABD = 7(22◦) − 46◦ = 108◦ m∠DBC = 180◦ − 108◦ = 72◦


Example 6: Are ∠CDA and ∠DAB a linear pair? Are they supplementary?
Solution: The two angles are not a linear pair because they do not have the same vertex. They are
supplementary, 120◦ + 60◦ = 180◦.


Vertical Angles
Vertical Angles: Two non-adjacent angles formed by intersecting lines.


∠1 and ∠3 are vertical angles


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∠2 and ∠4 are vertical angles
These angles are labeled with numbers. You can tell that these are labels because they do not have a
degree symbol.
Investigation 1-6: Vertical Angle Relationships


1. Draw two intersecting lines on your paper. Label the four angles created ∠1, ∠2, ∠3, and ∠4, just like
the picture above.


2. Use your protractor to find m∠1.
3. What is the angle relationship between ∠1 and ∠2 called? Find m∠2.
4. What is the angle relationship between ∠1 and ∠4 called? Find m∠4.
5. What is the angle relationship between ∠2 and ∠3 called? Find m∠3.
6. Are any angles congruent? If so, write them down.


From this investigation, you should find that ∠1 ∠3 and ∠2 ∠4.
Vertical Angles Theorem: If two angles are vertical angles, then they are congruent.
We can prove the Vertical Angles Theorem using the same process we used in the investigation. We will
not use any specific values for the angles.
From the picture above:


∠1 and ∠2 are a linear pair → m∠1 + m∠2 = 180◦ Equation 1
∠2 and ∠3 are a linear pair → m∠2 + m∠3 = 180◦ Equation 2
∠3 and ∠4 are a linear pair → m∠3 + m∠4 = 180◦ Equation 3


All of the equations = 180◦, so Equation 1 = Equation 2 and Equation 2 = Equation 3.


m∠1 + m∠2 = m∠2 + m∠3 and m∠2 + m∠3 = m∠3 + m∠4


Cancel out the like terms


m∠1 = m∠3 and m∠2 = m∠4


Recall that anytime the measures of two angles are equal, the angles are also congruent. So, ∠1 ∠3 and
∠2 ∠4 too.
Example 7: Find m∠1 and m∠2.


Solution: ∠1 is vertical angles with 18◦, so m∠1 = 18◦.
∠2 is a linear pair with ∠1 or 18◦, so 18◦ + m∠2 = 180◦.
m∠2 = 180◦ − 18◦ = 162◦.
Know What? Revisited The compass has several vertical angles and all of the smaller angles are
22.5◦, 180◦ ÷ 8. Directions that are opposite each other have the same angle measure, but of course, a


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different direction. All of the green directions have the same angle measure, 22.5◦, and the purple have the
same angle measure, 45◦. N, S , E and W all have different measures, even though they are all 90◦ apart.


Review Questions
• Questions 1 and 2 are similar to Examples 1, 2, and 3.
• Questions 3-8 are similar to Examples 3, 4, 6 and 7.
• Questions 9-16 use the definitions, postulates and theorems from this section.
• Questions 17-25 are similar to Example 5.


1. Find the measure of an angle that is complementary to ∠ABC if m∠ABC is
(a) 45◦
(b) 82◦
(c) 19◦
(d) z◦


2. Find the measure of an angle that is supplementary to ∠ABC if m∠ABC is
(a) 45◦
(b) 118◦
(c) 32◦
(d) x◦


Use the diagram below for exercises 3-7. Note that NK ⊥ ←→IL.


3. Name one pair of vertical angles.
4. Name one linear pair of angles.
5. Name two complementary angles.
6. Name two supplementary angles.


7. What is:
(a) m∠INL


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(b) m∠LNK


8. If m∠INJ = 63◦, find:


(a) m∠JNL
(b) m∠KNJ
(c) m∠MNL
(d) m∠MNI


For 9-16, determine if the statement is true or false.


9. Vertical angles are congruent.
10. Linear pairs are congruent.
11. Complementary angles add up to 180◦.
12. Supplementary angles add up to 180◦
13. Adjacent angles share a vertex.
14. Adjacent angles overlap.
15. Complementary angles are always 45◦.
16. Vertical angles have the same vertex.


For 17-25, find the value of x or y.


17.


18.


19.


20.


21.


22.


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23.


24. Find x.
25. Find y.


Review Queue Answers
1. x + 26 = 3x − 8


34 = 2x
17 = x


2. (7y + 6)◦ = 90◦
7y = 84◦
y = 12◦


3. z + 15 = 5z + 9
6 = 4z


1.5 = z


1.6 Classifying Polygons
Learning Objectives


• Define triangle and polygon.
• Classify triangles by their sides and angles.
• Understand the difference between convex and concave polygons.
• Classify polygons by number of sides.


Review Queue
1. Draw a triangle.
2. Where have you seen 4, 5, 6 or 8 - sided polygons in real life? List 3 examples.
3. Fill in the blank.


(a) Vertical angles are always _____________.
(b) Linear pairs are _____________.
(c) The parts of an angle are called _____________ and a _____________.


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Know What? The pentagon in Washington DC is a pentagon with congruent sides and angles. There is
a smaller pentagon inside of the building that houses an outdoor courtyard. Looking at the picture, the
building is divided up into 10 smaller sections. What are the shapes of these sections? Are any of these
division lines diagonals? How do you know?


Triangles
Triangle: Any closed figure made by three line segments intersecting at their endpoints.
Every triangle has three vertices (the points where the segments meet), three sides (the segments), and
three interior angles (formed at each vertex). All of the following shapes are triangles.


You might have also learned that the sum of the interior angles in a triangle is 180◦. Later we will prove
this, but for now you can use this fact to find missing angles.
Example 1: Which of the figures below are not triangles?


Solution: B is not a triangle because it has one curved side. D is not closed, so it is not a triangle either.
Example 2: How many triangles are in the diagram below?


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Solution: Start by counting the smallest triangles, 16.
Now count the triangles that are formed by 4 of the smaller triangles, 7.


Next, count the triangles that are formed by 9 of the smaller triangles, 3.


Finally, there is the one triangle formed by all 16 smaller triangles. Adding these numbers together, we
get 16 + 7 + 3 + 1 = 27.


Classifying by Angles


Angles can be grouped by their angles; acute, obtuse or right. In any triangle, two of the angles will always
be acute. The third angle can be acute, obtuse, or right. We classify each triangle by this angle.
Right Triangle: A triangle with one right angle.


Obtuse Triangle: A triangle with one obtuse angle.


Acute Triangle: A triangle where all three angles are acute.


Equiangular Triangle: When all the angles in a triangle are congruent.


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Example 3: Which term best describes 4RST below?


Solution: This triangle has one labeled obtuse angle of 92◦. Triangles can only have one obtuse angle, so
it is an obtuse triangle.


Classifying by Sides
You can also group triangles by their sides.
Scalene Triangle: A triangles where all three sides are different lengths.


Isosceles Triangle: A triangle with at least two congruent sides.


Equilateral Triangle: A triangle with three congruent sides.


From the definitions, an equilateral triangle is also an isosceles triangle.
Example 4: Classify the triangle by its sides and angles.


Solution: We see that there are two congruent sides, so it is isosceles. By the angles, they all look acute.


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We say this is an acute isosceles triangle.
Example 5: Classify the triangle by its sides and angles.


Solution: This triangle has a right angle and no sides are marked congruent. So, it is a right scalene
triangle.


Polygons
Polygon: Any closed, 2-dimensional figure that is made entirely of line segments that intersect at their
endpoints.
Polygons can have any number of sides and angles, but the sides can never be curved.
The segments are called the sides of the polygons, and the points where the segments intersect are called
vertices.
Example 6: Which of the figures below is a polygon?


Solution: The easiest way to identify the polygon is to identify which shapes are not polygons. B and C
each have at least one curved side, so they are not be polygons. D has all straight sides, but one of the
vertices is not at the endpoint, so it is not a polygon. A is the only polygon.
Example 7: Which of the figures below is not a polygon?


Solution: C is a three-dimensional shape, so it does not lie within one plane, so it is not a polygon.


Convex and Concave Polygons
Polygons can be either convex or concave. The term concave refers to a cave, or the polygon is “caving
in”. All stars are concave polygons.


A convex polygon does not do this. Convex polygons look like:


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Diagonals: Line segments that connect the vertices of a convex polygon that are not sides.


The red lines are all diagonals.
This pentagon has 5 diagonals.
Example 8: Determine if the shapes below are convex or concave.


Solution: To see if a polygon is concave, look at the polygons and see if any angle “caves in” to the
interior of the polygon. The first polygon does not do this, so it is convex. The other two do, so they are
concave.
Example 9: How many diagonals does a 7-sided polygon have?


Solution: Draw a 7-sided polygon, also called a heptagon.
Drawing in all the diagonals and counting them, we see there are 14.


Classifying Polygons
Whether a polygon is convex or concave, it is always named by the number of sides.


Table 1.11:


Polygon Name Number of Sides Number of Diago-
nals


Convex Example


Triangle 3 0


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Table 1.11: (continued)


Polygon Name Number of Sides Number of Diago-
nals


Convex Example


Quadrilateral 4 2


Pentagon 5 5


Hexagon 6 9


Heptagon 7 14


Octagon 8 ?


Nonagon 9 ?


Decagon 10 ?


Undecagon or hen-
decagon


11 ?


Dodecagon 12 ?


n-gon n (where n > 12) ?


Example 10: Name the three polygons below by their number of sides and if it is convex or concave.


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Solution: The pink polygon is a concave hexagon (6 sides).
The green polygon convex pentagon (5 sides).
The yellow polygon is a convex decagon (10 sides).
Know What? Revisited The pentagon is divided up into 10 sections, all quadrilaterals. None of these
dividing lines are diagonals because they are not drawn from vertices.


Review Questions
• Questions 1-8 are similar to Examples 3, 4 and 5.
• Questions 9-14 are similar to Examples 8 and 10
• Question 15 is similar to Example 6.
• Questions 16-19 are similar to Example 9 and the table.
• Questions 20-25 use the definitions, postulates and theorems in this section.


For questions 1-6, classify each triangle by its sides and by its angles.


1.


2.


3.


4.


5.


6.


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7. Can you draw a triangle with a right angle and an obtuse angle? Why or why not?
8. In an isosceles triangle, can the angles opposite the congruent sides be obtuse?


In problems 9-14, name each polygon in as much detail as possible.


9.


10.


11.


12.


13.


14.


15. Explain why the following figures are NOT polygons:


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16. How many diagonals can you draw from one vertex of a pentagon? Draw a sketch of your answer.
17. How many diagonals can you draw from one vertex of an octagon? Draw a sketch of your answer.
18. How many diagonals can you draw from one vertex of a dodecagon?
19. Determine the number of total diagonals for an octagon, nonagon, decagon, undecagon, and do-


decagon.


For 20-25, determine if the statement is true or false.


20. Obtuse triangles can be isosceles.
21. A polygon must be enclosed.
22. A star is a convex polygon.
23. A right triangle is acute.
24. An equilateral triangle is equiangular.
25. A quadrilateral is always a square.
26. A 5-point star is a decagon


Review Queue Answers
1.


2. Examples include: stop sign (8), table top (4), the Pentagon (5), snow crystals (6), bee hive combs
(6), soccer ball pieces (5 and 6)


3. (a) congruent or equal
(b) supplementary
(c) sides, vertex


1.7 Chapter 1 Review
Symbol Toolbox
←→AB,−→AB, AB - Line, ray, line segment
∠ABC - Angle with vertex B
mAB or AB - Distance between A and B
m∠ABC - Measure of ∠ABC
⊥ - Perpendicular
= - Equal
- Congruent
Markings


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Keywords, Postulates, and Theorems
Points, Lines, and Planes


• Geometry
• Point
• Line
• Plane
• Space
• Collinear
• Coplanar
• Endpoint
• Line Segment
• Ray
• Intersection
• Postulates
• Theorem
• Postulate 1-1
• Postulate 1-2
• Postulate 1-3
• Postulate 1-4
• Postulate 1-5


Segments and Distance


• Distance
• Measure
• Ruler Postulate
• Segment Addition Postulate


Angles and Measurement


• Angle
• Vertex
• Sides
• Protractor Postulate
• Straight Angle
• Right Angle
• Acute Angles
• Obtuse Angles
• Convex


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• Concave
• Polygon
• Perpendicular
• Construction
• Compass
• Angle Addition Postulate


Midpoints and Bisectors


• Congruent
• Midpoint.
• Midpoint Postulate
• Segment Bisector
• Perpendicular Bisector
• Perpendicular Bisector Postulate
• Angle Bisector
• Angle Bisector Postulate


Angle Pairs


• Complementary
• Supplementary
• Adjacent Angles
• Linear Pair
• Linear Pair Postulate
• Vertical Angles
• Vertical Angles Theorem


Classifying Polygons


• Triangle
• Right Triangle
• Obtuse Triangle
• Acute Triangle
• Equiangular Triangle
• Scalene Triangle
• Isosceles Triangle
• Equilateral Triangle
• Vertices
• Diagonals


Review
Match the definition or description with the correct word.


1. When three points lie on the same line. — A. Measure
2. All vertical angles are ________. — B. Congruent
3. Linear pairs add up to _______. — C. Angle Bisector


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4. The m in from of m∠ABC. — D. Ray
5. What you use to measure an angle. — E. Collinear
6. When two sides of a triangle are congruent. — F. Perpendicular
7. ⊥ — G. Line
8. A line that passes through the midpoint of another line. — H. Protractor
9. An angle that is greater than 90◦. — I. Segment Addition Postulate


10. The intersection of two planes is a ___________. — J. Obtuse
11. AB + BC = AC — K. Point
12. An exact location in space. — L. 180◦
13. A sunbeam, for example. — M. Isosceles
14. Every angle has exactly one. — N. Pentagon
15. A closed figure with 5 sides. — O. Hexagon


P. Bisector


Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities
designed to supplement the objectives for some of the lessons in this chapter. See http:
//www.ck12.org/flexr/chapter/9686.


1.8 Study Guide
Keywords: Define, write theorems, and/or draw a diagram for each word below.
1st Section: Points, Lines, and Planes
Geometry
Point
Line
Plane
Space
Collinear
Coplanar
Endpoint
Line Segment
Ray
Intersection
Postulates
Theorem
Postulate 1-1
Postulate 1-2
Postulate 1-3
Postulate 1-4


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Postulate 1-5
Use this picture to identify the geometric terms in this section.


Homework:
2nd Section: Segments and Distance
Distance
Measure
Ruler Postulate
Segment Addition Postulate


Homework:
3rd Section: Angles and Measurement
Angle
Vertex
Sides
Protractor Postulate
Straight Angle
Right Angle
Acute Angles
Obtuse Angles
Perpendicular
Construction
Compass
Angle Addition Postulate


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Homework:
4th Section: Midpoints and Bisectors
Congruent
Midpoint.
Midpoint Postulate
Segment Bisector
Perpendicular Bisector
Perpendicular Bisector Postulate
Angle Bisector
Angle Bisector Postulate


Homework:
5th Section: Angle Pairs
Complementary
Supplementary
Adjacent Angles
Linear Pair
Linear Pair Postulate
Vertical Angles
Vertical Angles Theorem


Homework:


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6th Section: Classifying Polygons
Draw your own pictures for this section
Triangle
Right Triangle
Obtuse Triangle
Acute Triangle
Equiangular Triangle
Scalene Triangle
Isosceles Triangle
Equilateral Triangle
Vertices
Sides
Polygon
Convex Polygon
Concave Polygon
Quadrilateral, Pentagon, Hexagon, Heptagon, Octagon, Nonagon, Decagon…
Diagonals
Homework:


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Chapter 2


Coordinate Geometry


2.1 Vocabulary Self-Rating
Table 2.1: Rating Guide: DK: I am sure I don’t know it K: I am sure I know it ?: I’m not
sure


Word Before Lesson/Unit After Lesson/Unit
Distance
Overbar
Estimation
Estimate
Ruler
Absolute value
x − y coordinate plane
x−coordinate
y−coordinate
Right triangle
Pythagorean Theorem
Distance Formula
Midpoint
Equidistant
Slope
Undefined
Parallel
Perpendicular
Slope-intercept form
y−intercept
Circle
Center
Radius
Leg
Hypotenuse
Concentric
Translation


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Table 2.1: (continued)


Word Before Lesson/Unit After Lesson/Unit
Prime
Preimage
Image
Isometry
Reflection
Perpendicular bisector
Rotation
Angle of rotation
Arc
Central angle
Diameter
Acute
Complementary


2.2 Distance and Midpoint
Learning Objectives


• Derive the Distance Formula using the Pythagorean Theorem.
• Use the Distance Formula to find the length of a line segment with known endpoints.
• Use the Midpoint Formula to calculate the coordinates of the midpoint of a line segment given both


endpoints, or to determine the coordinates of one endpoint given the midpoint and the other endpoint.


Measuring Distances
There are many different ways to identify measurements. This lesson will present some that may be
familiar, and probably a few that are new to you.
Before we begin to examine distances, however, it is important to identify the meaning of distance in the
context of geometry. The distance between two points is defined by the length of the line segment that
connects them.


• The distance between two points is the _________________________ of the line seg-
ment that connects them.


The most common way to measure distance is with a ruler. Also, distance can be estimated using scale
on a map.


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Notation Notes: When we name a segment we use the endpoints and an overbar (a bar or line above
the letters) with no arrows. For example, ”segment AB” is written AB. The length of a segment is named
by giving the endpoints without using an overbar. For example, the length of AB is written AB. In some
books you may also see mAB, or measure of AB, which means the same as AB, that is, it is the length of
the segment with endpoints A and B.
Example 1
Use the scale to estimate the distance between Aaron’s house and Bijal’s house. Assume that the first third
of the scale in black represents one inch.


You need to find the distance between the two houses in the map. The scale shows a sample distance.
Use the scale to estimate the distance. You will find that approximately 3 segments of the length of the
scale fit between the two points. Be careful — 3 is not the answer to this problem! As the scale shows 1
inch equal to 2 miles, you must multiply 3 units by 2 miles:


3 inches · 2 miles
1 inch


= 6 miles


The distance between the houses is about six miles.
You can also use estimation to identify measurements in other geometric figures. Remember to include
words like approximately, about, or estimation whenever you are finding an estimated answer.
The word “estimation” means using a non-exact guess of what a number is. Another similar word is
“approximation.”
Both of these words are nouns. The verb forms are: “to estimate” or “to approximate.”
We use these words when we are not sure of the exact measurement of a distance, length, or other number,
but when we can make an educated guess.


• To estimate (or to _________________________________ ) a number means to
give a non-exact but educated guess of what it is.


Rulers
You have probably been using rulers to measure distances for a long time and you know that a ruler is
a tool with measurement markings.


• A ruler is a tool with ___________________________________ markings.


Using a ruler: If you use a ruler to find the distance between two points, the distance will be the absolute
value of the difference between the numbers shown on the ruler.


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This means that you do not need to start measuring at the zero mark, as long as you use subtraction to
find the distance.
Note: We say absolute value here since distances in geometry must always be positive, and subtraction
can give a negative result.


• You do not need to measure from zero on a ruler; just ______________________ the start
number from the end number to find the distance!


• The distance on a ruler is the ___________________________ value of the difference
between the numbers.


• Absolute value is always a ____________________________ number.


Example 2
What distance is marked on the ruler in the diagram below? Assume that the scale is marked in centimeters.


The way to use the ruler is to find the absolute value of the difference between the numbers shown. This
means you subtract the numbers and then make sure your answer is positive. The line segment spans from
3 cm to 8 cm:


|3 − 8| = | − 5| = 5


The absolute value of the difference between the two numbers shown on the ruler above is 5 cm. So the
line segment is 5 cm long.
Remember, we use vertical bars around an expression to show absolute value: |x|


Distances on a Grid
In algebra you most likely worked with graphing lines in the x− y coordinate plane. Sometimes you can
find the distance between points on a coordinate plane using the values of the coordinates:


• If the two points line up horizontally, look at the change of value in the x-coordinates.
• If the two points line up vertically, look at the change of value in the y-coordinates.


The change in value will show the distance between the points. Remember to use absolute value, just
like you did with the ruler. Later you will learn how to calculate distance between points that do not line
up horizontally or vertically.


• When points line up horizontally, they have the same _______-coordinate. This means their
_______-coordinates are different so we take their difference to find the distance between the
points.


• When points line up vertically, they have the same _______-coordinate. This means their ___-
____-coordinates are different so we take their difference to find the distance between the points.


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Example 3
What is the distance between the two points shown below?


The two points shown on the grid are at (2, 9) and (2, 3). These points line up vertically (meaning they
have the same x−coordinate of 2), so we can look at the difference in their y−coordinates:


|9 − 3| = |6| = 6


So, the distance between the two points is 6 units.
Example 4
What is the distance between the two points shown below?


The two points shown on the grid are at (–4, 4) and (3, 4). These points line up horizontally (meaning they
have the same y−coordinate of 4), so we can look at the difference in their x−coordinates. Remember
to take the absolute value of the difference between the values to find the distance:


| − 4 − 3| = | − 7| = 7


The distance between the two points is 7 units.
Reading Check:
1. What is absolute value? Explain in your own words.


2. When 2 points line up vertically, what value do they have in common?


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3. When 2 points line up horizontally, what value do they have in common?


The Distance Formula
We have learned that a right triangle with sides of lengths a and b and hypotenuse of length c has a
special relationship called the Pythagorean Theorem. The sum of the squares of a and b is equal to
the square of c. Placing this in equation form we have:


If we put this triangle in a coordinate plane so A has coordinates of (x1, y1) and B has coordinates of
(x2, y2), we can find the lengths of the legs of the triangle using what we just learned about points that
line up horizontally or vertically:
the length of AC is |x2 − x1| and the length of BC is |y2 − y1|


We are finding the length, which means that we want a positive value; the absolute value bars guarantee


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that our answer is always positive. But in the final equation,


c2 = |x2 − x1|2 + |y2 − y1|2


the absolute value bars are not needed since we squared all three terms, and squared numbers are always
positive.
Getting the square root of both sides we have,


c =


(x2 − x1)2 + (y2 − y1)2


We say that c is the distance between the points A and B, and we call the formula above the Distance
Formula.
Reading Check:
1. On which famous theorem is the Distance Formula based?


2. How can you find the distance of one of the legs of a right triangle like the one in the diagram on the
previous page? Pick one leg and explain in your own words.


Segment Midpoints
Now that you understand congruent segments, there are a number of new terms and types of figures you
can explore.
A segment midpoint is a point on a line segment that divides the segment into two congruent segments.
So, each segment between the midpoint and an endpoint will have the same length.


• A midpoint divides a segment into two ___________________________ parts.


In the diagram below, point B is the midpoint of AC since AB is congruent to BC:


There is even a special postulate dedicated to midpoints:
Segment Midpoint Postulate
Any line segment will have exactly one midpoint—no more, and no less.
Example 5


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Nandi and Arshad measure and find that their houses are 10 miles apart. If they agree to meet at the
midpoint between their two houses, how far will each of them travel?


The easiest way to find the distance to the midpoint of the imagined segment connecting their houses is
to divide the length (which is 10 miles) by 2:


10 ÷ 2 = 5


Each person will travel five miles to meet at the midpoint between their houses.


The Midpoint Formula
The midpoint is the middle point of a line segment. It is equidistant (equal distances) from both
endpoints.
The formula for determining the midpoint of a segment in a coordinate plane is the average of the
x−coordinates and the y−coordinates. Remember, to find the average of 2 numbers, you take the sum
of the numbers and then divide by 2.
If a segment has endpoints (x1, y1) and (x2, y2):


• the average of the x−coordinates is: x1+x22
• and the average of the y-coordinates is: y1+y22


Therefore, the midpoint is at:


Reading Check:
1. What is an average? Explain in your own words.


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2. Where is a midpoint located on a line segment? Describe.


3. What does the word equidistant mean?


4. How many midpoints can a line segment have?


5. In the space below, draw a line segment. Then draw and label its midpoint.


2.3 Parallel and Perpendicular
Learning Objectives


• Identify and compute slope in the coordinate plane.
• Use the relationship between slopes of parallel lines.
• Use the relationship between slopes of perpendicular lines.
• Identify equations of parallel lines.
• Identify equations of perpendicular lines.


Slope in the Coordinate Plane
If you look at a graph of a line, you can think of the slope as the steepness of the line.


• Slope is the measure of the ______________________________ of a line.


Mathematically, you can calculate the slope using two different points on a line. Given two points (x1, y1)
and (x2, y2) the slope is:


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slope = y2 − y1
x2 − x1


You may have also learned that slope equals “rise over run.”
This means that:


• The numerator (top) of the fraction is the “rise,” or how many units the slope goes up (positive) or
down (negative).


→ Up or down is how the slope moves along the y−axis.


• The denominator (bottom) of the fraction is the “run,” or how many units the slope goes to the right
(positive) or left (negative).


→ Right or left is how the slope moves along the x−axis.
You can remember “rise” as up or down because an elevator “rises” up or down.
“Rise” (up/down) is in the y direction.
You can remember “run” as moving right or left because a person “runs” with her right and left feet.
“Run” (right/left) is in the x direction.


• The numerator of the slope represents the change in the _________ direction.
• The slope’s _______________________ represents the change in the x direction.


In other words, first calculate the distance that the line travels up (or down), and then divide that value
by the distance the line travels left to right.
A line that goes up from left to right has positive slope, and a line that goes down from left to right has
negative slope:


images from http://www.tutorvista.com/math/positive-and-negative-slope


• A line that goes up from left to right has a _________________________ slope.
• A line that goes down from left to right has a ________________________ slope.


Example 1
What is the slope of a line that goes through the points (2, 2) and (4, 6)?
You can use the slope formula on the previous page to find the slope of this line. When substituting
values, (x1, y1) is (2, 2) and (x2, y2) is (4, 6):


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x1 = , y1 = , and x2 = , y2 =


slope = y2 − y1
x2 − x1


slope = 6 − 2
4 − 2 =


4
2


= 2


The slope of this line is 2.
→ What does that mean graphically?
Look at the graph on the next page to see what the line looks like.
Notice: If the slope is positive, the line should go up from left to right. Does it?


You can see that the line “rises” 4 units as it “runs” 2 units to the right. So, the “rise” (numerator) is 4
units and the “run” (denominator) is 2 units. Since 4 ÷ 2 = 2, the slope of this line is 2.
As you read on the previous page, the slope of this line is 2, a positive number.


• Any line with a positive slope will go ____________________ from left to right.
• Any line with a negative slope will go ____________________ from left to right.


Example 2
What is the slope of the line that goes through the points (1, 9) and (3, 3)?
Again, use the formula to find the slope of this line:
x1 = , y1 = , and x2 = , y2 =


slope = y2 − y1
x2 − x1


slope = 3 − 9
3 − 1 =


−6
2


= −3


The slope of this line is –3.
Because the slope of the line in example 2 is negative, it will go down to the right. The points and the
line that connects them are shown below:


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Some types of lines have special slopes. Check out following examples to see what happens with horizontal
and vertical lines.
Example 3
What is the slope of a line that goes through the points (4, 4) and (8, 4)?
Use the formula to find the slope of this line:
x1 = , y1 = , and x2 = , y2 =


slope = y2 − y1
x2 − x1


slope = 4 − 4
8 − 4 =


0
8


= 0


The slope of this line is 0.
Every line with a slope of 0 is horizontal.


• A ____________________________________ line has a slope equal to zero.


Example 4
What is the slope of a line that goes through the points (3, 2) and (3, 6)?
Use the formula to find the slope of this line:
x1 = , y1 = , and x2 = , y2 =


slope = y2 − y1
x2 − x1


slope = 6 − 2
3 − 3 =


4
0


Zero is not allowed to be in the denominator of a fraction! Therefore, the slope of this line is undefined.
Every line with an undefined slope is vertical.


• All vertical lines have slopes that are ____________________.


The line in example 4 is vertical and its slope is undefined:


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In review, if you look at a graph of a line from left to right, then:


• Lines with positive slopes point up to the right.
• Lines with negative slopes point down to the right.
• Horizontal lines have a slope of zero.
• Vertical lines have undefined slope.


Reading Check:
1. On the coordinate plane below, draw a line with a positive slope.


2. On the coordinate plane below, draw a line with a negative slope.


3. On the coordinate plane below, draw a line with a slope of zero.


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4. On the coordinate plane below, draw a line with an undefined slope.


Slopes of Parallel Lines
Now that you know how to find the slope of lines using x− and y−coordinates, you can think about how
lines are related to their slopes.
If two lines in the coordinate plane are parallel, then they will have the same slope. Conversely, if two
lines in the coordinate plane have the same slope, then those lines are parallel.


• Parallel lines have the _____________________________ slope.


Example 5
Which of the following answers below could represent the slope of a line parallel to the one shown on the
graph?


A. – 4


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B. – 1
C. 14
D. 1
Since you are looking for the slope of a parallel line, it will have the same slope as the line on the graph.
First find the slope of the given line, and then choose the answer with that same slope. To do this, pick
any two points on the line and use the slope formula.
For example, for the points (–1, 5) and (3, 1) :
x1 = , y1 = , and x2 = , y2 =


slope = y2 − y1
x2 − x1


slope = 1 − 5
3 − (−1) =


−4
3 + 1


= −4
4


= −1


The slope of the line on the graph is –1. The answer is B.


Slopes of Perpendicular Lines
Parallel lines have the same slope. There is also a mathematical relationship for the slopes of perpen-
dicular lines.
The slopes of perpendicular lines will be the opposite reciprocal of each other.
Opposite here means the opposite sign.
If a slope is positive, then its opposite is negative.
If a slope is negative, then its opposite is positive.
A reciprocal is a fraction with its numerator and denominator flipped.
The reciprocal of 23 is 32 . The reciprocal of 12 is 2. The reciprocal of 4 is 14 .


• The opposite of 5 is _______________.
• The reciprocal of 5 is _______________.


The opposite reciprocal can be found in two steps:
1. First, find the reciprocal of the given slope. If the slope is a fraction, you can simply switch the
numbers in the numerator and the denominator to find the reciprocal. If the slope is not a fraction, you
can make it into a fraction by putting a 1 in the denominator. Then find the reciprocal by flipping the
numerator and denominator.
2. The second step is to find the opposite of the given number. If the value is positive, make it negative.
If the value is negative, make it positive.
The opposite reciprocal of 54 is −45 and the opposite reciprocal of - 3 is 13 .


• The opposite reciprocal of 5 is _______________.


Another way to check if lines are perpendicular is to multiply their slopes: if the slopes of two lines
multiply to be –1, then the two lines are perpendicular.


• The slopes of ____________________________ lines multiply to be –1.


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Example 6
Which of the following numbers could represent the slope of a line perpendicular to the one shown below?


A. −75
B. 75
C. −57
D. 57
Since you are looking for the slope of a perpendicular line, it will be the opposite reciprocal of the slope
of the line on the graph. First find the slope of the given line, then find its opposite reciprocal. You can
use the slope formula to find the original line’s slope. Pick two points on the line.
For example, for the points (–3, –2) and (4, 3) :
x1 = , y1 = , and x2 = , y2 =


slope = y2 − y1
x2 − x1


slope =
3 − (−2)
4 − (−3) =


3 + 2
4 + 3


= 5
7


The slope of the line on the graph is 57 . Now find the opposite reciprocal of that value. First switch the
numerator and denominator in the fraction, then find the opposite sign. The opposite reciprocal of 57 is −75 .
The answer is A.


Slope-Intercept Equations
The most common type of linear equation to study is called slope-intercept form, which uses both the
slope of the line and its y−intercept. A y−intercept is the point where the line crosses the vertical
y−axis. This is the value of y when x is equal to 0.


• Slope-intercept form is an equation that uses the _________________________ and
the ______________________________ of a line.


• The y−intercept is the point where the line intersects the ___________________.
• At the y−intercept, x equals ______________.


The formula for an equation in slope-intercept form is:


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y = mx + b


In this equation, y and x remain as variables, m is the slope of the line, and b is the y−intercept of the
line. For example, if you know that a line has a slope of 4 and it crosses the y−axis at (0, 8), then its
equation in slope-intercept form is: y = 4x + 8.


• In slope-intercept form, m represents the _____________.
• In slope-intercept form, b represents the _____________.


This form is especially useful for finding the equation of a line given its graph. You already know how to
calculate the slope by finding two points and using the slope formula. You can find the y−intercept by
seeing where the line crosses the y−axis on the graph. The value of b is the y−coordinate of this point.
Example 7
Write an equation in slope-intercept form that represents the following line:


First find the slope of the line. You already know how to do this using the slope formula. There are no
points given on the line, so you have to pick your own points. See where the line goes right through an
intersection (corner point) on the graph paper. You can use the two points (0, 3) and (2, 2) :
x1 = , y1 = , and x2 = , y2 =


slope = y2 − y1
x2 − x1


slope = 2 − 3
2 − 0 =


−1
2


= −1
2


The slope of the line is −12 . This will replace m in the slope-intercept equation.
Now you need to find the y−intercept. On the graph, find where the line intersects the y−axis. It crosses
the y−axis at (0, 3) so the y−intercept is 3. This will replace b in the slope-intercept equation, so now
you have all the information you need.
The equation for the line shown in the graph is: y = −12 x + 3.


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• In the slope-intercept equation, the slope is represented by the letter _________.
• In the slope-intercept equation, the y−intercept is the letter _________.


Equations of Parallel Lines
You studied parallel lines and their graphical relationships, so now you will learn how to easily identify
equations of parallel lines. When looking for parallel lines, look for equations that have the same slope.
As long as the y−intercepts are not the same and the slopes are equal, the lines are parallel. If the
y−intercept and the slope are both the same, then the two equations are for the same exact line, and a
line cannot be parallel to itself.


• Parallel lines have the _______________________ slope.


Reading Check:
1. True or false:
The reciprocal of a fraction is when you flip the numerator and the denominator.
2. Make up an example that supports the statement in #1 above.


3. What is the slope-intercept form of an equation?


4. What do the letters m and b stand for in the slope-intercept equation?
m :


b :


5. How are the slopes of parallel lines related?


Example 8
Juan drew the line below:


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Which of the following equations could represent a line parallel to the one Juan drew?
A. y = −12 x − 6
B. y = 12 x + 9
C. y = −2x − 18
D. y = 2x + 1
If you find the slope of the line in Juan’s graph, you can find the slope of a parallel line because it will
be the same. Pick two points on the graph and find the slope using the slope formula. Use the points (0,
5) and (1, 3) :
x1 = , y1 = , and x2 = , y2 =


slope = y2 − y1
x2 − x1


slope = 3 − 5
1 − 0 =


−2
1


= −2


The slope of Juan’s line is –2. Look at your four answer choices: which equation has a slope of –2? All
other parts of the equation do not matter. The only equation that has a slope of –2 is choice C, so that is
the correct answer.


Equations of Perpendicular Lines
You also studied perpendicular lines and their graphical relationships: remember that the slopes of
perpendicular lines are opposite reciprocals. To easily identify equations of perpendicular lines, look
for equations that have slopes that are opposite reciprocals of each other.
Here, it does not matter what the y−intercept is; as long as the slopes are opposite reciprocals, the lines
are perpendicular.
Example 9
Kara drew the line in this graph:


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Which of the following equations could represent a line perpendicular to the one Kara drew above?
A. y = 32 x + 10
B. y = −32 x + 6
C. y = 23 x − 4
D. y = −23 x − 1
First find the slope of the line in Kara’s graph. Then find the opposite reciprocal of this slope. To begin,
pick two points on the graph and calculate the slope using the slope formula. Use the points (0, 2) and
(3, 4) :
x1 = , y1 = , and x2 = , y2 =


slope = y2 − y1
x2 − x1


slope = 4 − 2
3 − 0 =


2
3


The slope of Kara’s line on the graph is 23 .
Find the opposite reciprocal: the reciprocal of 23 is 32 , and the opposite of 32 is −32 .
So, −32 is the opposite reciprocal of (or perpendicular slope to) 23 .
Now look in your answer choices for the equation that has a slope of −32 .
The only equation that has a slope of −32 is choice B, so that is the correct answer.
Reading Check:
1. How are the slopes of perpendicular lines related to each other?


2. In the context of perpendicular slope values, what does opposite mean?


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3. True or false: On a graph, perpendicular lines intersect at an angle of 45◦.
4. Correct the statement in #3 above to make it true.


2.4 Equation of a Circle
Learning Objectives


• Write the equation of a circle.


Equations and Graphs of Circles
A circle is defined as the set of all points that are the same distance from a single point called the center.
This definition can be used to find an equation of a circle in the coordinate plane.


• A circle is the set of all points equidistant from the ________________________.


Look at the circle shown below. As you can see, this circle has its center at the point (2, 2) and it has a
radius of 3.


All of the points (x, y) on the circle are a distance of 3 units away from the center of the circle.


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We can express this information as an equation with the help of the Pythagorean Theorem. The right
triangle shown above has legs of lengths (x − 2) and (y − 2), and hypotenuse of length 3. We can write:


(x − 2)2 + (y − 2)2 = 32 or
(x − 2)2 + (y − 2)2 = 9


We can generalize this equation for a circle with center at point (x0, y0) and radius r:


(x − x0)2 + (y − y0)2 = r2


Example 1
Find the center and radius of the following circles:
A. (x − 4)2 + (y − 1)2 = 25
B. (x + 1)2 + (y − 2)2 = 4
A. We rewrite the equation as: (x − 4)2 + (y − 1)2 = 52. Compare this to the standard equation. The
center of the circle is at the point (4, 1) and the radius is 5.
B. We rewrite the equation as: (x − (−1))2 + (y − 2)2 = 22. The center of the circle is at the point (–1,
2) and the radius is 2.
Example 2
Graph the following circles:
A. x2 + y2 = 9
B. (x + 2)2 + y2 = 1
In order to graph a circle, we first graph the center point and then draw points that are the length of the
radius away from the center in the directions up, down, right, and left. Then connect the outer points
in a smooth circle!
A. We rewrite the equation as: (x − 0)2 + (y − 0)2 = 32. The center of the circle is at the point (0, 0) and
the radius is 3.
Plot the center point and a point 3 units up at (0, 3), 3 units down at (0, –3), 3 units right at (3, 0) and
3 units left at (–3, 0) :


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B. We rewrite the equation as: (x − (−2))2 + (y − 0)2 = 12. The center of the circle is at the point (–2, 0)
and the radius is 1.
Plot the center point and a point 1 unit up at (–2, 1), 1 unit down at (–2, –1), 1 unit right at (–1, 0) and
1 unit left at (–3, 0) :


Reading Check:
1. In your own words, describe the radius of a circle.


2. In the general equation of a circle (x − x0)2 + (y − y0)2 = r2, the variables x0 and y0 stand for a special
point. What point is this?


3. How can you find the radius of a circle from its equation? What do you need to do to the right side of
the equation?


Example 3
Write the equation of the circle in the graph below:


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From the graph, we can see that the center of the circle is at the point (–2, 2) and the radius is 3 units
long, so we use these numbers in the standard circle equation:


(x + 2)2 + (y − 2)2 = 32


(x + 2)2 + (y − 2)2 = 9


Example 4
Determine if the point (1, 3) is on the circle given by the equation:


(x − 1)2 + (y + 1)2 = 16


In order to find the answer, we simply plug the point (1, 3) into the equation of the circle given.
Substitute the number _________ for x and the number _________ for y:


(1 − 1)2 + (3 + 1)2 = 16
(0)2 + (4)2 = 16


16 = 16


Since we end up with a true statement, the point (1, 3) satisfies the equation. Therefore, the point is on
the circle.


Concentric Circles
Concentric circles are circles of different radii that share the same center point.


• Circles with the same ___________________ but different _________________-
are called concentric circles.


Example 5


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Write the equations of the concentric circles shown in the graph:


All 4 circles have the same center point at (3, 2) so we know the equations will all be:


(x − 3)2 + (y − 2)2


Since the circles have different radius lengths, the right side of the equations will all be different numbers.
The smallest circle has a radius of 2:


(x − 3)2 + (y − 2)2 = 22 or
(x − 3)2 + (y − 2)2 = 4


The next larger circle has a radius of 3: (x − 3)2 + (y − 2)2 = 9
The next larger circle has a radius of 4: (x − 3)2 + (y − 2)2 = 16
The largest circle has a radius of 5: (x − 3)2 + (y − 2)2 = 25
Look at the word concentric:
In Spanish, the word “con” means “with.”
The second part of the word, “-centric” looks very similar to the word “center.”
When we put these two parts together, “concentric” means “with” the same “center.”
Reading Check:
1. If you are given a point and an equation of a circle, how can you tell if the given point is on the circle?
Describe what you would do.


2. What are concentric circles?


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3. If you are given two equations of two different circles, how can you tell if the circles are concentric?
Describe what the two equations would have to have in common.


2.5 Translating and Reflecting
Learning Objectives


• Graph a translation in a coordinate plane.
• Recognize that a translation is an isometry.
• Find the reflection of a point in a line on a coordinate plane.
• Verify that a reflection is an isometry.


Translations
A translation moves every point a given horizontal distance and/or a given vertical distance.


• When a point is moved a certain distance horizontally and/or vertically, the move is called a ___-
___________________________.


For example, if a translation moves point A(3, 7) 2 units to the right and 4 units up to A′(5, 11), then this
translation moves every point in a larger figure the same way.
The symbol next to the letter A′ above is called the prime symbol.
The prime symbol looks like an apostrophe like you may use to show possessive, such as, “that is my
brother’s book.”
(The apostrophe is before the s in brother’s)
In math, we use the prime symbol to show that two things are related.
In the translation above, the original point is related to the translated point, so instead of renaming the
translated point, we use the prime symbol to show this.
The original point (or figure) is called the preimage and the translated point (or figure) is called the
image. In the example given above, the preimage is point A(3, 7) and the image is point A′(5, 11). The
image is designated (or shown) with the prime symbol.


• Another name for the original point is the __________________________.
• Another name for the translated point is the __________________________.
• The translated point uses the _______________ symbol next to its naming letter.


Example 1
The point A(3, 7) in a translation becomes the point A′(2, 4). What is the image of B(−6, 1) in the same
translation?


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Point A moved 1 unit to the left and 3 units down to get to A′. Point B will also move 1 unit to the left
and 3 units down.
We subtract 1 from the x−coordinate and 3 from the y−coordinate of point B:


B′ = (−6 − 1, 1 − 3) = (−7,−2)


B′(−7,−2) is the image of B(−6, 1).
Using the Distance Formula, you can notice the following:


AB =


(−6 − 3)2 + (1 − 7)2 =


(−9)2 + (−6)2 =

117


A′B′ =


(−7 − 2)2 + (−2 − 4)2 =


(−9)2 + (−6)2 =

117


Since the endpoints of AB and A′B′ moved the same distance horizontally and vertically, both segments
have the same length.


Translation is an Isometry
An isometry is a transformation in which distance is “preserved.” This means that the distance between
any two points in the preimage (before the translation) is the same as the distance between the points
in the image (after the translation).


• An isometry is when ______________________________ is preserved from the preim-
age to the image.


As you saw in Example 1 above:
The preimage AB = the image A′B′ (since they are both equal to



117)


Would we get the same result for any other point in this translation? The answer is yes. It is clear that
for any point X, the distance from X to X′ will be



117. Every point moves



117 units to its image.


This is true in general:
Translation Isometry Theorem
Every translation in the coordinate plane is an isometry.


• Every translation in an x − y coordinate plane is an ________________________.


Reflection in a Line
A reflection in a line is as if the line were a mirror:


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• When an object is reflected in a line, the line is like a ______________________.


An object reflects in the mirror, and we see the image of the object.


• The image is the same distance behind the mirror line as the object is in front of the mirror line.
• The “line of sight” from the object to the mirror is perpendicular to the mirror line itself.
• The “line of sight” from the image to the mirror is also perpendicular to the mirror line.


Reflection of a Point in a Line
Point P′ is the reflection of point P in line k if and only if line k is the perpendicular bisector of PP′.


• The mirror line is a perpendicular ____________________________ of the line that
connects the object to its reflected image.


Reflections in Special Lines
In a coordinate plane there are some “special” lines for which it is relatively easy to create reflections:


• the x−axis
• the y−axis
• the line y = x (this line makes a 45◦ angle between the x−axis and the y−axis)
• The _________-axis, the _________-axis, and the line _________ = _________-


are “special” lines to use as mirrors when finding reflections of figures.


We can develop simple formulas for reflections in these lines.
Let P(x, y) be a point in the coordinate plane:


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We now have the following reflections of P(x, y) :


• Reflection of P in the x−axis is Q(x, –y)


[ the x−coordinate stays the same, and the y−coordinate is opposite]


• Reflection of P in the y−axis is R(−x, y)


[ the x−coordinate is opposite, and the y−coordinate stays the same ]


• Reflection of P in the line y = x is S (y, x)


[ switch the x−coordinate and the y−coordinate ]
Look at the graph above and you will be convinced of the first two reflections in the axes. We will prove
the third reflection in the line y = x on the next page.


• Reflections in the x−axis have the same ________-coordinate, but the y−coordinate has the
_________________________ value.


• Reflections in the y−axis have an ___________________________ x−coordinate, and
the y−coordinate stays the ____________________.


• For reflections in the y = x line, __________________ the x− and y−coordinates.


Example 2
Prove that the reflection of point P(h, k) in the line y = x is the point S (k, h).
Here is an “outline” proof:
First, we know the slope of the line y = x is 1 because y = 1x + 0.
Next, we will investigate the slope of the line that connects our two points, PS . Use the slope formula
and the values of the points’ coordinates given above:
Slope of PS is k−hh−k =


−1(h−k)
h−k = −1


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Therefore, we have just shown that PS and y = x are perpendicular because the product of their slopes
is –1.
Finally, we can show that y = x is the perpendicular bisector of PS by finding the midpoint of PS :
Midpoint of PS is


(


h+k
2 ,


h+k
2


)


We know the midpoint of PS is on the line y = x because the x−coordinate and the y−coordinate of the
midpoint are the same.
Therefore, the line y = x is the perpendicular bisector of PS .
Conclusion: The points P and S are reflections in the line y = x.
Example 3
Point P(5, 2) is reflected in the line y = x. The image is P′. P′ is then reflected in the y−axis. The image
is P′′. What are the coordinates of P′′?
We find one reflection at a time:


• Reflect P in the line y = x to find P′ :


For reflections in the line y = x we ___________________ coordinates.
Therefore, P′ is (2, 5).


• Reflect P′ in the y−axis:


For reflections in the y−axis, the x−coordinate is _____________________ and the y−coordinate
stays the _______________________.
Therefore, P′′ is (–2, 5).


Reflections Are Isometries
Like a translation, a reflection in a line is also an isometry. Distance between points is “preserved”
(stays the same).


• A reflection in a line is an __________________________, which means that distance
is preserved.


We will verify the isometry for reflection in the x−axis. The proof is very similar for reflection in the
y−axis.
The diagram below shows PQ and its reflection in the x−axis, P′Q′ :


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Use the Distance Formula:


PQ =


(m − h)2 + (n − k)2


P′Q′ =


(m − h)2 + (−n − (−k))2 =


(m − h)2 + (k − n)2


=


(m − h)2 + (n − k)2


So PQ = P′Q′


Conclusion: When a segment is reflected in the x−axis, the image segment has the same length as the
original preimage segment. This is the meaning of isometry. You can see that a similar argument would
apply to reflection in any line.
Reading Check:
1. True or false: Both translations and reflections are isometries.
2. What is the meaning of the statement in #1 above?


3. If a translation rule is (x + 3, y − 1), in which directions is a point moved?


4. When a point or figure is reflected in a line, that line acts as a mirror.
a. How does the x−axis change a point that is reflected? What do you do to the coordinates of the point in
this type of reflection?


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b. How does the y−axis change a point that is reflected? What do you do to the coordinates of the point in
this type of reflection?


c. How does the line y = x change a point that is reflected? What do you do to the coordinates of the point
in this type of reflection?


2.6 Rotating
Learning Objectives


• Find the image of a point in a rotation in a coordinate plane.
• Recognize that a rotation is an isometry.


Sample Rotations


In this lesson we will study rotations centered at the origin of a coordinate plane. We begin with some
specific examples of rotations. Later we will see how these rotations fit into a general formula.
We define a rotation as follows: In a rotation centered at the origin with an angle of rotation of n◦, a
point moves counterclockwise along an arc of a circle. The central angle of the circle measures n◦.
The original preimage point is one endpoint of the arc, and the image of the original point is the other
endpoint of the arc:


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• Rotations centered at the origin move points ______________________ along an arc of
a circle.


• For a rotation of n◦, the central angle of the circle measures ____________.
• The preimage point is one endpoint of the _________________ and the image is the other


endpoint.


180◦ Rotation
Our first example is rotation through an angle of 180◦:


In a 180◦ rotation, the image of P(h, k) is the point P′(−h,−k).
Notice:


• P and P′ are the endpoints of a diameter of a circle.


→ This means that the distance from the point P to the origin (or the distance from the point P′ to the
origin) is a radius of the circle.


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The distance from P to the origin equals the distance from ________ to the origin.


• The rotation is the same as a “reflection in the origin.”


→This means that if we use the origin as a mirror, the point P is directly across from the point P′.
A 180◦ _____________________________ is also a reflection in the origin.
In a rotation of 180◦, the x−coordinate and the y−coordinate of the __________________ become
the negative versions of the values in the image.
A 180◦ rotation is an isometry. The image of a segment is a congruent segment:


Use the Distance Formula:


PQ =


(k − t)2 + (h − r)2


P′Q′ =


(−k − (−t))2 + (−h − (−r))2 =


(−k + t)2 + (−h + r)2


=


(t − k)2 + (r − h)2


=


(k − t)2 + (h − r)2


So PQ = P′Q′


• A 180◦ rotation is an ___________________________, so distance is preserved.
• When a segment is rotated 180◦ (or reflected in the origin), its image is a ____________-


_____________________ segment.


90◦ Rotation
The next example is a rotation through an angle of 90◦. The rotation is in the counterclockwise direction:


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In a 90◦ rotation, the image of P(h, k) is the point P′(−k, h).
Notice:


• PO and P′O are both radii of the same circle, so PO = P′O.


If PO and P′O are both radii, then they are the same ______________________.


• ∠POP′ is a right angle.
• The acute angle formed by PO and the x−axis and the acute angle formed by P′O and the x−axis


are complementary angles.


Remember, complementary angles add up to ________________◦.
You can see by the coordinates of the preimage and image points, in a 90◦ rotation:


• the x− and y−coordinates are switched AND
• the x−coordinate is negative.


In a 90◦ rotation, switch the _________- and _________-coordinates and make the new x−coordinate
_________________________.
A 90◦ rotation is an isometry. The image of a segment is a congruent segment.


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Use the Distance Formula:


PQ =


(k − t)2 + (h − r)2


P′Q′ =


(h − r)2 + (−k − (−t))2 =


(h − r)2 + (t − k)2


=


(k − t)2 + (h − r)2


So PQ = P′Q′


Reading Check:
Which of the following are isometries? Circle all that apply:


30◦rotation 45◦rotation 60◦rotation
90◦rotation 150◦rotation 180◦rotation
Re f lection Translation Bisection


Example 1
What are the coordinates of the vertices of ∆ABC in a rotation of 90◦?


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Point A is (4, 6), B is (–4, 2), and C is (6, –2).
In a 90◦ rotation, the x−coordinate and the y−coordinate are switched AND the new x−coordinate is
made negative:


• A becomes A′ : switch x and y to (6, 4) and make x negative (–6, 4)
• B becomes B′ : switch x and y to (2, –4) and make x negative (–2, –4)
• C becomes C′ : switch x and y to (–2, 6) and make x negative (−(−2), 6) = (2, 6)


So the vertices of ∆A′B′C′ are (–6, 4), (–2, –4), and (2, 6).
Plot each of these points on the coordinate plane above and draw in each side of the new rotated triangle.
Can you see how ∆ABC is rotated 90◦ to ∆A′B′C′?
Reading Check:
1. True or false: A rotation is always in the counterclockwise direction.
2. On the coordinate plane below, create a point anywhere you like, and label it P.
Then draw a second point W that is the image of point P rotated 180◦.


3. On the coordinate plane above, draw a third point R that is the image of your original point P rotated
90◦.


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4. Is a 90◦ rotation an isometry? Explain.


5. Is a 180◦ rotation an isometry? Explain.


6. What type of rotation is the same as a reflection in the origin?


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Chapter 3


Triangles and Congruence


In this chapter, you will learn all about triangles. First, we will find out how many degrees are in a triangle
and other properties of the angles within a triangle. Second, we will use that information to determine if
two different triangles are congruent. Finally, we will investigate the properties of isosceles and equilateral
triangles.


3.1 Triangle Sums
Learning Objectives


• Understand the Triangle Sum Theorem.
• Identify interior and exterior angles in a triangle.
• Use the Exterior Angle Theorem.


Review Queue
Classify the triangles below by their angles and sides.


1.


2.


3.


4. Draw and label a straight angle, ∠ABC. Which point is the vertex? How many degrees does a straight
angle have?


Know What? To the right is the Bermuda Triangle. The myth of this triangle is that ships and planes
have passed through and mysteriously disappeared.


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The measurements of the sides of the triangle are in the picture. Classify the Bermuda triangle by its sides
and angles. Then, using a protractor, find the measure of each angle. What do they add up to?


Recall that a triangle can be classified by its sides...


and its angles...


Interior Angles: The angles inside of a polygon.
Vertex: The point where the sides of a polygon meet.


Triangles have three interior angles, three vertices, and three sides.
A triangle is labeled by its vertices with a 4. This triangle can be labeled 4ABC, 4ACB, 4BCA, 4BAC, 4CBA
or 4CAB.
Triangle Sum Theorem The interior angles in a polygon are measured in degrees. How many degrees
are there in a triangle?
Investigation 4-1: Triangle Tear-Up


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Tools Needed: paper, ruler, pencil, colored pencils


1. Draw a triangle on a piece of paper. Make all three angles different sizes. Color the three interior
angles three different colors and label each one, ∠1, ∠2, and ∠3.


2. Tear off the three colored angles, so you have three separate angles.


3. Line up the angles so the vertices points all match up. What happens? What measure do the three
angles add up to?


This investigation shows us that the sum of the angles in a triangle is 180◦ because the three angles fit
together to form a straight angle where all the vertices meet.
Triangle Sum Theorem: The interior angles of a triangle add up to 180◦.


m∠1 + m∠2 + m∠3 = 180◦


Example 1: What m∠T?


Solution: Set up an equation.


m∠M + m∠A + m∠T = 180◦


82◦ + 27◦ + m∠T = 180◦


109◦ + m∠T = 180◦


m∠T = 71◦


Even thought Investigation 4-1 is a way to show that the angles in a triangle add up to 180◦, it is not a
proof. Here is the proof of the Triangle Sum Theorem.


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Given: 4ABC with ←→AD||BC
Prove: m∠1 + m∠2 + m∠3 = 180◦


Table 3.1:


Statement Reason
1. 4ABC above with ←→AD||BC Given
2. ∠1 ∠4, ∠2 ∠5 Alternate Interior Angles Theorem
3. m∠1 = m∠4, m∠2 = m∠5 angles have = measures
4. m∠4 + m∠CAD = 180◦ Linear Pair Postulate
5. m∠3 + m∠5 = m∠CAD Angle Addition Postulate
6. m∠4 + m∠3 + m∠5 = 180◦ Substitution PoE
7. m∠1 + m∠3 + m∠2 = 180◦ Substitution PoE


Example 2: What is the measure of each angle in an equiangular triangle?


Solution: 4ABC is an equiangular triangle, where all three angles are equal. Write an equation.


m∠A + m∠B + m∠C = 180◦


m∠A + m∠A + m∠A = 180◦ S ubstitute, all angles are equal.
3m∠A = 180◦ Combine like terms.
m∠A = 60◦


If m∠A = 60◦, then m∠B = 60◦ and m∠C = 60◦.
Each angle in an equiangular triangle is 60◦.
Example 3: Find the measure of the missing angle.


Solution: m∠O = 41◦ and m∠G = 90◦ because it is a right angle.


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m∠D + m∠O + m∠G = 180◦


m∠D + 41◦ + 90◦ = 180◦


m∠D + 41◦ = 90◦


m∠D = 49◦


Notice that m∠D + m∠O = 90◦.
The acute angles in a right triangle are always complementary.


Exterior Angles
Exterior Angle: The angle formed by one side of a polygon and the extension of the adjacent side.
In all polygons, there are two sets of exterior angles, one that goes around clockwise and the other goes
around counterclockwise.


Notice that the interior angle and its adjacent exterior angle form a linear pair and add up to 180◦.


m∠1 + m∠2 = 180◦


Example 4: Find the measure of ∠RQS .


Solution: 112◦ is an exterior angle of 4RQS and is supplementary to ∠RQS .


112◦ + m∠RQS = 180◦


m∠RQS = 68◦


Example 5: Find the measure of the numbered interior and exterior angles in the triangle.


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Solution: m∠1 + 92◦ = 180◦ by the Linear Pair Postulate. m∠1 = 88◦


m∠2 + 123◦ = 180◦ by the Linear Pair Postulate. m∠2 = 57◦


m∠1 + m∠2 + m∠3 = 180◦ by the Triangle Sum Theorem.
88◦ + 57◦ + m∠3 = 180


m∠3 = 35◦


Lastly, m∠3 + m∠4 = 180◦ by the Linear Pair Postulate.
35◦ + m∠4 = 180◦


m∠4 = 145◦


In Example 5, the exterior angles are 92◦, 123◦, and 145◦. Adding these angles together, we get 92◦ +
123◦ + 145◦ = 360◦. This is true for any set of exterior angles for any polygon.
Exterior Angle Sum Theorem: The exterior angles of a polygon add up to 360◦.


m∠1 + m∠2 + m∠3 = 360◦


m∠4 + m∠5 + m∠6 = 360◦


Example 6: What is the value of p in the triangle below?


Solution: First, we need to find the missing exterior angle, let’s call it x. Set up an equation using the
Exterior Angle Sum Theorem.


130◦ + 110◦ + x = 360◦


x = 360◦ − 130◦ − 110◦


x = 120◦


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x and p add up to 180◦ because they are a linear pair.


x + p = 180◦


120◦ + p = 180◦


p = 60◦


Example 7: Find m∠A.


Solution:


m∠ACB + 115◦ = 180◦ because they are a linear pair
m∠ACB = 65◦


m∠A + 65◦ + 79◦ = 180◦ by the Triangle Sum Theorem
m∠A = 36◦


Remote Interior Angles: The two angles in a triangle that are not adjacent to the indicated exterior
angle.
In Example 7 above, ∠A and 79◦ are the remote interior angles relative to 115◦.
Exterior Angle Theorem From Example 7, we can find the sum of m∠A and m∠B, which is 36◦ + 79◦ =
115◦. This is equal to the exterior angle at C.
Exterior Angle Theorem: The sum of the remote interior angles is equal to the non-adjacent exterior
angle.


m∠A + m∠B = m∠ACD


Proof of the Exterior Angle Theorem
Given: Triangle with exterior ∠4
Prove: m∠1 + m∠2 = m∠4


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Table 3.2:


Statement Reason
1. Triangle with exterior ∠4 Given
2. m∠1 + m∠2 + m∠3 = 180◦ Triangle Sum Theorem
3. m∠3 + m∠4 = 180◦ Linear Pair Postulate
4. m∠1 + m∠2 + m∠3 = m∠3 + m∠4 Transitive PoE
5. m∠1 + m∠2 = m∠4 Subtraction PoE


Example 8: Find m∠C.


Solution: Using the Exterior Angle Theorem


m∠C + 16◦ = 121◦


m∠TCA = 105◦


If you forget the Exterior Angle Theorem, you can do this problem just like Example 7.
Example 9: Algebra Connection Find the value of x and the measure of each angle.


Solution: All the angles add up to 180◦.


(8x − 1)◦ + (3x + 9)◦ + (3x + 4)◦ = 180◦


(14x + 12)◦ = 180◦


14x = 168◦


x = 12◦


Substitute in 12◦ for x to find each angle.


3(12◦) + 9◦ = 45◦ 3(12◦) + 4◦ = 40◦ 8(12◦) − 1◦ = 95◦


Example 10: Algebra Connection Find the value of x and the measure of each angle.


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Solution: Set up an equation using the Exterior Angle Theorem.


(4x + 2)◦ + (2x − 9)◦ = (5x + 13)◦


↑ ↗ ↑
interior angles exterior angle


(6x − 7)◦ = (5x + 13)◦


x = 20◦


Substitute in 20◦ for x to find each angle.


4(20◦) + 2◦ = 82◦ 2(20◦) − 9◦ = 31◦ Exterior angle: 5(20◦) + 13◦ = 113◦


Know What? Revisited The Bermuda Triangle is an acute scalene triangle. The actual angle measures
are in the picture to the right. Your measured angles should be within a degree or two of these measures
and should add up to 180◦. However, because your measures are estimates using a protractor, they might
not exactly add up.


Review Questions
• Questions 1-16 are similar to Examples 1-8.
• Questions 17 and 18 use the definition of an Exterior Angle and the Exterior Angle Sum Theorem.
• Question 19 is similar to Example 3.
• Questions 20-27 are similar to Examples 9 and 10.


Determine m∠1.


1.


2.


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3.


4.


5.


6.


7.


8.


9.


10.


11.


12.


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13.


14.


15.


16. Find the lettered angles, a − f , in the picture to the right. Note that the two lines are parallel.


17. Draw both sets of exterior angles on the same triangle.


(a) What is m∠1 + m∠2 + m∠3?
(b) What is m∠4 + m∠5 + m∠6?
(c) What is m∠7 + m∠8 + m∠9?
(d) List all pairs of congruent angles.


18. Fill in the blanks in the proof below.
Given: The triangle to the right with interior angles and exterior angles.
Prove: m∠4 + m∠5 + m∠6 = 360◦


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Table 3.3:


Statement Reason
1. Triangle with interior and exterior angles. Given
2. m∠1 + m∠2 + m∠3 = 180◦
3. ∠3 and ∠4 are a linear pair, ∠2 and ∠5 are a
linear pair, and ∠1 and ∠6 are a linear pair
4. Linear Pair Postulate (do all 3)
5. m∠1 + m∠6 = 180◦
m∠2 + m∠5 = 180◦
m∠3 + m∠4 = 180◦
6. m∠1 + m∠6 + m∠2 + m∠5 + m∠3 + m∠4 = 540◦
7. m∠4 + m∠5 + m∠6 = 360◦


19. Fill in the blanks in the proof below.
Given: 4ABC with right angle B.
Prove: ∠A and ∠C are complementary.


Table 3.4:


Statement Reason
1. 4ABC with right angle B. Given
2. Definition of a right angle
3. m∠A + m∠B + m∠C = 180◦
4. m∠A + 90◦ + m∠C = 180◦
5.
6. ∠A and ∠C are complementary


Algebra Connection Solve for x.


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20.


21.


22.


23.


24.


25.


26.


27.


Review Queue Answers
1. acute isosceles
2. obtuse scalene
3. right scalene
4. B is the vertex, 180◦,


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3.2 Congruent Figures
Learning Objectives


• Define congruent triangles and use congruence statements.
• Understand the Third Angle Theorem.


Review Queue
What part of each pair of triangles are congruent? Write out each congruence statement for the marked
congruent sides and angles.


1.


2.


3. Determine the measure of x.


(a)


(b) What is the measure of each angle?
(c) What type of triangle is this?


Know What? Quilt patterns are very geometrical. The pattern to the right is made up of several
congruent figures. In order for these patterns to come together, the quilter rotates and flips each block (in
this case, a large triangle, smaller triangle, and a smaller square) to get new patterns and arrangements.
How many different sets of colored congruent triangles are there? How many triangles are in each set?
How do you know these triangles are congruent?


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Congruent Triangles
Two figures are congruent if they have exactly the same size and shape.
Congruent Triangles: Two triangles are congruent if the three corresponding angles and sides are
congruent.


4ABC and 4DEF are congruent because


AB DE ∠A ∠D


BC EF and ∠B ∠E
AC DF ∠C ∠F


When referring to corresponding congruent parts of congruent triangles it is called Corresponding Parts
of Congruent Triangles are Congruent, or CPCTC.
Example 1: Are the two triangles below congruent?


Solution: To determine if the triangles are congruent, match up sides with the same number of tic marks:
BC MN, AB LM, AC LN.


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Next match up the angles with the same markings:
∠A ∠L, ∠B ∠M, and ∠C ∠N.
Lastly, we need to make sure these are corresponding parts. To do this, check to see if the congruent angles
are opposite congruent sides. Here, ∠A is opposite BC and ∠L is opposite MN. Because ∠A ∠L and
BC MN, they are corresponding. Doing this check for the other sides and angles, we see that everything
matches up and the two triangles are congruent.


Creating Congruence Statements
In Example 1, we determined that 4ABC and 4LMN are congruent. When stating that two triangles are
congruent, the corresponding parts must be written in the same order. Using Example 1, we would have:


Notice that the congruent sides also line up within the congruence statement.
AB LM, BC MN, AC LN


We can also write this congruence statement five other ways, as long as the congruent angles match up.
For example, we can also write 4ABC 4LMN as:


4ACB 4LNM 4BCA 4MNL 4BAC 4MLN
4CBA 4NML 4CAB 4NLM


Example 2: Write a congruence statement for the two triangles below.


Solution: Line up the corresponding angles in the triangles:
∠R ∠F, ∠S ∠E, and ∠T ∠D.
4RST ∠FED
Example 3: If 4CAT 4DOG, what else do you know?
Solution: From this congruence statement, we know three pairs of angles and three pairs of sides are
congruent.


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Third Angle Theorem
Example 4: Find m∠C and m∠J.


Solution: The sum of the angles in a triangle is 180◦.


4ABC : 35◦ + 88◦ + m∠C = 180◦


m∠C = 57◦


4HIJ : 35◦ + 88◦ + m∠J = 180◦


m∠J = 57◦


Notice we were given m∠A = m∠H and m∠B = m∠I and we found out m∠C = m∠J. This can be generalized
into the Third Angle Theorem.
Third Angle Theorem: If two angles in one triangle are congruent to two angles in another triangle,
then the third pair of angles must also congruent.
If ∠A ∠D and ∠B ∠E, then ∠C ∠F.


Example 5: Determine the measure of the missing angles.


Solution: From the Third Angle Theorem, we know ∠C ∠F.


m∠A + m∠B + m∠C = 180◦


m∠D + m∠B + m∠C = 180◦


42◦ + 83◦ + m∠C = 180◦


m∠C = 55◦ = m∠F


Congruence Properties Recall the Properties of Congruence from Chapter 2. They will be very useful
in the upcoming sections.


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Reflexive Property of Congruence: AB AB or 4ABC 4ABC
Symmetric Property of Congruence: ∠EFG ∠XYZ and ∠XYZ ∠EFG


4ABC 4DEF and 4DEF 4ABC
Transitive Property of Congruence: 4ABC 4DEF and 4DEF 4GHI then


4ABC 4GHI


These three properties will be very important when you begin to prove that two triangles are congruent.
Example 6: In order to say that 4ABD 4ABC, you must show the three corresponding angles and sides
are congruent. Which pair of sides is congruent by the Reflexive Property?


Solution: The side AB is shared by both triangles. In a geometric proof, AB AB by the Reflexive
Property.
Know What? Revisited The 16 “A” triangles are congruent. The 16 “B” triangles are also congruent.
The quilt pattern is made from dividing up the entire square into smaller squares. Both the “A” and “B”
triangles are right triangles.


Review Questions
• Questions 1 and 2 are similar to Example 3.
• Questions 3-12 are a review and use the definitions and theorems explained in this section.
• Questions 13-17 are similar to Example 1 and 2.
• Question 18 the definitions and theorems explained in this section.
• Questions 19-22 are similar to Examples 4 and 5.
• Question 23 is a proof of the Third Angle Theorem.
• Questions 24-28 are similar to Example 6.
• Questions 29 and 30 are investigations using congruent triangles, a ruler and a protractor.


1. If 4RAT 4UGH, what is also congruent?
2. If 4BIG 4TOP, what is also congruent?


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For questions 3-7, use the picture to the right.


3. What theorem tells us that ∠FGH ∠FGI?
4. What is m∠FGI and m∠FGH? How do you know?
5. What property tells us that the third side of each triangle is congruent?
6. How does FG relate to ∠IFH?
7. Write the congruence statement for these two triangles.


For questions 8-12, use the picture to the right.


8. AB||DE, what angles are congruent? How do you know?
9. Why is ∠ACB ∠ECD? It is not the same reason as #8.


10. Are the two triangles congruent with the information you currently have? Why or why not?
11. If you are told that C is the midpoint of AE and BD, what segments are congruent?
12. Write a congruence statement.


For questions 13-16, determine if the triangles are congruent. If they are, write the congruence statement.


13.


14.


15.


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16.


17. Suppose the two triangles to the right are congruent. Write a congruence statement for these triangles.


18. Explain how we know that if the two triangles are congruent, then ∠B ∠Z.


For questions 19-22, determine the measure of all the angles in the each triangle.


19.


20.


21.


22.


23. Fill in the blanks in the Third Angle Theorem proof below.
Given: ∠A ∠D, ∠B ∠E
Prove: ∠C ∠F


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Table 3.5:


Statement Reason
1. ∠A ∠D, ∠B ∠E
2. angles have = measures
3. m∠A + m∠B + m∠C = 180◦
m∠D + m∠E + m∠F = 180◦
4. Substitution PoE
5. Substitution PoE
6. m∠C = m∠F
7. ∠C ∠F


For each of the following questions, determine if the Reflexive, Symmetric or Transitive Properties of
Congruence is used.


24. ∠A ∠B and ∠B ∠C, then ∠A ∠C
25. AB AB
26. 4XYZ 4LMN and 4LMN 4XYZ
27. 4ABC 4BAC
28. What type of triangle is 4ABC in #27? How do you know?


Review Queue Answers
1. ∠B ∠H, AB GH, BC HI
2. ∠C ∠M, BC LM
3. The angles add up to 180◦


(a) (5x + 2)◦ + (4x + 3)◦ + (3x − 5)◦ = 180◦
12x = 180◦


x = 15◦
(b) 77◦, 63◦, 40◦
(c) acute scalene


3.3 Triangle Congruence using SSS and SAS
Learning Objectives


• Use the distance formula to analyze triangles on the x − y plane.
• Apply the SSS and SAS Postulate to show two triangles are congruent.


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Review Queue
1. Use the distance formula,




(x2 − x1)2 + (y2 − y1)2 to find the distance between the two points.
(a) (-1, 5) and (4, 12)
(b) (-6, -15) and (-3, 8)


2.


(a) If we know that AB||CD, AD||BC, what angles are congruent? By which theorem?
(b) Which side is congruent by the Reflexive Property?
(c) Is this enough to say 4ADC 4CBA?


3.


(a) If we know that B is the midpoint of AC and DE, what segments are congruent?
(b) Are there any angles that are congruent by looking at the picture? Which ones and why?
(c) Is this enough to say 4ABE 4CBD?


Know What?
The “ideal” measurements in a kitchen from the sink, refrigerator and oven are as close to an equilateral
triangle as possible. Your parents are remodeling theirs to be as close to this as possible and the measure-
ments are in the picture at the left, below. Your neighbor’s kitchen has the measurements on the right.
Are the two triangles congruent? Why or why not?


SSS Postulate of Triangle Congruence
Consider the question: If I have three lengths: 3 in, 4 in, and 5 in, can I construct more than one triangle?
Investigation 4-2: Constructing a Triangle Given Three Sides
Tools Needed: compass, pencil, ruler, and paper
1. Draw the longest side (5 in) horizontally, halfway down the page.


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The drawings in this investigation are to scale.
2. Take the compass and, using the ruler, widen the compass to measure 4 in, the second side.


3. Using the measurement from Step 2, place the pointer of the compass on the left endpoint of the side
drawn in Step 1. Draw an arc mark above the line segment.


4. Repeat Step 2 with the third measurement, 3 in. Then, like Step 3, place the pointer of the compass
on the right endpoint of the side drawn in Step 1. Draw an arc mark above the line segment. Make sure
it intersects the arc mark drawn in Step 3.


5. Draw lines from each endpoint to the arc intersections. These segments are the other two sides of the
triangle.


An animation of this construction can be found at: http://www.mathsisfun.com/geometry/construct-ruler-compass-1.
html
Can another triangle be drawn with these measurements that look different? NO. Only one triangle
can be created from any given three lengths. You can rotate, flip, or move this triangle but
it will still be the same size.
Side-Side-Side (SSS) Triangle Congruence Postulate: If 3 sides in one triangle are congruent to 3
sides in another triangle, then the triangles are congruent.


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BC YZ, AB XY, and AC XZ then 4ABC 4XYZ.
The SSS Postulate is a shortcut. Before, you had to show 3 sides and 3 angles in one triangle were
congruent to 3 sides and 3 angles in another triangle. Now you only have to show 3 sides in one triangle
are congruent to 3 sides in another.
Example 1: Write a triangle congruence statement based on the picture below:


Solution: From the tic marks, we know AB LM, AC LK, BC MK. From the SSS Postulate, the
triangles are congruent. Lining up the corresponding sides, we have 4ABC 4LMK.
Don’t forget ORDER MATTERS when writing congruence statements. Line up the sides with the same
number of tic marks.
Example 2: Write a two-column proof to show that the two triangles are congruent.


Given: AB DE
C is the midpoint of AE and DB.
Prove: 4ACB 4ECD
Solution:


Table 3.6:


Statement Reason
1. AB DE
C is the midpoint of AE and DB


Given


2. AC CE, BC CD Definition of a midpoint
3. 4ACB 4ECD SSS Postulate


Prove Move: You must clearly state the three sets of sides are congruent BEFORE stating the triangles
are congruent.
Prove Move: Mark the picture with the information you are given as well as information that you see in


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the picture (vertical angles, information from parallel lines, midpoints, angle bisectors, right angles). This
information may be used in a proof.


SAS Triangle Congruence Postulate
SAS refers to Side-Angle-Side. The placement of the word Angle is important because it indicates that
the angle you are given is between the two sides.
Included Angle: When an angle is between two given sides of a polygon.


∠B would be the included angle for sides AB and BC.
Consider the question: If I have two sides of length 2 in and 5 in and the angle between them is 45◦, can I
construct one triangle?
Investigation 4-3: Constructing a Triangle Given Two Sides and Included Angle
Tools Needed: protractor, pencil, ruler, and paper
1. Draw the longest side (5 in) horizontally, halfway down the page.


The drawings in this investigation are to scale.
2. At the left endpoint of your line segment, use the protractor to measure a 45◦ angle. Mark this
measurement.


3. Connect your mark from Step 2 with the left endpoint. Make your line 2 in long, the length of the
second side.


4. Connect the two endpoints to draw the third side.


Can you draw another triangle, with these measurements that looks different? NO. Only one triangle
can be created from any two lengths and the INCLUDED angle.


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Side-Angle-Side (SAS) Triangle Congruence Postulate: If two sides and the included angle in one
triangle are congruent to two sides and the included angle in another triangle, then the two triangles are
congruent.


AC XZ, BC YZ, and ∠C ∠Z, then 4ABC 4XYZ.
Example 3: What additional piece of information do you need to show that these two triangles are
congruent using the SAS Postulate?


a) ∠ABC ∠LKM
b) AB LK
c) BC KM
d) ∠BAC ∠KLM
Solution: For the SAS Postulate, you need the side on the other side of the angle. In 4ABC, that is BC
and in 4LKM that is KM. The answer is c.
Example 4: Write a two-column proof to show that the two triangles are congruent.
Given: C is the midpoint of AE and DB
Prove: 4ACB 4ECD


Solution:
Table 3.7:


Statement Reason
1. C is the midpoint of AE and DB Given
2. AC CE, BC CD Definition of a midpoint
3. ∠ACB ∠DCE Vertical Angles Postulate
4. 4ACB 4ECD SAS Postulate


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SSS in the Coordinate Plane
The only way we will show two triangles are congruent in an x− y plane is using SSS. To do this, you need
to use the distance formula.


Example 5: Find the distances of all the line segments from both triangles to see if the two triangles are
congruent.
Solution: Begin with 4ABC and its sides.


AB =


(−6 − (−2))2 + (5 − 10)2


=


(−4)2 + (−5)2


=

16 + 25


=

41


BC =


(−2 − (−3))2 + (10 − 3)2


=


(1)2 + (7)2


=

1 + 49


=

50 = 5



2


AC =


(−6 − (−3))2 + (5 − 3)2


=


(−3)2 + (2)2


=

9 + 4


=

13


Now, find the distances of all the sides in 4DEF.


DE =


(1 − 5)2 + (−3 − 2)2


=


(−4)2 + (−5)2


=

16 + 25


=

41


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EF =


(5 − 4)2 + (2 − (−5))2


=


(1)2 + (7)2


=

1 + 49


=

50 = 5



2


DF =


(1 − 4)2 + (−3 − (−5))2


=


(−3)2 + (2)2


=

9 + 4


=

13


AB = DE, BC = EF, and AC = DF, so two triangles are congruent by SSS.
Example 6: Determine if the two triangles are congruent.


Solution: Start with 4ABC.


AB =


(−2 − (−8))2 + (−2 − (−6))2


=


(6)2 + (4)2


=

36 + 16


=

52 = 2



13


BC =


(−8 − (−6))2 + (−6 − (−9))2


=


(−2)2 + (3)2


=

4 + 9


=

13


AC =


(−2 − (−6))2 + (−2 − (−9))2


=


(4)2 + (7)2


=

16 + 49


=

65


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Now find the sides of 4DEF.


DE =


(3 − 6)2 + (9 − 4)2


=


(−3)2 + (5)2


=

9 + 25


=

34


EF =


(6 − 10)2 + (4 − 7)2


=


(−4)2 + (−3)2


=

16 + 9


=

25 = 5


DF =


(3 − 10)2 + (9 − 7)2


=


(−7)2 + (2)2


=

49 + 4


=

53


No sides have equal measures, so the triangles are not congruent.
Know What? Revisited From what we have learned in this section, the two triangles are not congruent
because the distance from the fridge to the stove in your house is 4 feet and in your neighbor’s it is 4.5
ft. The SSS Postulate tells us that all three sides have to be congruent in order for the triangles to be
congruent.


Review Questions
• Questions 1-10 are similar to Example 1.
• Questions 11-16 are similar to Example 3.
• Questions 17-23 are similar to Examples 2 and 4.
• Questions 24-27 are similar to Examples 5 and 6.


Are the pairs of triangles congruent? If so, write the congruence statement and why.


1.


2.


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3.


4.


5.


6.


7.


8.


9.


10.


State the additional piece of information needed to show that each pair of triangles is congruent.


11. Use SAS


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12. Use SSS


13. Use SAS


14. Use SAS


15. Use SSS


16. Use SAS


Fill in the blanks in the proofs below.


17. Given: AB DC, BE CE
Prove: 4ABE 4ACE


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Table 3.8:


Statement Reason
1. 1.
2. ∠AEB ∠DEC 2.
3. 4ABE 4ACE 3.


18. Given: AB DC, AC DB
Prove: 4ABC 4DCB


Table 3.9:


Statement Reason
1. 1.
2. 2. Reflexive PoC
3. 4ABC 4DCB 3.


19. Given: B is a midpoint of DC
AB ⊥ DC
Prove: 4ABD 4ABC


Table 3.10:


Statement Reason
1. B is a midpoint of DC, AB ⊥ DC 1.
2. 2. Definition of a midpoint
3. ∠ABD and ∠ABC are right angles 3.


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Table 3.10: (continued)


Statement Reason
4. 4. All right angles are
5. 5.
6. 4ABD 4ABC 6.


20. Given: AB is an angle bisector of ∠DAC
AD AC
Prove: 4ABD 4ABC


Table 3.11:


Statement Reason
1.
2. ∠DAB ∠BAC
3. Reflexive PoC
4. 4ABD 4ABC


21. Given: B is the midpoint of DC
AD AC
Prove: 4ABD 4ABC


Table 3.12:


Statement Reason
1.
2. Definition of a Midpoint
3. Reflexive PoC
4. 4ABD 4ABC


22. Given: B is the midpoint of DE and AC
∠ABE is a right angle


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Prove: 4ABE 4CBD


Table 3.13:


Statement Reason
1. Given
2. DB BE, AB BC
3. Definition of a Right Angle
4. Vertical Angle Theorem
5. 4ABE 4CBD


23. Given: DB is the angle bisector of ∠ADC
AD DC
Prove: 4ABD 4CBD


Table 3.14:


Statement Reason
1.
2. ∠ADB ∠BDC
3.
4. 4ABD 4CBD


Find the lengths of the sides of each triangle to see if the two triangles are congruent. Leave your answers
under the radical.


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24.


25.


26. 4ABC : A(−1, 5), B(−4, 2), C(2,−2) and 4DEF : D(7,−5), E(4, 2), F(8,−9)
27. 4ABC : A(−8,−3), B(−2,−4), C(−5,−9) and 4DEF : D(−7, 2), E(−1, 3), F(−4, 8)


Review Queue Answers
1. (a)



74


(b)

538


2. (a) ∠BAC ∠DCA, ∠DAC ∠BCA by the Alternate Interior Angles Theorem.
(b) AC AC
(c) Not yet, this would be ASA.


3. (a) DB BE, AB BC
(b) ∠DBC ∠ABE by the Vertical Angles Theorem.
(c) By the end of this section, yes, we will be able to show that these two triangles are congruent


by SAS.


3.4 Triangle Congruence using ASA, AAS, and
HL


Learning Objectives
• Use and understand the ASA, AAS, and HL Congruence Postulate.
• Complete two-column proofs using SSS, SAS, ASA and AAS.


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Review Queue
1.


(a) What sides are marked congruent?
(b) Is third side congruent? Why?
(c) Write the congruence statement for the two triangles. Why are they congruent?


2.


(a) From the parallel lines, what angles are congruent?
(b) How do you know the third angle is congruent?
(c) Are any sides congruent? How do you know?
(d) Are the two triangles congruent? Why or why not?


3. If 4DEF 4PQR, can it be assumed that:
(a) ∠F ∠R? Why or why not?
(b) EF PR? Why or why not?


Know What? Your parents changed their minds at the last second about their kitchen layout. Now, the
measurements are in the triangle on the left, below. Your neighbor’s kitchen is in blue on the right. Are
the kitchen triangles congruent now?


ASA Congruence
ASA refers to Angle-Side-Angle. The placement of the word Side is important because it indicates that
the side that you are given is between the two angles.
Consider the question: If I have two angles that are 45◦ and 60◦ and the side between them is 5 in, can I
construct only one triangle?
Investigation 4-4: Constructing a Triangle Given Two Angles and Included Side


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Tools Needed: protractor, pencil, ruler, and paper
1. Draw the side (5 in) horizontally, about halfway down the page.


The drawings in this investigation are to scale.
2. At the left endpoint of your line segment, use the protractor to measure the 45◦ angle. Mark this
measurement and draw a ray from the left endpoint through the 45◦ mark.


3. At the right endpoint of your line segment, use the protractor to measure the 60◦ angle. Mark this
measurement and draw a ray from the left endpoint through the 60◦ mark. Extend this ray so that it
crosses through the ray from Step 2.


4. Erase the extra parts of the rays from Steps 2 and 3 to leave only the triangle.
Can you draw another triangle, with these measurements that looks different? NO. Only one triangle
can be created from any given two angle measures and the INCLUDED side.
Angle-Side-Angle (ASA) Congruence Postulate: If two angles and the included side in one triangle
are congruent to two angles and the included side in another triangle, then the two triangles are congruent.
∠A ∠X, ∠B ∠Y, and AB XY, then 4ABC 4XYZ.


Example 1: What information do you need to prove that these two triangles are congruent using the
ASA Postulate?


a) AB UT
b) AC UV
c) BC TV
d) ∠B ∠T


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Solution: For ASA, we need the side between the two given angles, which is AC and UV. The answer is
b.
Example 2: Write a 2-column proof.
Given: ∠C ∠E, AC AE
Prove: 4ACF 4AEB


Solution:
Table 3.15:


Statement Reason
1. ∠C ∠E, AC AE Given
2. ∠A ∠A Reflexive PoC
3. 4ACF 4AEB ASA


AAS Congruence


A variation on ASA is AAS, which is Angle-Angle-Side. For ASA you need two angles and the side between
them. But, if you know two pairs of angles are congruent, the third pair will also be congruent by the 3rd
Angle Theorem. This means you can prove two triangles are congruent when you have any two pairs of
corresponding angles and a pair of sides.
ASA


AAS


Angle-Angle-Side (AAS) Congruence Theorem: If two angles and a non-included side in one triangle
are congruent to two angles and a non-included side in another triangle, then the triangles are congruent.


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Proof of AAS Theorem
Given: ∠A ∠Y, ∠B ∠Z, AC XY
Prove: 4ABC 4YZX


Table 3.16:


Statement Reason
1. ∠A ∠Y, ∠B ∠Z, AC XY Given
2. ∠C ∠X 3rd Angle Theorem
3. 4ABC 4YZX ASA


By proving 4ABC 4YZX with ASA, we have also proved that the AAS Theorem is true.
Example 3: What information do you need to prove that these two triangles are congruent using:
a) ASA?
b) AAS?
c) SAS?


Solution:
a) For ASA, we need the angles on the other side of EF and QR. ∠F ∠Q
b) For AAS, we would need the other angle. ∠G ∠P
c) For SAS, we need the side on the other side of ∠E and ∠R. EG RP
Example 4: Can you prove that the following triangles are congruent? Why or why not?


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Solution: We cannot show the triangles are congruent because KL and ST are not corresponding, even
though they are congruent. To determine if KL and ST are corresponding, look at the angles around them,
∠K and ∠L and ∠S and ∠T . ∠K has one arc and ∠L is unmarked. ∠S has two arcs and ∠T is unmarked. In
order to use AAS, ∠S needs to be congruent to ∠K.
Example 5: Write a 2-column proof.


Given: BD is an angle bisector of ∠CDA, ∠C ∠A
Prove: 4CBD ∠ABD
Solution:


Table 3.17:


Statement Reason
1. BD is an angle bisector of ∠CDA, ∠C ∠A Given
2. ∠CDB ∠ADB Definition of an Angle Bisector
3. DB DB Reflexive PoC
4. 4CBD 4ABD AAS


Hypotenuse-Leg
So far, the congruence postulates we have used will work for any triangle. The last congruence theorem
can only be used on right triangles. A right triangle has exactly one right angle. The two sides adjacent
to the right angle are called legs and the side opposite the right angle is called the hypotenuse.


You may or may not know the Pythagorean Theorem, which says, for any right triangle, this equation is
true:


(leg)2 + (leg)2 = (hypotenuse)2


What this means is that if you are given two sides of a right triangle, you can always find the third.
Therefore, if you have two sides of a right triangle are congruent to two sides of another right triangle; you
can conclude that third sides are also congruent.
The Hypotenuse-Leg (HL) Congruence Theorem is a shortcut of this process.


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HL Congruence Theorem: If the hypotenuse and leg in one right triangle are congruent to the hy-
potenuse and leg in another right triangle, then the two triangles are congruent.
4ABC and 4XYZ are both right triangles and AB XY and BC YZ then 4ABC 4XYZ.


Example 6: What information would you need to prove that these two triangles were congruent using
the:
a) HL Theorem?
b) SAS Theorem?


Solution:
a) For HL, you need the hypotenuses to be congruent. AC MN.
b) To use SAS, we would need the other legs to be congruent. AB ML.
AAA and SSA Relationships There are two other side-angle relationships that we have not discussed:
AAA and SSA.


AAA implies that all the angles are congruent.
As you can see, 4ABC and 4PRQ are not congruent, even though all the angles are.
SSA relationships do not prove congruence either. See 4ABC and 4DEF below.


Because ∠B and ∠D are not the included angles between the congruent sides, we cannot prove that these
two triangles are congruent.


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Recap


Table 3.18:


Side-Angle Relationship Picture Determine Congruence?
SSS Yes


4ABC 4XYZ


SAS Yes
4ABC 4XYZ


ASA Yes
4ABC 4XYZ


AAS (or SAA) Yes
4ABC 4YZX


HL Yes, Right Triangles Only
4ABC 4XYZ


SSA NO


AAA NO


Example 7: Write a 2-column proof.


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Given: AB||ED, ∠C ∠F, AB ED
Prove: AF CD
Solution:


Table 3.19:


Statement Reason
1. AB||ED, ∠C ∠F, AB ED Given
2. ∠ABE ∠DDB Alternate Interior Angles Theorem
3. 4ABF 4DEC ASA
4. AF CD CPCTC


Prove Move: At the beginning of this chapter we introduced CPCTC. Now, it can be used in a proof
once two triangles are proved congruent. It is used to prove the parts of congruent triangles are congruent.
Know What? Revisited Even though we do not know all of the angle measures in the two triangles,
we can find the missing angles by using the Third Angle Theorem. In your parents’ kitchen, the missing
angle is 39◦. The missing angle in your neighbor’s kitchen is 50◦. From this, we can conclude that the two
kitchens are now congruent, either by ASA or AAS.


Review Questions
• Questions 1-10 are similar to Examples 1, 3, 4, and 6.
• Questions 11-20 are review and use the definitions and theorems explained in this section.
• Question 21-26 are similar to Examples 1, 3, 4 and 6.
• Questions 27 and 28 are similar to Examples 2 and 5.
• Questions 29-31 are similar to Example 4 and Investigation 4-4.


For questions 1-10, determine if the triangles are congruent. If they are, write the congruence statement
and which congruence postulate or theorem you used.


1.


2.


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3.


4.


5.


6.


7.


8.


9.


10.


For questions 11-15, use the picture to the right and the given information below.


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Given: DB ⊥ AC, DB is the angle bisector of ∠CDA


11. From DB ⊥ AC, which angles are congruent and why?
12. Because DB is the angle bisector of ∠CDA, what two angles are congruent?
13. From looking at the picture, what additional piece of information are you given? Is this enough to


prove the two triangles are congruent?
14. Write a 2-column proof to prove 4CDB 4ADB, using #11-13.
15. What would be your reason for ∠C ∠A?


For questions 16-20, use the picture to the right and the given information.
Given: LP||NO, LP NO


16. From LP||NO, which angles are congruent and why?
17. From looking at the picture, what additional piece of information can you conclude?
18. Write a 2-column proof to prove 4LMP 4OMN.
19. What would be your reason for LM MO?
20. Fill in the blanks for the proof below. Use the given from above.


Prove: M is the midpoint of PN.


Table 3.20:


Statement Reason
1. LP||NO, LP NO Given
2. Alternate Interior Angles
3. ASA
4. LM MO
5. M is the midpoint of PN.


Determine the additional piece of information needed to show the two triangles are congruent by the given
postulate.


21. AAS


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22. ASA


23. ASA


24. AAS


25. HL


26. SAS


Fill in the blanks in the proofs below.


27. Given: SV ⊥ WU
T is the midpoint of SV and WU
Prove: WS UV


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Table 3.21:


Statement Reason
1.
2. ∠STW and ∠UTV are right angles
3.
4. ST TV , WT TU
5. 4STW 4UTV
6. WS UV


28. Given: ∠K ∠T , EI is the angle bisector of ∠KET
Prove: EI is the angle bisector of ∠KIT


Table 3.22:


Statement Reason
1.
2. Definition of an angle bisector
3. EI EI
4. 4KEI 4TEI
5. ∠KIE ∠T IE
6. EI is the angle bisector of ∠KIT


Construction Let’s see if we can construct two different triangles like 4KLM and 4STU from Example
4.


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29. Look at 4KLM.
(a) If m∠K = 70◦ and m∠M = 60◦, what is m∠L?
(b) If KL = 2 in, construct 4KLM using ∠L, ∠K, KL and Investigation 4-4 (ASA Triangle construc-


tion).
30. Look at 4STU.


(a) If m∠S = 60◦ and m∠U = 70◦, what is m∠T?
(b) If ST = 2 in, construct 4STU using ∠S , ∠T, ST and Investigation 4-4 (ASA Triangle construc-


tion).
31. Are the two triangles congruent?


Review Queue Answers
1. (a) AD DC, AB BC


(b) Yes, by the Reflexive Property
(c) 4DAB 4DCB by SSS


2. (a) ∠L ∠N and ∠M ∠P by the Alternate Interior Angles Theorem
(b) ∠PON ∠LOM by Vertical Angles or the 3rd Angle Theorem
(c) No, no markings or midpoints
(d) No, no congruent sides.


3. (a) Yes, CPCTC
(b) No, these sides do not line up in the congruence statement.


3.5 Isosceles and Equilateral Triangles
Learning Objectives


• Understand the properties of isosceles and equilateral triangles.
• Use the Base Angles Theorem and its converse.
• Understand that an equilateral triangle is also equiangular.


Review Queue
Find the value of x and/or y.


1.


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2.


3.


4. If a triangle is equiangular, what is the measure of each angle?


Know What? Your parents now want to redo the bathroom. To the right are 3 of the tiles they would
like to place in the shower. Each blue and green triangle is an equilateral triangle. What shape is each
dark blue polygon? Find the number of degrees in each of these figures?


Isosceles Triangle Properties


An isosceles triangle is a triangle that has at least two congruent sides. The congruent sides of the isosceles
triangle are called the legs. The other side is called the base. The angles between the base and the legs
are called base angles. The angle made by the two legs is called the vertex angle.
Investigation 4-5: Isosceles Triangle Construction
Tools Needed: pencil, paper, compass, ruler, protractor
1. Refer back to Investigation 4-2. Using your compass and ruler, draw an isosceles triangle with sides of
3 in, 5 in and 5 in. Draw the 3 in side (the base) horizontally at least 6 inches down the page.


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2. Now that you have an isosceles triangle, use your protractor to measure the base angles and the vertex
angle.
The base angles should each be 72.5◦ and the vertex angle should be 35◦.
We can generalize this investigation for all isosceles triangles.
Base Angles Theorem: The base angles of an isosceles triangle are congruent.
For 4DEF, if DE EF, then ∠D ∠F.


To prove the Base Angles Theorem, we need to draw the angle bisector (Investigation 1-5) of ∠E.
Given: Isosceles triangle 4DEF above, with DE EF.
Prove: ∠D ∠F


Table 3.23:


Statement Reason
1. Isosceles triangle 4DEF with DE EF Given
2. Construct angle bisector EG of ∠E Every angle has one angle bisector


3. ∠DEG ∠FEG Definition of an angle bisector
4. EG EG Reflexive PoC
5. 4DEG 4FEG SAS
6. ∠D ∠F CPCTC


Let’s take a further look at the picture from step 2 of our proof.


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Because 4DEG 4FEG, we know ∠EGD ∠EGF by CPCTC. These two angles are also a linear pair, so
90◦ each and EG ⊥ DF.
Additionally, DG GF by CPCTC, so G is the midpoint of DF. This means that EG is the perpendicular
bisector of DF.
Isosceles Triangle Theorem: The angle bisector of the vertex angle in an isosceles triangle is also the
perpendicular bisector of the base.
Note this is ONLY true of the vertex angle. We will prove this theorem in the review questions.
Example 1: Which two angles are congruent?


Solution: This is an isosceles triangle. The congruent angles, are opposite the congruent sides. From the
arrows we see that ∠S ∠U.


Example 2: If an isosceles triangle has base angles with measures of 47◦, what is the measure of the
vertex angle?


Solution: Draw a picture and set up an equation to solve for the vertex angle, v.


47◦ + 47◦ + v = 180◦


v = 180◦ − 47◦ − 47◦


v = 86◦


Example 3: If an isosceles triangle has a vertex angle with a measure of 116◦, what is the measure of
each base angle?


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Solution: Draw a picture and set up and equation to solve for the base angles, b.


116◦ + b + b = 180◦


2b = 64◦


b = 32◦


The converses of the Base Angles Theorem and the Isosceles Triangle Theorem are both true.
Base Angles Theorem Converse: If two angles in a triangle are congruent, then the opposite sides are
also congruent.
For 4DEF, if ∠D ∠F, then DE EF.


Isosceles Triangle Theorem Converse: The perpendicular bisector of the base of an isosceles triangle
is also the angle bisector of the vertex angle.
For isosceles 4DEF, if EG ⊥ DF and DG GF, then ∠DEG ∠FEG.


Equilateral Triangles By definition, all sides in an equilateral triangle have the same length.
Investigation 4-6: Constructing an Equilateral Triangle
Tools Needed: pencil, paper, compass, ruler, protractor
1. Because all the sides of an equilateral triangle are equal, pick one length to be all the sides of the
triangle. Measure this length and draw it horizontally on you paper.


2. Put the pointer of your compass on the left endpoint of the line you drew in Step 1. Open the compass
to be the same width as this line. Make an arc above the line. Repeat Step 2 on the right endpoint.


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4. Connect each endpoint with the arc intersections to make the equilateral triangle.


Use the protractor to measure each angle of your constructed equilateral triangle. What do you notice?
From the Base Angles Theorem, the angles opposite congruent sides in an isosceles triangle are congruent.
So, if all three sides of the triangle are congruent, then all of the angles are congruent, 60◦ each.
Equilateral Triangle Theorem: All equilateral triangles are also equiangular. Also, all equiangular
triangles are also equilateral.


If AB BC AC, then ∠A ∠B ∠C.
If ∠A ∠B ∠C, then AB BC AC.
Example 4: Algebra Connection Find the value of x.


Solution: Because this is an equilateral triangle 3x − 1 = 11. Solve for x.


3x − 1 = 11
3x = 12
x = 4


Example 5: Algebra Connection Find the value of x and the measure of each angle.


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Solution: Similar to Example 4, the two angles are equal, so set them equal to each other and solve for x.


(4x + 12)◦ = (5x − 3)◦


15◦ = x


Substitute x = 15◦; the base angles are 4(15◦) + 12, or 72◦. The vertex angle is 180◦ − 72◦ − 72◦ = 36◦.
Know What? Revisited Let’s focus on one tile. First, these triangles are all equilateral, so this is an
equilateral hexagon (6 sides). Second, we now know that every equilateral triangle is also equiangular, so
every triangle within this tile has 3 60◦ angles. This makes our equilateral hexagon also equiangular, with
each angle measuring 120◦. Because there are 6 angles, the sum of the angles in a hexagon are 6 · 120◦ or
720◦.


Review Questions
• Questions 1-5 are similar to Investigations 4-5 and 4-6.
• Questions 6-14 are similar to Examples 2-5.
• Question 15 uses the definition of an equilateral triangle.
• Questions 16-20 use the definition of an isosceles triangle.
• Question 21 is similar to Examples 2 and 3.
• Questions 22-25 are proofs and use definitions and theorems learned in this section.
• Questions 26-30 use the distance formula.


Constructions For questions 1-5, use your compass and ruler to:


1. Draw an isosceles triangle with sides 3.5 in, 3.5 in, and 6 in.
2. Draw an isosceles triangle that has a vertex angle of 100◦ and legs with length of 4 cm. (you will


also need your protractor for this one)
3. Draw an equilateral triangle with sides of length 7 cm.
4. Using what you know about constructing an equilateral triangle, construct (without a protractor) a


60◦ angle.
5. Draw an isosceles right triangle. What is the measure of the base angles?


For questions 6-14, find the measure of x and/or y.


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6.


7.


8.


9.


10.


11.


12.


13.


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14.


15. 4EQG is an equilateral triangle. If EU bisects ∠LEQ, find:


(a) m∠EUL
(b) m∠UEL
(c) m∠ELQ
(d) If EQ = 4, find LU.


Determine if the following statements are true or false.


16. Base angles of an isosceles triangle are congruent.
17. Base angles of an isosceles triangle are complementary.
18. Base angles of an isosceles triangle can be equal to the vertex angle.
19. Base angles of an isosceles triangle can be right angles.
20. Base angles of an isosceles triangle are acute.
21. In the diagram below, l1||l2. Find all of the lettered angles.


Fill in the blanks in the proofs below.


22. Given: Isosceles 4CIS , with base angles ∠C and ∠S
IO is the angle bisector of ∠CIS
Prove: IO is the perpendicular bisector of CS


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Table 3.24:


Statement Reason
1. Given
2. Base Angles Theorem
3. ∠CIO ∠S IO
4. Reflexive PoC
5. 4CIO 4S IO
6. CO OS
7. CPCTC
8. ∠IOC and ∠IOS are supplementary
9. Congruent Supplements Theorem
10. IO is the perpendicular bisector of CS


23. Given: Equilateral 4RST with RT ST RS
Prove: 4RST is equiangular


Table 3.25:


Statement Reason
1. Given
2. Base Angles Theorem
3. Base Angles Theorem
4. Transitive PoC
5. 4RST is equiangular


24. Given: Isosceles 4ICS with ∠C and ∠S
IO is the perpendicular bisector of CS
Prove: IO is the angle bisector of ∠CIS


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Table 3.26:


Statement Reason
1.
2. ∠C ∠S
3. CO OS
4. m∠IOC = m∠IOS = 90◦
5.
6. CPCTC
7. IO is the angle bisector of ∠CIS


25. Given: Isosceles 4ABC with base angles ∠B and ∠C
Isosceles 4XYZ with base angles ∠Y and ∠Z
∠C ∠Z, BC YZ
Prove: 4ABC 4XYZ


Table 3.27:


Statement Reason
1.
2. ∠B ∠C, ∠Y ∠Z
3. ∠B ∠Y
4. 4ABC 4XYZ


Coordinate Plane Geometry On the x − y plane, plot the coordinates and determine if the given three
points make a scalene or isosceles triangle.


26. (-2, 1), (1, -2), (-5, -2)
27. (-2, 5), (2, 4), (0, -1)
28. (6, 9), (12, 3), (3, -6)
29. (-10, -5), (-8, 5), (2, 3)
30. (-1, 2), (7, 2), (3, 9)


Review Queue Answers
1. (5x − 1)◦ + (8x + 5)◦ + (4x + 6)◦ = 180◦


17x + 10 = 180◦
17x = 170◦


x = 10◦
2. x = 40◦, y = 70◦


www.ck12.org 156




3. x − 3 = 8
x = 5


4. Each angle is 180◦3 , or 60◦


3.6 Chapter 4 Review
Symbols Toolbox
Congruent Triangles and their corresponding parts


Definitions, Postulates, and Theorems
Triangle Sums


• Interior Angles
• Vertex
• Triangle Sum Theorem
• Exterior Angle
• Exterior Angle Sum Theorem
• Remote Interior Angles
• Exterior Angle Theorem


Congruent Figures


• Congruent Triangles
• Congruence Statements
• Third Angle Theorem
• Reflexive Property of Congruence
• Symmetric Property of Congruence
• Transitive Property of Congruence


Triangle Congruence using SSS and SAS


• Side-Side-Side (SSS) Triangle Congruence Postulate
• Included Angle
• Side-Angle-Side (SAS) Triangle Congruence Postulate
• Distance Formula


Triangle Congruence using ASA, AAS, and HL


• Angle-Side-Angle (ASA) Congruence Postulate


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• Angle-Angle-Side (AAS) Congruence Theorem
• Hypotenuse
• Legs (of a right triangle)
• HL Congruence Theorem


Isosceles and Equilateral Triangles


• Base
• Base Angles
• Vertex Angle
• Legs (of an isosceles triangle)
• Base Angles Theorem
• Isosceles Triangle Theorem
• Base Angles Theorem Converse
• Isosceles Triangle Theorem Converse
• Equilateral Triangles Theorem


Review
For each pair of triangles, write what needs to be congruent in order for the triangles to be congruent.
Then, write the congruence statement for the triangles.


1. HL


2. ASA


3. AAS


4. SSS


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5. SAS


Using the pictures below, determine which theorem, postulate or definition that supports each statement
below.


6. m∠1 + m∠2 = 180◦
7. ∠5 ∠6
8. m∠1 + m∠4 + m∠3
9. m∠8 = 60◦


10. m∠5 + m∠6 + m∠7 = 180◦
11. ∠8 ∠9 ∠10
12. If m∠7 = 90◦, then m∠5 = m∠6 = 45◦


Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities
designed to supplement the objectives for some of the lessons in this chapter. See http:
//www.ck12.org/flexr/chapter/9689.


3.7 Study Guide
Keywords: Define, write theorems, and/or draw a diagram for each word below.
1st Section: Triangle Sums
Interior Angles
Vertex
Triangle Sum Theorem
Exterior Angle Exterior Angle Sum Theorem
Remote Interior Angles


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Exterior Angle Theorem


Homework:
2nd Section: Congruent Figures
Congruent Triangles
Congruence Statements
Third Angle Theorem
Reflexive Property of Congruence
Symmetric Property of Congruence
Transitive Property of Congruence
Homework:
3rd Section: Triangle Congruence using SSS and SAS
Side-Side-Side (SSS) Triangle Congruence Postulate
Included Angle
Side-Angle-Side (SAS) Triangle Congruence Postulate
Distance Formula


Homework:
4th Section: Triangle Congruence using ASA, AAS, and HL
Angle-Side-Angle (ASA) Congruence Postulate
Angle-Angle-Side (AAS) Congruence Theorem
Hypotenuse
Legs (of a right triangle)
HL Congruence Theorem


Homework:
5th Section: Isosceles and Equilateral Triangles


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Base
Base Angles
Vertex Angle
Legs (of an isosceles triangle)
Base Angles Theorem
Isosceles Triangle Theorem
Base Angles Theorem Converse
Isosceles Triangle Theorem Converse
Equilateral Triangle Theorem
Homework:


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Chapter 4


Right Triangle Trigonometry


Chapter 8 takes a look at right triangles. A right triangle is a triangle with exactly one right angle. In this
chapter, we will prove the Pythagorean Theorem and its converse. Then, we will introduce trigonometry
ratios. Finally, there is an extension about the Law of Sines and the Law of Cosines.


4.1 The Pythagorean Theorem
Learning Objectives


• Review simplifying and reducing radicals.
• Prove and use the Pythagorean Theorem.
• Use the Pythagorean Theorem to derive the distance formula.


Review Queue
1. Draw a right scalene triangle.
2. Draw an isosceles right triangle.
3. List all the factors of 75.
4. Write the prime factorization of 75.


Know What? For a 52” TV, 52” is the length of the diagonal. High Definition Televisions (HDTVs) have
sides in a ratio of 16:9. What are the length and width of a 52” HDTV?


Simplifying and Reducing Radicals
In algebra, you learned how to simplify radicals. Let’s review it here.
Example 1: Simplify the radical.


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a)

50


b)

27


c)

272


Solution: For each radical, find the square number(s) that are factors.
a)

50 =



25 · 2 = 5



2


b)

27 =



9 · 3 = 3



3


c)

272 =



16 · 17 = 4



17


When adding radicals, you can only combine radicals with the same number underneath it. For example,
2

5 + 3



6 cannot be combined, because 5 and 6 are not the same number.


Example 2: Simplify the radicals.
a) 2

10 +



160


b) 5

6 · 4

18


c)

8 · 12



2


d)
(


5

2
)2


Solution:
a) Simplify



160 before adding: 2



10 +



160 = 2



10 +



16 · 10 = 2



10 + 4



10 = 6



10


b) To multiply two radicals, multiply what is under the radicals and what is in front.
5

6 · 4

18 = 5 · 4



6 · 18 = 20



108 = 20



36 · 3 = 20 · 6



3 = 120



3


c)

8 · 12



2 = 12



8 · 2 = 12



16 = 12 · 4 = 48


d)
(


5

2
)2


= 52
(√


2
)2


= 25 · 2 = 50→ the √ and the 2 cancel each other out
Lastly, to divide radicals, you need to simplify the denominator, which means multiplying the top and
bottom of the fraction by the radical in the denominator.
Example 3: Divide and simplify the radicals.
a) 4

6 ÷

3


b)

30√
8


c) 8

2


6

7


Solution: Rewrite all division problems like a fraction.
a)


b)

30√
8
·

8√
8
=

240√
64


=

16·15
8 =


4

15


8 =

15
2


c) 8

2


6

7
·

7√
7
= 8



14


6·7 =
4

14


3·7 =
4

14


21


Notice, we do not really “divide” radicals, but get them out of the denominator of a fraction.


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The Pythagorean Theorem
We have used the Pythagorean Theorem already in this text, but have not proved it. Recall that the sides
of a right triangle are the legs (the sides of the right angle) and the hypotenuse (the side opposite the
right angle). For the Pythagorean Theorem, the legs are “a” and “b” and the hypotenuse is “c”.


Pythagorean Theorem: Given a right triangle with legs of lengths a and b and a hypotenuse of length
c, then a2 + b2 = c2.
Investigation 8-1: Proof of the Pythagorean Theorem
Tools Needed: pencil, 2 pieces of graph paper, ruler, scissors, colored pencils (optional)


1. On the graph paper, draw a 3 in. square, a 4 in. square, a 5 in. square and a right triangle with legs
of 3 in. and 4 in.


2. Cut out the triangle and square and arrange them like the picture on the right.


3. This theorem relies on area. Recall that the area of a square is side2. In this case, we have three
squares with sides 3 in., 4 in., and 5 in. What is the area of each square?


4. Now, we know that 9+16 = 25, or 32 +42 = 52. Cut the smaller squares to fit into the larger square,
thus proving the areas are equal.


For two more proofs, go to: http://www.mathsisfun.com/pythagoras.html and scroll down to “And
You Can Prove the Theorem Yourself.”
Using the Pythagorean Theorem
Here are several examples of the Pythagorean Theorem in action.
Example 4: Do 6, 7, and 8 make the sides of a right triangle?


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Solution: Plug in the three numbers to the Pythagorean Theorem. The largest length will always be
the hypotenuse. If 62 + 72 = 82, then they are the sides of a right triangle.


62 + 72 = 36 + 49 = 85
82 = 64 85 , 64, so the lengths are not the sides of a right triangle.


Example 5: Find the length of the hypotenuse.


Solution: Use the Pythagorean Theorem. Set a = 8 and b = 15. Solve for c.


82 + 152 = c2


64 + 225 = c2


289 = c2 Take the square root o f both sides.
17 = c


When you take the square root of an equation, the answer is 17 or -17. Length is never negative, which
makes 17 the answer.
Example 6: Find the missing side of the right triangle below.


Solution: Here, we are given the hypotenuse and a leg. Let’s solve for b.


72 + b2 = 142


49 + b2 = 196
b2 = 147


b =

147 =



49 · 3 = 7



3


Example 7: What is the diagonal of a rectangle with sides 10 and 16?


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Solution: For any square and rectangle, you can use the Pythagorean Theorem to find the length of a
diagonal. Plug in the sides to find d.


102 + 162 = d2


100 + 256 = d2


356 = d2


d =

356 = 2



89 ≈ 18.87


Pythagorean Triples
In Example 5, the sides of the triangle were 8, 15, and 17. This combination of numbers is called a
Pythagorean triple.
Pythagorean Triple: A set of three whole numbers that makes the Pythagorean Theorem true.


3, 4, 5 5, 12, 13 7, 24, 25 8, 15, 17 9, 12, 15 10, 24, 26


Any multiple of a Pythagorean triple is also considered a triple because it would still be three whole
numbers. Multiplying 3, 4, 5 by 2 gives 6, 8, 10, which is another triple. To see if a set of numbers makes
a triple, plug them into the Pythagorean Theorem.
Example 8: Is 20, 21, 29 a Pythagorean triple?
Solution: If 202 + 212 = 292, then the set is a Pythagorean triple.


202 + 212 = 400 + 441 = 841
292 = 841


Therefore, 20, 21, and 29 is a Pythagorean triple.


Height of an Isosceles Triangle
One way to use The Pythagorean Theorem is to find the height of an isosceles triangle.


Example 9: What is the height of the isosceles triangle?


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Solution: Draw the altitude from the vertex between the congruent sides, which bisect the base.


72 + h2 = 92


49 + h2 = 81
h2 = 32


h =

32 =



16 · 2 = 4



2


The Distance Formula
Another application of the Pythagorean Theorem is the Distance Formula. We will prove it here.


Let’s start with point A(x1, y1) and point B(x2, y2), to the left. We will call the distance between A and
B, d.
Draw the vertical and horizontal lengths to make a right triangle.


Now that we have a right triangle, we can use the Pythagorean Theorem to find the hypotenuse, d.


d2 = (x1 − x2)2 + (y1 − y2)2


d =


(x1 − x2)2 + (y1 − y2)2


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Distance Formula: The distance A(x1, y1) and B(x2, y2) is d =


(x1 − x2)2 + (y1 − y2)2.
Example 10: Find the distance between (1, 5) and (5, 2).
Solution: Make A(1, 5) and B(5, 2). Plug into the distance formula.


d =


(1 − 5)2 + (5 − 2)2


=


(−4)2 + (3)2


=

16 + 9 =



25 = 5


Just like the lengths of the sides of a triangle, distances are always positive.
Know What? Revisited To find the length and width of a 52” HDTV, plug in the ratios and 52 into
the Pythagorean Theorem. We know that the sides are going to be a multiple of 16 and 9, which we will
call n.


(16n)2 + (9n)2 = 522


256n2 + 81n2 = 2704
337n2 = 2704


n2 = 8.024
n = 2.83


The dimensions of the TV are 16(2.83”) × 9(2.83”), or 45.3” × 25.5”.


Review Questions
• Questions 1-9 are similar to Examples 1-3.
• Questions 10-15 are similar to Example 5 and 6.
• Questions 16-19 are similar to Example 7.
• Questions 20-25 are similar to Example 8.
• Questions 26-28 are similar to Example 9.
• Questions 29-31 are similar to Example 10.
• Questions 32 and 33 are similar to the Know What?
• Question 34 and 35 are a challenge and similar to Example 9.


Simplify the radicals.


1. 2

5 +

20


2.

24


3.
(


6

3
)2


4. 8

8 ·

10


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5.
(


2

30


)2


6.

320


7. 4

5√
6


8. 12√
10


9. 21

5


9

15


Find the length of the missing side. Simplify all radicals.


10.


11.


12.


13.


14.


15.


16. If the legs of a right triangle are 10 and 24, then the hypotenuse is __________.
17. If the sides of a rectangle are 12 and 15, then the diagonal is _____________.
18. If the sides of a square are 16, then the diagonal is ____________.
19. If the sides of a square are 9, then the diagonal is _____________.


Determine if the following sets of numbers are Pythagorean Triples.


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20. 12, 35, 37
21. 9, 17, 18
22. 10, 15, 21
23. 11, 60, 61
24. 15, 20, 25
25. 18, 73, 75


Find the height of each isosceles triangle below. Simplify all radicals.


26.


27.


28.


Find the length between each pair of points.


29. (-1, 6) and (7, 2)
30. (10, -3) and (-12, -6)
31. (1, 3) and (-8, 16)
32. What are the length and width of a 42” HDTV? Round your answer to the nearest tenth.
33. Standard definition TVs have a length and width ratio of 4:3. What are the length and width of a


42” Standard definition TV? Round your answer to the nearest tenth.
34. Challenge An equilateral triangle is an isosceles triangle. If all the sides of an equilateral triangle


are 8, find the height. Leave your answer in simplest radicalform.


35. If the sides are length s, what would the height be?


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Review Queue Answers
1.


2.


3. Factors of 75: 1, 3, 5, 15, 25, 75
4. Prime Factorization of 75: 3 · 5 · 5


4.2 Converse of the Pythagorean Theorem
Learning Objectives


• Understand the converse of the Pythagorean Theorem.
• Determine if a triangle is acute or obtuse from side measures.


Review Queue
1. Determine if the following sets of numbers are Pythagorean triples.


(a) 14, 48, 50
(b) 9, 40, 41
(c) 12, 43, 44
(d) 12, 35, 37


2. Simplify the radicals.


(a)
(


5

12


)2


(b) 14√
2


(c) 18√
3


Know What? A friend of yours is designing a building and wants it to be rectangular. One wall 65 ft.
long and the other is 72 ft. long. How can he ensure the walls are going to be perpendicular?


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Converse of the Pythagorean Theorem


Pythagorean Theorem Converse: If the square of the longest side of a triangle is equal to the sum of
the squares of the other two sides, then the triangle is a right triangle.
If a2 + b2 = c2, then 4ABC is a right triangle.


With this converse, you can use the Pythagorean Theorem to prove that a triangle is a right triangle, even
if you do not know any angle measures.
Example 1: Determine if the triangles below are right triangles.
a)


b)


Solution: Check to see if the three lengths satisfy the Pythagorean Theorem. Let the longest side represent
c.
a) a2 + b2 = c2
82 + 162 ?=


(


8

5
)2


64 + 256 ?= 64 · 5
320 = 320 Yes
b) a2 + b2 = c2
222 + 242 ?= 262


484 + 576 ?= 676
1060 , 676 No
Example 2: Do the following lengths make a right triangle?
a)

5, 3,



14


b) 6, 2

3, 8


c) 3

2, 4

2, 5

2


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Solution: Even though there is no picture, you can still use the Pythagorean Theorem. Again, the longest
length will be c.
a)


(√
5
)2


+ 32 =

142


5 + 9 = 14
Yes
b) 62 +


(


2

3
)2


= 82
36 + (4 · 3) = 64
36 + 12 , 64
c) This is a multiple of



2 of a 3, 4, 5 right triangle. Yes, this is a right triangle.


Identifying Acute and Obtuse Triangles
We can extend the converse of the Pythagorean Theorem to determine if a triangle is an obtuse or acute
triangle.
Theorem 8-3: If the sum of the squares of the two shorter sides in a right triangle is greater than the
square of the longest side, then the triangle is acute.
b < c and a < c
If a2 + b2 > c2, then the triangle is acute.


Theorem 8-4: If the sum of the squares of the two shorter sides in a right triangle is less than the square
of the longest side, then the triangle is obtuse.
b < c and a < c
If a2 + b2 < c2, then the triangle is obtuse.


Example 3: Determine if the following triangles are acute, right or obtuse.
a)


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b)


Solution: Set the longest side equal to c.
a) 62 +


(


3

5
)2


? 82
36 + 45 ? 64
81 > 64
The triangle is acute.
b) 152 + 142 ? 212
225 + 196 ? 441
421 < 441
The triangle is obtuse.
Example 4: Graph A(−4, 1), B(3, 8), and C(9, 6). Determine if 4ABC is acute, obtuse, or right.


Solution: Use the distance formula to find the length of each side.


AB =


(−4 − 3)2 + (1 − 8)2 =

49 + 49 =



98 = 7



2


BC =


(3 − 9)2 + (8 − 6)2 =

36 + 4 =



40 = 2



10


AC =


(−4 − 9)2 + (1 − 6)2 =

169 + 25 =



194


Plug these lengths into the Pythagorean Theorem.
(√


98
)2


+
(√


40
)2


?
(√


194
)2


98 + 40 ? 194
138 < 194


4ABC is an obtuse triangle.
Know What? Revisited Find the length of the diagonal.


www.ck12.org 174




652 + 722 = c2


4225 + 5184 = c2


9409 = c2


97 = c To make the building rectangular, both diagonals must be 97 feet.


Review Questions
• Questions 1-6 are similar to Examples 1 and 2.
• Questions 7-15 are similar to Example 3.
• Questions 16-20 are similar to Example 4.
• Questions 21-24 use the Pythagorean Theorem.
• Question 25 uses the definition of similar triangles.


Determine if the following lengths make a right triangle.


1. 7, 24, 25
2.

5, 2

10, 3



5


3. 2

3,

6, 8


4. 15, 20, 25
5. 20, 25, 30
6. 8



3, 6, 2



39


Determine if the following triangles are acute, right or obtuse.


7. 7, 8, 9
8. 14, 48, 50
9. 5, 12, 15


10. 13, 84, 85
11. 20, 20, 24
12. 35, 40, 51
13. 39, 80, 89
14. 20, 21, 38
15. 48, 55, 76


Graph each set of points and determine if 4ABC is acute, right, or obtuse, using the distance formula.


16. A(3,−5), B(−5,−8),C(−2, 7)
17. A(5, 3), B(2,−7),C(−1, 5)
18. A(1, 6), B(5, 2),C(−2, 3)
19. A(−6, 1), B(−4,−5),C(5,−2)
20. Show that #18 is a right triangle by using the slopes of the sides of the triangle.


The figure to the right is a rectangular prism. All sides (or faces) are either squares (the front and
back) or rectangles (the four around the middle). All faces are perpendicular.


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21. Find c.
22. Find d.


Now, the figure is a cube, where all the sides are squares. If all the sides have length 4, find:


23. Find c.
24. Find d.


25. Writing Explain why m∠A = 90◦.


Review Queue Answers
1. (a) Yes


(b) Yes
(c) No
(d) Yes


2. (a)
(


5

12


)2
= 52 · 12 = 25 · 12 = 300


(b) 14√
2
·

2√
2
= 14



2


2 = 7

2


(c) 18√
3
·

3√
3
= 18



3


3 = 6

3


4.3 Using Similar Right Triangles
Learning Objectives


• Identify similar triangles inscribed in a larger triangle.
• Use proportions in similar right triangles.


Review Queue
1. Solve the following ratios.


(a) 3x = x27
(b)



6
x =


x
9

6


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(c) x15 = 12x
2. If the legs of an isosceles right triangle are 4, find the length of the hypotenuse. Draw a picture and


simplify the radical.


Know What? The bridge to the right is called a truss bridge. It is a steel bridge with a series of right
triangles that are connected as support. All the red right triangles are similar. Can you find x, y and z?


Inscribed Similar Triangles
You may recall that if two objects are similar, corresponding angles are congruent and their sides are
proportional in length.
Theorem 8-5: If an altitude is drawn from the right angle of any right triangle, then the two triangles
formed are similar to the original triangle and all three triangles are similar to each other.


In 4ADB,m∠A = 90◦ and AC⊥DB, then 4ADB ∼ 4CDA ∼ 4CAB.


Example 1: Write the similarity statement for the triangles below.


Solution: Separate out the three triangles.


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Line up the congruent angles: 4IRE ∼ 4ITR ∼ 4RTE
We can also use the side proportions to find the length of the altitude.
Example 2: Find the value of x.


Solution: Separate the triangles to find the corresponding sides.


Set up a proportion.


shorter leg in 4EDG
shorter leg in 4DFG =


hypotenuse in 4EDG
hypotenuse in 4DFG


6
x


= 10
8


48 = 10x
4.8 = x


Example 3: Find the value of x.


Solution: Set up a proportion.


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shorter leg in 4SVT
shorter leg in 4RST =


hypotenuse in 4SVT
hypotenuse in 4RST


4
x


= x
20


x2 = 80


x =

80 = 4



5


Example 4: Find the value of y in 4RST above.
Solution: Use the Pythagorean Theorem.


y2 +
(


4

5
)2


= 202


y2 + 80 = 400
y2 = 320


y =

320 = 8



5


The Geometric Mean
Geometric Mean: The geometric mean of two positive numbers a and b is the positive number x, such
that ax = xb or x2 = ab and x =



ab.


Example 5: Find the geometric mean of 24 and 36.
Solution: x =



24 · 36 =



12 · 2 · 12 · 3 = 12



6


Example 6: Find the geometric mean of 18 and 54.
Solution: x =



18 · 54 =



18 · 18 · 3 = 18



3


In both of these examples, we did not multiply the numbers together. This makes it easier to simplify the
radical. A practical application of the geometric mean is to find the altitude of a right triangle.
Example 7: Find the value of x.


Solution: Set up a proportion.
shortest leg o f smallest 4
shortest leg o f middle 4 =


longer leg o f smallest 4
longer leg o f middle 4


9
x


= x
27


x2 = 243


x =

243 = 9



3


In Example 7, 9x = x27 is in the definition of the geometric mean. So, the altitude is the geometric mean of
the two segments that it divides the hypotenuse into. In other words, BCAC = ACDC . Two other true proportions
are BCAB = ABDB and DCAD = ADDB .


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Example 8: Find the value of x and y.


Solution: Separate the triangles. Write a proportion for x.


20
x


= x
35


x2 = 20 · 35
x =



20 · 35


x = 10

7


Set up a proportion for y. Or, you can use the Pythagorean Theorem to solve for y.
15
y


= y
35


(10

7)2 + y2 = 352


y2 = 15 · 35 700 + y2 = 1225
y =



15 · 35 y =



525 = 5



21


y = 5

21 Use the method you feel most comfortable with.


Know What? Revisited To find the hypotenuse of the smallest triangle, do the Pythagorean Theorem.


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452 + 282 = x2


2809 = x2


53 = x


Because the triangles are similar, find the scale factor of 7028 = 2.5.
y = 45 · 2.5 = 112.5 and z = 53 · 2.5 = 135.5


Review Questions
• Questions 1-4 use the ratios of similar right triangles.
• Questions 5-8 are similar to Example 1.
• Questions 9-11 are similar to Examples 2-4
• Questions 12-17 are similar to Examples 5 and 6.
• Questions 18-29 are similar to Examples 2, 3, 4, 7, and 8.
• Question 30 is a proof of theorem 8-5.


Fill in the blanks.


1. 4BAD ∼ 4 ∼ 4
2. BC? = ?CD
3. BCAB = AB?
4. ?AD = ADBD


Write the similarity statement for the right triangles in each diagram.


5.


6.


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7.


Use the diagram to answer questions 8-11.


8. Write the similarity statement for the three triangles in the diagram.
9. If JM = 12 and ML = 9, find KM.


10. Find JK.
11. Find KL.


Find the geometric mean between the following two numbers. Simplify all radicals.


12. 16 and 32
13. 45 and 35
14. 10 and 14
15. 28 and 42
16. 40 and 100
17. 51 and 8


Find the length of the missing variable(s). Simplify all radicals.


18.


19.


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20.


21.


22.


23.


24.


25.


26.


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27.


28.


29.


30. Fill in the blanks of the proof for Theorem 8-5.


Given: 4ABD with AC ⊥ DB and ∠DAB is a right angle.
Prove: 4ABD ∼ 4CBA ∼ 4CAD


Table 4.1:


Statement Reason
1. Given
2. ∠DCA and ∠ACB are right angles
3. ∠DAB ∠DCA ∠ACB
4. Reflexive PoC
5. AA Similarity Postulate
6. B ∠B
7. 4CBA 4ABD
8. 4CAD 4CBA


Review Queue Answers
1. (a) 3x = x27 → x2 = 81→ x = 9


(b)

6
x =


x
9

6
→ x2 = 54→ x =



54 =



9 · 6 = 3



6


(c) x15 = 12x → x2 = 180→ x =

180 =



4 · 9 · 5 = 2 · 3



5 = 6



5


2. 42 + 42 = h2
h =



32 = 4



2


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4.4 Special Right Triangles
Learning Objectives


• Learn and use the 45-45-90 triangle ratio.
• Learn and use the 30-60-90 triangle ratio.


Review Queue
Find the value of the missing variables. Simplify all radicals.


1.


2.


3.


4. Is 9, 12, and 15 a right triangle?
5. Is 3, 3



3, and 6 a right triangle?


Know What? A baseball diamond is a square with sides that are 90 feet long. Each base is a corner of
the square. What is the length between 1st and 3rd base and between 2nd base and home plate? (the red
dotted lines in the diagram).


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Isosceles Right Triangles
There are two special right triangles. The first is an isosceles right triangle.
Isosceles Right Triangle: A right triangle with congruent legs and acute angles. This triangle is also
called a 45-45-90 triangle (after the angle measures).


4ABC is a right triangle with:


m∠A = 90◦


AB AC
m∠B = m∠C = 45◦


Investigation 8-2: Properties of an Isosceles Right Triangle
Tools Needed: Pencil, paper, compass, ruler, protractor
1. Draw an isosceles right triangle with 2 inch legs and the 90◦ angle between them.


2. Find the measure of the hypotenuse, using the Pythagorean Theorem. Simplify the radical.


22 + 22 = c2


8 = c2


c =

8 =



4 · 2 = 2



2


What do you notice about the length of the legs and hypotenuse?


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3. Now, let’s say the legs are of length x and the hypotenuse is h. Use the Pythagorean Theorem to find
the hypotenuse. How is it similar to your answer in #2?


x2 + x2 = h2


2x2 = h2


x

2 = h


45-45-90 Theorem: If a right triangle is isosceles, then its sides are x : x : x

2.


For any isosceles right triangle, the legs are x and the hypotenuse is always x

2. Because the three


angles are always 45◦, 45◦, and 90◦, all isosceles right triangles are similar.
Example 1: Find the length of the missing sides.
a)


b)


Solution: Use the x : x : x

2 ratio.


a) TV = 6 because it is equal to ST . So, SV = 6 ·

2 = 6



2.


b) AB = 9

2 because it is equal to AC. So, BC = 9



2 ·

2 = 9 · 2 = 18.


Example 2: Find the length of x.
a)


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b)


Solution: Use the x : x : x

2 ratio.


a) 12

2 is the diagonal of the square. Remember that the diagonal of a square bisects each angle, so it


splits the square into two 45-45-90 triangles. 12

2 would be the hypotenuse, or equal to x



2.


12

2 = x



2


12 = x


b) Here, we are given the hypotenuse. Solve for x in the ratio.


x

2 = 16


x = 16√
2
·

2

2


= 16

2


2
= 8

2


In part b, we rationalized the denominator which we learned in the first section.


30-60-90 Triangles
The second special right triangle is called a 30-60-90 triangle, after the three angles. To draw a 30-60-90
triangle, start with an equilateral triangle.
Investigation 8-3: Properties of a 30-60-90 Triangle
Tools Needed: Pencil, paper, ruler, compass
1. Construct an equilateral triangle with 2 inch sides.


http://www.mathsisfun.com/geometry/construct-equitriangle.html
2. Draw or construct the altitude from the top vertex to form two congruent triangles.


3. Find the measure of the two angles at the top vertex and the length of the shorter leg.


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The top angles are each 30◦ and the shorter leg is 1 in because the altitude of an equilateral triangle is also
the angle and perpendicular bisector.
4. Find the length of the longer leg, using the Pythagorean Theorem. Simplify the radical.


12 + b2 = 22


1 + b2 = 4
b2 = 3


b =

3


5. Now, let’s say the shorter leg is length x and the hypotenuse is 2x. Use the Pythagorean Theorem to
find the longer leg. How is this similar to your answer in #4?


x2 + b2 = (2x)2


x2 + b2 = 4x2


b2 = 3x2


b = x

3


30-60-90 Theorem: If a triangle has angle measures 30◦, 60◦ and 90◦, then the sides are x : x

3 : 2x.


The shortest leg is always x, the longest leg is always x

3, and the hypotenuse is always 2x. If you ever


forget these theorems, you can still use the Pythagorean Theorem.
Example 3: Find the length of the missing sides.
a)


b)


Solution: In part a, we are given the shortest leg and in part b, we are given the hypotenuse.
a) If x = 5, then the longer leg, b = 5



3, and the hypotenuse, c = 2(5) = 10.


b) Now, 2x = 20, so the shorter leg, f = 202 = 10, and the longer leg, g = 10

3.


Example 4: Find the value of x and y.


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a)


b)


Solution: In part a, we are given the longer leg and in part b, we are given the hypotenuse.
a) x

3 = 12


x = 12√
3
·

3√
3
= 12



3


3 = 4

3


The hypotenuse is
y = 2(4



3) = 8



3


b) 2x = 16
x = 8
The longer leg is
y = 8 ·



3 = 8



3


Example 5: A rectangle has sides 4 and 4

3. What is the length of the diagonal?


Solution: If you are not given a picture, draw one.


The two lengths are x, x

3, so the diagonal would be 2x, or 2(4) = 8.


If you did not recognize this is a 30-60-90 triangle, you can use the Pythagorean Theorem too.


42 +
(


4

3
)2


= d2


16 + 48 = d2


d =

64 = 8


Example 6: A square has a diagonal with length 10, what are the sides?
Solution: Draw a picture.


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We know half of a square is a 45-45-90 triangle, so 10 = s

2.


s

2 = 10


s = 10√
2
·

2

2


= 10

2


2
= 5

2


Know What? Revisited The distance between 1st and 3rd base is one of the diagonals of the square.
So, it would be the same as the hypotenuse of a 45-45-90 triangle. Using our ratios, the distance is
90

2 ≈ 127.3 f t. The distance between 2nd base and home plate is the same length.


Review Questions
• Questions 1-4 are similar to Example 1-4.
• Questions 5-8 are similar to Examples 5 and 6.
• Questions 9-23 are similar to Examples 1-4.
• Questions 24 and 25 are a challenge.


1. In an isosceles right triangle, if a leg is 4, then the hypotenuse is __________.
2. In a 30-60-90 triangle, if the shorter leg is 5, then the longer leg is __________ and the hy-


potenuse is ___________.
3. In an isosceles right triangle, if a leg is x, then the hypotenuse is __________.
4. In a 30-60-90 triangle, if the shorter leg is x, then the longer leg is __________ and the hy-


potenuse is ___________.
5. A square has sides of length 15. What is the length of the diagonal?
6. A square’s diagonal is 22. What is the length of each side?
7. A rectangle has sides of length 6 and 6



3. What is the length of the diagonal?


8. Two (opposite) sides of a rectangle are 10 and the diagonal is 20. What is the length of the other
two sides?


For questions 9-23, find the lengths of the missing sides. Simplify all radicals.


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9.


10.


11.


12.


13.


14.


15.


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16.


17.


18.


19.


20.


21.


22.


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23.


Challenge For 24 and 25, find the value of y. You may need to draw in additional lines. Round all answers
to the nearest hundredth.


24.


25.


Review Queue Answers
1. 42 + 42 = x2


32 = x2
x = 4



2


2. 32 + y2 = 62
y2 = 27
y = 3



3


3. x2 + x2 =
(


10

2
)2


2x2 = 200
x2 = 100
x = 10


4. Yes, 92 + 122 = 152 → 81 + 144 = 225
5. Yes, 32 +


(


3

3
)2


= 62 → 9 + 27 = 36


4.5 Tangent, Sine and Cosine
Learning Objectives


• Use the tangent, sine and cosine ratios.
• Use a scientific calculator to find sine, cosine and tangent.
• Use trigonometric ratios in real-life situations.


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Review Queue
1. The legs of an isosceles right triangle have length 14. What is the hypotenuse?
2. Do the lengths 8, 16, 20 make a right triangle? If not, is the triangle obtuse or acute?
3. In a 30-60-90 triangle, what do the 30, 60, and 90 refer to?


Know What? A restaurant is building a wheelchair ramp. The angle of elevation for the ramp is 5◦. If
the vertical distance from the sidewalk to the front door is 4 feet, how long will the ramp be (x)? Round
your answers to the nearest hundredth.


What is Trigonometry?
In this lesson we will define three trigonometric (or trig) ratios. Once we have defined these ratios, we will
be able to solve problems like the Know What? above.
Trigonometry: The study of the relationships between the sides and angles of right triangles.
The legs are called adjacent or opposite depending on which acute angle is being used.


a is ad jacent to ∠B a is opposite ∠A
b is ad jacent to ∠A b is opposite ∠B


c is the hypotenuse


Sine, Cosine, and Tangent Ratios
The three basic trig ratios are called, sine, cosine and tangent. For now, we will only take the sine, cosine
and tangent of acute angles. However, you can use these ratios with obtuse angles as well.
For right triangle 4ABC, we have:
Sine Ratio: opposite leghypotenuse sin A = ac or sin B = bc
Cosine Ratio: ad jacent leghypotenuse cos A = bc or cos B = ac
Tangent Ratio: opposite legad jacent leg tan A = ab or tan B = ba


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An easy way to remember ratios is to use SOH-CAH-TOA.


Example 1: Find the sine, cosine and tangent ratios of ∠A.


Solution: First, we need to use the Pythagorean Theorem to find the length of the hypotenuse.


52 + 122 = h2


13 = h


sin A = leg opposite ∠A
hypotenuse


= 12
13


cos A = leg ad jacent to ∠A
hypotenuse


= 5
13


,


tan A = leg opposite ∠A
leg ad jacent to ∠A


= 12
5


A few important points:


• Always reduce ratios (fractions) when you can.
• Use the Pythagorean Theorem to find the missing side (if there is one).
• If there is a radical in the denominator, rationalize the denominator.


Example 2: Find the sine, cosine, and tangent of ∠B.


Solution: Find the length of the missing side.


AC2 + 52 = 152


AC2 = 200


AC = 10

2


sin B = 10

2


15
= 2

2


3
cos B = 5


15
= 1


3
tan B = 10



2


5
= 2

2


Example 3: Find the sine, cosine and tangent of 30◦.


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Solution: This is a 30-60-90 triangle. The short leg is 6, y = 6

3 and x = 12.


sin 30◦ = 6
12


= 1
2


cos 30◦ = 6

3


12
= 3


2
tan 30◦ = 6


6

3


= 1√
3
· 3√


3
=

3


3


Sine, Cosine, and Tangent with a Calculator
From Example 3, we can conclude that there is a fixed sine, cosine, and tangent value for every
angle, from 0◦ to 90◦. Your scientific (or graphing) calculator knows all the trigonometric values for any
angle. Your calculator, should have [SIN], [COS], and [TAN] buttons.
Example 4: Find the trigonometric value, using your calculator. Round to 4 decimal places.
a) sin 78◦


b) cos 60◦


c) tan 15◦


Solution: Depending on your calculator, you enter the degree and then press the trig button or the other
way around. Also, make sure the mode of your calculator is in DEGREES.
a) sin 78◦ = 0.9781
b) cos 60◦ = 0.5
c) tan 15◦ = 0.2679


Finding the Sides of a Triangle using Trig Ratios
One application of the trigonometric ratios is to use them to find the missing sides of a right triangle.
Example 5: Find the value of each variable. Round your answer to the nearest tenth.


Solution: We are given the hypotenuse. Use sine to find b, and cosine to find a.


sin 22◦ = b
30


cos 22◦ = a
30


30 · sin 22◦ = b 30 · cos 22◦ = a
b ≈ 11.2 a ≈ 27.8


Example 6: Find the value of each variable. Round your answer to the nearest tenth.


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Solution: We are given the adjacent leg to 42◦. To find c, use cosine and tangent to find d.


cos 42◦ = ad jacent
hypotenuse


= 9
c


tan 42◦ = opposite
ad jacent


= d
9


c · cos 42◦ = 9 9 · tan 42◦ = d


c = 9cos 42◦ ≈ 12.1 d ≈ 8.1


Anytime you use trigonometric ratios, only use the information that you are given. This will give the most
accurate answers.


Angles of Depression and Elevation
Another application of the trigonometric ratios is to find lengths that you cannot measure. Very frequently,
angles of depression and elevation are used in these types of problems.
Angle of Depression: The angle measured from the horizon or horizontal line, down.


Angle of Elevation: The angle measure from the horizon or horizontal line, up.
Example 7: A math student is standing 25 feet from the base of the Washington Monument. The angle
of elevation from her horizontal line of sight is 87.4◦. If her “eye height” is 5 ft, how tall is the monument?


Solution: We can find the height of the monument by using the tangent ratio.


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tan 87.4◦ = h
25


h = 25 · tan 87.4◦ = 550.54


Adding 5 ft, the total height of the Washington Monument is 555.54 ft.
Know What? Revisited To find the length of the ramp, we need to use sine.


sin 5◦ = 4
x


y = 2sin 5◦ = 22.95


Review Questions
• Questions 1-8 use the definitions of sine, cosine and tangent.
• Questions 9-16 are similar to Example 4.
• Questions 17-22 are similar to Examples 1-3.
• Questions 23-28 are similar to Examples 5 and 6.
• Questions 29 and 30 are similar to Example 7.


Use the diagram to fill in the blanks below.


1. tan D = ??
2. sin F = ??
3. tan F = ??
4. cos F = ??
5. sin D = ??
6. cos D = ??


From questions 1-6, we can conclude the following. Fill in the blanks.


7. cos = sin F and sin = cos F.
8. tan D and tan F are _________ of each other.


Use your calculator to find the value of each trig function below. Round to four decimal places.


9. sin 24◦
10. cos 45◦
11. tan 88◦
12. sin 43◦


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13. tan 12◦
14. cos 79◦
15. sin 82◦
16. tan 45◦


Find the sine, cosine and tangent of ∠A. Reduce all fractions and radicals.


17.


18.


19.


20.


21.


22.


Find the length of the missing sides. Round your answers to the nearest tenth.


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23.


24.


25.


26.


27.


28.


29. Kristin is swimming in the ocean and notices a coral reef below her. The angle of depression is 35◦
and the depth of the ocean, at that point is 250 feet. How far away is she from the reef?


30. The Leaning Tower of Piza currently “leans” at a 4◦ angle and has a vertical height of 55.86 meters.
How tall was the tower when it was originally built?


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Review Queue Answers
1. The hypotenuse is 14



2.


2. No, 82 + 162 < 202, the triangle is obtuse.
3. 30◦, 60◦, and 90◦ refer to the angle measures in the special right triangle.


4.6 Inverse Trigonometric Ratios
Learning Objectives


• Use the inverse trigonometric ratios to find an angle in a right triangle.
• Solve a right triangle.


Review Queue
Find the lengths of the missing sides. Round your answer to the nearest tenth.


1.


2.


3. Draw an isosceles right triangle with legs of length 3. What is the hypotenuse?
4. Use the triangle from #3, to find the sine, cosine, and tangent of 45◦.


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Know What? The longest escalator in North America is at the Wheaton Metro Station in Maryland. It
is 230 feet long and is 115 ft. high. What is the angle of elevation, x◦, of this escalator?


Inverse Trigonometric Ratios
In mathematics, the word inverse means “undo.” For example, addition and subtraction are inverses of
each other because one undoes the other. When we apply inverses to the trigonometric ratios, we can find
acute angle measures as long as we are given two sides.
Inverse Tangent: Labeled tan−1, the “-1” means inverse.


tan−1
(


b
a


)


= m∠B tan−1
(a
b


)


= m∠A


Inverse Sine: Labeled sin−1.


sin−1
(


b
c


)


= m∠B sin−1
(a
c


)


= m∠A


Inverse Cosine: Labeled cos−1.


cos−1
(a
c


)


= m∠B cos−1
(


b
c


)


= m∠A


In order to find the measure of the angles, you will need you use your calculator. On most scientific and
graphing calculators, the buttons look like [SIN−1], [COS−1], and [TAN−1]. You might also have to hit a
shift or 2nd button to access these functions.
Example 1: Use the sides of the triangle and your calculator to find the value of ∠A. Round your answer
to the nearest tenth of a degree.


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Solution: In reference to ∠A, we are given the opposite leg and the adjacent leg. This means we should
use the tangent ratio.
tan A = 2025 = 45 . So, tan−1 45 = m∠A. Now, use your calculator.
If you are using a TI-83 or 84, the keystrokes would be: [2nd][TAN]


(


4
5


)


[ENTER] and the screen looks like:


m∠A = 38.7◦


Example 2: ∠A is an acute angle in a right triangle. Find m∠A to the nearest tenth of a degree.
a) sin A = 0.68
b) cos A = 0.85
c) tan A = 0.34
Solution:
a) m∠A = sin−1 0.68 = 42.8◦


b) m∠A = cos−1 0.85 = 31.8◦


c) m∠A = tan−1 0.34 = 18.8◦


Solving Triangles
To solve a right triangle, you need to find all sides and angles in a right triangle, using sine, cosine or
tangent, inverse sine, inverse cosine, or inverse tangent, or the Pythagorean Theorem.
Example 3: Solve the right triangle.


Solution: To solve this right triangle, we need to find AB,m∠C and m∠B. Only use the values you are
given.
AB: Use the Pythagorean Theorem.


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242 + AB2 = 302


576 + AB2 = 900
AB2 = 324


AB =

324 = 18


m∠B: Use the inverse sine ratio.


sin B = 24
30


= 4
5


sin−1
(4
5


)


= 53.1◦ = m∠B


m∠C: Use the inverse cosine ratio.


cosC = 24
30


= 4
5
−→ cos−1


(4
5


)


= 36.9◦ = m∠C


Example 4: Solve the right triangle.


Solution: To solve this right triangle, we need to find AB, BC and m∠A.
AB: Use sine ratio.


sin 62◦ = 25
AB


AB = 25sin 62◦
AB ≈ 28.31


BC: Use tangent ratio.


tan 62◦ = 25
BC


BC = 25tan 62◦
BC ≈ 13.30


m∠A: Use Triangle Sum Theorem


62◦ + 90◦ + m∠A = 180◦


m∠A = 28◦


Example 5: Solve the right triangle.


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Solution: The two acute angles are congruent, making them both 45◦. This is a 45-45-90 triangle. You
can use the trigonometric ratios or the special right triangle ratios.
Trigonometric Ratios


tan 45◦ = 15
BC


sin 45◦ = 15
AC


BC = 15tan 45◦ = 15 AC =
15


sin 45◦ ≈ 21.21


45-45-90 Triangle Ratios


BC = AB = 15, AC = 15

2 ≈ 21.21


Real-Life Situations
Example 6: A 25 foot tall flagpole casts a 42 feet shadow. What is the angle that the sun hits the
flagpole?


Solution: Draw a picture. The angle that the sun hits the flagpole is x◦. We need to use the inverse
tangent ratio.


tan x = 42
25


tan−1 42
25
≈ 59.2◦ = x


Example 7: Elise is standing on top of a 50 foot building and sees her friend, Molly. If Molly is 35 feet
away from the base of the building, what is the angle of depression from Elise to Molly? Elise’s eye height
is 4.5 feet.
Solution: Because of parallel lines, the angle of depression is equal to the angle at Molly, or x◦. We can
use the inverse tangent ratio.


tan−1
(54.5


30


)


= 61.2◦ = x


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Know What? Revisited To find the escalator’s angle of elevation, use the inverse sine.


sin−1
(115
230


)


= 30◦ The angle of elevation is 30◦.


Review Questions
• Questions 1-6 are similar to Example 1.
• Questions 7-12 are similar to Example 2.
• Questions13-21 are similar to Examples 3 and 4.
• Questions 22-24 are similar to Examples 6 and 7.
• Questions 25-30 are a review of the trigonometric ratios.


Use your calculator to find m∠A to the nearest tenth of a degree.


1.


2.


3.


207 www.ck12.org




4.


5.


6.


Let ∠A be an acute angle in a right triangle. Find m∠A to the nearest tenth of a degree.


7. sin A = 0.5684
8. cos A = 0.1234
9. tan A = 2.78


10. cos−1 0.9845
11. tan−1 15.93
12. sin−1 0.7851


Solving the following right triangles. Find all missing sides and angles. Round any decimal answers to the
nearest tenth.


13.


14.


15.


www.ck12.org 208




16.


17.


18.


19.


20.


21.


Real-Life Situations Use what you know about right triangles to solve for the missing angle. If needed,
draw a picture. Round all answers to the nearest tenth of a degree.


22. A 75 foot building casts an 82 foot shadow. What is the angle that the sun hits the building?
23. Over 2 miles (horizontal), a road rises 300 feet (vertical). What is the angle of elevation?
24. A boat is sailing and spots a shipwreck 650 feet below the water. A diver jumps from the boat and


swims 935 feet to reach the wreck. What is the angle of depression from the boat to the shipwreck?


Examining Patterns Below is a table that shows the sine, cosine, and tangent values for eight different
angle measures. Answer the following questions.


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Table 4.2:


10◦ 20◦ 30◦ 40◦ 50◦ 60◦ 70◦ 80◦


Sine 0.1736 0.3420 0.5 0.6428 0.7660 0.8660 0.9397 0.9848
Cosine 0.9848 0.9397 0.8660 0.7660 0.6428 0.5 0.3420 0.1736
Tangent 0.1763 0.3640 0.5774 0.8391 1.1918 1.7321 2.7475 5.6713


25. What value is equal to sin 40◦?
26. What value is equal to cos 70◦?
27. Describe what happens to the sine values as the angle measures increase.
28. Describe what happens to the cosine values as the angle measures increase.
29. What two numbers are the sine and cosine values between?
30. Find tan 85◦, tan 89◦, and tan 89.5◦ using your calculator. Now, describe what happens to the tangent


values as the angle measures increase.


Review Queue Answers
1. sin 36◦ = y7 cos 36◦ = x7


y = 4.11 x = 5.66
2. cos 12.7◦ = 40x tan 12.7◦ =


y
40


x = 41.00 y = 9.01
3.


4. sin 45◦ = 3
3

2
=

2


2


cos 45◦ = 3
3

2
=

2


2
tan 45◦ = 33 = 1


4.7 Chapter 8 Review
Keywords & Theorems
The Pythagorean Theorem


• Pythagorean Theorem
• Pythagorean Triple
• Distance Formula


The Pythagorean Theorem Converse


• Pythagorean Theorem Converse
• Theorem 8-3


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• Theorem 8-4


Similar Right Triangles


• Theorem 8-5
• Geometric Mean


Special Right Triangles


• Isosceles Right (45-45-90) Triangle
• 30-60-90 Triangle
• 45-45-90 Theorem
• 30-60-90 Theorem


Tangent, Sine and Cosine Ratios


• Trigonometry
• Adjacent (Leg)
• Opposite (Leg)
• Sine Ratio
• Cosine Ratio
• Tangent Ratio
• Angle of Depression
• Angle of Elevation


Solving Right Triangles


• Inverse Tangent
• Inverse Sine
• Inverse Cosine


Review
Fill in the blanks using right triangle 4ABC.


1. a2 + 2 = c2
2. sin = bc
3. tan = fd
4. cos = bc
5. tan−1


( f
e


)


=


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6. sin−1
( f
b


)


=
7. 2 + d2 = b2
8. ?b = bc
9. e? = ?c


10. df =
f
?


Solve the following right triangles using the Pythagorean Theorem, the trigonometric ratios, and the inverse
trigonometric ratios. When possible, simplify the radical. If not, round all decimal answers to the nearest
tenth.


11.


12.


13.


14.


15.


16.


17.


www.ck12.org 212




18.


19.


Determine if the following lengths make an acute, right, or obtuse triangle. If they make a right triangle,
determine if the lengths are a Pythagorean triple.


20. 11, 12, 13
21. 16, 30, 34
22. 20, 25, 42
23. 10



6, 30, 10



15


24. 22, 25, 31
25. 47, 27, 35


Find the value of x.


26.


27.


28.


29. The angle of elevation from the base of a mountain to its peak is 76◦. If its height is 2500 feet, what
is the length to reach the top? Round the answer to the nearest tenth.


30. Taylor is taking an aerial tour of San Francisco in a helicopter. He spots AT&T Park (baseball
stadium) at a horizontal distance of 850 feet and down (vertical) 475 feet. What is the angle of
depression from the helicopter to the park? Round the answer to the nearest tenth.


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Texas Instruments Resources


In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities
designed to supplement the objectives for some of the lessons in this chapter. See http:
//www.ck12.org/flexr/chapter/9693.


4.8 Study Guide
Keywords: Define, write theorems, and/or draw a diagram for each word below.
1st Section: The Pythagorean Theorem
Pythagorean Theorem
Pythagorean Triple
Distance Formula


Homework:
2nd Section: The Pythagorean Theorem Converse
Pythagorean Theorem Converse
Theorem 8-3
Theorem 8-4


Homework:
3rd Section: Similar Right Triangles
Theorem 8-5
Geometric Mean


Homework:


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4th Section: Special Right Triangles
Isosceles Right (45-45-90) Triangle


30-60-90 Triangle
45-45-90 Theorem
30-60-90 Theorem


Homework:
5th Section: Tangent, Sine and Cosine Ratios
Trigonometry
Adjacent (Leg)
Opposite (Leg)
Sine Ratio
Cosine Ratio
Tangent Ratio
Angle of Depression
Angle of Elevation


Homework:
6th Section: Solving Right Triangles
Inverse Tangent


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Inverse Sine
Inverse Cosine
Solving Right Triangles


Homework:


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Chapter 5


Circles


First, we will define all the parts of circles and explore the properties of tangent lines, arcs, inscribed
angles, and chords. Next, we will learn about angles and segments that are formed by chords, tangents
and secants. Lastly, we will place circles in the coordinate plane and find the equation of and graph circles.


5.1 Parts of Circles & Tangent Lines
Learning Objectives


• Define the parts of a circle.
• Discover the properties of tangent lines.


Review Queue
1. Find the equation of the line with m = 25 and y−intercept of 4.
2. Find the equation of the line with m = −2 and passes through (4, -5).
3. Find the equation of the line that passes though (6, 2) and (-3, -1).
4. Find the equation of the line perpendicular to the line in #2 and passes through (-8, 11).


Know What? The clock to the right is an ancient astronomical clock in Prague. It has a large background
circle that tells the local time and the “ancient time” and the smaller circle rotates to show the current
astrological sign. The yellow point is the center of the larger clock. How does the orange line relate to the
small and large circle? How does the hand with the moon on it relate to both circles?


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Defining Terms


Circle: The set of all points that are the same distance away from a specific point, called the center.
The center of the circle is point A. We call this circle, “circle A,” and it is labeled




A.
Radii (the plural of radius) are line segments. There are infinitely many radii in any circle and they are
all equal.


Radius: The distance from the center to the circle.
Chord: A line segment whose endpoints are on a circle.
Diameter: A chord that passes through the center of the circle.
Secant: A line that intersects a circle in two points.


The tangent ray −→TP and tangent segment TP are also called tangents.
The length of a diameter is two times the length of a radius.
Tangent: A line that intersects a circle in exactly one point.
Point of Tangency: The point where the tangent line touches the circle.
Example 1: Find the parts of




A that best fit each description.


a) A radius
b) A chord


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c) A tangent line
d) A point of tangency
e) A diameter
f) A secant
Solution:
a) HA or AF
b) CD, HF, or DG
c) ←→BJ
d) Point H
e) HF
f) ←→BD


Coplanar Circles


Example 2: Draw an example of how two circles can intersect with no, one and two points of intersection.
You will make three separate drawings.
Solution:


Tangent Circles: When two circles intersect at one point.


Concentric Circles: When two circles have the same center, but different radii.
Congruent Circles: Two circles with the same radius, but different centers.
If two circles have different radii, they are similar. All circles are similar.
Example 3: Determine if any of the following circles are congruent.


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Solution: From each center, count the units to the circle. It is easiest to count vertically or horizontally.
Doing this, we have:


Radius of


A = 3 units


Radius of


B = 4 units


Radius of


C = 3 units


From these measurements, we see that


A


C.
Notice the circles are congruent. The lengths of the radii are equal.


Internally & Externally Tangent
If two circles are tangent to each other, then they are internally or externally tangent.
Internally Tangent Circles: When two circles are tangent and one is inside the other.
Externally Tangent Circles: When two circles are tangent and next to each other.
Internally Tangent


Externally Tangent


If circles are not tangent, they can still share a tangent line, called a common tangent.


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Common Internal Tangent: A line that is tangent to two circles and passes between the circles.
Common External Tangent: A line that is tangent to two circles and stays on the top or bottom of
both circles.
Common Internal Tangent


Common External Tangent


Tangents and Radii
Let’s investigate a tangent line and the radius drawn to the point of tangency.
Investigation 9-1: Tangent Line and Radius Property
Tools Needed: compass, ruler, pencil, paper, protractor


1. Using your compass, draw a circle. Locate the center and draw a radius. Label the radius AB, with
A as the center.


2. Draw a tangent line, ←→BC, where B is the point of tangency. To draw a tangent line, take your ruler
and line it up with point B. B must be the only point on the circle that the line passes through.


3. Find m∠ABC.


Tangent to a Circle Theorem: A line is tangent to a circle if and only if the line is perpendicular to
the radius drawn to the point of tangency.


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←→
BC is tangent at point B if and only if ←→BC ⊥ AB.
This theorem uses the words “if and only if,” making it a biconditional statement, which means the converse
of this theorem is also true.
Example 4: In




A, CB is tangent at point B. Find AC. Reduce any radicals.


Solution: CB is tangent, so AB ⊥ CB and 4ABC a right triangle. Use the Pythagorean Theorem to find
AC.


52 + 82 = AC2


25 + 64 = AC2


89 = AC2


AC =

89


Example 5: Find DC, in


A. Round your answer to the nearest hundredth.
Solution: DC = AC − AD
DC =



89 − 5 ≈ 4.43


Example 6: Determine if the triangle below is a right triangle.


Solution: Again, use the Pythagorean Theorem. 4

10 is the longest side, so it will be c.


82 + 102 ?
(


4

10


)2


64 + 100 , 160


4ABC is not a right triangle. From this, we also find that CB is not tangent to


A.
Example 7: Find AB in




A and


B. Reduce the radical.


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Solution: AD ⊥ DC and DC ⊥ CB. Draw in BE, so EDCB is a rectangle. Use the Pythagorean Theorem
to find AB.


52 + 552 = AC2


25 + 3025 = AC2


3050 = AC2


AC =

3050 = 5



122


Tangent Segments
Theorem 9-2: If two tangent segments are drawn from the same external point, then they are equal.
BC and DC have C as an endpoint and are tangent; BC DC.


Example 8: Find the perimeter of 4ABC.


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Solution: AE = AD, EB = BF, and CF = CD. Therefore, the perimeter of 4ABC = 6+6+4+4+7+7 = 34.


G is inscribed in 4ABC. A circle is inscribed in a polygon, if every side of the polygon is tangent to
the circle.
Example 9: If D and A are the centers and AE is tangent to both circles, find DC.


Solution: AE ⊥ DE and AE ⊥ AC and 4ABC ∼ 4DBE.
To find DB, use the Pythagorean Theorem.


102 + 242 = DB2


100 + 576 = 676


DB =

676 = 26


To find BC, use similar triangles. 510 = BC26 −→ BC = 13. DC = AB + BC = 26 + 13 = 39
Example 10: Algebra Connection Find the value of x.


Solution: AB CB by Theorem 9-2. Set AB = CB and solve for x.


4x − 9 = 15
4x = 24
x = 6


Know What? Revisited The orange line is a diameter of the smaller circle. Since this line passes
through the center of the larger circle (yellow point), it is part of one of its diameters. The “moon” hand
is a diameter of the larger circle, but a secant of the smaller circle.


Review Questions
• Questions 1-9 are similar to Example 1.
• Questions 10-12 are similar to Example 2.
• Questions 13-17 are similar to Example 3.
• Questions 18-20 are similar to Example 6.
• Questions 21-26 are similar to Example 4, 5, 7, and 10.
• Questions 27-31 are similar to Example 9.
• Questions 32-37 are similar to Example 8.


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• Question 38 and 39 use the proof of Theorem 9-2.
• Question 40 uses Theorem 9-2.


Determine which term best describes each of the following parts of


P.


1. KG
2. ←→FH
3. KH
4. E
5. ←→BK
6. ←→CF
7. A
8. JG
9. What is the longest chord in any circle?


Copy each pair of circles. Draw in all common tangents.


10.


11.


12.


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Coordinate Geometry Use the graph below to answer the following questions.


13. Find the radius of each circle.
14. Are any circles congruent? How do you know?
15. Find all the common tangents for




B and


C.
16.




C and


E are externally tangent. What is CE?
17. Find the equation of CE.


Determine whether the given segment is tangent to


K.


18.


19.


20.


Algebra Connection Find the value of the indicated length(s) in


C. A and B are points of tangency.
Simplify all radicals.


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21.


22.


23.


24.


25.


26.


A and B are points of tangency for


C and


D.


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27. Is 4AEC ∼ 4BED? Why?
28. Find CE.
29. Find BE.
30. Find ED.
31. Find BC and AD.




A is inscribed in BDFH.


32. Find the perimeter of BDFH.
33. What type of quadrilateral is BDFH? How do you know?
34. Draw a circle inscribed in a square. If the radius of the circle is 5, what is the perimeter of the


square?
35. Can a circle be inscribed in a rectangle? If so, draw it. If not, explain.
36. Draw a triangle with two sides tangent to a circle, but the third side is not.
37. Can a circle be inscribed in an obtuse triangle? If so, draw it. If not, explain.
38. Fill in the blanks in the proof of Theorem 9-2.


Given: AB and CB with points of tangency at A and C.
AD and DC are radii.
Prove: AB CB


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Table 5.1:


Statement Reason
1.
2. AD DC
3. DA ⊥ AB and DC ⊥ CB
4. Definition of perpendicular lines
5. Connecting two existing points
6. 4ADB and 4DCB are right triangles
7. DB DB
8. 4ABD 4CBD
9. AB CB


39. Fill in the blanks, using the proof from #38.
(a) ABCD is a _____________ (type of quadrilateral).
(b) The line that connects the ___________ and the external point B __________ ∠ABC.


40. Points A, B, and C are points of tangency for the three tangent circles. Explain why AT BT CT .


Review Queue Answers
1. y = 25 x + 4
2. y = −2x + 3
3. m = 2−(−1)6−(−3) = 39 = 13


y = 13 x + b→ plug in(6, 2)
2 = 13(6) + b
2 = 2 + b→ b = 0
y = 13 x


4. m⊥ = −3
11 = −3(−8) + b
11 = 24 + b→ b = −13
y = −3x − 13


5.2 Properties of Arcs
Learning Objectives


• Define and measure central angles, minor arcs, and major arcs.


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Review Queue


1. What kind of triangle is 4ABC?
2. How does BD relate to 4ABC?
3. Find m∠ABC and m∠ABD.
Round to the nearest tenth. Use the trig ratios.
4. Find AD.
5. Find AC.
Know What? The Ferris wheel to the right has equally spaced seats, such that the central angle is 20◦.
How many seats are on this ride? Why do you think it is important to have equally spaced seats on a
Ferris wheel?


Central Angles & Arcs
Recall that a straight angle is 180◦. If take two straight angles and put one on top of the other, we would
have a circle. This means that a circle has 360◦, 180◦ + 180◦. This also means that a semicircle, or half
circle, is 180◦.


Arc: A section of the circle.
Semicircle: An arc that measures 180◦.


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To label an arc, place a curve above the endpoints. You may want to use 3 points to clarify.
ÊHG and ̂EJG are semicircles mÊHG = 180◦


Central Angle: The angle formed by two radii and its vertex at the center of the circle.
Minor Arc: An arc that is less than 180◦


Major Arc: An arc that is greater than 180◦. Always use 3 letters to label a major arc.


The central angle is ∠BAC.
The minor arc is ̂BC.
The major arc is B̂DC.
Every central angle divides a circle into two arcs.
An arc can be measured in degrees or in a linear measure (cm, ft, etc.). In this chapter we will use degree
measure. The measure of the minor arc is the same as the measure of the central angle that
corresponds to it. The measure of the major arc is 360◦ minus the measure of the minor arc.
Example 1: Find m̂AB and mÂDB in




C.


Solution: m̂AB = m ̂ACB. So, m̂AB = 102◦.


mÂDB = 360◦ − m̂AB = 360◦ − 102◦ = 258◦


Example 2: Find the measures of the arcs in


A. EB is a diameter.


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Solution: Because EB is a diameter, m∠EAB = 180◦. Each arc is the same as its corresponding central
angle.


m̂BF = m∠FAB = 60◦


m̂EF = m∠EAF = 120◦ → 180◦ − 60◦


m̂ED = m∠EAD = 38◦ → 180◦ − 90◦ − 52◦


m̂DC = m∠DAC = 90◦


m̂BC = m∠BAC = 52◦


Congruent Arcs: Two arcs are congruent if their central angles are congruent.
Example 3: List the congruent arcs in




C below. AB and DE are diameters.


Solution: ∠ACD = ∠ECB because they are vertical angles. ∠DCB = ∠ACE because they are also vertical
angles.
̂AD ̂EB and ̂AE ̂DB
Example 4: Are the blue arcs congruent? Explain why or why not.
a)


b)


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Solution:
a) ̂AD ̂BC because they have the same central angle measure and in the same circle.
b) The two arcs have the same measure, but are not congruent because the circles have different radii.


Arc Addition Postulate


Just like the Angle Addition Postulate and the Segment Addition Postulate, there is an Arc Addition
Postulate.
Arc Addition Postulate: The measure of the arc formed by two adjacent arcs is the sum of the measures
of the two arcs.


m̂AD + m̂DB = mÂDB


Example 5: Find the measure of the arcs in


A. EB is a diameter.


a) mF̂ED
b) mĈDF
c) mD̂FC
Solution: Use the Arc Addition Postulate.
a) mF̂ED = m̂FE + m̂ED = 120◦ + 38◦ = 158◦


b) mĈDF = m̂CD + m̂DE + m̂EF = 90◦ + 38◦ + 120◦ = 248◦


c) mD̂FC = 38◦ + 120◦ + 60◦ + 52◦ = 270◦


Example 6: Algebra Connection Find the value of x for


C below.


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Solution:


m̂AB + m̂AD + m̂DB = 360◦


(4x + 15)◦ + 92◦ + (6x + 3)◦ = 360◦


10x + 110◦ = 360◦


10x = 250◦


x = 25◦


Know What? Revisited Because the seats are 20◦ apart, there will be 360◦20◦ = 18 seats. It is important
to have the seats evenly spaced for the balance of the Ferris wheel.


Review Questions
• Questions 1-6 use the definition of minor arc, major arc, and semicircle.
• Question 7 is similar to Example 3.
• Questions 8 and 9 are similar to Example 5.
• Questions 10-15 are similar to Example 1.
• Questions 16-18 are similar to Example 4.
• Questions 19-26 are similar to Example 2 and 5.
• Questions 27-29 are similar to Example 6.
• Question 30 is a challenge.


Determine if the arcs below are a minor arc, major arc, or semicircle of


G. EB is a diameter.


1. ̂AB
2. ÂBD
3. B̂CE
4. ĈAE
5. ̂ABC
6. ̂EAB


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7. Are there any congruent arcs? If so, list them.
8. If m̂BC = 48◦, find m̂CD.
9. Using #8, find mĈAE.


Find the measure of the minor arc and the major arc in each circle below.


10.


11.


12.


13.


14.


15.


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Determine if the blue arcs are congruent. If so, state why.


16.


17.


18.


Find the measure of the indicated arcs or central angles in


A. DG is a diameter.


19. ̂DE
20. ̂DC
21. ̂GAB
22. ̂FG
23. ÊDB
24. ̂EAB
25. D̂CF
26. D̂BE


Algebra Connection Find the measure of x in


P.


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27.


28.


29.


30. Challenge What can you conclude about


A and


B?


Review Queue Answers
1. isosceles
2. BD is the angle bisector of ∠ABC and the perpendicular bisector of AC.
3. m∠ABC = 40◦,m∠ABD = 25◦
4. cos 70◦ = AD9 → AD = 9 · cos 70◦ = 3.1
5. AC = 2 · AD = 2 · 3.1 = 6.2


5.3 Properties of Chords
Learning Objectives


• Find the lengths of chords in a circle.
• Discover properties of chords and arcs.


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Review Queue


1. Draw a chord in a circle.
2. Draw a diameter in the circle from #1. Is a diameter a chord?
3. 4ABC is an equilateral triangle in




A. Find m̂BC and mB̂DC.


4. 4ABC and 4ADE are equilateral triangles in


A. List a pair of congruent arcs and chords.


Know What? To the right is the Gran Teatro Falla, in Cadiz, Andalucía, Spain. Notice the five windows,
A − E.




A


E and


B


C


D. Each window is topped with a 240◦ arc. The gold chord in
each circle connects the rectangular portion of the window to the circle. Which chords are congruent?


Recall from the first section, a chord is a line segment whose endpoints are on a circle. A diameter is the
longest chord in a circle.


Congruent Chords & Congruent Arcs


From #4 in the Review Queue above, we noticed that BC DE and ̂BC ̂DE.
Theorem 9-3: In the same circle or congruent circles, minor arcs are congruent if and only if their
corresponding chords are congruent.


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In both of these pictures, BE CD and ̂BE ̂CD.
In the second circle, 4BAE 4CAD by SAS.
Example 1: Use




A to answer the following.


a) If m̂BD = 125◦, find m̂CD.
b) If m̂BC = 80◦, find m̂CD.
Solution:
a) BD = CD, which means the arcs are equal too. m̂CD = 125◦.
b) m̂CD m̂BD because BD = CD.


m̂BC + m̂CD + m̂BD = 360◦


80◦ + 2m̂CD = 360◦


2m̂CD = 280◦


m̂CD = 140◦


Investigation 9-2: Perpendicular Bisector of a Chord
Tools Needed: paper, pencil, compass, ruler
1. Draw a circle. Label the center A.


2. Draw a chord. Label it BC.


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3. Find the midpoint of BC using a ruler. Label it D.


4. Connect A and D to form a diameter. How does AD relate to BC?
Theorem 9-4: The perpendicular bisector of a chord is also a diameter.


If AD ⊥ BC and BD DC then EF is a diameter.
If EF ⊥ BC, then BD DC and ̂BE ̂EC.
Theorem 9-5: If a diameter is perpendicular to a chord, then the diameter bisects the chord and its
corresponding arc.
Example 2: Find the value of x and y.


Solution: The diameter perpendicular to the chord. From Theorem 9-5, x = 6 and y = 75◦.
Example 3: Is the converse of Theorem 9-4 true?


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Solution: The converse of Theorem 9-4 would be: A diameter is also the perpendicular bisector of a chord.
This is not true, a diameter cannot always be a perpendicular bisector to every chord. See the picture.
Example 4: Algebra Connection Find the value of x and y.


Solution: The diameter is perpendicular to the chord, which means it bisects the chord and the arc. Set
up an equation for x and y.


(3x − 4)◦ = (5x − 18)◦ y + 4 = 2y + 1
14◦ = 2x 3 = y
7◦ = x


Equidistant Congruent Chords
Investigation 9-3: Properties of Congruent Chords
Tools Needed: pencil, paper, compass, ruler


1. Draw a circle with a radius of 2 inches and two chords that are both 3 inches. Label like the picture
to the right. This diagram is drawn to scale.


2. From the center, draw the perpendicular segment to AB and CD. You can use Investigation 3-2


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3. Erase the arc marks and lines beyond the points of intersection, leaving FE and EG. Find the
measure of these segments. What do you notice?


Theorem 9-6: In the same circle or congruent circles, two chords are congruent if and only if they are
equidistant from the center.


The shortest distance from any point to a line is the perpendicular line between them.
If FE = EG and EF ⊥ EG, then AB and CD are equidistant to the center and AB CD.
Example 5: Algebra Connection Find the value of x.


Solution: Because the distance from the center to the chords is congruent and perpendicular to the chords,
the chords are equal.


6x − 7 = 35
6x = 42
x = 7


Example 6: BD = 12 and AC = 3 in


A. Find the radius.


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Solution: First find the radius. AB is a radius, so we can use the right triangle 4ABC, so AB is the
hypotenuse. From Theorem 9-5, BC = 6.


32 + 62 = AB2


9 + 36 = AB2


AB =

45 = 3



5


Example 7: Find m̂BD from Example 6.
Solution: First, find the corresponding central angle, ∠BAD. We can find m∠BAC using the tangent ratio.
Then, multiply m∠BAC by 2 for m∠BAD and m̂BD.


tan−1
(6
3


)


= m∠BAC


m∠BAC ≈ 63.43◦


m∠BAD ≈ 2 · 63.43◦ ≈ 126.86◦ ≈ m̂BD


Know What? Revisited In the picture, the chords from


A and


E are congruent and the chords
from




B,


C, and


D are also congruent. We know this from Theorem 9-3.


Review Questions
• Questions 1-3 use the theorems from this section and similar to Example 3.
• Questions 4-10 use the definitions and theorems from this section.
• Questions 11-16 are similar to Example 1 and 2.
• Questions 17-25 are similar to Examples 2, 4, 5, and 6.
• Questions 26 and 27 are similar to Example 7.
• Questions 28-30 use the theorems from this section.


1. Two chords in a circle are perpendicular and congruent. Does one of them have to be a diameter?
Why or why not?


2. Write the converse of Theorem 9-5. Is it true? If not, draw a counterexample.
3. Write the original and converse statements from Theorem 9-3 and Theorem 9-6.


Fill in the blanks.


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4. DF
5. ̂AC
6. ̂DJ
7. EJ
8. ∠AGH
9. ∠DGF


10. List all the congruent radii in


G.


Find the value of the indicated arc in


A.


11. m̂BC


12. m̂BD


13. m̂BC


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14. m̂BD


15. m̂BD


16. m̂BD


Algebra Connection Find the value of x and/or y.


17.


18.


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19.


20. AB = 32


21.


22.


23.


24.


25. AB = 20


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26. Find m̂AB in Question 20. Round your answer to the nearest tenth of a degree.
27. Find m̂AB in Question 25. Round your answer to the nearest tenth of a degree.


In problems 28-30, what can you conclude about the picture? State a theorem that justifies your answer.
You may assume that A is the center of the circle.


28.


29.


30.


Review Queue Answers
1 & 2. Answers will vary


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3. m̂BC = 60◦,mB̂DC = 300◦


4. BC DE and ̂BC ̂DE


5.4 Inscribed Angles
Learning Objectives


• Find the measure of inscribed angles and the arcs they intercept.


Review Queue
We are going to use #14 from the homework in the previous section.


1. What is the measure of each angle in the triangle? How do you know?
2. What do you know about the three arcs?
3. What is the measure of each arc?


Know What? The closest you can get to the White House are the walking trails on the far right. You
want to get as close as you can (on the trail) to the fence to take a picture (you were not allowed to walk
on the grass). Where else can you take a picture from to get the same frame of the White House? Your
line of sight in the camera is marked in the picture as the grey lines.


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Inscribed Angles
In addition to central angles, we will now learn about inscribed angles in circles.
Inscribed Angle: An angle with its vertex on the circle and sides are chords.


Intercepted Arc: The arc that is inside the inscribed angle and endpoints are on the angle.
The vertex of an inscribed angle can be anywhere on the circle as long as its sides intersect the circle to
form an intercepted arc.
Investigation 9-4: Measuring an Inscribed Angle
Tools Needed: pencil, paper, compass, ruler, protractor


1. Draw three circles with three different inscribed angles. Try to make all the angles different sizes.


2. Using your ruler, draw in the corresponding central angle for each angle and label each set of end-
points.


3. Using your protractor measure the six angles and determine if there is a relationship between the
central angle, the inscribed angle, and the intercepted arc.


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m∠LAM = m∠NBP = m∠QCR =
m̂LM = m̂NP = m̂QR =
m∠LKM = m∠NOP = m∠QSR =


Inscribed Angle Theorem: The measure of an inscribed angle is half the measure of its intercepted arc.


m∠ADC = 1
2
m̂AC


m̂AC = 2m∠ADC


Example 1: Find m̂DC and m∠ADB.


Solution: From the Inscribed Angle Theorem:


m̂DC = 2 · 45◦ = 90◦


m∠ADB = 1
2
· 76◦ = 38◦


Example 2: Find m∠ADB and m∠ACB.


Solution: The intercepted arc for both angles is ̂AB. Therefore,


m∠ADB = 1
2
· 124◦ = 62◦


m∠ACB = 1
2
· 124◦ = 62◦


This example leads us to our next theorem.


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Theorem 9-8: Inscribed angles that intercept the same arc are congruent.


∠ADB and ∠ACB intercept ̂AB, so m∠ADB = m∠ACB.
∠DAC and ∠DBC intercept ̂DC, so m∠DAC = m∠DBC.
Example 3: Find m∠DAB in




C.


Solution: C is the center, so DB is a diameter. ∠DAB endpoints are on the diameter, so the central angle
is 180◦.


m∠DAB = 1
2
· 180◦ = 90◦.


Theorem 9-9: An angle intercepts a semicircle if an only if it is a right angle.


∠DAB intercepts a semicircle, so m∠DAB = 90◦.
∠DAB is a right angle, so ̂DB is a semicircle.
Anytime a right angle is inscribed in a circle, the endpoints of the angle are the endpoints
of a diameter and the diameter is the hypotenuse.
Example 4: Find m∠PMN, m̂PN, m∠MNP, and m∠LNP.


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Solution:


m∠PMN = m∠PLN = 68◦ by Theorem 9 − 8.
m̂PN = 2 · 68◦ = 136◦ from the Inscribed Angle Theorem.


m∠MNP = 90◦ by Theorem 9 − 9.


m∠LNP = 1
2
· 92◦ = 46◦ from the Inscribed Angle Theorem.


Inscribed Quadrilaterals
Inscribed Polygon: A polygon where every vertex is on a circle.


Investigation 9-5: Inscribing Quadrilaterals
Tools Needed: pencil, paper, compass, ruler, colored pencils, scissors


1. Draw a circle. Mark the center point A.


2. Place four points on the circle. Connect them to form a quadrilateral. Color in the 4 angles.


3. Cut out the quadrilateral. Then cut the diagonal CE, making two triangles.


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4. Line up ∠B and ∠D so that they are next to each other. What do you notice?


By cutting the quadrilateral in half, we are able to show that ∠B and ∠D form a linear pair when they are
placed next to each other, making ∠B and ∠D supplementary.
Theorem 9-10: A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary.


If ABCD is inscribed in


E, then m∠A + m∠C = 180◦ and m∠B + m∠D = 180◦.
If m∠A + m∠C = 180◦ and m∠B + m∠D = 180◦, then ABCD is inscribed in




E.
Example 5: Find the value of the missing variables.
a)


b)


Solution:


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a) x + 80◦ = 180◦ y + 71◦ = 180◦
x = 100◦ y = 109◦


b) z + 93◦ = 180◦ x = 12(58◦ + 106◦) y + 82◦ = 180◦
z = 87◦ x = 82◦ y = 98◦


Example 6: Algebra Connection Find x and y in the picture below.


Solution:


(7x + 1)◦ + 105◦ = 180◦ (4y + 14)◦ + (7y + 1)◦ = 180◦


7x + 106◦ = 180◦ 11y + 15◦ = 180◦


7x = 84◦ 11y = 165◦


x = 12◦ y = 15◦


Know What? Revisited You can take the picture from anywhere on the semicircular walking path, the
frame will be the same.


Review Questions
• Questions 1-8 use the vocabulary and theorems learned in this section.
• Questions 9-27 are similar to Examples 1-5.
• Questions 28-33 are similar to Example 6.
• Question 34 is a proof of the Inscribed Angle Theorem.


Fill in the blanks.


1. A(n) _______________ polygon has all its vertices on a circle.
2. An inscribed angle is ____________ the measure of the intercepted arc.
3. A central angle is ________________ the measure of the intercepted arc.
4. An angle inscribed in a ________________ is 90◦.
5. Two inscribed angles that intercept the same arc are _______________.


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6. The _____________ angles of an inscribed quadrilateral are ________________.
7. The sides of an inscribed angle are ___________________.
8. Draw inscribed angle ∠JKL in




M. Then draw central angle ∠JML. How do the two angles relate?


Quadrilateral ABCD is inscribed in


E. Find:


9. m∠DBC
10. m̂BC
11. m̂AB
12. m∠ACD
13. m∠ADC
14. m∠ACB


Quadrilateral ABCD is inscribed in


E. Find:


15. m∠A
16. m∠B
17. m∠C
18. m∠D


Find the value of x and/or y in


A.


19.


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20.


21.


22.


23.


24.


25.


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26.


27.


Algebra Connection Solve for x.


28.


29.


30.


31.


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32.


33.


34. Fill in the blanks of the Inscribed Angle Theorem proof.


Given: Inscribed ∠ABC and diameter BD
Prove: m∠ABC = 12m̂AC


Table 5.2:


Statement Reason
1. Inscribed ∠ABC and diameter BD
m∠ABE = x◦ and m∠CBE = y◦


2. x◦ + y◦ = m∠ABC
3. All radii are congruent
4. Definition of an isosceles triangle
5. m∠EAB = x◦ and m∠ECB = y◦
6. m∠AED = 2x◦ and m∠CED = 2y◦
7. m̂AD = 2x◦ and m̂DC = 2y◦
8. Arc Addition Postulate
9. m̂AC = 2x◦ + 2y◦
10. Distributive PoE
11. m̂AC = 2m∠ABC
12. m∠ABC = 12m̂AC


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Review Queue Answers
1. 60◦, it is an equilateral triangle.
2. They are congruent because the chords are congruent.
3. 360◦3 = 120◦


5.5 Angles of Chords, Secants, and Tangents
Learning Objectives


• Find the measures of angles formed by chords, secants, and tangents.


Review Queue


1. What is m∠OML and m∠OPL? How do you know?
2. Find m∠MLP.
3. Find mM̂NP.


Know What? The sun’s rays hit the Earth such that the tangent rays determine when daytime and night
time are. If the arc that is exposed to sunlight is 178◦, what is the angle at which the sun’s rays hit the
earth (x◦)?


Angle on a Circle
When an angle is on a circle, the vertex is on the edge of the circle. One type of angle on a circle is the
inscribed angle, from the previous section. Another type of angle on a circle is one formed by a tangent
and a chord.
Investigation 9-6: The Measure of an Angle formed by a Tangent and a Chord
Tools Needed: pencil, paper, ruler, compass, protractor


1. Draw


A with chord BC and tangent line ←→ED with point of tangency C.


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2. Draw in central angle ∠CAB. Find m∠CAB and m∠BCE.


3. Find m̂BC. How does the measure of this arc relate to m∠BCE?


Theorem 9-11: The measure of an angle formed by a chord and a tangent that intersect on the circle is
half the measure of the intercepted arc.


m∠DBA = 1
2
m̂AB


We now know that there are two types of angles that are half the measure of the intercepted arc; an
inscribed angle and an angle formed by a chord and a tangent.
Example 1: Find:
a) m∠BAD


b) m ̂AEB


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Solution: Use Theorem 9-11.
a) m∠BAD = 12m̂AB = 12 · 124◦ = 62◦


b) m ̂AEB = 2 · m∠DAB = 2 · 133◦ = 266◦


Example 2: Find a, b, and c.


Solution:
50◦ + 45◦ + m∠a = 180◦ straight angle


m∠a = 85◦


m∠b = 1
2
· m̂AC


m̂AC = 2 · m∠EAC = 2 · 45◦ = 90◦


m∠b = 1
2
· 90◦ = 45◦


85◦ + 45◦ + m∠c = 180◦ Triangle Sum Theorem
m∠c = 50◦


From this example, we see that Theorem 9-8 is true for angles formed by a tangent and chord with the
vertex on the circle. If two angles, with their vertices on the circle, intercept the same arc then
the angles are congruent.


Angles inside a Circle
An angle is inside a circle when the vertex anywhere inside the circle, but not on the center.
Investigation 9-7: Find the Measure of an Angle inside a Circle
Tools Needed: pencil, paper, compass, ruler, protractor, colored pencils (optional)


1. Draw


A with chord BC and DE. Label the point of intersection P.


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2. Draw central angles ∠DAB and ∠CAE. Use colored pencils, if desired.


3. Find m∠DPB, m∠DAB, and m∠CAE. Find m̂DB and m̂CE.
4. Find m̂DB+m̂CE2 .
5. What do you notice?


Theorem 9-12: The measure of the angle formed by two chords that intersect inside a circle is the average
of the measure of the intercepted arcs.


m∠SVR = 1
2


(


m̂SR + m̂TQ
)


= m
̂SR + m̂TQ


2
= m∠TVQ


m∠SVT = 1
2


(


m̂ST + m̂RQ
)


= m
̂ST + m̂RQ


2
= m∠RVQ


Example 3: Find x.
a)


b)


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c)


Solution: Use Theorem 9-12 to write an equation.
a) x = 129◦+71◦2 = 200




2 = 100


b) 40◦ = 52◦+x2
80◦ = 52◦ + x
28◦ = x
c) x is supplementary to the angle that the average of the given intercepted arcs, y.


y = 19
◦ + 107◦


2
= 126




2
= 63◦ x + 63◦ = 180◦; x = 117◦


Angles outside a Circle
An angle is outside a circle if the vertex of the angle is outside the circle and the sides are tangents or
secants. The possibilities are: an angle formed by two tangents, an angle formed by a tangent and a secant,
and an angle formed by two secants.
Investigation 9-8: Find the Measure of an Angle outside a Circle
Tools Needed: pencil, paper, ruler, compass, protractor, colored pencils (optional)


1. Draw three circles and label the centers A, B, and C. In


A draw two secant rays with the same
endpoint. In




B, draw two tangent rays with the same endpoint. In


C, draw a tangent ray and
a secant ray with the same endpoint. Label the points like the pictures below.


2. Draw in all the central angles. Using a protractor, measure the central angles and find the measures
of each intercepted arc.


3. Find m∠EDF, m∠MLN, and m∠RQS .
4. Find m̂EF−m̂GH2 , mM̂PN−m


̂MN
2 , and m


̂RS−m̂RT
2 . What do you notice?


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Theorem 9-13: The measure of an angle formed by two secants, two tangents, or a secant and a tangent
from a point outside the circle is half the difference of the measures of the intercepted arcs.


m∠D = m
̂EF − m̂GH


2


m∠L = mM̂PN − m
̂MN


2


m∠Q = m
̂RS − m̂RT


2


Example 4: Find the measure of x.
a)


b)


c)


Solution: For all of the above problems we can use Theorem 9-13.
a) x = 125◦−27◦2 = 98




2 = 49


b) 40◦ is not the intercepted arc. The intercepted arc is 120◦, (360◦−200◦−40◦). x = 200◦−120◦2 = 80


2 = 40


c) Find the other intercepted arc, 360◦ − 265◦ = 95◦ x = 265◦−95◦2 = 170


2 = 85


Know What? Revisited From Theorem 9-13, we know x = 182◦−178◦2 = 4


2 = 2
◦.


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Review Questions
• Questions 1-3 use the definitions of tangent and secant lines.
• Questions 4-7 use the definition and theorems learned in this section.
• Questions 8-25 are similar to Examples 1-4.
• Questions 26 and 27 are similar to Example 4, but also a challenge.
• Questions 28 and 29 are fill-in-the-blank proofs of Theorems 9-12 and 9-13.


1. Draw two secants that intersect:
(a) inside a circle.
(b) on a circle.
(c) outside a circle.


2. Can two tangent lines intersect inside a circle? Why or why not?
3. Draw a tangent and a secant that intersect:


(a) on a circle.
(b) outside a circle.


Fill in the blanks.


4. If the vertex of an angle is on the _______________ of a circle, then its measure is ____-
___________ to the intercepted arc.


5. If the vertex of an angle is _______________ a circle, then its measure is the average of the
__________________ arcs.


6. If the vertex of an angle is ________ a circle, then its measure is ______________ the
intercepted arc.


7. If the vertex of an angle is ____________ a circle, then its measure is ___________ the
difference of the intercepted arcs.


For questions 8-25, find the value of the missing variable(s).


8.


9.


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10.


11.


12.


13.


14.


15.


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16.


17.


18.


19.


20.


21.


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22.


23. y , 60◦


24.


25.


Challenge Solve for x.


26.


27.


28. Fill in the blanks of the proof for Theorem 9-12.


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Given: Intersecting chords AC and BD.
Prove: m∠a = 12


(


m̂DC + m̂AB
)


Table 5.3:


Statement Reason
1. Intersecting chords AC and BD.
2. Draw BC Construction


3. m∠DBC = 12m̂DC
m∠ACB = 12m̂AB
4. m∠a = m∠DBC + m∠ACB
5. m∠a = 12m̂DC + 12m̂AB


29. Fill in the blanks of the proof for Theorem 9-13.


Given: Secant rays −→AB and −→AC
Prove: m∠a = 12


(


m̂BC − m̂DE
)


Table 5.4:


Statement Reason
1. Intersecting secants −→AB and −→AC.
2. Draw BE. Construction


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Table 5.4: (continued)


Statement Reason
3. m∠BEC = 12m̂BC
m∠DBE = 12m̂DE
5. m∠a + m∠DBE = m∠BEC
6. Subtraction PoE
7. Substitution
8. m∠a = 12


(


m̂BC − m̂DE
)


Review Queue Answers
1. m∠OML = m∠OPL = 90◦ because a tangent line and a radius drawn to the point of tangency are


perpendicular.
2. 165◦ + m∠OML + m∠OPL + m∠MLP = 360◦


165◦ + 90◦ + 90◦ + m∠MLP = 360◦
m∠MLP = 15◦


3. mM̂NP = 360◦ − 165◦ = 195◦


5.6 Segments of Chords, Secants, and Tangents
Learning Objectives


• Find the lengths of segments within circles.


Review Queue


1. What do you know about m∠DAC and m∠DBC? Why?
2. What do you know about m∠AED and m∠BEC? Why?
3. Is 4AED ∼ 4BEC? How do you know?
4. If AE = 8, ED = 7, and BE = 6, find EC.


Know What? At a particular time during its orbit, the moon is 238,857 miles from Beijing, China. On
the same line, Yukon is 12,451 miles from Beijing. Drawing another line from the moon to Cape Horn we
see that Jakarta, Indonesia is collinear. If the distance from the moon to Jakarta is 240,128 miles, what is
the distance from Cape Horn to Jakarta?


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Segments from Chords
In the Review Queue above, we have two chords that intersect inside a circle. The two triangles are similar,
making the sides in each triangle proportional.
Theorem 9-14: If two chords intersect inside a circle so that one is divided into segments of length a and
b and the other into segments of length c and d then ab = cd.


The product of the segments of one chord is equal to the product of segments of the second
chord.
ab = cd
Example 1: Find x in each diagram below.
a)


b)


Solution: Use the ratio from Theorem 9-14.
a) 12 · 8 = 10 · x
96 = 10x
9.6 = x
b) x · 15 = 5 · 9


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15x = 45
x = 3
Example 2: Algebra Connection Solve for x.
a)


b)


Solution: Use Theorem 9-13.
a) 8 · 24 = (3x + 1) · 12
192 = 36x + 12
180 = 36x
5 = x
b) (x − 5)21 = (x − 9)24
21x − 105 = 24x − 216
111 = 3x
37 = x


Segments from Secants
In addition to forming an angle outside of a circle, the circle can divide the secants into segments that are
proportional with each other.
Theorem 9-15: If two secants are drawn from a common point outside a circle and the segments are
labeled as below, then a(a + b) = c(c + d).


The product of the outer segment and the whole of one secant equals the product of the outer
segment and the whole of the other secant.


a(a + b) = c(c + d)


Example 3: Find the value of the missing variable.


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a)


b)


Solution: Use Theorem 9-15 to set up an equation.
a) 18 · (18 + x) = 16 · (16 + 24)
324 + 18x = 256 + 384
18x = 316
x = 1759
b) x · (x + x) = 9 · 32
2x2 = 288
x2 = 144
x = 12, x , −12 (length is not negative)


Segments from Secants and Tangents
If a tangent and secant meet at a common point outside a circle, the segments created have a similar
relationship to that of two secant rays in Example 3.
Theorem 9-16: If a tangent and a secant are drawn from a common point outside the circle (and the
segments are labeled like the picture below), then a2 = b(b + c).


The product of the outside segment of the secant and the whole is equal to the square of the
tangent.


a2 = b(b + c)


Example 4: Find the value of the missing segment.


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a)


b)


Solution: Use Theorem 9-16.
a) x2 = 4(4 + 12)
x2 = 4 · 16 = 64
x = 8
b) 202 = y(y + 30)
400 = y2 + 30y
0 = y2 + 30y − 400
0 = (y + 40)(y − 10)
y =XX−40, 10
When you have to factor a quadratic equation to find an answer, always eliminate the negative answer
because length is never negative.
Example 5: Ishmael found a broken piece of a CD in his car. He places a ruler across two points on the
rim, and the length of the chord is 9.5 cm. The distance from the midpoint of this chord to the nearest
point on the rim is 1.75 cm. Find the diameter of the CD.


Solution: Think of this as two chords intersecting each other. If we were to extend the 1.75 cm segment,
it would be a diameter. So, if we find x, in the diagram to the left, and add it to 1.75 cm, we would find
the diameter.


4.25 · 4.25 = 1.75 · x
18.0625 = 1.75x


x ≈ 10.3 cm, making the diameter 12 cm, which is the
actual diameter of a CD.


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Know What? Revisited The given information is to the left. Let’s set up an equation using Theorem
9-15.


238857 · 251308 = 240128(240128 + x)
60026674956 = 57661456380 + 240128x
2365218572 = 240128x


x ≈ 9849.8 miles


Review Questions
• Questions 1-25 are similar to Examples 1, 3, and 4.
• Questions 26-28 are similar to Example 2.
• Questions 29 is similar to Example 5.
• Questions 30 and 31 are proofs of Theorem 9-14 and 9-15.


Fill in the blanks for each problem below. Then, solve for the missing segment.


1.


· 4 = · x


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2.


3( + ) = 2(2 + 7)


3.


20x =


4.


x · = 8( + )


5.


= (4 + 5)


6.


102 = x( + )
Find x in each diagram below. Simplify any radicals.


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7.


8.


9.


10.


11.


12.


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13.


14.


15.


16.


17.


18.


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19.


20.


21.


22.


23.


24.


25. Error Analysis Describe and correct the error in finding y.


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10 · 10 = y · 15y
100 = 15y2


20
3


= y2


2

15


3
= y ←− y in not correct


Algebra Connection Find the value of x.


26.


27.


28.


29. Suzie found a piece of a broken plate. She places a ruler across two points on the rim, and the length
of the chord is 6 inches. The distance from the midpoint of this chord to the nearest point on the
rim is 1 inch. Find the diameter of the plate.


30. Fill in the blanks of the proof of Theorem 9-14.


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Given: Intersecting chords AC and BE.
Prove: ab = cd


Table 5.5:


Statement Reason
1. Intersecting chords AC and BE with segments
a, b, c, and d.
2. Theorem 9-8
3. 4ADE ∼ 4BDC
4. Corresponding parts of similar triangles are pro-


portional
5. ab = cd


31. Fill in the blanks of the proof of Theorem 9-15.


Given: Secants PR and RT
Prove: a(a + b) = c(c + d)


Table 5.6:


Statement Reason
1. Secants PR and RT with segments a, b, c, and
d.


given


2. ∠R ∠R Reflexive PoC
3. ∠QPS ∠STQ Theorem 9-8
4. 4RPS ∼ 4RTQ AA Similarity Postulate
5. ac+d = ca+b Corresponding parts of similar triangles are pro-


portional
6. a(a + b) = c(c + d) Cross multiplication


Review Queue Answers
1. m∠DAC = m∠DBC by Theorem 9-8, they are inscribed angles and intercept the same arc.
2. m∠AED = m∠BEC by the Vertical Angles Theorem.
3. Yes, by AA Similarity Postulate.
4. 86 = 7EC


8 · EC = 42
EC = 214 = 5.25


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5.7 Extension: Writing and Graphing the Equa-
tions of Circles


Learning Objectives
• Graph a circle.
• Find the equation of a circle in the x − y plane.
• Find the radius and center, given the equation of a circle and vice versa.
• Find the equation of a circle, given the center and a point on the circle.


Graphing a Circle in the Coordinate Plane
Recall that the definition of a circle is the set of all points that are the same distance from the center. This
definition can be used to find an equation of a circle in the coordinate plane.


Let’s start with the circle centered at (0, 0). If (x, y) is a point on the circle, then the distance from the
center to this point would be the radius, r. x is the horizontal distance y is the vertical distance. This
forms a right triangle. From the Pythagorean Theorem, the equation of a circle, centered at the origin is
x2 + y2 = r2.
Example 1: Graph x2 + y2 = 9.
Solution: The center is (0, 0). It’s radius is the square root of 9, or 3. Plot the center, and then go out 3
units in every direction and connect them to form a circle.


The center does not always have to be on (0, 0). If it is not, then we label the center (h, k) and would use
the distance formula to find the length of the radius.


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r =


(x − h)2 + (y − k)2


If you square both sides of this equation, then we would have the standard equation of a circle.
Standard Equation of a Circle: The standard equation of a circle with center (h, k) and radius r is
r2 = (x − h)2 + (y − k)2.
Example 2: Find the center and radius of the following circles.
a) (x − 3)2 + (y − 1)2 = 25
b) (x + 2)2 + (y − 5)2 = 49
Solution:
a) Rewrite the equation as (x − 3)2 + (y − 1)2 = 52. The center is (3, 1) and r = 5.
b) Rewrite the equation as (x − (−2))2 + (y − 5)2 = 72. The center is (-2, 5) and r = 7.
When finding the center of a circle always take the opposite sign of what the value is in the equation.
Example 3: Find the equation of the circle below.


Solution: First locate the center. Draw in the horizontal and vertical diameters to see where they intersect.


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From this, we see that the center is (-3, 3). If we count the units from the center to the circle on either of
these diameters, we find r = 6. Plugging this into the equation of a circle, we get: (x−(−3))2+(y−3)2 = 62
or (x + 3)2 + (y − 3)2 = 36.


Finding the Equation of a Circle
Example 4: Determine if the following points are on (x + 1)2 + (y − 5)2 = 50.
a) (8, -3)
b) (-2, -2)
Solution: Plug in the points for x and y in (x + 1)2 + (y − 5)2 = 50.
a) (8 + 1)2 + (−3 − 5)2 = 50
92 + (−8)2 = 50
81 + 64 , 50
(8, -3) is not on the circle
b) (−2 + 1)2 + (−2 − 5)2 = 50
(−1)2 + (−7)2 = 50
1 + 49 = 50
(-2, -2) is on the circle
Example 5: Find the equation of the circle with center (4, -1) and passes through (-1, 2).
Solution: First plug in the center to the standard equation.


(x − 4)2 + (y − (−1))2 = r2


(x − 4)2 + (y + 1)2 = r2


Now, plug in (-1, 2) for x and y and solve for r.


(−1 − 4)2 + (2 + 1)2 = r2


(−5)2 + (3)2 = r2


25 + 9 = r2


34 = r2


Substituting in 34 for r2, the equation is (x − 4)2 + (y + 1)2 = 34.


Review Questions
• Questions 1-4 are similar to Examples 1 and 2.
• Questions 5-8 are similar to Example 3.
• Questions 9-11 are similar to Example 4.
• Questions 12-15 are similar to Example 5.


Find the center and radius of each circle. Then, graph each circle.


1. (x + 5)2 + (y − 3)2 = 16
2. x2 + (y + 8)2 = 4


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3. (x − 7)2 + (y − 10)2 = 20
4. (x + 2)2 + y2 = 8


Find the equation of the circles below.


5.


6.


7.


8.


9. Is (-7, 3) on (x + 1)2 + (y − 6)2 = 45?
10. Is (9, -1) on (x − 2)2 + (y − 2)2 = 60?


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11. Is (-4, -3) on (x + 3)2 + (y − 3)2 = 37?
12. Is (5, -3) on (x + 1)2 + (y − 6)2 = 45?


Find the equation of the circle with the given center and point on the circle.


13. center: (2, 3), point: (-4, -1)
14. center: (10, 0), point: (5, 2)
15. center: (-3, 8), point: (7, -2)
16. center: (6, -6), point: (-9, 4)


5.8 Chapter 9 Review
Keywords & Theorems
Parts of Circles & Tangent Lines


• Circle
• Center
• Radius
• Chord
• Diameter
• Secant
• Tangent
• Point of Tangency
• Congruent Circles
• Concentric Circles
• Externally Tangent Circles
• Internally Tangent Circles
• Common Internal Tangent
• Common External Tangent
• Tangent to a Circle Theorem
• Theorem 9-2


Properties of Arcs


• Central Angle
• Arc
• Semicircle
• Minor Arc
• Major Arc
• Congruent Arcs
• Arc Addition Postulate


Properties of Chords


• Theorem 9-3
• Theorem 9-4


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• Theorem 9-5
• Theorem 9-6


Inscribed Angles


• Inscribed Angle
• Intercepted Arc
• Inscribed Angle Theorem
• Theorem 9-8
• Theorem 9-9
• Inscribed Polygon
• Theorem 9-10


Angles from Chords, Secants and Tangents


• Theorem 9-11
• Theorem 9-12
• Theorem 9-13


Segments from Secants and Tangents


• Theorem 9-14
• Theorem 9-15
• Theorem 9-16


Extension: Equations of Circles


• Standard Equation of a Circle


Vocabulary


Match the description with the correct label.


1. minor arc - A. CD
2. chord - B. AD
3. tangent line - C. ←→CB
4. central angle - D. ←→EF


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5. secant - E. A
6. radius - F. D
7. inscribed angle - G. ∠BAD
8. center - H. ∠BCD
9. major arc - I. ̂BD


10. point of tangency - J. B̂CD


Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities
designed to supplement the objectives for some of the lessons in this chapter. See http:
//www.ck12.org/flexr/chapter/9694.


5.9 Study Guide
Keywords: Define, write theorems, and/or draw a diagram for each word below.
1st Section: Parts of Circles & Tangent Lines
Circle


Radius
Diameter
Tangent
Congruent Circles
Concentric Circles
Externally Tangent Circles
Internally Tangent Circles
Common Internal Tangent


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Common External Tangent
Tangent to a Circle Theorem
Theorem 9-2
Center
Chord
Secant
Point of Tangency
Homework:
2nd Section: Properties of Arcs
Central Angle


Arc
Semicircle
Minor Arc


Major Arc
Congruent Arcs
Arc Addition Postulate
Homework:
3rd Section: Properties of Chords
Theorem 9-3


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Theorem 9-4
Theorem 9-5


Theorem 9-6
Homework:
4th Section: Inscribed Angles
Inscribed Angle
Intercepted Arc
Inscribed Angle Theorem
Theorem 9-8


Theorem 9-9
Inscribed Polygon


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Theorem 9-10
Homework:
5th Section: Angles from Chords, Secants and Tangents
Theorem 9-11


Theorem 9-12


Theorem 9-13


Homework:
6th Section: Segments from Secants and Tangents
Theorem 9-14


Theorem 9-15


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Theorem 9-16


Homework:
Extension: Equations of Circles
Standard Equation of a Circle


Homework:


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Chapter 6


Perimeter and Area


Now that we have explored triangles, quadrilaterals, polygons, and circles, we are going to learn how to
find the perimeter and area of each.


6.1 Triangles and Parallelograms
Learning Objectives


• Understand the basic concepts of area.
• Use formulas to find the area of triangles and parallelograms.


Review Queue
1. Define perimeter and area, in your own words.
2. Solve the equations below. Simplify any radicals.


(a) x2 = 121
(b) 4x + 6 = 80
(c) x2 − 6x + 8 = 0
(d) 12 x − 3 = 5
(e) x2 + 2x − 15 = 0
(f) x2 − x − 12 = 0


Know What? Ed’s parents are getting him a new king bed. Upon further research, Ed discovered there
are two types of king beds, and Eastern (or standard) King and a California King. The Eastern King has
76” × 80” dimensions, while the California King is 72” × 84” (both dimensions are width × length). Which
bed has a larger area to lie on?


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Areas and Perimeters of Squares and Rectangles
Perimeter: The distance around a shape.
The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given
(feet, inches, centimeters, etc), write “units.”
Example 1: Find the perimeter of the figure to the left.
Solution: Here, we can use the grid as our units. Count around the figure to find the perimeter.


5 + 1 + 1 + 1 + 5 + 1 + 3 + 1 + 1 + 1 + 1 + 2 + 4 + 7 = 34 units


You are probably familiar with the area of squares and rectangles from a previous math class. Recall that
you must always establish a unit of measure for area. Area is always measured in square units, square feet
( f t.2), square inches (in.2). square centimeters (cm.2), etc. If no specific units are given, write “units2.”
Example 2: Find the area of the figure from Example 1.
Solution: Count the number of squares within the figure. If we start on the left and count each column.
5 + 6 + 1 + 4 + 3 + 4 + 4 = 27 units2


Area of a Rectangle: A = bh, where b is the base (width) and h is the height (length).


Example 3: Find the area and perimeter of a rectangle with sides 4 cm by 9 cm.


Solution: The perimeter is 4 + 9 + 4 + 9 = 36 cm. The area is A = 9 · 4 = 26 cm2.
Perimeter of a Rectangle: P = 2b + 2h.
If a rectangle is a square, with sides of length s, the formulas are as follows:
Perimeter of a Square: Psquare = 2s + 2s = 4s


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Area of a Square: Asqaure = s · s = s2


Example 4: The area of a square is 75 in2. Find the perimeter.
Solution: To find the perimeter, we need to find the length of the sides.


A = s2 = 75 in2


s =

75 = 5



3 in


From this, P = 4
(


5

3
)


= 20

3 in.


Area Postulates
Congruent Areas Postulate: If two figures are congruent, they have the same area.


Example 5: Draw two different rectangles with an area of 36 cm2.
Solution: Think of all the different factors of 36. These can all be dimensions of the different rectangles.


Other possibilities could be 6 × 6, 2 × 18, and 1 × 36.
Example 5 shows two rectangles with the same area and are not congruent. This tells us that the converse
of the Congruent Areas Postulate is not true.
Area Addition Postulate: If a figure is composed of two or more parts that do not overlap each other,
then the area of the figure is the sum of the areas of the parts.
Example 6: Find the area of the figure below. You may assume all sides are perpendicular.


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Solution: Split the shape into two rectangles and find the area of each.


Atop rectangle = 6 · 2 = 12 f t2


Abottom square = 3 · 3 = 9 f t2


The total area is 12 + 9 = 21 f t2.


Area of a Parallelogram
Recall that a parallelogram is a quadrilateral whose opposite sides are parallel.


To find the area of a parallelogram, make it into a rectangle.


From this, we see that the area of a parallelogram is the same as the area of a rectangle.
Area of a Parallelogram: The area of a parallelogram is A = bh.
The height of a parallelogram is always perpendicular to the base. This means that the sides are not the
height.


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Example 7: Find the area of the parallelogram.


Solution: A = 15 · 8 = 120 in2


Example 8: If the area of a parallelogram is 56 units2 and the base is 4 units, what is the height?
Solution: Solve for the height in A = bh.


56 = 4h
14 = h


Area of a Triangle


If we take parallelogram and cut it in half, along a diagonal, we would have two congruent triangles. The
formula for the area of a triangle is half the area of a parallelogram.
Area of a Triangle: A = 12 bh or A = bh2 .


Example 9: Find the area of the triangle.


Solution: To find the area, we need to find the height of the triangle. We are given the two sides of the


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small right triangle, where the hypotenuse is also the short side of the obtuse triangle.


32 + h2 = 52


9 + h2 = 25
h2 = 16
h = 4


A = 1
2
(4)(7) = 14 units2


Example 10: Find the perimeter of the triangle in Example 9.


Solution: To find the perimeter, we need to find the longest side of the obtuse triangle. If we used the
black lines in the picture, we would see that the longest side is also the hypotenuse of the right triangle
with legs 4 and 10.


42 + 102 = c2


16 + 100 = c2


c =

116 ≈ 10.77


The perimeter is 7 + 5 + 10.77 = 22.77 units
Example 11: Find the area of the figure below.


Solution: Divide the figure into a triangle and a rectangle with a small rectangle cut out of the lower
right-hand corner.


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A = Atop triangle + Arectangle − Asmall triangle


A =
(1
2
· 6 · 9


)


+ (9 · 15) +
(1
2
· 3 · 6


)


A = 27 + 135 + 9
A = 171 units2


Know What? Revisited The area of an Eastern King is 6080 in2 and the California King is 6048 in2.


Review Questions
• Questions 1-12 are similar to Examples 3-5, 7-9.
• Questions 13-18 are similar to Examples 9 and 10.
• Questions 19-24 are similar to Examples 7 and 9.
• Questions 25-30 are similar to Examples 6 and 11.
• Questions 31-36 use the formula for the area of a triangle.


1. Find the area and perimeter of a square with sides of length 12 in.
2. Find the area and perimeter of a rectangle with height of 9 cm and base of 16 cm.
3. Find the area of a parallelogram with height of 20 m and base of 18 m.
4. Find the area and perimeter of a rectangle if the height is 8 and the base is 14.
5. Find the area and perimeter of a square if the sides are 18 ft.
6. If the area of a square is 81 f t2, find the perimeter.
7. If the perimeter of a square is 24 in, find the area.
8. Find the area of a triangle with base of length 28 cm and height of 15 cm.
9. What is the height of a triangle with area 144 m2 and a base of 24 m?


10. The perimeter of a rectangle is 32. Find two different dimensions that the rectangle could be.
11. Draw two different rectangles that haven an area of 90 mm2.
12. Write the converse of the Congruent Areas Postulate. Determine if it is a true statement. If not,


write a counterexample. If it is true, explain why.


Use the triangle to answer the following questions.


13. Find the height of the triangle by using the geometric mean.
14. Find the perimeter.
15. Find the area.


Use the triangle to answer the following questions.


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16. Find the height of the triangle.
17. Find the perimeter.
18. Find the area.


Find the area of the following shapes.


19.


20.


21.


22.


23.


24.


25. (a) Divide the shape into two triangles and one rectangle.
(b) Find the area of the two triangles and rectangle.
(c) Find the area of the entire shape.


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26. (a) Divide the shape into two rectangles and one triangle.
(b) Find the area of the two rectangles and triangle.
(c) Find the area of the entire shape (you will need to subtract the area of the small triangle in the


lower right-hand corner).


Use the picture below for questions 27-30. Both figures are squares.


27. Find the area of the outer square.
28. Find the area of one grey triangle.
29. Find the area of all four grey triangles.
30. Find the area of the inner square.


In questions 31-36 we are going to derive a formula for the area of an equilateral triangle.


31. What kind of triangle is 4ABD? Find AD and BD.
32. Find the area of 4ABC.
33. If each side is x, what is AD and BD?
34. If each side is x, find the area of 4ABC.
35. Using your formula from #34, find the area of an equilateral triangle with 12 inch sides.
36. Using your formula from #34, find the area of an equilateral triangle with 5 inch sides.


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Review Queue Answers


1. Possible Answers
Perimeter: The distance around a shape.
Area: The space inside a shape.
2. (a) x = ±11
(b) x = 18.5
(c) x = 4, 2
(d) x = 16
(e) x = 3,−5
(f) x = 4,−3


6.2 Trapezoids, Rhombi, and Kites
Learning Objectives


• Derive and use the area formulas for trapezoids, rhombi, and kites.


Review Queue


Find the area of the shaded regions in the figures below.


1.


2. ABCD is a square.


3. ABCD is a square.


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Know What? The Brazilian flag is to the right. The flag has dimensions of 20×14 (units vary depending
on the size, so we will not use any here). The vertices of the yellow rhombus in the middle are 1.7 units
from the midpoint of each side.


Find the area of the rhombus (including the circle). Do not round your answer.


Area of a Trapezoid
Recall that a trapezoid is a quadrilateral with one pair of parallel sides. The lengths of the parallel sides
are the bases and the perpendicular distance between the parallel sides is the height of the trapezoid.


To find the area of the trapezoid, make a copy of the trapezoid and then rotate the copy 180◦. Now, this
is a parallelogram with height h and base b1 + b2. The area of this shape is A = h(b1 + b2).


Because the area of this parallelogram is two congruent trapezoids, the area of one trapezoid would be
A = 12h(b1 + b2).
Area of a Trapezoid: A = 12h(b1 + b2)
h is always perpendicular to the bases.


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You could also say the area of a trapezoid is the average of the bases times the height.
Example 1: Find the area of the trapezoids below.
a)


b)


Solution:
a) A = 12(11)(14 + 8)
A = 12(11)(22)
A = 121 units2


b) A = 12(9)(15 + 23)
A = 12(9)(38)
A = 171 units2


Example 2: Find the perimeter and area of the trapezoid.


Solution: Even though we are not told the length of the second base, we can find it using special right
triangles. Both triangles at the ends of this trapezoid are isosceles right triangles, so the hypotenuses are
4

2 and the other legs are of length 4.


P = 8 + 4

2 + 16 + 4



2 A = 1


2
(4)(8 + 16)


P = 24 + 8

2 ≈ 35.3 units A = 48 units2


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Area of a Rhombus and Kite


Recall that a rhombus is an equilateral quadrilateral and a kite has adjacent congruent sides.
Both of these quadrilaterals have perpendicular diagonals, which is how we are going to find their areas.


Notice that the diagonals divide each quadrilateral into 4 triangles. If we move the two triangles on the
bottom of each quadrilateral so that they match up with the triangles above the horizontal diagonal, we
would have two rectangles.


So, the height of these rectangles is half of one of the diagonals and the base is the length of the other
diagonal.


Area of a Rhombus: A = 12d1d2
The area is half the product of the diagonals.


Area of a Kite: A = 12d1d2
Example 3: Find the perimeter and area of the rhombi below.
a)


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b)


Solution: In a rhombus, all four triangles created by the diagonals are congruent.
a) To find the perimeter, you must find the length of each side, which would be the hypotenuse of one of
the four triangles. Use the Pythagorean Theorem.


122 + 82 = side2 A = 1
2
· 16 · 24


144 + 64 = side2 A = 192


side =

208 = 4



13


P = 4
(


4

13


)


= 16

13


b) Here, each triangle is a 30-60-90 triangle with a hypotenuse of 14. From the special right triangle ratios
the short leg is 7 and the long leg is 7



3.


P = 4 · 14 = 56 A = 1
2
· 14 · 14



3 = 98



3


Example 4: Find the perimeter and area of the kites below.
a)


b)


Solution: In a kite, there are two pairs of congruent triangles. Use the Pythagorean Theorem in both
problems to find the length of sides or diagonals.
a) Shorter sides of kite Longer sides of kite


62 + 52 = s21 12
2 + 52 = s22


36 + 25 = s21 144 + 25 = s
2
2


s1 =

61 s2 =



169 = 13


P = 2
(√


61
)


+ 2(13) = 2

61 + 26 ≈ 41.6 A = 1


2
(10)(18) = 90


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b) Smaller diagonal portion Larger diagonal portion
202 + d2s = 252 202 + d2l = 35


2


d2s = 225 d2l = 825


ds = 15 dl = 5

33


A = 1
2


(


15 + 5

33


)


(40) ≈ 874.5 P = 2(25) + 2(35) = 120


Example 5: The vertices of a quadrilateral are A(2, 8), B(7, 9),C(11, 2), and D(3, 3). Show ABCD is a kite
and find its area.
Solution: After plotting the points, it looks like a kite. AB = AD and BC = DC. The diagonals are
perpendicular if the slopes are opposite signs and flipped.


mAC =
2 − 8
11 − 2 = −


6
9


= −2
3


mBD =
9 − 3
7 − 3 =


6
4


= 3
2


The diagonals are perpendicular, so ABCD is a kite. To find the area, we need to find the length of the
diagonals.


d1 =


(2 − 11)2 + (8 − 2)2 d2 =


(7 − 3)2 + (9 − 3)2


=


(−9)2 + 62 =


42 + 62


=

81 + 36 =



117 = 3



13 =



16 + 36 =



52 = 2



13


Plug these lengths into the area formula for a kite. A = 12
(


3

13


) (


2

13


)


= 39 units2


Know What? Revisited To find the area of the rhombus, we need to find the length of the diagonals.
One diagonal is 20− 1.7− 1.7 = 16.6 and the other is 14− 1.7− 1.7 = 10.6. The area is A = 12(16.6)(10.6) =
87.98 units2.


Review Questions
• Question 1 uses the formula of the area of a kite and rhombus.
• Questions 2-16 are similar to Examples 1-4.


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• Questions 17-23 are similar to Example 5.
• Questions 24-27 use the area formula for a kite and rhombus and factors.
• Questions 28-30 are similar to Example 4.


1. Do you think all rhombi and kites with the same diagonal lengths have the same area? Explain your
answer.


Find the area of the following shapes. Round your answers to the nearest hundredth.


2.


3.


4.


5.


6.


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7.


8.


9.


10.


Find the area and perimeter of the following shapes. Round your answers to the nearest hundredth.


11.


12.


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13.


14.


15.


16.


Quadrilateral ABCD has vertices A(−2, 0), B(0, 2),C(4, 2), and D(0,−2). Leave your answers in simplest
radical form.


17. Find the slopes of AB and DC. What type of quadrilateral is this? Plotting the points will help you
find the answer.


18. Find the slope of AD. Is it perpendicular to AB and DC?
19. Find AB, AD, and DC.
20. Use #19 to find the area of the shape.


Quadrilateral EFGH has vertices E(2,−1), F(6,−4),G(2,−7), and H(−2,−4).


21. Find the slopes of all the sides and diagonals. What type of quadrilateral is this? Plotting the points
will help you find the answer.


22. Find HF and EG.
23. Use #22 to find the area of the shape.


For Questions 24 and 25, the area of a rhombus is 32 units2.


24. What would the product of the diagonals have to be for the area to be 32 units2?
25. List two possibilities for the length of the diagonals, based on your answer from #24.


For Questions 26 and 27, the area of a kite is 54 units2.


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26. What would the product of the diagonals have to be for the area to be 54 units2?
27. List two possibilities for the length of the diagonals, based on your answer from #26.


Sherry designed the logo for a new company, made up of 3 congruent kites.


28. What are the lengths of the diagonals for one kite?
29. Find the area of one kite.
30. Find the area of the entire logo.


Review Queue Answers
1. A = 9(8) +


[


1
2(9)(8)


]


= 72 + 36 = 108 units2


2. A = 12(6)(12)2 = 72 units2
3. A = 4


[


1
2(6)(3)


]


= 36 units2


6.3 Areas of Similar Polygons
Learning Objectives


• Understand the relationship between the scale factor of similar polygons and their areas.
• Apply scale factors to solve problems about areas of similar polygons.


Review Queue
1. Are two squares similar? Are two rectangles?


2. Find the scale factor of the sides of the similar shapes. Both figures are squares.
3. Find the area of each square.
4. Find the ratio of the smaller square’s area to the larger square’s area. Reduce it.


Know What? One use of scale factors and areas is scale drawings. This technique takes a small object,
like the handprint to the right, divides it up into smaller squares and then blows up the individual squares.


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In this Know What? you are going to make a scale drawing of your own hand. Trace your hand on a piece
of paper. Then, divide your hand into 9 squares, like the one to the right, 2 in×2 in. Take a larger piece of
paper and blow up each square to be 6 in×6 in (you will need at least an 18 in square piece of paper). Once
you have your 6 in× 6 in squares drawn, use the proportions and area to draw in your enlarged handprint.


Areas of Similar Polygons
In Chapter 7, we learned about similar polygons. Polygons are similar when the corresponding angles are
equal and the corresponding sides are in the same proportion.
Example 1: The two rectangles below are similar. Find the scale factor and the ratio of the perimeters.


Solution: The scale factor is 1624 = 23 .


Psmall = 2(10) + 2(16) = 52 units
Plarge = 2(15) + 2(24) = 78 units


The ratio of the perimeters is 5278 = 23 .
The ratio of the perimeters is the same as the scale factor. In fact, the ratio of any part of two
similar shapes (diagonals, medians, midsegments, altitudes, etc.) is the same as the scale factor.
Example 2: Find the area of each rectangle from Example 1. Then, find the ratio of the areas.
Solution:


Asmall = 10 · 16 = 160 units2


Alarge = 15 · 24 = 360 units2


The ratio of the areas would be 160360 = 49 .
The ratio of the sides, or scale factor was 23 and the ratio of the areas is 49 . Notice that the ratio of the
areas is the square of the scale factor.


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Area of Similar Polygons Theorem: If the scale factor of the sides of two similar polygons is mn , then
the ratio of the areas would be


(


m
n


)2.


If the scale factor is mn , then the ratio of the areas is
(


m
n


)2.


Example 3: Find the ratio of the areas of the rhombi below. The rhombi are similar.


Solution: Find the ratio of the sides and square it.
(3
5


)2
= 9


25


Example 4: Two trapezoids are similar. If the scale factor is 34 and the area of the smaller trapezoid is
81 cm2, what is the area of the larger trapezoid?
Solution: First, the ratio of the areas would be


(


3
4


)2
= 916 . Now, we need the area of the larger trapezoid.


To find this, set up a proportion using the area ratio.
9
16


= 81
A
→ 9A = 1296


A = 144 cm2


Example 5: Two triangles are similar. The ratio of the areas is 2564 . What is the scale factor?


Solution: The scale factor is


25
64 =


5
8 .


Example 6: Using the ratios from Example 5, find the length of the base of the smaller triangle if the
length of the base of the larger triangle is 24 units.
Solution: Set up a proportion using the scale factor.


5
8


= b
24
→ 8b = 120


b = 15 units


Know What? Revisited You should end up with an 18 in × 18 in drawing of your handprint.


Review Questions
• Questions 1-4 are similar to Example 3.


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• Questions 5-8 are similar to Example 5.
• Questions 9-18 are similar to Examples 1-3, and 5.
• Questions 19-22 are similar to Examples 4 and 6.
• Questions 23-26 are similar to Examples 5 and 6.


Determine the ratio of the areas, given the ratio of the sides of a polygon.


1. 35
2. 14
3. 72
4. 611


Determine the ratio of the sides of a polygon, given the ratio of the areas.


5. 136
6. 481
7. 499
8. 25144


This is an equilateral triangle made up of 4 congruent equilateral triangles.


9. What is the ratio of the areas of the large triangle to one of the small triangles?


10. What is the scale factor of large to small triangle?
11. If the area of the large triangle is 20 units2, what is the area of a small triangle?
12. If the length of the altitude of a small triangle is 2



3, find the perimeter of the large triangle.


13. Find the perimeter of the large square and the blue square.
14. Find the scale factor of the blue square and large square.
15. Find the ratio of their perimeters.
16. Find the area of the blue and large squares.
17. Find the ratio of their areas.
18. Find the length of the diagonals of the blue and large squares. Put them into a ratio. Which ratio


is this the same as?
19. Two rectangles are similar with a scale factor of 47 . If the area of the larger rectangle is 294 in2, find


the area of the smaller rectangle.
20. Two triangles are similar with a scale factor of 13 . If the area of the smaller triangle is 22 f t2, find


the area of the larger triangle.


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21. The ratio of the areas of two similar squares is 1681 . If the length of a side of the smaller square is 24
units, find the length of a side in the larger square.


22. The ratio of the areas of two right triangles is 49 . If the length of the hypotenuse of the larger triangle
is 48 units, find the length of the smaller triangle’s hypotenuse.


Questions 23-26 build off of each other. You may assume the problems are connected.


23. Two similar rhombi have areas of 72 units2 and 162 units2. Find the ratio of the areas.
24. Find the scale factor.
25. The diagonals in these rhombi are congruent. Find the length of the diagonals and the sides.
26. What type of rhombi are these quadrilaterals?


Review Queue Answers
1. Two squares are always similar. Two rectangles can be similar as long as the sides are in the same


proportion.
2. 1025 = 25
3. Asmall = 100, Alarge = 625


6.4 Circumference and Arc Length
Learning Objectives


• Find the circumference of a circle.
• Define the length of an arc and find arc length.


Review Queue
1. Find a central angle in that intercepts ̂CE


2. Find an inscribed angle that intercepts ̂CE.
3. How many degrees are in a circle? Find mÊCD.
4. If m̂CE = 26◦, find m̂CD and m∠CBE.


Know What? A typical large pizza has a diameter of 14 inches and is cut into 8 pieces. Think of the
crust as the circumference of the pizza. Find the length of the crust for the entire pizza. Then, find the
length of the crust for one piece of pizza if the entire pizza is cut into 8 pieces.


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Circumference of a Circle
Circumference: The distance around a circle.
The circumference can also be called the perimeter of a circle. However, we use the term circumference for
circles because they are round. In order to find the circumference of a circle, we need to explore pi (pi).
Investigation 10-1: Finding pi (pi)
Tools Needed: paper, pencil, compass, ruler, string, and scissors


1. Draw three circles with radii of 2 in, 3 in, and 4 in. Label the centers of each A, B, and C.
2. Draw in the diameters and determine their lengths.


3. Take the string and outline each circle with it. Cut the string so that it perfectly outlines the circle.
Then, lay it out straight and measure it in inches. Round your answer to the nearest 18 -inch. Repeat
this for the other two circles.


4. Find circum f erencediameter for each circle. Record your answers to the nearest thousandth.


You should see that circum f erencediameter approaches 3.14159... We call this number pi, the Greek letter “pi.” When
finding the circumference and area of circles, we must use pi.
pi, or “pi”: The ratio of the circumference of a circle to its diameter. It is approximately equal to
3.14159265358979323846...
To see more digits of pi, go to http://www.eveandersson.com/pi/digits/.
From Investigation 10-1, we found that circum f erencediameter = pi. Let’s solve for the circumference, C.


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C
d


= pi


C = pid


We can also say C = 2pir because d = 2r.
Circumference Formula: C = pid or C = 2pir


d = 2r
Example 1: Find the circumference of a circle with a radius of 7 cm.
Solution: Plug the radius into the formula.


C = 2pi(7) = 14pi ≈ 44 cm


Example 2: The circumference of a circle is 64pi. Find the diameter.
Solution: Again, you can plug in what you know into the circumference formula and solve for d.


64pi = pid
64 = d


Example 3: A circle is inscribed in a square with 10 in. sides. What is the circumference of the circle?
Leave your answer in terms of pi.


Solution: From the picture, we can see that the diameter of the circle is equal to the length of a side.
C = 10pi in.
Example 4: Find the perimeter of the square. Is it more or less than the circumference of the circle?
Why?
Solution: The perimeter is P = 4(10) = 40 in. In order to compare the perimeter with the circumference
we should change the circumference into a decimal.
C = 10pi ≈ 31.42 in. This is less than the perimeter of the square, which makes sense because the circle is
inside the square.


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Arc Length
In Chapter 9, we measured arcs in degrees. This was called the “arc measure” or “degree measure.” Arcs
can also be measured in length, as a portion of the circumference.
Arc Length: The length of an arc or a portion of a circle’s circumference.
The arc length is directly related to the degree arc measure.


Example 5: Find the length of ̂PQ. Leave your answer in terms of pi.
Solution: In the picture, the central angle that corresponds with ̂PQ is 60◦. This means that m̂PQ = 60◦.
Think of the arc length as a portion of the circumference. There are 360◦ in a circle, so 60◦ would be 16 of
that


(


60◦
360◦ =


1
6


)


. Therefore, the length of ̂PQ is 16 of the circumference. length of ̂PQ = 16 · 2pi(9) = 3pi


Arc Length Formula: The length of ̂AB = m̂AB360◦ · pid or m
̂AB


360◦ · 2pir.
Another way to write this could be x◦360◦ · 2pir, where x is the central angle.


Example 6: The arc length of ̂AB = 6pi and is 14 the circumference. Find the radius of the circle.
Solution: If 6pi is 14 the circumference, then the total circumference is 4(6pi) = 24pi. To find the radius,
plug this into the circumference formula and solve for r.


24pi = 2pir
12 = r


Example 7: Find the measure of the central angle or ̂PQ.


Solution: Let’s plug in what we know to the Arc Length Formula.


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15pi = m
̂PQ


360◦
· 2pi(18)


15 = m
̂PQ


10◦
150◦ = m̂PQ


Example 8: The tires on a compact car are 18 inches in diameter. How far does the car travel after the
tires turn once? How far does the car travel after 2500 rotations of the tires?


Solution: One turn of the tire is the circumference. This would be C = 18pi ≈ 56.55 in. 2500 rotations
would be 2500 · 56.55 in = 141371.67 in, 11781 ft, or 2.23 miles.
Know What? Revisited The entire length of the crust, or the circumference of the pizza is 14pi ≈ 44 in.
In 18 of the pizza, one piece would have 448 ≈ 5.5 inches of crust.


Review Questions
• Questions 1-10 are similar to Examples 1 and 2.
• Questions 11-14 are similar to Examples 3 and 4.
• Questions 15-20 are similar to Example 5.
• Questions 21-23 are similar to Example 6.
• Questions 24-26 are similar to Example 7.
• Questions 27-30 are similar to Example 8.


Fill in the following table. Leave all answers in terms of pi.


Table 6.1:


diameter radius circumference
1. 15
2. 4
3. 6
4. 84pi
5. 9
6. 25pi
7. 2pi
8. 36


9. Find the radius of circle with circumference 88 in.


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10. Find the circumference of a circle with d = 20
pi


cm.


Square PQSR is inscribed in


T . RS = 8

2.


11. Find the length of the diameter of


T .
12. How does the diameter relate to PQSR?
13. Find the perimeter of PQSR.
14. Find the circumference of




T .


Find the arc length of ̂PQ in


A. Leave your answers in terms of pi.


15.


16.


17.


18.


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19.


20.


Find PA (the radius) in


A. Leave your answer in terms of pi.


21.


22.


23.


Find the central angle or m̂PQ in


A. Round any decimal answers to the nearest tenth.


24.


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25.


26.


For questions 27-30, a truck has tires with a 26 in diameter.


27. How far does the truck travel every time a tire turns exactly once? What is this the same as?
28. How many times will the tire turn after the truck travels 1 mile? (1 mile = 5280 feet)
29. The truck has travelled 4072 tire rotations. How many miles is this?
30. The average recommendation for the life of a tire is 30,000 miles. How many rotations is this?


Review Queue Answers
1. ∠CAE
2. ∠CBE
3. 360◦, 180◦
4. m̂CD = 180◦ − 26◦ = 154◦,m∠CBE = 13◦


6.5 Areas of Circles and Sectors
Learning Objectives


• Find the area of circles, sectors, and segments.


Review Queue
1. Find the area of both squares.
2. Find the area of the shaded region.


3. The triangle to the right is an equilateral triangle.


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(a) Find the height of the triangle.
(b) Find the area of the triangle.


Know What? Back to the pizza. In the previous section, we found the length of the crust for a 14
in pizza. However, crust typically takes up some area on a pizza. Round your answers to the nearest
hundredth.


a) Find the area of the crust of a deep-dish 16 in pizza. A typical deep-dish pizza has 1 in of crust around
the toppings.
b) A thin crust pizza has 12 -in of crust around the edge of the pizza. Find the area of a thin crust 16 in
pizza.


Area of a Circle
Take a circle and divide it into several wedges. Then, unfold the wedges so they are in a line, with the
points at the top.


The height of the wedges is the radius and the length is the circumference of the circle. Now, take half of
these wedges and flip them upside-down and place them so they all fit together.


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Now our circle looks like a parallelogram. The area of this parallelogram is A = bh = pir · r = pir2.
To see an animation of this derivation, see http://www.rkm.com.au/ANIMATIONS/animation-Circle-Area-Derivation.
html, by Russell Knightley.
Area of a Circle: If r is the radius of a circle, then A = pir2.
Example 1: Find the area of a circle with a diameter of 12 cm.
Solution: If d = 12 cm, then r = 6 cm. The area is A = pi


(


62
)


= 36pi cm2.
Example 2: If the area of a circle is 20pi, what is the radius?
Solution: Plug in the area and solve for the radius.


20pi = pir2


20 = r2


r =

20 = 2



5


Just like the circumference, we will leave our answers in terms of pi, unless otherwise specified.
Example 3: A circle is inscribed in a square. Each side of the square is 10 cm long. What is the area of
the circle?


Solution: The diameter of the circle is the same as the length of a side of the square. Therefore, the
radius is 5 cm.


A = pi52 = 25pi cm2


Example 4: Find the area of the shaded region.
Solution: The area of the shaded region would be the area of the square minus the area of the circle.


A = 102 − 25pi = 100 − 25pi ≈ 21.46 cm2


Area of a Sector
Sector of a Circle: The area bounded by two radii and the arc between the endpoints of the radii.


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Area of a Sector: If r is the radius and ̂AB is the arc bounding a sector, then A = m̂AB360◦ · pir2.
Example 5: Find the area of the blue sector. Leave your answer in terms of pi.


Solution: In the picture, the central angle that corresponds with the sector is 60◦. 60◦ would be 16 of
360◦, so this sector is 16 of the total area. area o f blue sector = 16 · pi82 = 323 pi
Another way to write the sector formula is A = central angle360◦ · pir2.
Example 6: The area of a sector is 8pi and the radius of the circle is 12. What is the central angle?
Solution: Plug in what you know to the sector area formula and then solve for the central angle, we will
call it x.


8pi = x
360◦


· pi122


8pi = x
360◦


· 144pi


8 = 2x
5◦


x = 8 · 5


2
= 20◦


Example 7: The area of a sector is 135pi and the arc measure is 216◦. What is the radius of the circle?


Solution: Plug in what you know to the sector area formula and solve for r.


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135pi = 216


360◦
· pir2


135 = 3
5
· r2


5
3
· 135 = r2


225 = r2 → r =

225 = 15


Example 8: Find the area of the shaded region. The quadrilateral is a square.


Solution: The radius of the circle is 16, which is also half of the diagonal of the square. So, the diagonal
is 32 and the sides would be 32√


2
·

2√
2
= 16



2 because each half of a square is a 45-45-90 triangle.


Acircle = 162pi = 256pi


Asquare =
(


16

2
)2


= 256 · 2 = 512


The area of the shaded region is 256pi − 512 ≈ 292.25


Segments of a Circle
The last part of a circle that we can find the area of is called a segment, not to be confused with a line
segment.
Segment of a Circle: The area of a circle that is bounded by a chord and the arc with the same endpoints
as the chord.


Asegment = Asector − A4ABC


Example 9: Find the area of the blue segment below.


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Solution: The area of the segment is the area of the sector minus the area of the isosceles triangle made
by the radii. If we split the isosceles triangle in half, each half is a 30-60-90 triangle, where the radius is
the hypotenuse. The height of 4ABC is 12 and the base would be 2


(


12

3
)


= 24

3.


Asector =
120
360


pi · 242 A4 =
1
2


(


24

3
)


(12)


= 192pi = 144

3


The area of the segment is A = 192pi − 144

3 ≈ 353.8.


Know What? Revisited The area of the crust for a deep-dish pizza is 82pi − 72pi = 15pi. The area of the
crust of the thin crust pizza is 82pi − 7.52pi = 314 pi.


Review Questions
• Questions 1-10 are similar to Examples 1 and 2.
• Questions 11-16 are similar to Example 5.
• Questions 17-19 are similar to Example 7.
• Questions 20-22 are similar to Example 6.
• Questions 23-25 are similar to Examples 3, 4, and 8.
• Questions 26-31 are similar to Example 9.


Fill in the following table. Leave all answers in terms of pi.


Table 6.2:


radius Area circumference
1. 2
2. 16pi
3. 10pi
4. 24pi
5. 9
6. 90pi
7. 35pi
8. 7


pi


9. 60
10. 36


Find the area of the blue sector or segment in


A. Leave your answers in terms of pi. Round any decimal
answers to the nearest hundredth.


11.


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12.


13.


14.


15.


16.


Find the radius of the circle. Leave your answer in terms of pi.


17.


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18.


19.


Find the central angle of each blue sector. Round any decimal answers to the nearest tenth.


20.


21.


22.


Find the area of the shaded region. Round your answer to the nearest hundredth.


23.


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24.


25.


26. Find the area of the sector in


A. Leave your answer in terms of pi.


27. Find the area of the equilateral triangle.
28. Find the area of the segment. Round your answer to the nearest hundredth.
29. Find the area of the sector in




A. Leave your answer in terms of pi.
30. Find the area of the right triangle.


31. Find the area of the segment. Round your answer to the nearest hundredth.


Review Queue Answers


1. 82 − 42 = 64 − 16 = 48
2. 6(10) − 12(7)(3) = 60 − 10.5 = 49.5
3. 12(6)


(


3

3
)


= 9

3


4. 12(s)
(


1
2 s

3
)


= 14 s
2 √3


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6.6 Chapter 10 Review
Keywords, Theorems and Formulas
Triangles and Parallelograms


• Perimeter
• Area of a Rectangle: A = bh
• Perimeter of a Rectangle P = 2b + 2h
• Perimeter of a Square: P = 4s
• Area of a Square: A = s2
• Congruent Areas Postulate
• Area Addition Postulate
• Area of a Parallelogram: A = bh
• Area of a Triangle: A = 12 bh or A = bh2


Trapezoids, Rhombi, and Kites


• Area of a Trapezoid: A = 12h(b1 + b2)
• Area of a Rhombus: A = 12d1d2
• Area of a Kite: A = 12d1d2


Area of Similar Polygons


• Area of Similar Polygons Theorem


Circumference and Arc Length


• pi
• Circumference: C = pid or C = 2pir
• Arc Length
• Arc Length Formula: length of ̂AB = m̂AB360◦ · pid or m


̂AB
360◦ · 2pir


Area of Circles and Sectors


• Area of a Circle: A = pir2
• Sector
• Area of a Sector: A = m̂AB360◦ · pir2
• Segment of a Circle


Review Questions
Find the area and perimeter of the following figures. Round your answers to the nearest hundredth.


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1. square


2. rectangle


3. rhombus


4. equilateral triangle


5. parallelogram


6. kite


Find the area of the following figures. Leave your answers in simplest radical form.


7. triangle


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8. kite


9. isosceles trapezoid


10. Find the area and circumference of a circle with radius 17.
11. Find the area and circumference of a circle with diameter 30.
12. Two similar rectangles have a scale factor 43 . If the area of the larger rectangle is 96 units2, find the


area of the smaller rectangle.


Find the area of the following figures. Round your answers to the nearest hundredth.


13.


14.


15. find the shaded area


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(figure is a rhombus)


Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities
designed to supplement the objectives for some of the lessons in this chapter. See http:
//www.ck12.org/flexr/chapter/9695.


6.7 Study Guide
Keywords: Define, write theorems, and/or draw a diagram for each word below.
1st Section: Triangles and Parallelograms
Perimeter


Area of a Rectangle: A = bh
Perimeter of a Rectangle P = 2b + 2h
Perimeter of a Square: P = 4s


Area of a Square: A = s2


Congruent Areas Postulate
Area Addition Postulate
Area of a Parallelogram: A = bh


Area of a Triangle: A = 12 bh or A = bh2


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Homework:
2nd Section: Trapezoids, Rhombi, and Kites
Area of a Trapezoid: A = 12h(b1 + b2)


Area of a Rhombus: A = 12d1d2
Area of a Kite: A = 12d1d2


Homework:
3rd Section: Area of Similar Polygons
Area of Similar Polygons Theorem


Homework:
4th Section: Circumference and Arc Length
pi


Circumference: C = pid or C = 2pir
Arc Length
Arc Length Formula: length of ̂AB = m̂AB360◦ · pid or m


̂AB
360◦ · 2pir


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Homework:
5th Section: Area of Circles and Sectors
Area of a Circle: A = pir2


Sector
Area of a Sector: A = m̂AB360◦ · pir2


Segment of a Circle
Homework:


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Chapter 7


Surface Area and Volume


In this chapter we extend what we know about two-dimensional figures to three-dimensional shapes. First,
we will define the different types of 3D shapes and their parts. Then, we will find the surface area and
volume of prisms, cylinders, pyramids, cones, and spheres.


7.1 Exploring Solids
Learning Objectives


• Identify different types of solids and their parts.
• Use Euler’s formula and nets.


Review Queue
1. Draw an octagon and identify the edges and vertices of the octagon. How many of each are there?
2. Find the area of a square with 5 cm sides.
3. Draw the following polygons.


(a) A convex pentagon.
(b) A concave nonagon.


Know What? Until now, we have only talked about two-dimensional, or flat, shapes. Copy the equilateral
triangle to the right onto a piece of paper and cut it out. Fold on the dotted lines. What shape do these
four equilateral triangles make?


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Polyhedrons


Polyhedron: A 3-dimensional figure that is formed by polygons that enclose a region in space.
Each polygon in a polyhedron is a face.
The line segment where two faces intersect is an edge.
The point of intersection of two edges is a vertex.


Examples of polyhedrons include a cube, prism, or pyramid. Non-polyhedrons are cones, spheres, and
cylinders because they have sides that are not polygons.
Prism: A polyhedron with two congruent bases, in parallel planes, and the lateral sides are rectangles.


Pyramid: A polyhedron with one base and all the lateral sides meet at a common vertex.


All prisms and pyramids are named by their bases. So, the first prism would be a triangular prism and
the first pyramid would be a hexagonal pyramid.
Example 1: Determine if the following solids are polyhedrons. If the solid is a polyhedron, name it and
find the number of faces, edges and vertices each has.
a)


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b)


c)


Solution:
a) The base is a triangle and all the sides are triangles, so this is a triangular pyramid. There are 4 faces,
6 edges and 4 vertices.
b) This solid is also a polyhedron. The bases are both pentagons, so it is a pentagonal prism. There are 7
faces, 15 edges, and 10 vertices.
c) The bases that are circles. Circles are not polygons, so it is not a polyhedron.


Euler’s Theorem
Let’s put our results from Example 1 into a table.


Table 7.1:


Faces Vertices Edges
Triangular Pyramid 4 4 6
Pentagonal Prism 7 10 15


Notice that faces + vertices is two more that the number of edges. This is called Euler’s Theorem, after
the Swiss mathematician Leonhard Euler.
Euler’s Theorem: F + V = E + 2.


Faces + Vertices = Edges + 2
5 + 6 = 9 + 2


Example 2: Find the number of faces, vertices, and edges in the octagonal prism.


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Solution: There are 10 faces and 16 vertices. Use Euler’s Theorem, to solve for E.


F + V = E + 2
10 + 16 = E + 2


24 = E


Example 3: In a six-faced polyhedron, there are 10 edges. How many vertices does the polyhedron have?
Solution: Solve for V in Euler’s Theorem.


F + V = E + 2
6 + V = 10 + 2


V = 6


Example 4: A three-dimensional figure has 10 vertices, 5 faces, and 12 edges. Is it a polyhedron?
Solution: Plug in all three numbers into Euler’s Theorem.


F + V = E + 2
5 + 10 = 12 + 2


15 , 14


Because the two sides are not equal, this figure is not a polyhedron.


Regular Polyhedra
Regular Polyhedron: A polyhedron where all the faces are congruent regular polygons.
All regular polyhedron are convex.
A concave polyhedron “caves in.”


There are only five regular polyhedra, called the Platonic solids.
Regular Tetrahedron: A 4-faced polyhedron and all the faces are equilateral triangles.
Cube: A 6-faced polyhedron and all the faces are squares.
Regular Octahedron: An 8-faced polyhedron and all the faces are equilateral triangles.


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Regular Dodecahedron: A 12-faced polyhedron and all the faces are regular pentagons.
Regular Icosahedron: A 20-faced polyhedron and all the faces are equilateral triangles.


Cross-Sections
One way to “view” a three-dimensional figure in a two-dimensional plane, like in this text, is to use
cross-sections.
Cross-Section: The intersection of a plane with a solid.
The cross-section of the peach plane and the tetrahedron is a triangle.


Example 5: What is the shape formed by the intersection of the plane and the regular octahedron?
a)


b)


c)


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Solution:
a) Square
b) Rhombus
c) Hexagon


Nets
Net: An unfolded, flat representation of the sides of a three-dimensional shape.


Example 6: What kind of figure does this net create?


Solution: The net creates a rectangular prism.


Example 7: Draw a net of the right triangular prism below.


Solution: The net will have two triangles and three rectangles. The rectangles are different sizes and the
two triangles are the same.


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There are several different nets of any polyhedron. For example, this net could have the triangles anywhere
along the top or bottom of the three rectangles. Click the site http://www.cs.mcgill.ca/~sqrt/unfold/
unfolding.html to see a few animations of other nets.
Know What? Revisited The net of the shape is a regular tetrahedron.


Review Questions
• Questions 1-8 are similar to Examples 2-4.
• Questions 9-14 are similar to Example 1.
• Questions 15-17 are similar to Example 5.
• Questions 18-23 are similar to Example 7.
• Questions 24-29 are similar to Example 6.
• Question 30 uses Euler’s Theorem.


Complete the table using Euler’s Theorem.


Table 7.2:


Name Faces Edges Vertices
1. Rectangular Prism 6 12
2. Octagonal Pyra-


mid
16 9


3. Regular Icosahe-
dron


20 12


4. Cube 12 8
5. Triangular Pyra-


mid
4 4


6. Octahedron 8 12
7. Heptagonal Prism 21 14
8. Triangular Prism 5 9


Determine if the following figures are polyhedra. If so, name the figure and find the number of faces, edges,
and vertices.


9.


10.


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11.


12.


13.


14.


Describe the cross section formed by the intersection of the plane and the solid.


15.


16.


17.


Draw the net for the following solids.


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18.


19.


20.


21.


22.


23.


Determine what shape is formed by the following nets.


24.


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25.


26.


27.


28.


29.


30. A truncated icosahedron is a polyhedron with 12 regular pentagonal faces and 20 regular hexagonal
faces and 90 edges. This icosahedron closely resembles a soccer ball. How many vertices does it have?
Explain your reasoning.


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Review Queue Answers
1. There are 8 vertices and 8 edges in an octagon.


2. 52 = 25 cm2
3. (a)


(b)


7.2 Surface Area of Prisms and Cylinders
Learning Objectives


• Find the surface area of a prism and cylinder.


Review Queue
1. Find the area of a rectangle with sides:


(a) 6 and 9
(b) 11 and 4
(c) 5



2 and 8



6


2. If the area of a square is 36 units2, what are the lengths of the sides?
3. If the area of a square is 45 units2, what are the lengths of the sides?


Know What? Your parents decide they want to put a pool in the backyard. The shallow end will be 4
ft. and the deep end will be 8 ft. The pool will be 10 ft. by 25 ft. How much siding do they need to cover
the sides and bottom of the pool?


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Parts of a Prism
Prism: A 3-dimensional figure with 2 congruent bases, in parallel planes, and the other faces are rectangles.


The non-base faces are lateral faces.
The edges between the lateral faces are lateral edges.
This is a pentagonal prism.
Right Prism: A prism where all the lateral faces are perpendicular to the bases.
Oblique Prism: A prism that leans to one side and the height is outside the prism.


Surface Area of a Prism
Surface Area: The sum of the areas of the faces.


S ur f ace Area = B1 + B2 + L1 + L2 + L3
Lateral Area = L1 + L2 + L3


Lateral Area: The sum of the areas of the lateral faces.


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Example 1: Find the surface area of the prism below.


Solution: Draw the net of the prism.
Using the net, we have:


S Aprism = 2(4)(10) + 2(10)(17) + 2(17)(4)
= 80 + 340 + 136
= 556 cm2


Surface Area of a Right Prism: The surface area of a right prism is the sum of the area of the bases
and the area of each rectangular lateral face.
Example 2: Find the surface area of the prism below.


Solution: This is a right triangular prism. To find the surface area, we need to find the length of the
hypotenuse of the base because it is the width of one of the lateral faces.


72 + 242 = c2


49 + 576 = c2


625 = c2 c = 25


Looking at the net, the surface area is:


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S A = 28(7) + 28(24) + 28(25) + 2
(1
2
· 7 · 24


)


S A = 196 + 672 + 700 + 168 = 1736 units2


Cylinders
Cylinder: A solid with congruent circular bases that are in parallel planes. The space between the circles
is enclosed.
A cylinder has a radius and a height.
A cylinder can also be oblique, like the one on the far right.


Surface Area of a Right Cylinder
Let’s find the net of a right cylinder. One way to do this is to take the label off of a soup can. The label
is a rectangle where the height is the height of the cylinder and the base is the circumference of the circle.


Surface Area of a Right Cylinder: S A = 2pir2 + 2pirh.


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2pir2
︸︷︷︸


+ 2pir
︸︷︷︸


h


area of length
both of


circles rectangle


To see an animation of the surface area, click http://www.rkm.com.au/ANIMATIONS/animation-Cylinder-Surface-Area-Derivation.
html, by Russell Knightley.
Example 3: Find the surface area of the cylinder.


Solution: r = 4 and h = 12.


S A = 2pi(4)2 + 2pi(4)(12)
= 32pi + 96pi
= 128pi units2


Example 4: The circumference of the base of a cylinder is 16pi and the height is 21. Find the surface area
of the cylinder.
Solution: We need to solve for the radius, using the circumference.


2pir = 16pi
r = 8


Now, we can find the surface area.


S A = 2pi(8)2 + (16pi)(21)
= 128pi + 336pi
= 464pi units2


Example 5: Algebra Connection The total surface area of the triangular prism is 540 units2. What is
x?


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Solution: The total surface area is equal to:


A2 triangles + A3 rectangles = 540


The hypotenuse of the triangle bases is 13,

52 + 122. Let’s fill in what we know.


A2 triangles = 2
(1
2
· 5 · 12


)


= 60


A3 triangles = 5x + 12x + 13x = 30x
60 + 30x = 540


30x = 480
x = 16 units The height is 16 units.


Know What? Revisited To the right is the net of the pool (minus the top). From this, we can see that
your parents would need 670 square feet of siding.


Review Questions
• Questions 1-9 are similar to Examples 1 and 2.
• Question 10 uses the definition of lateral and total surface area.
• Questions 11-18 are similar to Examples 1-3.
• Questions 19-21 are similar to Example 5.
• Questions 22-24 are similar to Example 4.
• Questions 25-30 use the Pythagorean Theorem and are similar to Examples 1-3.


1. What type of prism is this?


2. Draw the net of this prism.
3. Find the area of the bases.
4. Find the area of lateral faces, or the lateral surface area.
5. Find the total surface area of the prism.


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Use the right triangular prism to answer questions 6-9.


6. What shape are the bases of this prism? What are their areas?
7. What are the dimensions of each of the lateral faces? What are their areas?
8. Find the lateral surface area of the prism.
9. Find the total surface area of the prism.


10. Writing Describe the difference between lateral surface area and total surface area.
11. Fuzzy dice are cubes with 4 inch sides.


(a) What is the surface area of one die?
(b) Typically, the dice are sold in pairs. What is the surface area of two dice?


12. A right cylinder has a 7 cm radius and a height of 18 cm. Find the surface area.


Find the surface area of the following solids. Round your answer to the nearest hundredth.


13. bases are isosceles trapezoids


14.


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15.


16.


17.


18.


Algebra Connection Find the value of x, given the surface area.


19. S A = 432 units2


20. S A = 1536pi units2


21. S A = 1568 units2


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22. The area of the base of a cylinder is 25pi in2 and the height is 6 in. Find the lateral surface area.
23. The circumference of the base of a cylinder is 80pi cm and the height is 36 cm. Find the total surface


area.
24. The lateral surface area of a cylinder is 30pi m2 and the height is 5m. What is the radius?


Use the diagram below for questions 25-30. The barn is shaped like a pentagonal prism with dimensions
shown in feet.


25. What is the width of the roof? (HINT: Use the Pythagorean Theorem)
26. What is the area of the roof? (Both sides)
27. What is the floor area of the barn?
28. What is the area of the rectangular sides of the barn?
29. What is the area of the two pentagon sides of the barn? (HINT: Find the area of two congruent


trapezoids for each side)
30. Find the total surface area of the barn (Roof and sides).


Review Queue Answers
1. (a) 54


(b) 44
(c) 80



3


2. s = 6
3. s = 3



5


7.3 Surface Area of Pyramids and Cones
Learning Objectives


• Find the surface area of pyramids and cones.


Review Queue
1. A rectangular prism has sides of 5 cm, 6 cm, and 7 cm. What is the surface area?
2. A cylinder has a diameter of 10 in and a height of 25 in. What is the surface area?
3. A cylinder has a circumference of 72pi f t. and a height of 24 ft. What is the surface area?
4. Draw the net of a square pyramid.


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Know What? A typical waffle cone is 6 inches tall and has a diameter of 2 inches. What is the surface
area of the waffle cone? (You may assume that the cone is straight across at the top)


Parts of a Pyramid
Pyramid: A solid with one base and the lateral faces meet at a common vertex.
The edges between the lateral faces are lateral edges.
The edges between the base and the lateral faces are base edges.


Regular Pyramid: A pyramid where the base is a regular polygon.
All regular pyramids also have a slant height which is the height of a lateral face. A non-regular pyramid
does not have a slant height.


Example 1: Find the slant height of the square pyramid.


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Solution: The slant height is the hypotenuse of the right triangle formed by the height and half the base
length. Use the Pythagorean Theorem.


82 + 242 = l2


640 = l2


l =

640 = 8



10


Surface Area of a Regular Pyramid
Using the slant height, which is labeled l, the area of each triangular face is A = 12bl.
Example 2: Find the surface area of the pyramid from Example 1.


Solution: The four triangular faces are 4
(


1
2bl


)


= 2(16)
(


8

10


)


= 256

10. To find the total surface area, we


also need the area of the base, which is 162 = 256. The total surface area is 256

10+256 ≈ 1065.54 units2.


From this example, we see that the formula for a square pyramid is:


S A = (area o f the base) + 4(area o f triangular f aces)


S A = B + n
(1
2
bl


)


B is the area of the base and n is the number of triangles.
Surface Area of a Regular Pyramid: If B is the area of the base, then S A = B + 12nbl.
The net shows the surface area of a pyramid. If you ever forget the formula, use the net.


Example 3: Find the area of the regular triangular pyramid.


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Solution: “Regular” tells us the base is an equilateral triangle. Let’s draw it and find its area.


B = 12 · 8 · 4

3 = 16



3


The surface area is:
S A = 16



3 + 12 · 3 · 8 · 18 = 16



3 + 216 ≈ 243.71


Example 4: If the lateral surface area of a square pyramid is 72 f t2 and the base edge is equal to the
slant height. What is the length of the base edge?
Solution: In the formula for surface area, the lateral surface area is 12nbl. We know that n = 4 and b = l.
Let’s solve for b.


1
2
nbl = 72 f t2


1
2
(4)b2 = 72


2b2 = 72
b2 = 36
b = 6


Surface Area of a Cone


Cone: A solid with a circular base and sides taper up towards a vertex.
A cone has a slant height, just like a pyramid.


A cone is generated from rotating a right triangle, around one leg, in a circle.
Surface Area of a Right Cone: S A = pir2 + pirl.


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Area of the base: pir2


Area of the sides: pirl
Example 5: What is the surface area of the cone?


Solution: First, we need to find the slant height. Use the Pythagorean Theorem.


l2 = 92 + 212


= 81 + 441


l =

522 ≈ 22.85


The surface area would be S A = pi92 + pi(9)(22.85) ≈ 900.54 units2.
Example 6: The surface area of a cone is 36pi and the radius is 4 units. What is the slant height?
Solution: Plug in what you know into the formula for the surface area of a cone and solve for l.


36pi = pi42 + pi4l
36 = 16 + 4l When each term has a pi, they cancel out.
20 = 4l
5 = l


Know What? Revisited The standard cone has a surface area of pi +

35pi ≈ 21.73 in2.


Review Questions
• Questions 1-10 use the definitions of pyramids and cones.
• Questions 11-19 are similar to Example 1.
• Questions 20-26 are similar to Examples 2, 3, and 5.
• Questions 27-31 are similar to Examples 4 and 6.
• Questions 32-25 are similar to Example 5.


Fill in the blanks about the diagram to the left.


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1. x is the ___________.
2. The slant height is ________.
3. y is the ___________.
4. The height is ________.
5. The base is _______.
6. The base edge is ________.


Use the cone to fill in the blanks.


7. v is the ___________.
8. The height of the cone is ______.
9. x is a __________ and it is the ___________ of the cone.


10. w is the _____________ ____________.


For questions 11-13, sketch each of the following solids and answer the question. Your drawings should be
to scale, but not one-to-one. Leave your answer in simplest radical form.


11. Draw a right cone with a radius of 5 cm and a height of 15 cm. What is the slant height?
12. Draw a square pyramid with an edge length of 9 in and a 12 in height. Find the slant height.
13. Draw an equilateral triangle pyramid with an edge length of 6 cm and a height of 6 cm. What is the


height of the base?


Find the slant height, l, of one lateral face in each pyramid or of the cone. Round your answer to the
nearest hundredth.


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14.


15.


16.


Find the area of a lateral face of the regular pyramid. Round your answers to the nearest hundredth.


17.


18.


19.


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Find the surface area of the regular pyramids and right cones. Round your answers to 2 decimal places.


20.


21.


22.


23.


24.


25.


26. A regular tetrahedron has four equilateral triangles as its faces.
(a) Find the height of one of the faces if the edge length is 6 units.
(b) Find the area of one face.
(c) Find the total surface area of the regular tetrahedron.


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27. If the lateral surface area of a cone is 30pi cm2 and the radius is 5 cm, what is the slant height?
28. If the surface area of a cone is 105pi cm2 and the slant height is 8 cm, what is the radius?
29. If the surface area of a square pyramid is 40 f t2 and the base edge is 4 ft, what is the slant height?
30. If the lateral area of a square pyramid is 800 in2 and the slant height is 16 in, what is the length of


the base edge?
31. If the lateral area of a regular triangle pyramid is 252 in2 and the base edge is 8 in, what is the slant


height?


The traffic cone is cut off at the top and the base is a square with 24 in sides. Round answers to the
nearest hundredth.


32. Find the area of the entire square. Then, subtract the area of the base of the cone.
33. Find the lateral area of the cone portion (include the 4 inch cut off top of the cone).
34. Subtract the cut-off top of the cone, to only have the lateral area of the cone portion of the traffic


cone.
35. Combine your answers from #27 and #30 to find the entire surface area of the traffic cone.


Review Queue Answers
1. 2(5 · 6) + 2(5 · 7) + 2(6 · 7) = 214 cm2
2. 2(15 · 18) + 2(15 · 21) + 2(18 · 21) = 1926 cm2
3. 2 · 25pi + 250pi = 300pi in2
4. 362(2pi) + 72pi(24) = 4320pi f t2
5.


7.4 Volume of Prisms and Cylinders
Learning Objectives


• Find the volume of prisms and cylinders.


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Review Queue
1. Define volume in your own words.
2. What is the surface area of a cube with 3 inch sides?
3. A regular octahedron has 8 congruent equilateral triangles as the faces.


(a) If each edge is 4 cm, what is the slant height for one face?
(b) What is the surface area of one face?
(c) What is the total surface area?


Know What? Let’s fill the pool it with water. The shallow end is 4 ft. and the deep end is 8 ft. The
pool is 10 ft. wide by 25 ft. long. How many cubic feet of water is needed to fill the pool?


Volume of a Rectangular Prism
Volume: The measure of how much space a three-dimensional figure occupies.
Another way to define volume would be how much a three-dimensional figure can hold. The basic unit of
volume is the cubic unit: cubic centimeter (cm3), cubic inch (in3), cubic meter (m3), cubic foot ( f t3).
Volume of a Cube Postulate: V = s3.
V = s · s · s = s3


What this postulate tells us is that every solid can be broken down into cubes. For example, if we wanted
to find the volume of a cube with 9 inch sides, it would be 93 = 729 in3.
Volume Congruence Postulate: If two solids are congruent, then their volumes are congruent.
These prisms are congruent, so their volumes are congruent.


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Example 1: Find the volume of the right rectangular prism below.


Solution: Count the cubes. The bottom layer has 20 cubes, or 4× 5, and there are 3 layers. There are 60
cubes. The volume is also 60 units3.
Each layer in Example 1 is the same as the area of the base and the number of layers is the same as the
height. This is the formula for volume.
Volume of a Rectangular Prism: V = l · w · h.


Example 2: A typical shoe box is 8 in by 14 in by 6 in. What is the volume of the box?
Solution: We can assume that a shoe box is a rectangular prism.
V = (8)(14)(6) = 672 in2


Volume of any Prism
Notcie that l · w is equal to the area of the base of the prism, which we will re-label B.
Volume of a Prism: V = B · h.


“B” is not always going to be the same. So, to find the volume of a prism, you would first find the area of
the base and then multiply it by the height.
Example 3: You have a small, triangular prism shaped tent. How much volume does it have, once it is
set up?


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Solution: First, we need to find the area of the base.


B = 1
2
(3)(4) = 6 f t2.


V = Bh = 6(7) = 42 f t3


Even though the height in this problem does not look like a “height,” it is, according to the formula.
Usually, the height of a prism is going to be the last length you need to use.


Oblique Prisms
Recall that oblique prisms are prisms that lean to one side and the height is outside the prism. What
would be the volume of an oblique prism? Consider to piles of books below.


Both piles have 15 books, which means they will have the same volume. Cavalieri’s Principle says that
leaning does not matter, the volumes are the same.
Cavalieri’s Principle: If two solids have the same height and the same cross-sectional area at every level,
then they will have the same volume.
If an oblique prism and a right prism have the same base area and height, then they will have the same
volume.


Example 4: Find the area of the oblique prism below.


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Solution: This is an oblique right trapezoidal prism. Find the area of the trapezoid.


B = 1
2
(9)(8 + 4) = 9(6) = 54 cm2


V = 54(15) = 810 cm3


Volume of a Cylinder
If we use the formula for the volume of a prism, V = Bh, we can find the volume of a cylinder. In the case
of a cylinder, the base is the area of a circle. Like a prism, Cavalieri’s Principle holds.
Volume of a Cylinder: V = pir2h.


Example 5: Find the volume of the cylinder.


Solution: If the diameter is 16, then the radius is 8.
V = pi82(21) = 1344pi units3


Example 6: Find the volume of the cylinder.


Solution: V = pi62(15) = 540pi units3


Example 7: If the volume of a cylinder is 484pi in3 and the height is 4 in, what is the radius?
Solution: Solve for r.


484pi = pir2(4)
121 = r2


11 = r


Example 8: Find the volume of the solid below.


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Solution: This solid is a parallelogram-based prism with a cylinder cut out of the middle.


Vprism = (25 · 25)30 = 18750 cm3


Vcylinder = pi(4)2(30) = 480pi cm3


The total volume is 18750 − 480pi ≈ 17242.04 cm3.
Know What? Revisited Even though it doesn’t look like it, the trapezoid is the base of this prism. The
area of the trapezoids are 12(4 + 8)25 = 150 f t2. V = 150(10) = 1500 f t3


Review Questions
• Question 1 uses the volume formula for a cylinder.
• Questions 2-4 are similar to Example 1.
• Questions 5-18 are similar to Examples 2-6.
• Questions 19-24 are similar to Example 7.
• Questions 25-30 are similar to Example 8.


1. Two cylinders have the same surface area. Do they have the same volume? How do you know?
2. How many one-inch cubes can fit into a box that is 8 inches wide, 10 inches long, and 12 inches tall?


Is this the same as the volume of the box?
3. A cereal box in 2 inches wide, 10 inches long and 14 inches tall. How much cereal does the box hold?
4. A can of soda is 4 inches tall and has a diameter of 2 inches. How much soda does the can hold?


Round your answer to the nearest hundredth.
5. A cube holds 216 in3. What is the length of each edge?
6. A cube has sides that are 8 inches. What is the volume?
7. A cylinder has r = h and the radius is 4 cm. What is the volume?
8. A cylinder has a volume of 486pi f t.3. If the height is 6 ft., what is the diameter?


Use the right triangular prism to answer questions 9 and 10.


9. What is the length of the third base edge?
10. Find the volume of the prism.


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11. Fuzzy dice are cubes with 4 inch sides.


(a) What is the volume of one die?
(b) What is the volume of both dice?


12. A right cylinder has a 7 cm radius and a height of 18 cm. Find the volume.


Find the volume of the following solids. Round your answers to the nearest hundredth.


13.


14.


15.


16.


17.


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18.


Algebra Connection Find the value of x, given the surface area.


19. V = 504 units3


20. V = 6144pi units3


21. V = 2688 units3


22. The area of the base of a cylinder is 49pi in2 and the height is 6 in. Find the volume.
23. The circumference of the base of a cylinder is 80pi cm and the height is 15 cm. Find the volume.
24. The lateral surface area of a cylinder is 30pi m2 and the circumference is 10pi m. What is the volume


of the cylinder?


The bases of the prism are squares and a cylinder is cut out of the center.


25. Find the volume of the prism.


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26. Find the volume of the cylinder in the center.
27. Find the volume of the figure.


This is a prism with half a cylinder on the top.


28. Find the volume of the prism.
29. Find the volume of the half-cylinder.
30. Find the volume of the entire figure.


Review Queue Answers
1. The amount a three-dimensional figure can hold.
2. 54 in2
3. (a) 2



3


(b) 12 · 4 · 2

3 = 4



3


(c) 8 · 4

3 = 32



3


7.5 Volume of Pyramids and Cones
Learning Objectives


• Find the volume of pyramids and cones.


Review Queue
1. Find the volume of a square prism with 8 inch base edges and a 12 inch height.
2. Find the volume of a cylinder with a diameter of 8 inches and a height of 12 inches.
3. Find the surface area of a square pyramid with 10 inch base edges and a height of 12 inches.


Know What? The Khafre Pyramid is a pyramid in Giza, Egypt. It is a square pyramid with a base edge
of 706 feet and an original height of 407.5 feet. What was the original volume of the Khafre Pyramid?


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Volume of a Pyramid
The volume of a pyramid is closely related to the volume of a prism with the same sized base.
Investigation 11-1: Finding the Volume of a Pyramid
Tools needed: pencil, paper, scissors, tape, ruler, dry rice.


1. Make an open net (omit one base) of a cube, with 2 inch sides.


2. Cut out the net and tape up the sides to form an open cube.


3. Make an open net (no base) of a square pyramid, with lateral edges of 2.5 inches and base edges of
2 inches.


4. Cut out the net and tape up the sides to form an open pyramid.


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5. Fill the pyramid with dry rice and dump the rice into the open cube. Repeat this until you have
filled the cube?


Volume of a Pyramid: V = 13Bh.


Example 1: Find the volume of the pyramid.


Solution: V = 13(122)12 = 576 units3


Example 2a: Find the height of the pyramid.


Solution: In this example, we are given the slant height. Use the Pythagorean Theorem.


72 + h2 = 252


h2 = 576
h = 24


Example 2b: Find the volume of the pyramid in Example 2a.
Solution: V = 13(142)(24) = 1568 units3.
Example 3: Find the volume of the pyramid.


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Solution: The base of the pyramid is a right triangle. The area of the base is 12(14)(8) = 56 units2.
V = 13(56)(17) ≈ 317.33 units3


Example 4: A rectangular pyramid has a base area of 56 cm2 and a volume of 224 cm3. What is the
height of the pyramid?
Solution:


V = 1
3
Bh


224 = 1
3
· 56h


12 = h


Volume of a Cone
Volume of a Cone: V = 13pir2h.
This is the the same relationship as a pyramid’s volume with a prism’s volume.


Example 5: Find the volume of the cone.


Solution: First, we need the height. Use the Pythagorean Theorem.


52 + h2 = 152


h =

200 = 10



2


V = 1
3
(52)


(


10

2
)


pi ≈ 370.24


Example 6: Find the volume of the cone.


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Solution: We can use the same volume formula. Find the radius.


V = 1
3
pi(32)(6) = 18pi ≈ 56.55


Example 7: The volume of a cone is 484pi cm3 and the height is 12 cm. What is the radius?
Solution: Plug in what you know to the volume formula.


484pi = 1
3
pir2(12)


121 = r2


11 = r


Composite Solids
Example 8: Find the volume of the composite solid. All bases are squares.


Solution: This is a square prism with a square pyramid on top. First, we need the height of the pyramid
portion. Using the Pythagorean Theorem, we have, h =



252 − 242 = 7.


Vprism = (48)(48)(18) = 41472 cm3


Vpyramid =
1
3
(482)(7) = 5376 cm3


The total volume is 41472 + 5376 = 46, 848 cm3.
Know What? Revisited The original volume of the pyramid is 13(7062)(407.5) ≈ 67, 704, 223.33 f t3.


Review Questions
• Questions 1-13 are similar to Examples 1-3, 5 and 6.
• Questions 14-22 are similar to Examples 4 and 7.


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• Questions 23-31 are similar to Example 8.


Find the volume of each regular pyramid and right cone. Round any decimal answers to the nearest
hundredth. The bases of these pyramids are either squares or equilateral triangles.


1.


2.


3.


4.


5.


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6.


7.


Find the volume of the following non-regular pyramids and cones. Round any decimal answers to the
nearest hundredth.


8.


9.


10.


11. base is a rectangle


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12.


13.


A regular tetrahedron has four equilateral triangles as its faces. Use the diagram to answer questions
14-16. Round your answers to the nearest hundredth.


14. What is the area of the base of this regular tetrahedron?
15. What is the height of this figure? Be careful!
16. Find the volume.


A regular octahedron has eight equilateral triangles as its faces. Use the diagram to answer questions
17-21. Round your answers to the nearest hundredth.


17. Describe how you would find the volume of this figure.
18. Find the volume.


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19. The volume of a square pyramid is 72 square inches and the base edge is 4 inches. What is the
height?


20. If the volume of a cone is 30pi cm3 and the radius is 5 cm, what is the height?
21. If the volume of a cone is 105pi cm3 and the height is 35 cm, what is the radius?
22. The volume of a triangle pyramid is 170 in3 and the base area is 34 in2. What is the height of the


pyramid?


For questions 23-31, round your answer to the nearest hundredth.


23. Find the volume of the base prism.


24. Find the volume of the pyramid.
25. Find the volume of the entire solid.


The solid to the right is a cube with a cone cut out.


26. Find the volume of the cube.
27. Find the volume of the cone.
28. Find the volume of the entire solid.


The solid to the left is a cylinder with a cone on top.


29. Find the volume of the cylinder.
30. Find the volume of the cone.
31. Find the volume of the entire solid.


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Review Queue Answers
1. (82)(12) = 768 in3
2. (42)(12)pi = 192pi ≈ 603.19
3. Find slant height, l = 13. S A = 100 + 12(40)(13) = 360 in2


7.6 Surface Area and Volume of Spheres
Learning Objectives


• Find the surface area of a sphere.
• Find the volume of a sphere.


Review Queue
1. List three spheres you would see in real life.
2. Find the area of a circle with a 6 cm radius.
3. Find the volume of a cylinder with the circle from #2 as the base and a height of 5 cm.


Know What? A regulation bowling ball is a sphere with a circumference of 27 inches. Find the radius
of a bowling ball, its surface area and volume. You may assume the bowling ball does not have any finger
holes. Round your answers to the nearest hundredth.


Defining a Sphere
A sphere is the last of the three-dimensional shapes that we will find the surface area and volume of. Think
of a sphere as a three-dimensional circle.
Sphere: The set of all points, in three-dimensional space, which are equidistant from a point.
The radius has an endpoint on the sphere and the other endpoint is the center.


The diameter must contain the center.


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Great Circle: A cross section of a sphere that contains the diameter.
A great circle is the largest circle cross section in a sphere. The circumference of a sphere is the
circumference of a great circle.
Every great circle divides a sphere into two congruent hemispheres.


Example 1: The circumference of a sphere is 26pi f eet. What is the radius of the sphere?
Solution: The circumference is referring to the circumference of a great circle. Use C = 2pir.


2pir = 26pi
r = 13 f t.


Surface Area of a Sphere
The best way to understand the surface area of a sphere is to watch the link by Russell Knightley, http:
//www.rkm.com.au/ANIMATIONS/animation-Sphere-Surface-Area-Derivation.html.
Surface Area of a Sphere: S A = 4pir2.


Example 2: Find the surface area of a sphere with a radius of 14 feet.
Solution:


S A = 4pi(14)2


= 784pi f t2


Example 3: Find the surface area of the figure below.


Solution: Be careful when finding the surface area of a hemisphere because you need to include the area
of the base.


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S A = pir2 + 1
2
4pir2


= pi(62) + 2pi(62)
= 36pi + 72pi = 108pi cm2


Example 4: The surface area of a sphere is 100pi in2. What is the radius?
Solution:


S A = 4pir2


100pi = 4pir2


25 = r2


5 = r


Example 5: Find the surface area of the following solid.


Solution: This solid is a cylinder with a hemisphere on top. It is one solid, so do not include the bottom
of the hemisphere or the top of the cylinder.


S A = LAcylinder + LAhemisphere + Abase circle


= pirh + 1
2
4pir2 + pir2


= pi(6)(13) + 2pi62 + pi62


= 78pi + 72pi + 36pi
= 186pi in2 “LA” stands for lateral area.


Volume of a Sphere
To see an animation of the volume of a sphere, see http://www.rkm.com.au/ANIMATIONS/animation-Sphere-Volume-Derivation.
html by Russell Knightley.
Volume of a Sphere: V = 43pir3.


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Example 6: Find the volume of a sphere with a radius of 9 m.
Solution:


V = 4
3
pi63


= 4
3
pi(216)


= 288pi m3


Example 7: A sphere has a volume of 14137.167 f t3, what is the radius?
Solution:


V = 4
3
pir3


14137.167 = 4
3
pir3


3
4pi
· 14137.167 = r3


3375 = r3


At this point, you will need to take the cubed root of 3375. Ask your teacher how to do this on your
calculator.


3√3375 = 15 = r


Example 8: Find the volume of the following solid.


Solution:


Vcylinder = pi62(13) = 78pi


Vhemisphere =
1
2


(4
3
pi63


)


= 36pi


Vtotal = 78pi + 36pi = 114pi in3


Know What? Revisited The radius would be 27 = 2pir, or r = 4.30 inches. The surface area would be
4pi4.32 ≈ 232.35 in2, and the volume would be 43pi4.33 ≈ 333.04 in3.


Review Questions
• Questions 1-3 look at the definition of a sphere.


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• Questions 4-17 are similar to Examples 1, 2, 4, 6 and 7.
• Questions 18-21 are similar to Example 3 and 5.
• Questions 22-25 are similar to Example 8.
• Question 26 is a challenge.
• Questions 27-29 are similar to Example 8.
• Question 30 analyzes the formula for the surface area of a sphere.


1. Are there any cross-sections of a sphere that are not a circle? Explain your answer.
2. List all the parts of a sphere that are the same as a circle.
3. List any parts of a sphere that a circle does not have.


Find the surface area and volume of a sphere with: (Leave your answer in terms of pi)


4. a radius of 8 in.
5. a diameter of 18 cm.
6. a radius of 20 ft.
7. a diameter of 4 m.
8. a radius of 15 ft.
9. a diameter of 32 in.


10. a circumference of 26pi cm.
11. a circumference of 50pi yds.
12. The surface area of a sphere is 121pi in2. What is the radius?
13. The volume of a sphere is 47916pi m3. What is the radius?
14. The surface area of a sphere is 4pi f t2. What is the volume?
15. The volume of a sphere is 36pi mi3. What is the surface area?
16. Find the radius of the sphere that has a volume of 335 cm3. Round your answer to the nearest


hundredth.
17. Find the radius of the sphere that has a surface area 225pi f t2.


Find the surface area of the following shapes. Leave your answers in terms of pi.


18.


19.


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20.


21. You may assume the bottom is open.


Find the volume of the following shapes. Round your answers to the nearest hundredth.


22.


23.


24.


25.


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26. A sphere has a radius of 5 cm. A right cylinder has the same radius and volume. Find the height of
the cylinder.


Tennis balls with a 3 inch diameter are sold in cans of three. The can is a cylinder. Round your answers
to the nearest hundredth.


27. What is the volume of one tennis ball?
28. What is the volume of the cylinder?
29. Assume the balls touch the can on the sides, top and bottom. What is the volume of the space not


occupied by the tennis balls?
30. How does the formula of the surface area of a sphere relate to the area of a circle?


Review Queue Answers
1. Answers will vary. Possibilities are any type of ball, certain lights, or the 76/Unical orb.
2. 36pi
3. 180pi


7.7 Extension: Exploring Similar Solids
Learning Objectives


• Find the relationship between similar solids and their surface areas and volumes.


Similar Solids
Recall that two shapes are similar if all the corresponding angles are congruent and the corresponding sides
are proportional.
Similar Solids: Two solids are similar if they are the same type of solid and their corresponding radii,
heights, base lengths, widths, etc. are proportional.


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Example 1: Are the two rectangular prisms similar? How do you know?


Solution: Match up the corresponding heights, widths, and lengths.


small prism
large prism


= 3
4.5


= 4
6


= 5
7.5


The congruent ratios tell us the two prisms are similar.
Example 2: Determine if the two triangular pyramids similar.


Solution: Just like Example 1, let’s match up the corresponding parts.
6
8 =


3
4 =


12
16 however, 812 = 23 .


These triangle pyramids are not similar.


Surface Areas of Similar Solids
If two shapes are similar, then the ratio of the area is a square of the scale factor.


For example, the two rectangles are similar because their sides are in a ratio of 5:8. The area of the larger
rectangle is 8(16) = 128 units2. The area of the smaller rectangle is 5(10) = 50 units2.
Comparing the areas in a ratio, it is 50 : 128 = 25 : 64 = 52 = 82.
So, what happens with the surface areas of two similar solids?
Example 3: Find the surface area of the two similar rectangular prisms.


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Solution:


S Asmaller = 2(4 · 3) + 2(4 · 5) + 2(3 · 5)
= 24 + 40 + 30 = 94 units2


S Alarger = 2(6 · 4.5) + 2(4.5 · 7.5) + 2(6 · 7.5)
= 54 + 67.5 + 90 = 211.5 units2


Now, find the ratio of the areas. 94211.5 = 49 = 2
2


32 . The sides are in a ratio of 46 = 23 , so the surface areas are
in a ratio of 2232 .
Surface Area Ratio: If two solids are similar with a scale factor of ab , then the surface areas are in a
ratio of


(


a
b


)2.
Example 4: Two similar cylinders are below. If the ratio of the areas is 16:25, what is the height of the
taller cylinder?


Solution: First, we need to take the square root of the area ratio to find the scale factor,


16
25 =


4
5 . Set


up a proportion to find h.


4
5


= 24
h


4h = 120
h = 30


Example 5: Using the cylinders from Example 4, if the area of the smaller cylinder is 1536pi cm2, what is
the area of the larger cylinder?
Solution: Set up a proportion using the ratio of the areas, 16:25.


16
25


= 1536pi
A


16A = 38400pi
A = 2400pi cm2


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Volumes of Similar Solids
Let’s look at what we know about similar solids so far.


Table 7.3:


Ratios Units
Scale Factor ab in, ft, cm, m, etc.
Ratio of the Surface Areas


(


a
b


)2
in2, f t2, cm2,m2, etc.


Ratio of the Volumes ?? in3, f t3, cm3,m3, etc.


If the ratio of the volumes follows the pattern from above, it should be the cube of the scale factor.
Example 6: Find the volume of the following rectangular prisms. Then, find the ratio of the volumes.


Solution:


Vsmaller = 3(4)(5) = 60
Vlarger = 4.5(6)(7.5) = 202.5


The ratio is 60202.5 , which reduces to 827 = 2
3


33 .


Volume Ratio: If two solids are similar with a scale factor of ab , then the volumes are in a ratio of
(


a
b


)3.
Example 7: Two spheres have radii in a ratio of 3:4. What is the ratio of their volumes?
Solution: If we cube 3 and 4, we will have the ratio of the volumes. 33 : 43 = 27 : 64.
Example 8: If the ratio of the volumes of two similar prisms is 125:8, what is the scale factor?
Solution: Take the cubed root of 125 and 8 to find the scale factor.
3√125 : 3



8 = 5 : 2


Example 9: Two similar right triangle prisms are below. If the ratio of the volumes is 343:125, find the
missing sides in both triangles.


Solution: The scale factor is 7:5, the cubed root. With the scale factor, we can now set up several
proportions.


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7
5


= 7
y


7
5


= x
10


7
5


= 35
w


72 + x2 = z2 7
5


= z
v


y = 5 x = 14 w = 25 72 + 142 = z2


z =

245 = 7



5 7


5
= 7

5


v
→ v = 5



5


Example 10: The ratio of the surface areas of two similar cylinders is 16:81. What is the ratio of the
volumes?
Solution: First, find the scale factor. If we take the square root of both numbers, the ratio is 4:9. Now,
cube this to find the ratio of the volumes, 43 : 93 = 64 : 729.


Review Questions
• Questions 1-4 are similar to Examples 1 and 2.
• Questions 5-14 are similar to Examples 3-8 and 10.
• Questions 15-18 are similar to Example 9.
• Questions 19 and 20 are similar to Example 1.


Determine if each pair of right solids are similar.


1.


2.


3.


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4.


5. Are all cubes similar? Why or why not?
6. Two prisms have a scale factor of 1:4. What is the ratio of their surface areas?
7. Two pyramids have a scale factor of 2:7. What is the ratio of their volumes?
8. Two spheres have radii of 5 and 9. What is the ratio of their volumes?
9. The surface area of two similar cones is in a ratio of 64:121. What is the scale factor?


10. The volume of two hemispheres is in a ratio of 125:1728. What is the scale factor?
11. A cone has a volume of 15pi and is similar to another larger cone. If the scale factor is 5:9, what is


the volume of the larger cone?
12. The ratio of the volumes of two similar pyramids is 8:27. What is the ratio of their total surface


areas?
13. The ratio of the volumes of two tetrahedrons is 1000:1. The smaller tetrahedron has a side of length


6 cm. What is the side length of the larger tetrahedron?
14. The ratio of the surface areas of two cubes is 64:225. What is the ratio of the volumes?


Below are two similar square pyramids with a volume ratio of 8:27. The base lengths are equal to the
heights. Use this to answer questions 15-18.


18. What is the scale factor?
19. What is the ratio of the surface areas?
20. Find h, x and y.
21. Find the volume of both pyramids.


Use the hemispheres below to answer questions 19-20.


1. Are the two hemispheres similar? How do you know?
2. Find the ratio of the surface areas and volumes.


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7.8 Chapter 11 Review
Keywords, Theorems, & Formulas
Exploring Solids


• Polyhedron
• Face
• Edge
• Vertex
• Prism
• Pyramid
• Euler’s Theorem
• Regular Polyhedron
• Regular Tetrahedron
• Cube
• Regular Octahedron
• Regular Dodecahedron
• Regular Icosahedron
• Cross-Section
• Net


Surface Area of Prisms & Cylinders


• Lateral Face
• Lateral Edge
• Base Edge
• Right Prism
• Oblique Prism
• Surface Area
• Lateral Area
• Surface Area of a Right Prism
• Cylinder
• Surface Area of a Right Cylinder


Surface Area of Pyramids & Cones


• Surface Area of a Regular Pyramid
• Cone
• Slant Height
• Surface Area of a Right Cone


Volume of Prisms & Cylinders


• Volume
• Volume of a Cube Postulate
• Volume Congruence Postulate
• Volume of a Rectangular Prism


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• Volume of a Prism
• Cavalieri’s Principle
• Volume of a Cylinder


Volume of Pyramids & Cones


• Volume of a Pyramid
• Volume of a Cone


Surface Area and Volume of Spheres


• Sphere
• Great Circle
• Surface Area of a Sphere
• Volume of a Sphere


Extension: Similar Solids


• Similar Solids
• Surface Area Ratio
• Volume Ratio


Review Questions
Match the shape with the correct name.


1. Triangular Prism
2. Icosahedron
3. Cylinder
4. Cone


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5. Tetrahedron
6. Pentagonal Prism
7. Octahedron
8. Hexagonal Pyramid
9. Octagonal Prism


10. Sphere
11. Cube
12. Dodecahedron


Match the formula with its description.


13. Volume of a Prism - A. 13pir2h
14. Volume of a Pyramid - B. pir2h
15. Volume of a Cone - C. 4pir2
16. Volume of a Cylinder - D. 43pir3
17. Volume of a Sphere - E. pir2 + pirl
18. Surface Area of a Prism - F. 2pir2 + 2pirh
19. Surface Area of a Pyramid - G. 13Bh
20. Surface Area of a Cone - H. Bh
21. Surface Area of a Cylinder - I. B + 12Pl
22. Surface Area of a Sphere - J. The sum of the area of the bases and the area of each rectangular lateral


face.


Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities
designed to supplement the objectives for some of the lessons in this chapter. See http:
//www.ck12.org/flexr/chapter/9696.


7.9 Study Guide
Keywords: Define, write theorems, and/or draw a diagram for each word below.
1st Section: Exploring Solids
Polyhedron
Face
Edge
Vertex


Prism


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Pyramid
Euler’s Theorem
Regular Polyhedron


Regular Tetrahedron
Cube
Regular Octahedron
Regular Dodecahedron
Regular Icosahedron
Cross-Section
Net
Homework:
2nd Section: Surface Area of Prisms & Cylinders
Lateral Face
Lateral Edge
Base Edge


Right Prism
Oblique Prism
Surface Area
Lateral Area


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Surface Area of a Right Prism
Cylinder
Surface Area of a Right Cylinder
Homework:
3rd Section: Surface Area of Pyramids & Cones
Surface Area of a Regular Pyramid


Cone
Slant Height
Surface Area of a Right Cone
Homework:
4th Section: Volume of Prisms & Cylinders
Volume
Volume of a Cube Postulate


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Volume Congruence Postulate
Volume Addition Postulate
Volume of a Rectangular Prism
Volume of a Prism
Cavalieri’s Principle
Volume of a Cylinder
Homework:
5th Section: Volume of Pyramids & Cones
Volume of a Pyramid
Volume of a Cone
Homework:
6th Section: Surface Area and Volume of Spheres
Sphere


Great Circle
Surface Area of a Sphere
Volume of a Sphere
Homework:
Extension: Similar Solids
Similar Solids
Surface Area Ratio
Volume Ratio
Homework:


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Chapter 8


Analyzing Conic Sections


8.1 Introduction to Conic Sections
Learning Objectives


• Consider the results when two simple mathematical objects are intersected.
• Be comfortable working with an infinite two-sided cone.
• Know the basic types of figures that result from intersecting a plane and a cone.
• Know some of the history of the study of conic sections.


Introduction: Intersections of Figures


Some of the best mathematical shapes come from intersecting two other important shapes. Two spheres
intersect to form a circle:


Two planes intersect to form a line:


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While simple and beautiful, for these two examples of intersections there isn’t much else to investigate.
For the spheres, no matter how we put them together, their intersection is always either nothing, a point
(when they just touch), a circle of various sizes, or a sphere if they happen to be exactly the same size and
coincide. Once we’ve exhausted this list, the inquiry is over. Planes are even simpler: the intersection of
two distinct planes is either nothing (if the planes are parallel) or a line.
But some intersections yield more complex results. For instance a plane can intersect with a cube in
numerous ways. Below a plane intersects a cube to form an equilateral triangle.


Here a plane intersects a cube and forms a regular hexagon.


Review Questions
1. Describe all the types of shapes that can be produced by the intersection of a plane and a cube.
2. What is the side-length of the regular hexagon that is produced in the above diagram when the cube


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has side-length 1?


Review Answers
1. Square, rectangle, pentagon, others.
2. length =



2


2


Intersections with Cones
One class of intersections is of particular interest: The intersection of a plane and a cone. These intersections
are called conic sections and the first person known to have studied them extensively is the Ancient Greek
mathematician Menaechmus in the 3rd Century B. C. E. Part of his interest in the conic sections came
from his work on a classic Greek problem called “doubling the cube,” and we will describe this problem
and Menaechmus’ approach that uses conic sections in section three. The intersections of the cone and the
plane are so rich that the resulting shapes have continued to be of interest and generate new ideas from
Menaechmus’ time until the present.
Before we really delve into what we mean by a plane and a cone, we can look at an intuitive example.
Suppose by a cone we just mean an ice-cream cone. And by a plane we mean a piece of paper. Well, if
you sliced through an upright ice-cream cone with a horizontal piece of paper you would find that the two
objects intersect at a circle.


That’s very nice. But unlike the intersection of two spheres, which also resulted in a circle, that’s not all
we get. If we tilt the paper (or the ice cream cone) things start to get tricky.


First things first, we better make sure we know what we mean by “plane” and “cone”. Let’s use the simplest
definitions possible. So by “plane”, we mean the infinitely thin flat geometric object that extends forever
in all directions. Even though infinity is a tricky concept, this plane is in some sense simpler than one that
ends arbitrarily. There is no boundary to think about with the infinite plane. And what do we mean by
a cone? An ice-cream cone is a good start. In fact, it’s very similar to how the ancient Greeks defined a
cone, as a right triangle rotated about one if its legs.


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Like we do with the infinite geometric plane, we want to idealize this object a bit too. As with the plane,
to avoid having to deal with a boundary, let’s suppose it continues infinitely in the direction of its open
end.
But the Greek mathematician Apollonius noticed that it helps even more to have it go to infinity in the
other direction. This way, a cone can be thought of as an infinite collection of lines, and since geometric
lines go on forever in both directions, a cone also extends to infinity in both directions. Here is a picture
of what we will call a cone in this chapter (remember it extends to infinity in both directions).


A cone can be formally defined as a three-dimensional collection of lines, all forming an equal angle with
a central line or axis. In the above picture, the central line is vertical.


Review Questions
3. What other mathematical objects can be generated by a collection of lines?


Review Answers
3. Cylinder (infinite), plane, three-sided infinite pyramids, others.


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Intersections of an Infinite Cone and Plane


Like with the ice cream cone and the piece of paper, some intersections of this infinite cone with an infinite
plane yield a circle.


Other intersections yield something a bit less tidy. In the picture below, the plane is parallel to the axis of
the cone.


But it turns out that the set of intersections of a cone and a plane forms a beautiful, mathematically
consistent, set of shapes that have interesting properties. So in this chapter we embark on a study of these
intersections called conic sections.
Note: Although we will not prove it here, it doesn’t matter if the cone is asymmetrical or tilted to one
side (also called “oblique”). If you include “tilted cones” the same conic sections result. So, for simplicity’s
sake let’s stick with non-tilted, or “right” cones and focus on what happens


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Let’s begin by doing a rough tally of the kinds of shapes it seems we can generate by slicing planes through
our cone. First of all, we have the circle, as we discovered with the ice cream cone above. A circle is
formed when the plane is perpendicular to the line in the middle of the cone.
If we tilt the plane a little, but not so much that it intersects both cones, we get something more oval
shaped. This is called an ellipse. Later you will learn many of the fascinating properties of the ellipse.


If really tilt the plane more so that it only intersects one side of the cone, but we get a big infinite “U”.


This shape is called a parabola and like the ellipse it has a number of surprising properties.
Then if we tilt it even further, we intersect both sides of the cone and get two big “U’s” going in opposite


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directions.


This pair of objects is called a hyperbola, and, like the parabola and ellipse, the hyperbola has a number
of interesting properties that we will discuss.
Finally, in a few cases we get what are called “degenerate” conics. For instance, if we line up a vertical
plane with the vertex of the cone, we get two crossing lines.


Review Questions
4. Are there any other types of intersections between a plane and a cone besides the ones illustrated


above?
5. Is it possible for a plane to miss a cone entirely?


Review Answers
4. Yes, the plane could meet the vertex, resulting in an intersection of a single point.
5. It is not possible for a plane to miss a cone entirely since both of the objects extend infinitely.


Applications and Importance
The intersections of cones and planes produce an interesting set of shapes, which we will study in the
upcoming sections. We will also see numerous applications of these shapes to the physical world. Why is
it that the intersection of a cone and a plane would produce so many applications? We don’t seem to see


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a lot of cones or planes in our daily life. But at closer inspection they are everywhere. Point sources of
light, approximated by such sources as a flashlight or the sun, shine in cone-shaped array, so for instance
the image of a flashlight against a slanted wall is an ellipse.
The geometric properties of conics, such as the focal property of the ellipse, turn out to have many physical
ramifications, such as the design of telescopes. And, when we inspect the algebraic representations of conic
sections, we will see that there are similarities with the law of gravity, which in turn has ramifications for
planetary motion.


Lesson Summary
In summary, here are some of the ways that a plane can intersect a cone.


A circle


an “oval” called the ellipse


A big infinite “U” called a parabola.


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Two big, infinite “U”s called a hyperbola.


Strange “degenerate” shapes like two crossing lines, as well as other examples you may have found in
Review Question 1.
This array of shapes has a surprising amount of mathematical coherence, as well as a large number of
interesting properties.


Vocabulary
Cone A three-dimensional collection of lines, all forming an equal angle with a central line or axis.


Conic section The points of intersection between a cone and a plane.


8.2 Circles and Ellipses
Learning Objectives


• Understand the difference between an “oval” and an ellipse.
• Recognize and work with equations for ellipses.
• Derive the focal property of ellipses.


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• Understand the equivalence of different definitions of ellipses.
• Reconstruct Dandelin’s sphere construction.
• Know some of the different ways people have approached ellipses throughout history.
• Understand some of the important applications of ellipses.


Introduction
Let’s begin with the first class of shapes discussed in the last section. When the plane makes a finite
intersection with one side of the cone, we get either a circle or the “oval-shaped” object illustrated in the
previous section. It turns out that this is no ordinary oval, but something called an ellipse, a shape with
special properties.
Like parallelograms, or any other shape with lots of interesting properties, ellipses can be defined by some
of these properties, and then the other properties necessarily follow from the definition. For example,
a parallelogram is typically defined as a quadrilateral with each pair of opposite pairs of sides parallel.
Once you define it this way, it follows that the opposite sides must also be equal in length, and that the
diagonals must bisect each other. Well, if you instead started by defining a parallelogram by one of these
other properties, for instance opposite sides having equal lengths, then you would end up with the same
class of shapes. The same thing happens for ellipses. One way to define an ellipse is as a “stretched out
circle”. It’s the shape you would get if you sketched a circle on a deflated balloon and then stretched out
the balloon evenly in two opposite directions:


It’s also the shape of the surface of water that results when you tilt a round glass:


Or an ellipse could be thought of as the shape of a circle drawn on a piece of paper when it is viewed at
an angle.


Equations of Ellipses
This “stretching” can be represented algebraically. For simplicity, take the circle of radius 1 centered at
the origin (0,0). The distance formula tells us that this is the set of points (x, y) that is a distance 1 unit
away from the origin.


D =


(x1 − x2)2 + (y1 − y2)2


1 =


(x − 0)2 + (y − 0)2


1 = x2 + y2


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This equation, x2 + y2 = 1, can be altered to stretch the circle in the horizontal (i.e. x−axis) direction by
dividing the x variable by a constant a > 1,


( x
a


)2
+ y2 = 1


Why does this stretch the circle horizontally? Well, the effect of dividing x by a is that for each y−value
in an ordered pair (x, y) that satisfies the original equation, the corresponding x value must be multiplied
by a in order for the pair to make a solution to the altered equation. So solutions (x, y) of the circle are in
one-to-one correspondence with solutions (ax, y) of the altered equation, hence stretching the corresponding
graph to the left and right by a factor of a. For example, here is the graph of


(


x
2


)2
+ y2 = 1:


Generalizing the equation by allowing a stretch in the vertical direction, we get the following.
( x
a


)2
+


( y
b


)2
= 1


The factor a stretches the circle in the horizontal direction and the factor b stretches the circle in the
vertical direction. If a = b, this is just a circle. When a , b, this equation represents an ellipse. The ellipse
is stretched in the horizontal direction if b < a and it is stretched in the vertical direction if a < b. Often
the above equation is written as follows.


x2


a2
+ y


2


b2
= 1


This is called the standard form of the equation of an ellipse, assuming that the ellipse is centered at (0,0).
To sketch a graph of an ellipse with the equation x2a2 +


y2
b2 = 1, start by plotting the four axes intercepts,


which are easy to find by plugging in 0 for x and then for y. Then sketch the ellipse freehand, or with a
graphing program or calculator.


Example 1 Sketch the graph of x24 +
y2
9 = 1.


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Solution: This equation can be rewritten as x222 +
y2
32 = 1. After setting x = 0 and y = 0 to find the y− and


x−intercepts, (0,3), (0,-3), (2,0), and (-2,0), we sketch the ellipse about these points:


Example 2 Sketch the graph of x216 + y2 = 1.
Solution: This can be rewritten as x242 +


y2
12 = 1. After finding the intercepts and sketching the graph, we


have:


The segment spanning the long direction of the ellipse is called the major axis, and the segment spanning
the short direction of the ellipse is called the minor axis. So in the last example the major axis is the
segment from (-4,0) to (4,0) and the minor axis is the segment from (0,-1) to (0,1).
The major and minor axes are examples of what are sometimes called reference lines. Apollonius, the
Ancient Greek mathematician who wrote an early treatise on conics, used these and other reference lines to
orient conic sections. Though the Greeks did not use a coordinate plane to discuss geometry, these reference
lines offer a framing perspective that is similar to the Cartesian plane that we use today. Apollonius’ way of
framing conics with reference lines was the closest mathematics came to the system of coordinate geometry
that you know so well until Descartes’ and Fermat’s systematic work in the seventeenth century.
Example 3 Not all equations for ellipses start off in the standard form above. For example, 25x2+9y2 = 225
is an ellipse. Put it in the proper form and graph it.
Solution: First, divide both sides by 225, to get: x29 +


y2
25 = 1, or x


2


32 +
y2
52 = 1. Finding the intercepts and


graphing, we have:


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Review Questions


1. It was mentioned above that when a round glass of water is tilted, the surface of the water is an
ellipse. Using our working definition of an ellipse as “stretched out circle”, explain why you think
the water takes this shape.


2. Sketch the following ellipse: 36x2 + 25y2 = 900
3. Now try sketching this ellipse where the numbers don’t turn out to be so neat: 3x2 + 4y2 = 12


Review Answers


1. Answers may vary, but should explain why the shape that results stretches a circle in one direction
because the width of the glass is constant.


2.


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3.


The Focal Property


In every ellipse there are two special points called the foci (foci is plural, focus is singular), which lie
inside the ellipse and which can be used to define the shape. For an ellipse centered at (0,0) that is wider
than it is tall, its major axis is horizontal and its foci are at


(√
a2 − b2, 0


)


and
(




a2 − b2, 0


)


.


What is the significance of these points? The ellipse has a geometric property relating to these points that
is similar to a circle’s relationship with its center. Remember a circle can be thought of as the set of points
in a plane that are a certain distance from the center point. In fact, that is typically the definition of a
circle. Well, the foci act like the center except that there are two of them. An ellipse is the set of points
where the sum of the distance between each point on the ellipse and each of the two foci is a constant
number. In the diagram below, for any point P on the ellipse, F1P + F2P = d, where F1 and F2 are the
foci and d is a constant.


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Technology
This definition using the sum of the focal distance gives us a great way to draw ellipses. Sure, you can just
graph them on your calculator. But why not utilize a simpler technology that does the job just as well?
As you know, a circle could be drawn by fixing a string to a piece of paper, tying the other end to a pencil,
and then drawing the curve that keeps the string taut.
Similarly, an ellipse can be drawn by taking a string that is longer than the distance between two points,
fixing the two ends of the string to the two points. Then drawing all points that can be drawn when the
string is taut and the pencil is touching it. In the diagram above, the dotted line represents the string of
fixed length, which is attached at the foci F1 and F2. The string is looped around the pencil at P, and
then the pencil is moved, keeping the string taut, drawing all the points whose sum of distances between
F1 and F2 is the length of the string.


Review Questions
4. Use string and tacks to draw ellipses that


(a) are nearly circles
(b) are very different than circles


For each of these, what can you say about how the distance between the foci and the length of the string
compare?


Review Answers
4. Drawings may vary. For ellipses that are nearly circles, the distance between the foci is small


compared to the length of string.


The foci can also be used to measure how far an ellipse is “stretched” from a circle. The symbol ε stands
for the eccentricity of an ellipse, and it is defined by the distance between the foci divided by the length
of the major axis, or



a2−b2
a for horizontally oriented ellipses and



b2−a2
b for vertically oriented ellipses.


Since a circle is an ellipse where a = b, circles have an eccentricity of 0.


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Review Questions


5. What is the full range of the eccentricity of an ellipse? What does it look like near the extremes of
this range?


Review Answers


5. The interval of possible values is ε ∈ [0, 1). At ε = 0, the ellipse is a circle; as the eccentricity
approaches 1 it becomes more and more elongated.


Turning the Definition of Ellipses on its Head
Often, this focal property is not though of as a property of ellipses, but rather a defining feature. To see
that these are equivalent, we have some to work to do. Let’s start by proving that stretched out circles
actually have this focal property. We want to prove that for a “stretched out circle” defined by the equation
x2
a2 +


y2
b2 = 1, where a and b are two positive numbers, the sum of the distance between every point on


the stretched circle and the two foci
(√


a2 − b2, 0
)


and
(




a2 − b2, 0


)


is the same. So we need to prove
that for every point on the shape defined by x2a2 +


y2
b2 = 1, the sum of the distances to


(√
a2 − b2, 0


)


and
(




a2 − b2, 0


)


is the same number.


Proof: Suppose a point (x, y) is on the curve defined by x2a2 +
y2
b2 = 1. Then we can solve for y in the


equation and express the point in terms of x:


y2


b2
= 1 − x


2


a2


y2 = a
2b2 − b2x2


a2


y =


a2b2 − b2x2
a2


Now, using the distance formula to compute the sum of the distance between the pairs of points:
(


x,


a2b2−b2x2
a2


)


and
(√


a2 − b2, 0
)


; and,
(


x,


a2b2−b2x2
a2


)


and
(




a2 − b2, 0


)


We have:




(


x −

a2 − b2


)2
+ a


2b2 − b2x2
a2


+


(


x +

a2 − b2


)2
+ a


2b2 − b2x2
a2


This algebraic equation looks daunting, but simplifying the expressions inside these square roots results in
a surprising result. This becomes


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a2
(


x −

a2 − b2


)2
+ a2b2 − b2x2


a2
+




a2
(


x +

a2 − b2


)2
+ a2b2 − b2x2


a2


= 1
a


















a2
(


x −

a2 − b2


)2
+ a2b2 − b2x2 +




a2
(


x +

a2 − b2


)2
+ a2b2 − b2x2
















= 1
a


(




a2x2 − 2a2x

a2 − b2 + a4 − a2b2 + a2b2 − b2x2 +




a2x2 − 2a2x

a2 − b2 + a4 − a2b2 + a2b2 − b2x2


)


= 1
a


(




a2x2 − 2a2x

a2 − b2 + a4 − b2x2 +




a2x2 − 2a2x

a2 − b2 + a4 − b2x2


)


= 1
a


















(


a2 − x

a2 − b2


)2
+




(


a2 + x

a2 − b2


)2














= 1
a


(


a2 − x

a2 − b2 + a2 + x



a2 − b2


)


= 1
a
(2a2)


= 2a


What a miraculous collapse! One of the gnarliest algebraic expressions that I have ever encountered turned
into the simple expression 2a. Most importantly, this reduced expression is merely a constant—it doesn’t
depend on x or y. This is exactly what we were hoping for. The sum of the distances between every point
on the ellipse and the two foci is always 2a. If you every find yourself needing to remember what this sum
is, the number 2a can be remembered most easily by computing it from an easy point such as one of the
shape’s x−intercepts, say (a, 0).


Review Questions
6. Compute the distance from the x−intercept (a, 0) and the two foci and show that it is in fact 2a.
7. What is the sum of the distances to the foci of the points on a vertically-oriented ellipse?


Review Answers


6. The distance between the x−intercept (a, 0) and
(√


a2 − b2, 0
)


is: a−

a2 − b2. The distance between


the x−intercept (a, 0) and
(




a2 − b2, 0


)


is: a +

a2 − b2.


Together these add to: a −

a2 − b2 + a +



a2 − b2 = 2a


7. 2b


Defining an Ellipse by Focal Distance
So all this computing simply means that “stretched-out circles”—what we’ve been calling ellipses—satisfy
the focal property. What would be great is if we could define ellipses by the focal property. This would be
a nice generalization of the way we define circles.
Recall that circles are defined as the set of points in a plane that are a constant distance from a center
point. Analogously, ellipses could be defined as a set of points in a plane for which the sum of the distances


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to two focus points is a constant. In other words, this definition would yield the exact same set of shapes
as the “stretched out circle” definition that we started with.
Before proceeding we need to make sure of one thing. We’ve already proved that stretched out circles
satisfy the focal property, but how do we know that any shape satisfying the focal property is in turn a
stretched out circle? Well, the calculation above can simply be read backwards. In other words, suppose
you have a set of points that satisfy the focal property, that each point whose sum of the distances to the
points ( f , 0) and (− f , 0) is a fixed distance d. Now note that for any two positive numbers d and f with
2 f < d, there exist positive numbers a and b such that d = 2a and f =



a2 − b2. (this fact is a bit subtle


and is part of an exercise below.) So now we have a shape where every point (x, y) has a sum of distances
to the points


(√
a2 − b2, 0


)


and
(




a2 − b2, 0


)


which equals 2a. Using these expressions, the algebraic steps
of the proof above can simply be read backwards.


Review Questions
8. Explain why for any two positive numbers d and f with 2 f < d, there exist positive numbers a and


b such that d = 2a and f =

a2 − b2.


9. We just told you that the above proof could be read backwards. But you need to be careful when
following algebraic steps backwards, especially ones involving squares or square roots.
(a) For example, what happens when you follow this argument backwards?x = −2


x2 = 4


(b) Write a convincing argument that it is okay to follow the steps backwards in the above proof
that every stretched circle has the focal property.


Review Answers
8. Set a = d2 . Then we need to show that for f satisfying 2 f < d, there exists a number b such that


f =

a2 − b2. Since 2 f < d, 2 f < 2a by the definition of a (using the assumption d > 0). So f < a.


We can find b geometrically. Since f < a, there is a right triangle with one leg having length f and
hypotenuse a. Call the other leg of the triangle b.


Then the Pythagorean Theorem tells us that f 2 + b2 = a2, or equivalently f =

a2 − b2. So a and b


satisfy the necessary requirements.
9. (a) Taking the square root of both sides of x2 = 4 yields two solutions, x = ±2, instead of the one


value we already know (x = −2). The problem is that the operation of squaring a number is not
a one-to-one function. Both (−2)2 and 22 yield the same number. So some information is lost
during this step, and it cannot be perfectly “undone”, like other algebraic maneuvers.


(b) The student’s reason should include the fact no information is lost (through squaring both sides
or other operations) in any of these steps, so that each step is completely reversible.


Equation of an Ellipse Not Centered at the Origin
All the ellipses we’ve looked at so far are centered around the origin (0,0). To find an equation for ellipses
centered around another point, say (h, k), simply replace x with x − h and y with y − k. This will shift all


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the points of the ellipse to the right h units (or left if h < 0) and to up k units (or down if k < 0). So the
general form for a horizontally- or vertically-oriented ellipse is:


(x − h)2
a2


+
(y − k)2


b2
= 1


It is centered about the point (h, k). If b < a, the ellipse is horizontally oriented and has foci
(


h +

a2 − b2, k


)


and
(


h −

a2 − b2, k


)


on its horizontal major axis. If a < b, it is vertically oriented and has foci
(


h, k +

a2 − b2


)


and
(


h, k −

a2 − b2


)


on its vertical major axis.
Example 4
Graph the equation 4x2 + 8x + 9y2 − 36y + 4 = 0.
Solution : We need to get the equation into the form of general equation above. The first step is to group
all the x terms and y terms, factor our the leading coefficients of x2 and y2, and move the constants to the
other side of the equation:


4(x2 + 2x) + 9(y2 − 4y) = −4


Now, we “complete the square” by adding the appropriate terms to the x expressions and the y expressions
to make a perfect square. (See http://authors.ck12.org/wiki/index.php/Algebra_I-Chapter-10#
Solving_Quadratic_Equations_by_Completing_the_Square for more on completing the square.)


4(x2 + 2x + 1) + 9(y2 − 4y + 4) = −4 + 4 + 36


Now we factor and divide by the coefficients to get:


(x + 1)2


9
+


(y − 2)2
4


= 1


And there we have it. Once it’s in this form, we see this is an ellipse is centered around the point (-1,2), it
has a horizontal major axis of length 3 and a vertical minor axis of length 2, and from this we can make a
sketch of the ellipse:


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Review Questions
10. Explain why subtracting h from the x−term and k from the y−term in the equation for an ellipse


shifter the ellipse h horizontally and k vertically.
11. Graph this ellipse. x2 − 6x + 5y2 − 10y − 66 = 0
12. Now try this one that doesn’t have such nice numbers!


16x2 − 48x + 125y2 + 150y + 61 = 0


13. Now try this one. 3x2 − 12x + 5y2 + 10y − 3 = 0. What goes wrong? Explain what you think the
graph of this equation might look like.


14. What about this one. 5x2 − 15x− 2y2 +8y− 50 = 0? What goes wrong here? Explain what you think
the graph of this equation might look like.


Review Answers


10. If (x, y) is a solution to (x−h)2a2 +
(y−k)2


b2 = 1 then (x + h, y + k) is a solution to
(x−h)2


a2 +
(y−k)2


b2 = 1. This
produces a graph that is shifted horizontally by h and vertically by k.


11.


12.


13. After completing the square, we have the sum of positive numbers equaling a negative number. This
is an impossibility, so the equation has no solutions.


14. After completing the square, the x term and the y term are opposite signs. If you plot some points you
will see that the graph has two disconnected sections. This class of conic sections will be discussed
in the next section.


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Difference Between an Ellipse and an Oval or Proving the Sliced
Cone Definition of an Ellipse


There is still one critical step missing in our exploration of ellipses. We showed that “stretched-out circles”
satisfy the focal property, and that any shape satisfying this property is in fact a “stretched-out circle”.
So these are actually the same class of shapes, and they are called ellipses. But not any oval-shaped curve
is an ellipse. Draw a random oval and you’re not likely to be able to find two points that satisfy the focal
property. In particular, when we cut a cone with a tilted plane, how do we know that the oval-shaped
curve that results is a “stretched out circle” satisfying the focal property?


Amazingly, the Ancient Greeks had an argument for this fact over two millennia ago. While it is impressive
that this problem was solved so long ago, the argument itself involves an intricate construction that isn’t as
illuminating as a more modern the one I’m going to show you instead. Most mathematicians prefer to use
a more modern argument that is simply stunning in its simplicity. This modern argument isn’t fancy—the
Greeks had all the tools they need to understand it—they just didn’t happen to think of it. It wasn’t until
1822 that the French mathematician Germinal Dandelin thought of this very clever construction. Dandelin
found a way to find the foci and prove the focal property in one fell swoop. Here’s what he said.
Take the conic section in question. Then choose a sphere that is just the right size so that when it’s dropped
into the conic, it touches the intersecting plane, as well as being snug against the cone on all sides. If you
prefer, you can think of the sphere as a perfectly round balloon that is blown up until it “just fits” inside
the cone, still touching the plane. Then do the same on the other side of the plane. fter we’ve drawn both
of these spheres we have this picture:


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or


These spheres are often called “Dandelin spheres”, named after their discoverer. It turns out that not only
is our shape an ellipse (which, like all ellipses satisfies the focal property). But these spheres touch the
ellipse exactly at the two foci. To see this, consider this geometric argument.
The first thing to notice is that the circles C1 and C2 shown on the diagram below, where each sphere lies
snug against the cone, lie in parallel planes to one another. In particular, each line passing through these
circles and the vertex of the cone, such as the line l drawn below, cuts off equal segments between the
two circles. Let’s call d the shortest distance along the cone between circles C1 and C2. This can also be
thought of as the shortest distance between C1 and C2 that passes through the vertex of the cone.


The next thing to remember is a property of tangents to spheres that you may have learned in geometry.
If two segments are drawn between a point and a sphere, and if the line containing each segment is tangent
to the sphere, then the two segments are equal. In the diagram below, AB = AC. (This follows from the
fact that tangents are perpendicular to the radii of a sphere and that two congruent triangles are formed
in this configuration. See this description for more about this property.)


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Now consider the point P on the ellipse drawn below. Let QR be the segment of length d between C1 and
C2 that passes through P. The distances between the two foci are marked d1 and d2. But d1 = RP and
d2 = PQ by the property of tangents to spheres discussed above. So d1 + d2 = RP + PQ = QR = d. And
this sum will always equal d, no matter what point P on the ellipse is chosen. So this proves the focal
property of ellipses: that the sum of the distances between any point on the ellipse and the two foci is
constant.


Review Questions
15. What do the Dandelin spheres look like in the case of a circle?
16. What is the area of an ellipse with the equation (x−h)2a2 +


(y−k)2
b2 = 1 ? (Hint: use a geometric argument


starting with the area of a circle.)
17. What is the perimeter of an ellipse with the equation (x−h)2a2 +


(y−k)2
b2 = 1 ?


Review Answers
15. The Dandelin spheres for a circle lie directly above one another, and both touch the circle at the


center point.
16. The area of an ellipse is abpi. To see why this is true, start with a circle of radius 1, which has an


area of pi. Then imagine an approximation with rectangles of the circle. Then stretch the rectangles
by a factor of a in the x−direction and by a factor of b in the y−direction to obtain an approximation
of the ellipse. This makes the rectangles a times wider and b times taller, giving an area that is ab
multiplied by the area of the approximation of the circle. Since this is true of any approximation of


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the circle, the area of the ellipse must be abpi.
17. This is actually a much more difficult question than the previous one. You’re on your own! Even


the great Indian mathematician Ramanujan could only come up with an approximation: p ≈
pi


[


3(a + b) −


(3a + b)(a + 3b)
]


.


Applications
The number of places ellipses appear in the natural world is immense. Consider, for instance, how often
the simplest kind of ellipse, the circle, appears in your life. You see circles emanating as waves when you
throw a rock into a pond. You see circular pupils when you look at a set of eyes. You see what is roughly
a circle as the image of the sun or moon in the sky. The path of an object swung around on a string is a
circle. When any of these circles are viewed at an angle you see an elongated circle, or an ellipse. So it is
important to keep in mind that the discussion that follows covers only a few of the instances of ellipses in
our daily lives.


Planetary Motion
When a planet orbits the sun (or when any object orbits any other), it takes an elliptical path and the
sun lies at one of the two foci of the ellipse. Johannes Kepler first proposed this at the beginning of the
seventeenth century as one of three laws of planetary motions, after analyzing observational data of Tycho
Brahe. His law is accurate enough to produce modern computations which are still used to predict the
motion of artificial satellites. A century later, Newton’s law of gravity offers an explanation of why this
law might be true.


Figure 8.1


Review Questions
18. Though planets take an elliptical path around the sun, these ellipses often have a very low eccentricity,


meaning they are close to being circles. The diagram above exaggerates the elliptical shape of a
planet’s orbit. The Earth’s orbit has an eccentricity of 0.0167. Its minimum distance from the sun
is 146 million km. What is its maximum distance from the sun? If the sun’s diameter is 1.4 million
kilometers. Do both foci of the Earth’s orbit lie within the sun? Recall that the eccentricity of an
ellipse is ε =



a2−b2
a .


19. While the elliptical paths of planets are ellipses that are closely approximated by circles, comets
and asteroids often have orbits that are ellipses with very high eccentricity. Halley’s comet has an


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eccentricity of 0.967, and comes within 54.6 million miles of the sun at its closest point, or “perihelion”.
What is the furthest point it reaches from the sun?


Review Answers


18. Assume that the orbit of the sun is an ellipse centered at (0,0). Then we can use the distance from the
origin to the focus



a2 − b2 to set up the equations 146 + 146 + 2



a2 − b2 = 2a and 0.167 =



a2−b2
a .


Solving we get a = 175.270, b = 175.245, and the distance from (0,0) to the foci, c = 2.927 (all units
are in millions of km). Finally the maximum distance from the earth to the sun is approximately 152
million km. From Kepler’s law, we know one of the foci of its orbit is at the center of the sun. The
other foci is 2(2.927) = 5.854 million kilometers away, so it is outside the sun (but not by very far!).


19. ∼ 3.25 billion miles.


Echo Rooms


The National Statuary Hall in the United States Capital Building is an example of an ellipse-shaped room,
sometimes called an “echo room”, which provide an interesting application to a property of ellipses. If a
person whispers very quietly at one of the foci, the sound echoes in a way such that a person at the other
focus can often hear them very clearly. Rumor has it that John Quincy Adams took advantage of this
property to eavesdrop on conversations in this room.
The property of ellipses that makes echo rooms work is called the “optical property.” So why echoes, if this
is an optical property? Well, light rays and sound waves bounce around in similar ways. In particular,
they both bounce off walls at equal angles. In the diagram below, α = β.


For a curved wall, they bounce at equal angles to the tangent line at that point:


So the “optical property” of ellipses is that lines between a point on the ellipse and the two foci form equal
angles to the tangent at that point, or in other words, whispers coming from one foci bounce directly to
the other foci. In the diagram below, for each Q on the ellipse, ∠α ∠β.


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This seems reasonable, given the symmetry of the ellipse, but how do we know it is true? First, let’s prove
an important property that is a bit more general. Suppose you have two points, P1 and P2, and a line l,
and you are interested in the shortest possible path between two P1 and P2, that intersects with l.


A nice way to find this is to reflect P2 across L, obtaining the point P′2 in the diagram below:


Since l lies between P1 and P′2, it’s much easier to find the shortest distance between P1 and P′2 that passes
through l. It’s simply the shortest path between P1 and P′2: a straight line! But for every path between
P1 and P2, we can reflect the part from l to P2 to get a path equal in length between P1 and P′2. So the
shortest path between P1 and P2 is simply the shortest path between P1 and P′2 (the straight line), with
the part between l and P′2 reflected to get a path between l and P2. So the shortest path from P1 to P2
that intersects with l is P1Q followed by QP2.
The part of the above diagram that is going to help us prove the optical property is that since ∠1 ∠2
(vertical angles) and ∠2 ∠3 (reflected angles), then ∠1 ∠3 (transitive property). Thus the shortest path
from P1 to P2 that intersects with l consists of two segments that meet the line at equal angles.
Now to prove the optical property of the ellipse, apply the above situation to the ellipse. In the picture
below, P1 and P2 are foci of the ellipse and Q is the intersection with tangent line l.


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The optical property states that ∠α ∠β. This is true because all points other than Q on line l lie outside
the ellipse. Points outside the ellipse have a combined distance to the two foci that is greater than points
on the ellipse. So Q is the point on line l with the smallest combined distance to points P1 to P2. Thus,
like we showed in general above, ∠α ∠β. So the optical property has been proved.


Review Questions
20. Design the largest possible echo room with the following constraints: You would like to spy on


someone who will be 3 m from the tip of the ellipse. The room cannot be more than 100 m wide in
any direction. How far from the person you’re spying on will you be standing.


Review Answers
20. The echo room has a major axis of 100 m and a minor axis of 34.12 m. Situating the room in the


coordinate plane, the room can be represented by the equation: x22500 +
y2
291 = 1. You will be 94 m


from the person you are spying on.


Sundials:


Figure 8.2


Conic sections help us solve the problem of making a sundial. Depending on the season, the sun shines at a
different angle. However, due to the elliptical nature of the Earth’s orbit about the sun, a shadow-casting
stick can be placed in such a way that the shadow always tells the correct time of day, no matter what the
time of year, as long as the stick is lined up with the earth’s pole of rotation.


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Review Questions
21. No matter what the orientation of a stick, if you trace out the path that the shadow of the tip makes


on a flat surface, you will find it is an ellipse. Describe why this is true.(HINT: for simplicity, you
can assume that you are making the measurements throughout the course of one day and that with
the exception of the earth rotating about a pole, the sun and the earth are fixed with respect to one
another.)


22. It was mentioned earlier in the chapter that when a round glass of water is tilted, the surface of the
water is an ellipse. Or, in other words, this statement is claiming that the cross section of a cylinder
is an ellipse. Prove that this is true. (Hint: to prove that this is an ellipse all you need to do is show
that for any cross section of a cylinder there exists a cone that has the same cross section.)


23. From the exercise above, it appears that there is some overlap between “conic sections” and “cylin-
drical sections”. Are any of the classes of conic sections we found in the last section not cylindrical
sections? Are there any cylindrical sections that are not conic sections?


Review Answers
21. Answers may vary.
22. Answers may vary.
23. Answers may vary.


Vocabulary
Ellipse A conic section that can be equivalently defined as: 1) any finite conic section, 2) a circle which


has been dilated (or “stretched”) in one direction, 3) the set of points in which the sum of distances
to two special points called the foci is constant.


Major axis The segment spanning an ellipse in the longest direction.


Minor axis The segment spanning an ellipse in the shortest direction.


Focus One of two points that defines an ellipse in the above definition.


Eccentricity A measure of how “stretched out” an ellipse is. Formally, it is the distance between the two
foci divided by the length of the major axis. The eccentricity ranges from 0 (a circle) to points close
to 1, which are very elongated ellipses.


8.3 Parabolas
Learning Objectives


• Understand what results when a cone is intersected by a plane parallel to one side of the cone.
• Understand the focal-directrix property for parabolas.
• Recognize and work with equations for parabolas.
• Understand that all parabolas are self-similar.
• Understand the equivalence of different definitions of parabolas.


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Introduction
We’ve examined ellipses and circles, the two cases when a plane intersects only one side of the cone and
creates a finite cross-section. Is it possible for a plane to intersect only one side of the cone, but create
an infinite cross-section? It turns out that this is possible if and only if the plane is parallel to one of the
lines making up the cone. Or, in other words, the angle θ between the plane and the horizon, is equal to
the angle formed by a side of the cone and the horizontal plane.


In the image above, if you till the plane a little bit to the left it will cut off a finite ellipse (possibly a very
large one if you only tilt it a little.) Tilt the plane to the right and it will intersect both sides of the cone,
making a two-part conic section called a hyperbola, which will be discussed in the next section.
When the plane is parallel to from the side of the cone, the infinite shape that results from the intersection
of the plane and the cone is called a parabola. Like the ellipse, it has a number of interesting geometric
properties.


Focus-Directrix Property
Like the ellipse, the parabola has a focal property. And, also like the ellipse, a construction similar to
Dandelin’s with the spheres can show us what it is. Dandelin himself didn’t prove the focal property for
parabolas that we are about to discuss, but Pierce Morton used a sphere construction similar to Dandelin’s
to prove the focal property of parabolas in 1829. We’ll look at Morton’s argument here.
In contrast with the argument we made for the ellipse, for a parabola we can only fit one tangent sphere
inside the cone. That is, only one sphere can be tangent to both the cone and the cutting plane. In the
diagram below, the sphere fits underneath the cutting plane, but there is no room for a sphere to lie on
top of the cutting plane and still be tangent to the cone.
As with the ellipse, the point where the sphere intersects the plane is called a focus. But because there is
only one sphere in this construction, and this is related to the fact that a parabola has only one focus. The
other geometric object of interest is called the directrix. This is the line that results from the intersection
between the cutting plane and the plane that contains the circle of contact between the sphere and the
cone. In the diagram below, the directrix is labeled l and is found by intersecting the plane defined by


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circle C and the cutting plane (the planes are shown in dashed lines for clarity). Finally, we will call the
angle between the planes θ.


In the above diagram, we have labeled the point where the sphere contacts the cutting plane with F, and
we’ll call that point the focus of the parabola. Suppose P is an arbitrarily chosen point on the parabola.
Then, let Q be the point on circle c such that PQ is tangent to the sphere. In other words Q is chosen
so that PQ lies on the cone itself. Let L be the point on the directrix l such that PL is perpendicular to
l. Then PF = PQ since both segments are tangents to the sphere from the same point P. We can also
show that PQ = PL. This follows from the fact that the cutting plane is parallel to one side of the cone.
Consider the point P’ that is the projection of P onto the plane containing circle C. Then PP′Q and
PP′L are both right angles by the definition of a projection. PQP′ and PLP′ are both equal to the
angle 90 − θ , where θ is the angle defined above, because the cutting plane and the cone both have an
angle of θ with the horizon. Since they also share a side, triangles 4PQP′ and 4PLP′ are congruent by
AAS . So the corresponding sides PQ and PL are congruent. By the transitive property we have PF = PL,
so the distance between the point P on the parabola to the focus is the same as the distance between P
and the directrix l. We have just proven the focus-directrix property of parabolas.


Equations and Graphs of Parabolas
The equation of a parabola is simpler than that of the ellipse. We will arrive at the equation for a parabola
in two ways.


Method 1: Using the Distance formula
The first method arises directly from the focus-directrix property discussed in the previous section. Suppose
we have a line and a point not on that line in a plane, and we want to find the equation of the set of points
in the plane that is equidistant to these two objects. Without losing any generality, we can orient the line


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horizontally and the point on the y−axis, with the origin halfway between them. Since the parabola is the
set of points equidistant from the line and the point, The parabola passes through the origin, (0,0). The
picture below shows this configuration. The point directly between the directrix and the focus (the origin
in this case) is called the vertex of the parabola. Suppose the focus is located at (0, b). Then the directrix
must be y = −b.


Thus, the parabola is the set of points (x, y) equidistant from the line y = −b and the focus point (0, b). The
distance to the line is the vertical segment from (x, y) down to (0,−b), which has length y − (−b) = y + b.
The distance from (x, y) to the focus (0, b) is distance =




(x − 0)2 + (y − b)2 by the distance formula. So
the equation of the parabola is the set of points where these two distances equal.


y + b =


(x − 0)2 + (y − b)2


Since distances are always positive, we can square both sides without losing any information, obtaining
the following.


y2 + 2by + b2 = x2 + y2 − 2by + b2


2by = x2 − 2by
4by = x2


y = 1
4b


x2


But b was chosen arbitrarily and could have been any positive number. And for any positive number, a,
there always exists a number b such that a = 14b , so we can rewrite this equation as:


y = ax2


where a is any constant.
This is the general form of a parabola with a horizontal directrix, with a focus lying above it, and with a
vertex at the origin. If a is negative, the parabola is reflected about the x−axis, resulting in a parabola
with a horizontal directrix, with a focus lying below it, and with a vertex at the origin. The equation can
be shifted horizontally or vertically by moving the vertex, resulting in the general form of a parabola with
a horizontal directrix and passing through a vertex of (h, k):


y − k = a(x − h)2


Switching x and y, the equation for a parabola with a vertical directrix and with a vertex at (h, k) is:


x − h = a(y − k)2


Example 1


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Sketch a graph of the parabola y = 3x2 + 12x + 17.
Solution: First, we need to factor out the coefficient of the x2 term and complete the square:


y = 3(x2 + 4x) + 17
y = 3(x2 + 4x + 4) + 17 − 12
y = 3(x + 2)2 + 5


Now we write it in the form x − h = a(y − k)2:


y − 5 = 3(x + 2)2


So the vertex is at (-2,5) and plotting a few x−values on either side of x = −2, we can draw the following
sketch of the parabola:


Review Questions


1. Sketch a graph of the following parabola: y = 2x2 − 2x − 3
2. Sketch a graph of the following parabola: 3x2 + 12x + 11 − y = 0
3. Sketch a graph of the following parabola: 0 = x2 − y + 3x + 5
4. Identify which of the following equations are parabolas: y − 5x + x2 = 3, x2 − 3y2 + 3y − 2x + 15 = 0,


x − 6y2 + 20x − 100 = 0
5. Draw a sketch of the following parabola. Also identify its directrix and focus. 3x2 + 6x − y = 0
6. Find the equation for a parabola with directrix y = −2 and focus (3,8).


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Review Answers


1.


2.


3.


4. y − 5x + x2 = 3 and x − 6y2 + 20x − 100 = 0


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5.


6. y − 3 = 120(x − 3)2


Method 2: Using Three Dimensional Analytical Geometry
Alternatively, we could have arrived at the above equation for a parabola without the argument of Pierce
Morton or the use of Dandelin’s spheres. The following argument is closer to Apollonius’ approach to
parabolas in Ancient Greece, though the closed equation forms and the coordinate grid that I will use are
modern conventions.
Consider a cone oriented in space as pictured below:


If the cone opens at an angle such that at any point its radius to height ratio is a, then the cone could
be defined as the set of points such that the distance from the z−axis is a times the z−coordinate. Or, in
other words, the set of points (x, y, z) satisfying:




(x − 0)2 + (y − 0)2 = az


Or:


x2 + y2 = a2z2


This equation works for negative values of x, y, and z, giving the general equation for a two-sided cone.


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To consider the intersection of this cone with a plane that is parallel to the line marked l marked in the
diagram above, it is most convenient to rotate the entire cone about the y axis until the left side of the
cone is vertical, then intersect it with a vertical plane perpendicular to the x−axis. Such a rotation leaves
the y−variable unchanged. To see what it does to the x and z variables, let’s see what happens to the point
(x, z) on the xz−plane when it is rotated by an angle of θ.


In the above diagram, P(x, z) is rotated by an angle of θ to the point P′. We have marked the side lengths
QP = x and SQ = z. By the Pythagorean Theorem, S P =



x2 + z2. We also have S P′ =



x2 + z2, since


rotation leaves the distance from the origin unchanged. To find the x−coordinates of our rotated point P′,
we can use the fact that cos(90 − α − θ) = SQ′√


x2+z2
. But by properties of cosine we have:


cos(90 − α − θ) = sin(α + θ),


and substituting with the sine addition formula gives us:
SQ′

x2 + z2


= sin(α) cos(θ) + cos(α) sin(θ),


which we can use our diagram to change to:
SQ′

x2 + z2


= x√
x2 + z2


cos(θ) + z√
x2 + z2


sin(θ)


which simplifies to:


SQ′ = x cos(θ) + z sin(θ)


To find the x−coordinates of our rotated point P′, we can use the fact that sin(90 − α − θ) = P′Q′√
x2+z2


. But
by properties of sine we have:


sin(90 − α − θ) = cos(α + θ)


and substituting with the cosine addition formula gives us:
P′Q′

x2 + z2


= cos(α) cos(θ) − sin(α) sin(θ),


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which we can use our diagram to change to:
P′Q′

x2 + z2


= z√
x2 + z2


cos(θ) − x√
x2 + z2


sin(θ)


which simplifies to:


P′Q′ = z cos(θ) − x sin(θ)


Looking back at the picture, this means that the coordinates of P′ are (x cos(θ)+z sin(θ), z cos(θ)− x sin(θ)).
In other words, in rotating from P to P′, the x−coordinate changes to x cos(θ)+z sin(θ) and the z− coordinate
changes to z cos(θ) − x sin(θ).
If this rotation happens to every point on the cone, we can substitute x cos(θ)+ z sin(θ) for x and z cos(θ)−
x sin(θ) for z into our equation of the cone, resulting in a new equation for the cone after rotating by θ.


(x cos(θ) + z sin(θ))2 + y2 = a2(z cos(θ) − x sin(θ))2


x2 cos2(θ) + 2xz cos(θ) sin(θ) + z2 sin2(θ) + y2 = a2(x2 sin2(θ) − 2xz cos(θ) sin(θ) + z2 cos2(θ))
x2 cos2(θ) + 2xz cos(θ) sin(θ) + z2 sin2(θ) + y2 = a2x2 sin2(θ) − 2a2xz cos(θ) sin(θ) + a2z2 cos2(θ))
x2 cos2(θ) + 2xz cos(θ) sin(θ) + z2 sin2(θ) + y2 = a2x2 sin2(θ) − 2a2xz cos(θ) sin(θ) + a2z2 cos2(θ))


Now in the case of the tilted cone, we want to tilt the cone such that the left side becomes vertical. Since
the factor a determines how tilted the cone is, we can see from the triangle below that sin(θ) = a√


1+a2
and


cos(θ) = 1√
1+a2


.


So the equation becomes:


x2
1


1 + a2
+ 2xz a


1 + a2
+ z2 a


2


1 + a2
+ y2 = a2x2 a


2


1 + a2
− 2a2xz a


1 + a2
+ a2z2 1


1 + a2
x2 + 2xza + z2a2 + y2(1 + a2) = a4x2 − 2a3xz + a2z2


x2 + 2xza + y2(1 + a2) = a4x2 − 2a3xz


Now that we have tilted our cone, to take a cross section that is parallel to the left side of the cone,
we can simply cut it with a vertical plane. The equation of a vertical plane going through (b, 0, 0) and


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perpendicular to the x−axis is x = b. Therefore, setting x equal to the constant in the equation above will
give us the intersection of the tilted cone and a plane parallel to one side of the cone. b. Here is a picture
of the rotation and the cross-section, which lies in an xz−plane.


Setting x equal to the constant b, we have:


b2 + 2abz + y2(1 + a2) = a4b2 − 2a3bz
z(2ab + 2a3b) = −y2(1 + a2) + a4b2 − b2


z =
(


−1 − a2
2ab + 2a3b


)


y2 + (a4b2 − b2)


Although this coefficient and constant term seem complicated, a and b can be chosen so that the coefficient
of the y2 term can be equal to any number (you will explore this fact in an exercise). The constant term
can be ignored since any parabola can be shifted vertically by any amount.
So the general form of a parabola is:


z = Ay2


where A is any constant.
Or, using the more standard x− and y−coordinates the form of a parabola is


y = ax2


As before, this equation can be adapted to produce the shifted and horizontally oriented forms.


Review Questions
7. Explain why cos(90 − α − θ) = sin(α + θ) in the above argument.
8. Show that for any A there exist constants a and b such that A = −1−a22ab+2a3b .


Review Answers
7. There are many arguments that work. One route is to use the fact that cos(α) = cos(−α) for any α,


and then the fact that cos(α − 90) = sin(α) for any α.


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8. Solving for b in terms of A and a, we have:


A(2ab + 2a3b) = −1 − a2


2Aab(1 + a2) = −(1 + a2)
2Aab = −1
2Aab = −1


b = − 1
2Aa


So we can set b = 2Aa and the relationship will hold.


All Parabolas are the Same
There is a subtle point in the above two arguments that reveals a very interesting property of parabolas.
This is the fact that all parabolas have the same shape. Or, in the language of geometry, any two parabolas
are similar to one another. This means that one parabola can be scaled in or out to produce another
parabola of exactly the same shape. This may come across as surprising, because parabolas where x2 has
a large coefficient certainly look much “steeper” than parabolas with a small coefficient when examined
over the same domain, as shown in the graphs below.


But when one of the parabolas is scaled appropriately, these parabolas are identical:


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This fact about parabolas can be seen in the first argument simply from the fact that all parabolas are
generated from a line and a point not on that line. This configuration of generating objects, a line and
a point, is always the same shape. Any other line and point looks exactly the same—simply zoom in or
out until the line and point are the same distance from one another. So the shapes that any two such
configurations generate must also be the same shape.
In the second argument, there were two factors that might affect the shape of the parabola. The first is
the distance between the cutting plane and the apex of the cone. But cones have the same proportions at
any scale, so no matter what this distance, the picture can be reduced or enlarged, affecting this distance
but not the shape of the cone or plane. So this parameter does not actually change the shape of the conic
section that results. The other factor is the shape of the actual cone. This is its steepness, defined by the
angle at the apex, or equivalently by the ratio between the radius and the height at any point. This is a
bit trickier. It’s not at all obvious that short, squat cones and tall, skinny cones would produce parabolas
of the same shape.


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According to what we found, any parabola produced by slicing any cone resulted in an equation of this
form:


y = ax2


We want to show that if we generate two such parabolas, that they actually have the same shape. So
suppose we use two cone constructions and come up with these parabolas: y = a1x2 and y = a2x2. We
want to show there is some scale factor, call it f , that shrinks or enlarges y = a1x2 into y = a2x2. To keep
a shape the same, the scale factor needs to affect both the x− and y−variables. So we need to find an f
such that ( f y) = a1( f x)2 is equivalent to y = a2x2. The first equation can be written y = (a1 f )x2, which is
equivalent to the second equation when a1 f = a2, or when f = a2a1 . Such an f always exists for non-negativenumbers a1 and a2. So the parabolas are indeed the same shape. If a is less than zero, then the parabola
can be reflected vertically to produce a parabola of the same shape and with positive coefficient a.


Review Questions
9. What other classes of shapes have this property that all members of the class are similar to each


other?
10. Explain why not all ellipses are similar. While enlarging or shrinking doesn’t work to make two


ellipses identical, how can you change the view of two ellipses that have different shapes so that they
look the same?


Review Answers


9. A few examples are: circles, squares, finite sections of one-sided cones of the same angle.
10. The eccentricity of ellipses defines the shape, so when the eccentricity is different for two ellipses, the


ellipses are not similar to one another. Viewing one of the ellipses at an angle, however, changes the
perceived eccentricity of that ellipse, and the angle can be chosen to match the perceived eccentricity
to the eccentricity of the other ellipse, producing an image that is similar to the other ellipse.


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Applications
“Burning Mirrors”
Diocles (∼ 240 − 180 BCE) was a mathematician from Ancient Greece about whom we know very little.
However, we know enough from a few scant documents that he thought about an important application
of parabolas. It comes from what is sometimes called the “optical property” of parabolas. Remember the
optical property of ellipses: lines from one focus “bounce off” the side of the ellipse to hit the other focus.
For parabolas, since parabolas have only one focus, the directrix plays a role. For the parabola, the optical
property is that lines perpendicular to the directrix “bounce off” the parabola and converge at the focus.
Or, alternatively, lines from the focus “bounce off” off the parabola and continue perpendicular to the
directrix. As with the ellipse, “bouncing off” means that the two lines meet the parabola at equal angles
to the tangent.


In the above diagram, the optical property states that α β. The proof is similar to the proof of the
optical property of ellipses. In the above diagram P is the focus and Q is a point on the parabola. Let
R′ be the point on the directrix that is obtained by extending PQ. Then PR′, the straight line, is clearly
the shortest distance between P and R′ that passes through the tangent line. Let R be on the line lying
directly above Q such that QR = QR′. The R can be thought of as R′, reflected across the tangent line.
Then α γ (vertical angles) and γ β (reflected angles), and so α β (transitive property).
The optical property has some interesting applications. Diocles described one potential application in his
document “On Burning Mirrors”. He envisioned a parabolic-shaped mirror (basically a parabola rotated
about its line of symmetry) which would collect light from the sun and focus it on the focal point, creating
enough of a concentration of light to start a fire at that point. Some claim that Archimedes attempted to
make such a contraption with copper plates to fight the Romans in Syracuse.


Figure 8.3


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Review Questions
11. If nothing was used to deflect light before it entered the mirror, where would the sun have to be in


relationship to you and the place you want to start a fire? Why is this a constraint? Design a way
to circumvent this constraint.


12. You live in Ancient Greece and are defending your city against Roman invaders. Design a “burning
mirror” that you plan to use to destroy an army that is approaching at a distance of three miles.


Review Answers
11. The fire-locale must lie on the segment between you and the sun. This is a problem because to start


a ground fire, you would have to wait until evening when the sun is not as bright. A lens or mirror
that changes the angle of the suns rays could help you work around this constraint.


12. Answers will vary. The distance from the focus to the vertex should be 3 miles.


Headlights
The optical property is also responsible for parabola-shaped unidirectional lights, such as car headlights.
If a bulb is placed at the focus of a parabolic mirror, the light rays reflect off the mirror parallel to each
other, making a focused beam of light.


Figure 8.4


Review Questions
13. In the above diagram of a car headlight, the lens directs the beams of light downwards, to keep them


out of the eyes of oncoming drivers. But if that was the only purpose of the lens, alternatively the
lens could be omitted and the headlight could just be angled down slightly. But there is another
purpose to the lens. What is it?


Review Answers
13. The lens also expands the array of light which is why it is called “dispersed light.” Without the lens,


the headlight would only illuminate a strip the width of the headlight itself, which would not be very
useful for driving.


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Cassegrain Telescopes
Satellite telescopes take advantage of the optical property of parabolas to collect as much light from a
distant star as possible. The dish of the satellite below is parabolic in shape and reflects light to the point
in the middle.


Figure 8.5


This image at this point is then focused with a lens into the telescope as shown in the diagram below.
(from:http://commons.wikimedia.org/wiki/File:Telescope_cassegrain_principe.png) The


Figure 8.6


Technology
As with ellipses, you can examine the graphs of parabolas on a graphing calculator or computer program.
But as with ellipses, there is a much more low-technology tool that can draw a parabola. It is a generalized
version of the circle-drawing compass. Below is a model of a conic section producing compass drawn by
Leonardo da Vinci. Like a compass for making circles, this tool assists a pencil in swiveling about a vertex.
Unlike a compass, the pencil is held loosely in a shaft so that it can slide up and down, and the fixed side
is held at a constant angle. As this angle changes, different conic sections result. This essentially turns the
pencil shaft into the cone and the paper into the cutting plane.


Review Questions
14. Describe how the above compass should be set up to produce a parabola.


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Figure 8.7


Review Answers
14. The pencil should be parallel to the paper at its most extended point.


Vocabulary
Parabola A conic section resulting from intersecting a cone with a plane that is parallel to one of the


lines in the cone.


Focus The point to which all points on a parabola are the same distance as they are to a line, called the
directrix.


Directrix A line to which all points on a parabola are the same distance as they are to a point called the
focus.


8.4 Hyperbolas
Learning Objectives


• Understand what results when a cone is intersected by a plane that intersects both sides of the cone.
• Understand the focal-directrix property for hyperbolas.
• Recognize and work with equations for hyperbolas.
• Understand how a hyperbola can be described in terms of its asymptotes and foci.
• Understand the equivalence of different definitions of hyperbolas.


Two-part Conic Sections
The final way of cutting a cone with a plane appears at first to be much messier. Not only does it result
in an infinite shape, but there are two pieces that aren’t even connected! When the plane slices through
two parts of the cone, the two infinite “U”-shaped parts are together called a hyperbola.


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In this section we will see that this sprawling shape actually has some beautiful properties that make it as
noble as its cousins.


The Focal Property
Even though this shape seems much harder to conceive of than an ellipse, the hyperbola has a defining
focal property that is as simple as the ellipse’s. Remember, an ellipse has two foci and the shape can be
defined as the set of points in a plane whose distances to these two foci have a fixed sum.


Hyperbolas also have two foci, and they can be defined as the set of points in a plane whose distances to
these two points have the same difference. So in the picture below, for every point P on the hyperbola,
|d2 − d1| = C for some constant C.


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Review Questions
1. The hyperbola is infinite in size. In mathematics this is called unbounded, which means no circle,


no matter how large, can enclose the shape. Explain why a focal property involving a difference
results in an unbounded shape, while a focal property involving a sum results in a bounded shape.


Review Answers
1. Answer should include the following concept: In the case of an ellipse, we had two distances summing


to a constant. Since the distances are both positive then there is a limit to the size of the numbers.
In the case of hyperbolas, two very large positive numbers can have a much smaller difference.


To prove the focal property of hyperbolas, we examine Dandelin’s sphere construction. Unlike the con-
struction for ellipses, which used two spheres on one side of the cone, and the sphere construction for
parabolas, which used one sphere on one side of the cone, this construction uses two spheres, one on each
side of the cone. As with the ellipse construction, each sphere touches the plane at one of the foci of the
hyperbola. And as with the argument for the elliptical focal property, the argument uses the fact that
tangents from a common point to a sphere are equal.


In the above diagram, suppose P is an arbitrary point on the hyperbola. We would like to examine the
difference PF2 − PF1. Let C1 be the point on the upper sphere that lies on the line between P and the
vertex of the cone. Let C2 be the point on the upper sphere when this line is extended (so P, C1, and


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C2 are all on the same line and PC1 + C1C2 = PC2 and the cone By the common tangent property,
PF1 = PC1 and PF2 = PC2 for some points C1 and C2 on the circles where the spheres meet the cone. So
PF2−PF1 = PC2−PC1 = (PC1+C1C2)−PC1 = C1C2. But C1C2 is the distance along the cone between the
two circles of tangency and is constant regardless of the choice of C1 and C2. So the difference PF2 − PF1
is constant.


Equations for Hyperbolas


To derive the equation for a hyperbola, we can’t do exactly what we did for ellipses. Remember, when we
first derived the equations for ellipses, we were defining them as “stretched out circles.” Hyperbolas aren’t
as obviously a simple dilation of a shape as basic as a circle. Instead, we’ll use the focal property to derive
their equation. This was what we said could be done for ellipses when I said “simply read the proof of the
focal property backwards.” To derive the equation form of hyperbolas from the focal property, I’ll actually
show you the steps in the correct order.
Suppose we have a curve (actually a pair of curves), satisfying the focal property for hyperbolas. Let’s
orient the hyperbola so that the two foci are on the y−axis, and equidistant from the origin. In the diagram
below the foci are labeled with the points (0, c), and (0,−c).


The focal property states that the difference in distances between an arbitrary point P = (x, y) on the
hyperbola and (0, c), and (0,−c) is a constant. In particular, we know that this constant can be computed
from any point on the hyperbola, for instance the point on the y−axis marked (0, a). The distance between
(0, a) and (0,−c) is a + c and the distance between (0, a) and (0, c) is c − a. The difference between these
two quantities is (c+ a)− (c− a) = 2a. So 2a, the distance between the two y−intercepts of the hyperbola,
is the constant in the focal property for the hyperbola. Using this distance formula and the focal property,
we have:


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(x − 0)2 + (y − (−c))2 −


(x − 0)2 + (y − c)2 = 2a


x2 + (y + c)2 − 2


x2 + (y + c)2


x2 + (y − c)2 + x2 + (y − c)2 = 4a2


2x2 + (y + c)2 + (y − c)2 − 4a2 = 2


x2 + (y + c)2


x2 + (y − c)2


(2x2 + 2y2 + 2c2 − 4a2)2 = 4(x2 + (y + c)2)(x2 + (y − c)2)
4x4 + 8x2y2 + 8c2x2 + 8c2y2 + 4y4 − 16a2x2 + 4c4 − 16a2c2 − 16a2y2 + 16a4 = 4x4 + 4x2(y + c)2 + 4x2(y − c)2 + 4(y + c)2(y − c)2


4x4 + 8x2y2 + 8c2x2 + 8c2y2 + 4y4 − 16a2x2 + 4c4 − 16a2c2 − 16a2y2 + 16a4 = 4x4 + 4x2(y + c)2 + 4x2(y − c)2 + 4(y2 − c2)2


4x4 + 8x2y2 + 8c2x2 + 8c2y2 + 4y4 − 16a2x2 + 4c4 − 16a2c2 − 16a2y2 + 16a4 = 4x4 + 8x2y2 + 8c2x2 + 4y4 − 8c2y2 + 4c4


16c2y2 − 16a2x2 − 16a2c2 − 16a2y2 + 16a4 = 0
c2y2 − a2x2 − a2c2 − a2y2 + a4 = 0


c2y2 − a2y2 − a2x2 = a2c2 − a4


y2(c2 − a2) − x2(a2) = a2(c2 − a2)
y2


a2
− x


2


c2 − a2 = 1


Like the collapse we witnessed when investigating the focal property for ellipses, this equation became
pretty simple. We can even make it simpler. For any positive number b and a, there exists a c such that
b2 = c2 − a2. Thus the general form for a hyperbola that open upwards and downwards and whose foci lie
on the y−axis is:


y2


a2
− x


2


b2
= 1


Switching x and y we have hyperbolas that open rightwards and leftwards and whose foci lie on the x−axis.
x2


a2
− y


2


b2
= 1


These equations have so far been hyperbolas that are centered about the origin. For a hyperbola that is
centered around the point (h, k) we have the shifted equations:


(y − k)2
a2


− (x − h)
2


b2
= 1


for a hyperbola opening up and down, and
(x − h)2


a2
− (y − k)


2


b2
= 1


For a hyperbola opening left and right.
Example 1
Show that the following equation is a hyperbola. Graph it, and show its foci.


144x2 − 576x − 25y2 − 150y − 3249 = 0


Solution: The positive leading coefficient for the x2 term and the negative leading coefficient for the y2
term indicate that this is a hyperbola that is horizontally oriented. Grouping and completing the square,
we have:


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144(x2 − 4x) − 25(y2 + 6y) = 3249
144(x2 − 4x + 4) − 25(y2 + 6y + 9) = 3249 + 576 − 225


144(x − 2)2 − 25(y + 3)2 = 3600
(x − 2)2


52
− (y + 3)


2


122
= 1


So our hyperbola is centered at (2,-3). Its vertices are 5 units to the right and left of (2,-3), or at the points
(7,-3) and (-3,-3). It opens to the right and left from these vertices. It’s foci are c units to the left and
right of (2,-3), where c =



a2 + b2 =



52 + 122 = 13. So it’s foci are at (15,-3) and (-11,-3). Plotting a


few points near (7,-3) and (-3,-3), the graph looks like:


Review Questions


2. Explain why for any positive number b and a, there exists a c such that b2 = c2 − a2.
3. Graph the following hyperbola and mark its foci: 16x2 + 64x − 9y2 + 90y − 305 = 0
4. Graph the following hyperbola and mark its foci: 9y2 + 18y − x2 + 4x − 4 = 0
5. Graph the following hyperbola and mark its foci: 25x2 + 150x − 4y2 + 24y + 89 = 0
6. Find the equation for the following hyperbola:


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Review Answers


2. Let c =

a2 + b2. Since a2 +b2 is always positive for positive a and b, this number is always defined.


Geometrically, let c be the hypotenuse of a right triangle with side lengths a and b.
3.


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4.


5.


6. (x−4)24 −
(y+2)2


45 = 1


Asymptotes
In addition to their focal property, hyperbolas also have another interesting geometric property. Unlike a
parabola, a hyperbola becomes infinitesimally close to a certain line as the x− or y−coordinates approach
infinity. Such a line is called an asymptote.


Before we try to prove this property of hyperbolas, we have to figure out what we mean by “infinitesimally
close.” Here we mean two things: 1) The further you go along the curve, the closer you get to the asymptote,
and 2) If you name a distance, no matter how small, eventually the curve will be that close to the asymptote.
Or, using the language of limits, as we go further from the vertex of the hyperbola the limit of the distance


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between the hyperbola and the asymptote is 0.
So now we need to prove such a line (or lines) exist for the hyperbola. It turns out there are two of them,
and that they cross at the point at which the hyperbola is centered:


It also turns out that for a hyperbola of the form x2a2 −
y2
b2 = 1, the asymptotes are the lines y = ba x and


y = −ba x. (for a hyperbola of the form
y2
a2 −


x2
b2 = 1 the asymptotes are the lines y = ab x and y = −ab x. For a


shifted hyperbola, the asymptotes shift accordingly.)
To prove that the hyperbola actually gets infinitesimally close to these lines as it goes to infinity, let’s focus
on the point P on the upper right leg of the hyperbola in the picture below.


Now consider the point P′, the point on the asymptote lying directly above P. If we can show that P
and P′ become infinitesimally close, then that’s good enough (this subtlety will be explored in the next
exercise.) Well, the point P is on the hyperbola x2a2 −


y2
b2 = 1, so if it’s x−value is x, it’s y− value can be


found by solving for y in the equation. So


x2


a2
− 1 = y


2


b2


y2 = b
2


a2
x2 − b2


y =


b2


a2
x2 − b2


This expression simplifies further, but we’ll leave it for now and work on P′. Since P′ lies directly above
P, x is also the x− coordinate of P′. Similarly as with P, we can find the y−coordinate from the equation of
the asymptote. Since y = ba x is the equation of the asymptote, we have ba x as the y−coordinate of P′. So
P =


(


x,


b2
a2 x


2 − b2
)


, and P′ =
(


x, ba x
)


. Since these two points are vertically aligned, the distance between
them points is simply the difference between the y−coordinate:


Distance between P and P′ = ba x −


b2
a2 x


2 − b2


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= b
a
x −




b2


a2
x2 − b


2


a2
a2


= b
a
x −




b2


a2
(x2 − a2)


= b
a
x − b


a




(x2 − a2)


= b
a


(


x −


(x2 − a2)
)


So the distance is a constant multiplied by a quantity that depends on x. What happens to this quantity
when x approaches infinity? This kind of question involves working with limits in a way you will study
soon. For now, we can use a nice geometric picture to see what happens. In the picture below, we see that
a, x, and




(x2 − a2) can be thought of as side-lengths of a right triangle, where x is the hypotenuse. This
can be verified with the Pythagorean Theorem.


In this picture, what happens to difference in lengths between the side of length x and the side of length


(x2 − a2) when x approaches infinity and a remains fixed? The two side lengths become closer in length,
so their difference approaches zero.
Going back to the previous diagram where we defined P and P′, this means that P and P′ become infinites-
imally close as x approaches infinity.
This same argument can be repeated to show that the other three legs of the hyperbola approach their
respective asymptotes, and to show that the same holds in vertically-oriented and/or shifted parabolas.
Example: Graph the following hyperbola, drawing its foci and asymptotes and using them to create a
better drawing: 9x2 − 36x − 4y2 − 16y − 16 = 0.
Solution: First, we put the hyperbola into the standard form:


9(x2 − 4x) − 4(y2 + 4y) = 16
9(x2 − 4x + 4) − 4(y2 + 4y + 4) = 36


(x − 2)2
4


− (y + 2)
2


9
= 1


So a = 2, b = 3 and c = √4 + 9 =

13. The hyperbola is horizontally oriented, centered at the point (2,-


2), with foci at
(


2 +

13,−2


)


and
(


2 −

13,−2


)


. After taking shifting into consideration, the asymptotes
are the lines: y + 2 = 32(x − 2) and y + 2 = −32(x − 2). So graphing the vertices and a few points on either
side, we see the hyperbola looks something like this:


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Review Questions


7. In the diagram above, which was used in the proof that hyperbolas have asymptotes, why is it true
that if P and P′ become infinitesimally close, that this implies the hyperbola and the line become
infinitesimally close. (HINT: what exactly do we mean by the “closeness” of a point on a hyperbola
and a line?)


8. Graph the following hyperbola, drawing its foci and asymptotes and using them to create a better
drawing: 16x2 − 96x − 9y2 − 36y − 84 = 0.


9. Graph the following hyperbola, drawing its foci and asymptotes and using them to create a better
drawing: y2 − 14y − 25x2 − 200x − 376 = 0.


10. Find the equation for a hyperbola with asymptotes of slopes 512 and − 512 , and foci at points (2,11)
and (2,1).


11. A hyperbolas with perpendicular asymptotes is called perpendicular. What does the equation of
a perpendicular hyperbola look like?


Review Answers


7. The distance between a point and a line is the shortest segment between the point and a point on
the line. We have shown that some distance—not necessarily the shortest—between P and a point
on the asymptote becomes infinitesimally smaller. This means that the shortest distance between P
and the asymptote must also become shorter.


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8.


9.


10. (y−6)225 −
(x−2)2
144 = 1


11. The slopes of perpendicular lines are negative reciprocals of each other. This means that ab = ba ,
which, for positive a and b means a = b.


Applications
Vexing Questions from Ancient Greece
This application of hyperbolas is the reason Menaechmus, one of the first serious thinkers about conic
sections, got into them in the first place.
It is thought that conic sections first attracted the attention of the Ancient Greeks when Menaechmus used
two conics to find a solution to another famous Greek problem. A famous class of Greek problems involves
constructing a length with the simplest tools possible. One of these problems is often referred to as the
problem of “doubling a cube.” The problem is to make a cube that is twice the volume of a given cube.


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This, of course, does not simply mean making a cube whose sides are double in length. This would produce
a cube with eight times the volume. If the first cube has side length one unit then it has volume 13 = 1.
In order for a cube to have double that volume, 2, it must have a side length of 3



2. So the problem boils


down to producing a segment whose length is 3

2. Once this is done, the cube can be constructed out of


segments of this length.
So Menaechmus was working on constructing a segment of length 3



2. Now you are probably thinking,


“Easy, just plug 3

2 into your calculator.” But Menaechmus didn’t have a calculator into which he could


type 2∧(1/3). An even more important point is that even if he did have such a calculator, this wouldn’t
solve the problem in a way Menaechmus would have been pleased with. He was trying to use the simplest
technology possible, and a calculator—even if he had one—wouldn’t fit this bill. Plus, a calculator merely
finds an estimate of this number, and the Greeks aimed to find a mathematical situation in which the exact
number would be generated.
The very simplest tools the Greeks considered for making these kinds of constructions were an idealized
compass and straightedge. Unknown to the Greeks at the time, a compass and straightedge don’t do the
job for this length, though they didn’t know it at the time—it wasn’t discovered until 1837, when Pierre
Laurent Wantzel demonstrated that this construction is impossible.
According to some historians (Heath [footnote]), Menaechmus did find a solution to the problem and he
used conic sections to do so. Before his time, the problem had already been reduced to finding a specific
set of “mean proportionals”, or a chain of numbers with equivalent subsequent ratios. In particular, to
find numbers x and y such that 1 is to x as x is to y as y is to 2, or in fraction form:


1
x


= x
y


= y
2


How does this chain of proportions help find the number 3

2? Well, looking at the first equality and cross-


multiplying, we have y = x2. Looking at the second equality, we have y2 = 2x. Using the first equation to
substitute x2 for y in the second equation, we have (x2)2 = 2x, which reduces to x3 = 2, the solution of
which is 3



2.


But solving the above chain of ratios is also equivalent to finding the intersection of the parabola y = x2
and the hyperbola xy = 2. You’ll examine why this is true in an upcoming review questions.
But wait! Hold the presses! Did I say the hyperbola xy = 2 ? But that doesn’t look like the form of
any of the hyperbolas we have looked at! This is because it is oriented with its foci on a diagonal, rather
than horizontally or vertically. In the next section we will look at how rotating conic sections from their
standard positions affects their forms. For now, you can check that xy = 2 indeed looks like a hyperbola
by plugging in a few points near the origin—which you will do in the exercises below.


Review Questions
12. Draw a sketch of xy = 2 by plotting a few points near the origin.
13. What are the asymptotes of the hyperbola xy = 2 ? What are the foci?
14. Explain why solving the mean proportional 1x = xy =


y
2 is equivalent to finding the intersection of the


parabola y = x2 and the hyperbola xy = 2.
15. In the language of “compass and straightedge,” though Menaechmus’ construction can’t be done with


a traditional compass, it can be done with the generalized compass discussed in the previous section.


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Explain how this tool could be used to “double” an existing cube.


Review Answers
12.


13. The asymptotes are the x− and y−axes. The foci are (2,2) and (-2,-2) (these are relatively hard to
find, but it is relatively easy to show they are the foci once they are found.)


14. These two equations are obtained by looking at the first equality and cross multiplying, as well as
setting the first term equal to the third term and cross multiplying. These two equalities hold exactly
the same amount of information as the chain of equalities.


15. Using a compass and straightedge, a coordinate grid with unit lengths can be drawn. See http://
en.wikipedia.org/wiki/Compass_and_straightedge_constructions for more on using a compass
and straightedge to make geometric constructions. Upon this, the two shapes can be drawn using
the generalized compass as discussed in the last session. The distance between the y−axis and the
intersection point is of length 3



2.


Vocabulary
Hyperbola A conic section where the cutting plane intersects both sides of the cone, resulting in two


infinite “U”-shapes curves.


Unbounded A shape which is so large that no circle, no matter how large, can enclose the shape.


Asymptote A line which a curve approaches as the curve and the line approach infinity, eventually
becoming closer than any given positive number.


Perpendicular Hyperbola A hyperbola where the asymptotes are perpendicular.


8.5 General Algebraic Forms
Learning Objectives


• Understand how the cross sections of a cone relate to degree two polynomial equations.
• Understand what happens to the algebraic forms of conic sections when they are neither horizontally


or vertically oriented.
• Recognize the algebraic form of different types of conic sections.


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The Cross Sections of the Cone Are Degree Two Polynomial
Equations
Let’s examine all the equations of the conic sections we’ve studied in this chapter.
Ellipses: x2a2 +


y2
b2 = 1, where a and b are any positive numbers (the circle is the specific case when a = b).


Parabolas: y = ax2 or x = ay2 where a is any non-zero number.
Hyperbolas: x2a2 −


y2
b2 = 1 or


y2
a2 −


x2
b2 = 1, where a and b are any positive numbers.


All these equations have in common that they are degree-2 polynomials, meaning the highest exponent of
any variable—or sum of exponents of products of variables—is two. So for example, here are some degree
two polynomial equations in a more general form:


2x2 + 5y2 − 3y + 4 = 0
x2 − 3y2 + x − y + 3 = 0


10x2 − y − 5 = 0
xy = 2


Some of these probably already look like conic sections to you. For example, in the first equation, we can
complete the square to remove the −3y term and we will see that we have an ellipse. In the second equation
we can complete the square twice to remove both the x and −y terms and we will have a hyperbola. This
is a hyperbola, not an ellipse, because the coefficient of the x2 and y2 terms have opposite signs.
The third equation is a parabola since there is an x2 term and y term but not a y2 term. Do you see how
you can solve for y, putting the equation in the standard form for a vertically oriented parabola?
But what about the fourth equation? Like the others, it is a degree-2 polynomial, since the exponents of
the x and y term sum to 2. But the fourth equation looks nothing like any of the forms for conic sections
that we’ve examined so far. Nonetheless, as we saw in the last section, xy = 2 appears to be a hyperbola
with foci (2,2) and (-2,-2). The reason it doesn’t fit either of the standard forms for hyperbolas is because
it is diagonally oriented, rather than horizontally or vertically oriented (do you see how its two foci lie on
a diagonal line, rather than a horizontal or vertical line?)
In order to see how such differently-oriented conic sections fit into our standard forms, we need to rotate
them so that they are either horizontally or vertically oriented.


Rotation of Conics
Remember in the section on parabolas we discussed rotating objects in the plane. In that section we showed
that if we take a point P = (x, y) and rotate it θ degrees, it changes the x−coordinate to x′ cos(θ)− y′ sin(θ)
and the y−coordinate changes to x′ sin(θ) + y′ cos(θ). (Note: in that section, because we were looking at
a plane embedded in space, we happened to be looking at the xz−plane. We also we’re using the “prime”
symbols x′ and y′. Also, in the particular case of that section we were rotating the cone clockwise. It
is more standard to rotate counter-clockwise, so the signs on the sine functions have changed. So you
can simply substitute x′ for x and y′ for z into the formulas in that section, switch the signs on the sine
functions, and come up with these rotation rules for the xy−plane.)
Now suppose we have an equation of degree two polynomials, such as the xy = 2 example discussed above.
In order to put it into the more recognizable form of a ellipse, parabola, or a hyperbola, we need to rotate
it in such a way so that rotated version has no xy term. So we need to find an appropriate angle θ such
that changing the x−coordinate to x′ cos(θ)− y′ sin(θ) and the y−coordinate to x′ sin(θ)+ y′ cos(θ) results in


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an equation with no xy term. We need to investigate what happens to a degree-two polynomial equation.
Such an equation can be written in the form:


Ax2 + By2 +Cxy + Dx + Ey + F = 0


If C = 0, such as in the first three examples at the beginning of this section, we are done, as there is no
xy term and we already know how to classify these into conics. If C , 0, we need to rotate the curve so
that we have an equation with no xy term. When x is replaced by x′ cos(θ) − y′ sin(θ) and y is replaced by
x′ sin(θ) + y′ cos(θ), only the first three terms of this equation are in danger of producing an xy term. To
see if we can determine if an appropriate angle θ can always be found, let’s substitute our new variables in
for x and y into the first three terms of the equation:


A(x′ cos(θ) − y′ sin(θ))2 + B(x′ sin(θ) + y′ cos(θ))2 +C(x′ cos(θ) − y′ sin(θ))(x′ sin(θ) + y′ cos(θ))


Then, let’s multiply this expression out, but only examine the terms that are a multiple of x′y′, since that
is what we’re trying to eliminate.


−2Ax′y′ cos(θ) sin(θ) + 2Bx′y′ cos(θ) sin(θ) +Cx′y′(cos2(θ) − sin2(θ))


This reduces to:


(−2A cos(θ) sin(θ) + 2B cos(θ) sin(θ)) +C(cos2(θ) − sin2(θ)))x′y′


So we are interested in whether or not there is an angle θ such that the above coefficient of x′y′ is zero:


(−2A cos(θ) sin(θ) + 2B cos(θ) sin(θ)) +C(cos2(θ) − sin2(θ)) = 0
C(cos2(θ) − sin2(θ)) = (2A cos(θ) sin(θ) − 2B cos(θ) sin(θ))
C(cos2(θ) − sin2(θ)) = 2(A − B)(cos(θ) sin(θ))


Separating the θ terms from the A, B, and C terms, we have:
2 cos(θ) sin(θ)


cos2(θ) − sin2(θ) =
C


A − B


But remember the double angle formulas sin(2θ) = 2 cos(θ) sin(θ) and cos(2θ) = cos2(θ)−sin2(θ) (see http:
//authors.ck12.org/wiki/index.php/Trigonometric_Identities#Double-Angle_Identities for more
on the double angles). This means that we need to find an angle such that:


sin(2θ)
cos(2θ) =


C
A − B


or:


tan(2θ) = C
A − B


Can we find such an angle? Well remember the graph of tangent (see http://authors.ck12.org/wiki/
index.php/Circular_Functions#y_.3D_tan.28x.29.2C_The_Tangent_Graph for more on the tangent
graph). It spans all values, so no matter what C, A, and B equal, we can find the appropriate tangent
value. The only time this expression gives us trouble is if the denominator is zero, or A = B. But in this
case there’s an easy solution that you will find in the exercise below.


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For the other cases, you can use the inverse tan function on your calculator or computer program to find
out the value of 2θ, and then divide by 2 to find the value of θ.
Example: Rotate the following conic section so that it is oriented either horizontally or vertically, and then
analyze the result:


79x2 + 37y2 + 42

3xy +


(


−200

3 − 8


)


x +
(


8

3 − 200


)


y + 4 = 0


Solution: To find how much to rotate this conic we need to solve for θ in the equation tan(2θ) = CA−B . We
have C = 42



3, A = 79, and B = 37, so we have:


tan(2θ) = 42

3


79 − 37
tan(2θ) =



3


The angle that solves this equation is θ = pi6 (or 30◦).
So we replace x with x′ cos(30) − y′ sin(30) =



3


2 x
′ − 12y′, and y with x′ sin(30) + y′ cos(30) = 12 x′ +



3


2 y
′,


giving us:


79













3


2
x′ − 1


2
y′












2


+ 37












1
2
x′ +



3


2
y′












2


+ 42

3













3


2
x′ − 1


2
y′
























1
2
x′ +



3


2
y′












+


(


−200

3 − 8


)















3


2
x′ − 1


2
y′












+
(


8

3 − 200


)














1
2
x′ +



3


2
y′












+ 4 = 0


Multiplying through by 4 we have:


79
(√


3x′ − y′
)2


+ 37
(


x′ +

3y′


)2
+ 42



3
(√


3x′ − y′
) (


x′ +

3y′


)


+
(


−200

3 − 8


) (


2

3x′ − 2y′


)


+
(


8

3 − 200


) (


2x′ + 2

3y′


)


+ 16 = 0


Multiplying out we have:


237(x′)2 − 158

3x′y′ + 79(y′)2 + 37(x′)2 + 74



3x′y′ + 111(y′)2 + 126(x′)2 + 84



3x′y′ − 126(y′)2


− 1200x′ − 16

3x′ + 400



3y′ + 16y′ + 16



3x′ − 400x′ + 48y′ − 400



3y′ + 16 = 0


Which simplifies to:


400(x′)2 + 64(y′)2 − 1600x′ + 64y′ + 16 = 0


Which, divided by 16 is:


25(x′)2 + 4(y′)2 − 100x′ + 4y′ + 1 = 0


Grouping the x′ and y′ terms and completing the squares, we have:


25((x′)2 − 4x′ + 4) − 100 + 4
(


(y′)2 + y′ + 1
4


)


− 1 + 1 = 0


25((x′)2 − 2)2 + 4
(


(y′)2 + 1
2


)2
= 100


((x′)2 − 2)2
4


+


(


(y′)2 + 12
)2


25
= 1


We recognize this as an ellipse, centered at the point
(


2,−12
)


.


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Review Questions
1. What’s the easy solution to the above problem when A = B?
2. Use the above answer or a calculator or computer program to determine how much you would have


to rotate the following conic to eliminate the xy term and orient it either horizontally or vertically:
x2 + y2 + 6x − 3y + 10xy + 100 = 0.


3. Use the above answer or a calculator or computer program to determine how much you would have
to rotate the following conic to eliminate the xy term and orient it either horizontally or vertically:
5x2 − 3x + 4xy + 21 = 0.


4. Rotate the following conic section so that it is oriented either horizontally or vertically, and then
analyze the result: 3x2 + 3y2 +


(


4

2 + 2


)


xy + 4

2x + 4



2y − 4 = 0


Review Answers
1. θ = pi2
2. θ = pi2
3. θ ≈ 31.72◦
4. To figure out the angle of rotation, since A = B we have θ = pi4 (or 45◦). After shifting the equation


by this amount, we have a relation, 4x2 +8x− y2 +8y− 4 = 0. Completing the square, this results in
a shifted hyperbola: (x+1)21 −


(y−2)2
4 = 1.


General Algebraic Forms
How can we look at a degree-2 polynomial equation and determine which conic section it depicts?


Ax2 + By2 +Cxy + Dx + Ey + F = 0


When C = 0 we have already discussed how to determine which conic section the equation refers to. In
summary, if A and B are both positive, the conic section is an ellipse. This is also true of A and B are
both negative, as the entire equation can be multiplied by -1 without changing the solution set. If A and
B differ in sign, the equation is a hyperbola, and if A or B equals zero the equation is a parabola.
There are a few new, more general, rules I will show you that give more information about the case when
C , 0 and hence the conic section needs to be rotated to achieve horizontal or vertical orientation.
If C2 < 4AB, the equation is an ellipse (note when C = 0 this holds whenever A and B are the same sign,
which is consistent with our simpler rule stated above.)
If C2 > 4AB, the equation is an hyperbola (note when C = 0 this holds whenever A and B are the opposite
sign, which is consistent with our simpler rule stated above.)
If C2 = 4AB, the equation is a parabola (note when C = 0, either A or B equals zero, which is consistent
with our simpler rule stated above.)


Review Questions
5. State what type of conic section is represented by the following equation: 5x2+6y2+2x−5y+ xy = 0.
6. State what type of conic section is represented by the following equation: x2 + 3y − 20xy + 20 = 0.
7. The rules above do not account for “degenerate” conic sections, that is the conic section that looks


like an X made by the intersection of a plane containing the line at the center of the cone. Explain
the conditions on the coefficients that lead to the degenerate conic sections.


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Review Answers
5. Ellipse
6. Hyperbola
7. C = 0, and either A = 0 or B = 0 (or both).


Image Sources
(1) http://upload.wikimedia.org/wikipedia/commons/d/d8/Focke_Garten_0004.jpg. Creative


Commons Attribution 2.5.


(2) http://commons.wikimedia.org/wiki/File:Leonardo_parabolic_compass.JPG. Public Domain.


(3) http://commons.wikimedia.org/wiki/File:Parabolic_reflection_1.svg. GNU-FDL.


(4) http://commons.wikimedia.org/wiki/File:Cassegrain_telescope.png. GNU-FDL.


(5) http://commons.wikimedia.org/wiki/File:DSN_Antenna_details.jpg. Public Domain.


(6) http://commons.wikimedia.org/wiki/File:Headlight_lens_optics_schematic.png.
GNU-FDL.


(7) http://upload.wikimedia.org/wikipedia/commons/8/87/Kepler-first-law.png. CC by SA
3.0.


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