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School District of Palm Beach County Summer Packet Post-Geometry Answers

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Page 1: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

School District of 

Palm Beach County 

 

Summer Packet 

Post-Geometry

Answers

 

Page 2: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

Prentice Hall Geometry • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

9

Name Class Date

A net is a two-dimensional fl at diagram that represents a three-dimensional fi gure.

It shows all of the shapes that make up the faces of a solid.

Stepping through the process of building a three-dimensional fi gure from a net

will help you improve your ability to visualize the process. Here are the steps you

would take to build a square pyramid.

Step 1 Step 2 Step 3

Start with the net. Fold up on the Tape the adjacent

dotted lines. triangle sides

together.

Here are other examples of nets that also fold up into a square pyramid.

Problem

How can you be sure that none of the nets shown above are the same?

Make sure you cannot rotate or fl ip any net and place it on top of any other net.

Exercise

1. What is a possible net for the fi gure shown at the right?

A. B. C.

1-1 ReteachingNets and Drawings for Visualizing Geometry

A

Page 3: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

Prentice Hall Geometry • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

10

Name Class Date

1-1 Reteaching (continued)

Nets and Drawings for Visualizing Geometry

An isometric drawing is a corner-view drawing of a three-dimensional

fi gure. It shows the top, front, and side views.

To make an isometric drawing, you can start by visualizing picking up a

block structure and turning it so that you are looking directly at one face.

Draw the edges of that face. Th en visualize and draw the two other faces.

Follow the steps below to make an isometric drawing of the block structure

at the right.

Step 1 Step 2 Step 3

Draw the edges that Start at a vertex of the front Draw the back and top

surround the front face. face and draw an edge for the edges to connect the

right side view. Edges only fi gure. Draw all

occur at bends and folds. parallel edges.

Problem

What is an isometric drawing of the block structure at the right?

Isometric drawing:

Exercises 2. Draw a possible net for this fi gure. 3. Make the isometric drawing for

this structure.

FrontRig

ht

FrontRig

ht

Front Right

Page 4: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

Prentice Hall Geometry • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

19

Name Class Date

1-2 Reteaching Points, Lines, and Planes

Review these important geometric terms.

Remember:

1. When you name a ray, an arrowhead is not drawn over the beginning point.

2. When you name a plane with three points, choose no more than two collinear

points.

3. An arrow indicates the direction of a path that extends without end.

4. A plane is represented by a parallelogram. However, the plane actually has no

edges. It is fl at and extends forever in all directions.

Exercises

Identify each fi gure as a point, segment, ray, line, or plane, and name each.

1. 2. 3.

4. 5. 6.

Term Examples of Labels Diagram

Point Italicized capital letter: D

Line

Line Segment

Ray

Plane

Three capital letters: ABF, AFB, BAF, BFA,

FAB, or FBA

One italicized capital letter: W

D

FA

BW

B

A

BmA

AB

Two capital letters (called endpoints)

with a segment drawn over them: AB or BA

Two capital letters with a line drawn

over them: AB or BA

One italicized lowercase letter: m

Two capital letters with a ray symbol

drawn over them: AB

LL N

M CD

P

O

E F G T S

point; point L line; *

CD)

ray; ST)segment; PO

plane; Answers may vary. Sample: plane LMN

line; answers may vary. Sample: *

EG)

Page 5: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

Prentice Hall Geometry • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

20

Name Class Date

1-2 Reteaching (continued)

Points, Lines, and Planes

A postulate is a statement that is accepted as true.

Postulate 1–4 states that through any three

noncollinear points, there is only one plane.

Noncollinear points are points that do not all lie

on the same line.

In the fi gure at the right, points D, E, and F are

noncollinear. Th ese points all lie in one plane.

Th ree noncollinear points lie in only one plane. Th ree

points that are collinear can be contained by more

than one plane. In the fi gure at the right, points P, Q, and R are collinear, and lie in both plane O and

plane N.

Exercises

Identify the plane containing the given points as front, back, left side, right side, top, or bottom.

F G

I H

Z

W

Y

X

Top

Right

Front

7. F, G, and X 8. F, G, and H

9. H, I, and Z 10. F, W, and X

11. I, W, and Z 12. Z, X, and Y

13. H, G, and X 14. W, Y, and Z

Use the fi gure at the right to determine how many planes contain the given group of points. Note that

*

GF)

pierces the plane at R, *

GF)

is not coplanar with X, and

*

GF)

does not intersect *

CE)

.

15. C, D, and E 16. D, E, and F

17. C, G, E, and F 18. C and F

D F E

QR

P

N

O

C D

F

E

G

R X

back

back

left side

right side

infi nite number of planes

infi nite number of planes

bottom

bottom

1 plane

0 planes

front

top

Page 6: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

Prentice Hall Geometry • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

29

Name Class Date

Th e Segment Addition Postulate allows you to use known segment lengths to fi nd unknown segment lengths. If three points, A, B, and C, are on the same line, and point B is between points A and C, then the distance AC is the sum of the distances AB and BC.

A B C

AC 5 AB 1 BC

Problem

If QS 5 7 and QR 5 3, what is RS?

Q R S

QS 5 QR 1 RS Segment Addition Postulate

QS 2 QR 5 RS Subtract QR from each side.

7 2 3 5 RS Substitute.

4 5 RS Simplify.

Exercises

For Exercises 1–5, use the fi gure at the right.

1. If PN 5 29 cm and MN 5 13 cm, then PM 5 z z. 2. If PN 5 34 cm and MN 5 19 cm, then PM 5 z z. 3. If PM 5 19 and MN 5 23, then PN 5 z z. 4. If MN 5 82 and PN 5 105, then PM 5 z z. 5. If PM 5 100 and MN 5 100, then PN 5 z z.For Exercises 6–8, use the fi gure at the right.

6. If UW 5 13 cm and UX 5 46 cm, then WX 5 z z. 7. UW 5 2 and UX 5 y. Write an expression for WX.

8. UW 5 m and WX 5 y 1 14. Write an expression for UX.

On a number line, the coordinates of A, B, C, and D are 26, 22, 3, and 7, respectively. Find the lengths of the two segments. Th en tell whether they are congruent.

9. AB and CD 10. AC and BD 11. BC and AD

P M N

U W X

1-3 Reteaching Measuring Segments

16 cm

42

23

200

15 cm

33 cm

4; 4; yes 9; 9; yes 5; 13; no

m 1 y 1 14

y 2 2

Page 7: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

Prentice Hall Geometry • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

30

Name Class Date

1-3 Reteaching (continued)

Measuring Segments

Th e midpoint of a line segment divides the segment into two segments that are equal in length. If you know the distance between the midpoint and an endpoint of a segment, you can fi nd the length of the segment. If you know the length of a segment, you can fi nd the distance between its endpoint and midpoint.

W X Y

X is the midpoint of WY . XW 5 XY , so XW and XY are congruent.

Problem

C is the midpoint of BE. If BC 5 t 1 1, and CE 5 15 2 t , what is BE?

B C E

BC 5 CE Defi nition of midpoint

t 1 1 5 15 2 t Substitute.

t 1 t 1 1 5 15 2 t 1 t Add t to each side.

2t 1 1 5 15 Simplify.

2t 1 1 2 1 5 15 2 1 Subtract 1 from each side.

2t 5 14 Simplify.

t 5 7 Divide each side by 2.

BC 5 t 1 1 Given.

BC 5 7 1 1 Substitute.

BC 5 8 Simplify.

BE 5 2(BC) Defi nition of midpoint.

BE 5 2(8) Substitute.

BE 5 16 Simplify.

Exercises

12. W is the midpoint of UV . If UW 5 x 1 23, and WV 5 2x 1 8, what is x?

13. W is the midpoint of UV . If UW 5 x 1 23, and WV 5 2x 1 8, what is WU?

14. W is the midpoint of UV . If UW 5 x 1 23, and WV 5 2x 1 8, what is UV?

15. Z is the midpoint of YA. If YZ 5 x 1 12, and ZA 5 6x 2 13, what is YA?

15

38

76

34

Page 8: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

Prentice Hall Geometry • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

39

Name Class Date

1-4 Reteaching Measuring Angles

Th e vertex of an angle is the common endpoint of the rays that form the angle. An angle may be named by its vertex. It may also be named by a number or by a point on each ray and the vertex (in the middle).

Th is is /Z, /XZY , /YZX , or /1.

It is not /ZYX , /XYZ, /YXZ, or /ZXY .

Angles are measured in degrees, and the measure of an angle is used to classify it.

Exercises

Use the fi gure at the right for Exercises 1 and 2.

1. What are three other names for /S?

2. What type of angle is /S?

3. Name the vertex of each angle. a. /LGH b. /MBX

Classify the following angles as acute, right, obtuse, or straight.

4. m/LGH 5 14 5. m/SRT 5 114

6. m/SLI 5 90 7. m/1 5 139

8. m/L 5 179 9. m/P 5 73

Use the diagram below for Exercises 10–18. Find the measure of each angle.

10. /ADB 11. /FDE

12. /BDC 13. /CDE

14. /ADC 15. /FDC

16. /BDE 17. /ADE

18. /BDF

Y

X

1Z

The measure ofan acute angleis between 0 and 90.

The measure of aright angle is 90.

The measure of anobtuse angle is between90 and 180.

The measure of astraight angle is 180.

3

R

S T

A

B

C

D

E

F

9090

80

100

10080

11070

12060 13050 14040 15030 16020

17010

1800

70

11060

12050

130

40140

30150

2016

0

10 170

0 180

1800

right

right

point G

25

55

80

120

155

35

65

100

145

acute

acute

obtuse

obtuse

obtuse

point B

lRST , lTSR, l3

Page 9: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

Prentice Hall Geometry • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

40

Name Class Date

1-4 Reteaching (continued)

Measuring Angles

Th e Angle Addition Postulate allows you to use a known angle measure to fi nd an unknown angle measure. If point B is in the interior of /AXC, the sum of m/AXB and m/BXC is equal to m/AXC.

m/AXB 1 m/BXC 5 m/AXC

Problem

If m/LYN 5 125, what are m/LYM and m/MYN ?

Step 1 Solve for p.

m/LYN 5 m/LYM 1 m/MYN Angle Addition Postulate

125 5 (4p 1 7) 1 (2p 2 2) Substitute.

125 5 6p 1 5 Simplify.

120 5 6p Subtract 5 from each side.

20 5 p Divide each side by 6.

Step 2 Use the value of p to fi nd the measures of the angles.

m/LYM 5 4p 1 7 Given

m/LYM 5 4(20) 1 7 Substitute.

m/LYM 5 87 Simplify.

m/MYN 5 2p 2 2 Given

m/MYN 5 2(20) 2 2 Substitute.

m/MYN 5 38 Simplify.

Exercises

19. X is in the interior of /LIN . m/LIN 5 100, m/LIX 5 14t , and m/XIN 5 t 1 10.

a. What is the value of t? b. What are m/LIX and m/XIN ?

20. Z is in the interior of /GHI . m/GHI 5 170, m/GHZ 5 3s 2 5, and m/ZHI 5 2s 1 25.

a. What is the value of s? b. What are m/GHZ and m/ZHI ?

A

B C

X

L Y

M

(4p 7)

(2p 2)N

6 84, 16

30 85, 85

Page 10: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

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49

Name Class Date

1-5 ReteachingExploring Angle Pairs

Adjacent Angles and Vertical AnglesAdjacent means “next to.” Angles are adjacent if they lie next to each other. In other

words, the angles have the same vertex and they share a side without overlapping.

Adjacent Angles Overlapping Angles

Vertical means “related to the vertex.” So, angles are vertical if they share a vertex,

but not just any vertex. Th ey share a vertex formed by the intersection of two

straight lines. Vertical angles are always congruent.

Vertical Angles Non-vertical Angles

Exercises

1. Use the diagram at the right.

a. Name an angle that is adjacent to /ABE.

b. Name an angle that overlaps /ABE.

2. Use the diagram at the right.

a. Mark /DOE and its vertical angle as congruent angles.

b. Mark /AOE and its vertical angle as congruent angles.

21 21

2121

A

D E

F

B C

B

C

D

O

A

E

lEBF or lEBC

lAOB

lDOB

Answers may vary. Sample: lDBF or lDBC

Page 11: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

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50

Name Class Date

1-5 Reteaching (continued)

Exploring Angle Pairs

Supplementary Angles and Complementary AnglesTwo angles that form a line are supplementary angles. Another term for these

angles is a linear pair. However, any two angles with measures that sum to 180 are

also considered supplementary angles. In both fi gures below, m/1 5 120 and

m/2 5 60, so /1 and /2 are supplementary.

Two angles that form a right angle are complementary angles. However, any two

angles with measures that sum to 90 are also considered complementary angles.

In both fi gures below, m/1 5 60 and m/2 5 30, so /1 and /2 are

complementary.

Exercises

3. Copy the diagram at the right.

a. Label /ABD as /1.

b. Label an angle that is supplementary

to /ABD as /2.

c. Label as /3 an angle that is adjacent

and complementary to /ABD.

d. Label as /4 a second angle that is

complementary to /ABD.

e. Name an angle that is supplementary to /ABE.

f. Name an angle that is complementary to /EBF .

1 2

1

2

1 2 1 2

A B

DE

F

C60

60

Label lABD as l1.

Label lDBC as l2.

lCBE

lCBF

Label lDBE as l3.

Label lFBC as l4.

Page 12: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

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59

Name Class Date

1-6 Reteaching Basic Constructions

Before you start the construction, THINK about the goal. Th en SKETCH the segment or angle you are trying to construct. Next EXPLAIN the purpose of each step in the construction as you complete it.

Problem

Construct AB so that AB is congruent to XY .

X

Y

THINK: Can you describe what the problem is asking for in your own words? You want to draw a line segment that has the same length as one you are given.

SKETCH: Sketch a segment and label it as AB. Why do you start with a sketch? A sketch helps you to see what you need to construct.

EXPLAIN: First, draw a ray. What is the purpose of drawing a ray? Th e ray is a part of a line on which you can mark off the correct

length of AB.

Second, measure the length of XY , using the compass. What is the purpose of measuring the length of XY ? You need this measure to mark the same length on the ray.

Finally, mark the length of XY on the ray, using the compass. Why do you mark the length of XY on your ray?

Th is completes your construction of AB, which is congruent to XY .

Exercises

Analyze the construction of a congruent angle and bisectors.

1. Analyze the construction of a perpendicular bisector. First draw XY . a. THINK about what it means to construct a perpendicular bisector. What is

your goal?

b. SKETCH a perpendicular bisector to XY .

c. EXPLAIN the fi rst two steps. What is the purpose of drawing an arc from each endpoint?

d. EXPLAIN the last step. What is the purpose of drawing the segment between the intersections of the arcs?

Answers may vary. Sample: to draw a segment that is perpendicular to XY and that divides it into two equal pieces

Check students’ work.

Answers may vary. Sample: to form two intersections to become two points of the perpendicular bisector

Answers may vary. Sample: to draw the perpendicular bisector

Page 13: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

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60

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1-6 Reteaching (continued)

Basic Constructions

2. Analyze the construction of a congruent angle. First draw/X . a. THINK about what it means to construct a congruent angle. What is

your goal?

b. SKETCH /Y congruent to /X . Your goal is to construct an angle the same size as /Y .

c. EXPLAIN the fi rst step (drawing a ray). What is the purpose of the fi rst step?

d. EXPLAIN the second step (drawing an arc). What is the purpose of the second step?

e. EXPLAIN the third step (drawing an arc). What is the purpose of the third step?

f. EXPLAIN the fourth step (drawing an arc). What is the purpose of the fourth step?

g. EXPLAIN the fi fth step (drawing a segment). What is the purpose of the fi fth step?

3. Analyze the construction of an angle bisector. First draw /W . a. THINK about what it means to construct an angle bisector. What is

your goal?

b. SKETCH a ray that makes /V congruent to 12 /W , where /V shares a ray

with /W and has its other ray inside /W . Your goal is to construct a ray that bisects an angle.

c. EXPLAIN the fi rst step (drawing an arc). What is the purpose of the fi rst step?

d. EXPLAIN the second step (drawing two arcs). What is the purpose of the second step?

e. EXPLAIN the third step (drawing a ray). What is the purpose of the third step?

Answers may vary. Sample: to draw an angle of the same size as the given angle

Answers may vary. Sample: to divide a given angle into two equal angles

Answers may vary. Sample: to divide the given angle into two equal smaller angles

Check students’ work.

Answers may vary. Sample: to measure the opening of the given angle

Answers may vary. Sample: to mark the size of the given angle on the ray

Answers may vary. Sample: to draw the congruent angle

Answers may vary. Sample: to make a ray to serve as one side of the new congruent angle

Answers may vary. Sample: to locate two points of intersection to use to draw the other endpoint of the angle bisector

Answers may vary. Sample: to locate the point of intersection that will be another point on the angle bisector

Answers may vary. Sample: to fi nd the point of intersection so you can draw the other side of the angle

Check students’ work.

Page 14: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

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1-7 Reteaching Midpoint and Distance in the Coordinate Plane

Average the x-coordinates of the endpoints to fi nd the x-coordinate of the

midpoint. Average the y-coordinates of the endpoints to fi nd the y-coordinate

of the midpoint.

Problem

What is the midpoint of AB if the endpoints are A(1, 7) and B(5, 9)?

Find the average of the x-coordinates.

1 1 5

2 5 3

Repeat to fi nd the y-coordinate of the midpoint.

7 1 9

2 5 8

So, the midpoint of AB is (3, 8).

Remember the Midpoint Formula: ax1 1 x2

2 , y1 1 y2

2 b .

Th e formula gives a point whose coordinates are the average of the x-coordinates

and the y-coordinates.

So, the midpoint is halfway between the two points, and has coordinates that are

the average of the coordinates of the two points.

To fi nd an unknown endpoint, subtract the coordinates of the known endpoint

from the coordinates of the midpoint. Add that number to the coordinates of

the midpoint.

Problem

Th e midpoint of XY is M(7, 6). One endpoint is X(3, 5). What are the coordinates

of the other endpoint Y?

2

4

6

64 82

8

x

y

A(1, 7)M(3, 8)

B(5, 9)

O

7 3 4 7 4 11

X(3, 5) M(7 ,6) Y(11, 7)

6 5 1 6 1 7

Add the difference to the midpoint.Subtract.

Page 15: School District of Palm Beach County Post-Ge… · School District of Palm Beach County ... Nets and Drawings for Visualizing Geometry An isometric drawing is a corner-view drawing

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1-7 Reteaching (continued)

Midpoint and Distance in the Coordinate Plane

Exercises

Find the coordinates of the midpoint of AB by fi nding the averages of the coordinates.

1. A(4, 3) Mau,ub B(8, 8)

2. A(7, 2) Mau,ub B(1, 5)

3. A(25, 6) Mau,ub B(1, 23)

4. A(27, 21) Mau,ub B(25, 29)

M is the midpoint of XY . Find the coordinates of Y.

5. X(3, 4) and M(6, 10) 6. X(25, 1) and M(3, 25)

To help fi nd the distance between two points, make a sketch on graph paper.

Problem

What is the distance between A(2, 6) and B(6, 9)?

Step 1: Sketch the points on graph paper.

Step 2: Draw a right triangle along the gridlines.

Step 3: Find the length of each leg.

Step 4: Find the distance between the points.

Exercises

Find the distance between points A and B. If necessary, round to the nearest tenth.

7. A(1, 4) and B(6, 16) 8. A(23, 2) and B(1, 6)

9. A(21, 28) and B(1, 23) 10. A(25, 25) and B(7, 11)

Find the midpoint between each pair of points. Th en, fi nd the distance between each pair of points. If necessary, round to the nearest tenth.

11. C(3, 8) and D(0, 3) 12. H(22, 4) and I(4, 22)

13. K(1, 25) and L(23, 29) 14. M(7, 0) and N(23, 4)

15. O(25, 21) and P(22, 3) 16. R(0, 26) and S(28, 0)

B(6, 9)

A(2, 6) (6, 6)

Length of side is9 6 3.

Length of side is6 2 4.

Distance is32 42 5.

(9, 16) (11, 211)

13

20

5.7

5.4

(1.5, 5.5); 5.8

(21, 27); 5.7

(23.5, 1); 5 (24, 23); 10

(1, 1); 8.5

(2, 2); 10.8

5.56

4 3.5

1.522

2526

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1-8 Reteaching Perimeter, Circumference, and Area

Th e perimeter of a rectangle is the sum of the lengths of its sides. So, the perimeter is the distance around its outside. Th e formula for the perimeter of a rectangle is P 5 2(b 1 h).

Th e area of a rectangle is the number of square units contained within the rectangle. Th e formula for the area of a rectangle is A 5 bh.

Exercises

1. Fill in the missing information for each rectangle in the table below.

2. How does the perimeter vary as you move down the table? How does the area vary as you move down the table?

3. What pattern in the dimensions of the rectangles explains your answer to Exercise 2?

4. Fill in the missing information for each rectangle in the table below.

5. How does the perimeter vary as you move down the table? How does the area vary as you move down the table?

6. What pattern in the dimensions of the rectangles explains your answer to Exercise 5?

Dimensions Area, A bhPerimeter, P 2(b h)

1 ft 9 ft

2 ft 8 ft

3 ft 7 ft

4 ft 6 ft

2(1 ft 9 ft) 20 ft 1 ft 9 ft 9 ft2

Dimensions Area, A bhPerimeter, P 2(b h)

1 ft 24 ft

2 ft 12 ft

3 ft 8 ft

4 ft 6 ft

The perimeter stays the same. The area increases.

The perimeter decreases. The area stays the same.

Answers may vary. Sample: The product bh is always the same, while the sum b 1 h decreases.

Answers may vary. Sample: The sum b 1 h is always the same, while the product bh increases.

2(2 ft 1 8 ft) 5 20 ft

2(1 ft 1 24 ft) 5 50 ft

2(2 ft 1 12 ft) 5 28 ft

2(3 ft 1 8 ft) 5 22 ft

2(4 ft 1 6 ft) 5 20 ft

2(3 ft 1 7 ft) 5 20 ft

2(4 ft 1 6 ft) 5 20 ft

2 ft 3 8 ft 5 16 ft2

1 ft 3 24 ft 5 24 ft2

2 ft 3 12 ft 5 24 ft2

3 ft 3 8 ft 5 24 ft2

4 ft 3 6 ft 5 24 ft2

3 ft 3 7 ft 5 21 ft2

4 ft 3 6 ft 5 24 ft2

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1-8 Reteaching (continued)

Perimeter, Circumference, and Area

A square is a rectangle that has four sides of the same length. Because the perimeter is s 1 s 1 s 1 s, the formula for the perimeter of a square is P 5 4s.

Th e formula for the area of a square is A 5 s2.

Th e circumference of a circle is the distance around the circle. Th e formula for the circumference of a circle is C 5 pd or C 5 2pr. Th e area of a circle is the number of square units contained within the circle. Th e formula for the area of a circle is

A 5 pr 2.

Exercises

7. Fill in the missing information for each square in the table below.

8. Fill in the missing information for each circle in the table below.

9. A rectangle has a length of 5 cm and an area of 20 cm2. What is its width?

10. What is the perimeter of a square whose area is 81 ft2?

11. Can you fi nd the perimeter of a rectangle if you only know its area? What about a square? Explain.

12. Can you fi nd the area of a circle if you only know its circumference? Explain.

Side Area, A s2Perimeter, P 4s

3 cm

4 cm

4 3 cm 12 cm

4 5 cm 20 cm

(3 cm)2 9 cm2

(10 cm)2 100 cm2

Radius Area, A r2Diameter, D 2r Circumference, C 2 r

2 in.

3 in.

2 2 4 in.

2 5 10 in.

2 2 4 in.

2 8 16 in.

2 2 4 in.2

10 10 100 in.2

4 cm

36 ft

Answers may vary. Sample: It is impossible to determine the area of a rectangle if only the perimeter is known, because rectangles with the same area may have different perimeters. However, because all squares with the same area have the same perimeter, it is possible to determine the area if only the perimeter is known.

Answers may vary. Sample: Yes; all circles with the same circumference have the same area.

4 3 4 cm 5 16 cm

4 3 10 cm 5 40 cm

(4 cm)2 5 16 cm2

(5 cm)2 5 25 cm2

2 3 3 5 6 in. 2π 3 3 5 6π in.

2π 3 5 5 10π in.

2π 3 10 5 20π in.

π 3 3 3 3 5 9π in.2

π 3 5 3 5 5 25π in.2

π 3 8 3 8 5 64π in.22 3 8 5 16 in.

2 3 10 5 20 in.

5 cm

5 in.

8 in.

10 in.

10 cm

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2-1 ReteachingPatterns and Inductive Reasoning

Inductive reasoning is a type of reasoning in which you look at a pattern and then make some type of prediction based on the pattern. Th ese predictions are also called conjectures. A conjecture is a statement about what you think will happen based on the pattern you observed.

Problem

Which conjectures below are reasonable? Which are not?

Pattern Conjecture

Pattern 1: Every day for the two weeks that Alba

visited Cairo, the weather was hot and dry.

Alba thought to herself, “The weather

is always hot and dry in this city.”

Pattern 2: 5, 10, 15, 20, 25, 30, 35, 40,

45, 50, . . .

Each number increases by 5. The next

two numbers are most likely 55 and 60.

Pattern 3: Dani and Liz both examined this

pattern of letters: A, BB, CCC, . . .

Dani was sure that the next two terms

of the pattern would be DDDD and

EEEEE. Liz wasn’t so sure. She thought

the pattern might repeat and the next

two elements would be A and BB.

Pattern 1: Th e conjecture for Pattern 1 is probably not correct. In most cities, the weather will not be hot and dry all the time.

Pattern 2: You are given enough numbers in the pattern to assume that the numbers continue to increase by fi ve. Th e conjecture is probably correct.

Pattern 3: Only three terms of the pattern are shown. Th is makes it diffi cult to determine what rule the pattern follows. Either Dani or Liz could be correct, or they could both be incorrect.

Remember that conjectures are never the fi nal goal in a complete reasoning process. Th ey are simply the fi rst step to fi guring out a problem.

ExercisesMake a conjecture about the rule these patterns follow.

1. 3, 6, 9, 12, 15, c 2. 9, 3, 1, 13, 19, c

3. A, C, E, G, I, K, M, c 4. 0, 5, 22, 3, 24, 1, 26, c

5. 4, 8, 16, 32, 64, 128, c 6. 0.1, 0.01, 0.001, 0.0001, c

The numbers increase by 3. The numbers are divided by 3.

Add 5, then subtract 7.

The numbers are multiplied by 2. The numbers are divided by 10.

One letter of the alphabet is skipped between each term.

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2-1 Reteaching (continued)

Patterns and Inductive Reasoning

Draw the next fi gure in each sequence.

7.

8.

One way to show that a conjecture is not true is to fi nd a counterexample. A counterexample is an instance in which the conjectured pattern does not work. Only one counterexample is needed to prove a conjecture false. For example, one rainy or cool day in Cairo would prove to Alba that it is not always hot and dry there.

Exercises

Find one counterexample to show that each conjecture is false.

9. All vehicles on the highway have exactly four wheels.

10. All states in the United States share a border with another state.

11. All plurals end with the letter s.

12. Th e diff erence between two integers is always positive.

13. All pentagons have exactly fi ve congruent sides.

14. All numbers that are divisible by 3 are also divisible by 6.

15. All whole numbers are greater than their opposites.

16. All prime numbers are odd integers.

Sample: Motorcycles have two wheels.

Sample: Hawaii

Sample: 23 2 2 5 25

Sample: geese

Sample: an irregular pentagon

Sample: 15

0

2

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2-2 ReteachingConditional Statements

A conditional is a statement with an “if” clause and a “then” clause. Here are some

examples of conditional statements:

If the fl ag is extended all the way outward, then it is very windy.

If a cup is in pieces, then it is broken.

A conditional is divided into two parts—the hypothesis and the conclusion.

If Alex decides to eat dessert, then Alex will eat apple pie.

the conditionsnecessary to reachthe conclusion

Hypothesis: the result if theconditions from thehypothesis are present

Conclusion:

In the above conditional, the hypothesis is: Alex decides to eat dessert. Th e

conclusion is: Alex will eat apple pie.

Conditionals can be changed to form related statements. Besides the

original conditional, there are also the converse, inverse, and contrapositive

of a conditional.

Problem

Write the converse, inverse, and contrapositive of the following statement:

If the weather is rainy, then the sidewalks will be wet.

To write the converse of a conditional, write the conclusion as the hypothesis and

the hypothesis as the conclusion.

Conditional: If the weather is rainy, then the sidewalks are wet.

Converse: If the sidewalks are wet, then the weather is rainy.

To write the inverse of a conditional, negate the original hypothesis

and conclusion.

Conditional: If the weather is rainy, then the sidewalks are wet.

Inverse: If the weather is not rainy, then the sidewalks are not wet.

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2-2 Reteaching (continued)

Conditional Statements

To write the contrapositive of a conditional, negate the original hypothesis and

conclusion and also switch the hypothesis and conclusion.

Conditional: If the weather is rainy, then the sidewalks are wet.

Contrapositive: If the sidewalks are not wet, then the weather is not rainy.

Th e conditional and its contrapositive are always both true or both false, so they

have the same truth value. Th e converse and the inverse also have the same truth

value. However, the conditional and contrapositive may have a diff erent truth

value than the converse and the inverse.

ExercisesIdentify the hypothesis and conclusion of each conditional.

1. If a number is a multiple of 2, then the number is even.

2. If something is thrown up into the air, then it must come back down.

3. Two angles are supplementary if the angles form a linear pair.

4. If the shoe fi ts, then you should wear it.

State whether each conditional is true or false. Write the converse for the conditional and state whether the converse is true or false.

5. If the recipe uses 3 teaspoons of sugar, then it uses 1 tablespoon of sugar.

6. If the milk has passed its expiration date, then the milk should not be consumed.

State whether each conditional is true or false. Write the inverse for the conditional and state whether the inverse is true or false.

7. If the animal is a fi sh, then it lives in water.

8. If your car tires are not properly infl ated, then you will get lower gas mileage.

State whether each conditional is true or false. Write the contrapositive for the conditional and state whether the contrapositive is true or false.

9. If you ride on a roller coaster, then you will experience sudden drops.

10. If you only have $15, then you can buy a meal that costs $15.65.

Hypothesis: A number is a multiple of 2. Conclusion: The number is even.

Hypothesis: Something is thrown up into the air. Conclusion: It must come back down.

Hypothesis: Two angles form a linear pair. Conclusion: The angles are supplementary.

Hypothesis: The shoe fi ts. Conclusion: You should wear it.

True; if the recipe uses 1 tablespoon of sugar, then it uses 3 teaspoons of sugar; true.True;

if the milk should not be consumed, then the milk has passed its expiration date; false.

True; if the animal is not a fi sh, then the animal does not live in water; false.

True; if your car tires are infl ated properly, then you will not get lower gas mileage; true.

True; if you do not experience sudden drops, then you are not riding a roller coaster; true.

False; if you cannot buy a meal that costs $15.65, then you do not only have $15; false.

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2-3 ReteachingBiconditionals and Defi nitions

If a conditional statement and its converse are both true, we say the original

conditional is reversible. It works both ways because both pu q and qu p are true.

If a conditional is reversible, you can write it as a biconditional. A biconditional

uses the words “if and only if.” A biconditional can be written as p 4 q.

Conditional: If a triangle has three congruent sides, then the

triangle is equilateral.

Converse: If a triangle is equilateral, then the triangle has three

congruent sides.

Th e conditional is true. Th e converse is also true. Since the conditional and its

converse are both true, the original statement is “reversible” and the biconditional

will be true.

Biconditional: A triangle has three congruent sides if and only if it is an

equilateral triangle.

Exercises

Test each statement below to see if it is reversible. If it is reversible, write it as a true biconditional. If not, write not reversible.

1. If a whole number is a multiple of 2, then the whole number is even.

2. Rabbits are animals that eat carrots.

3. Two lines that intersect to form four 908 angles are perpendicular.

4. Mammals are warm-blooded animals.

Write the two conditionals that form each biconditional.

5. A parallelogram is a rectangle if and only if the diagonals are congruent.

6. An animal is a giraff e if and only if the animal’s scientifi c name is Giraff a camelopardalis.

Reversible; a whole number is a multiple of 2 if and only if the whole number is even.

not reversible

Reversible; two lines intersect to form four 908 angles if and only if the two lines are perpendicular.

not reversible

If a parallelogram is a rectangle, then the diagonals are congruent. If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle.

If an animal is a giraffe, then the animal’s scientifi c name is Giraffa camelopardalis. If the animal’s scientifi c name is Giraffa camelopardalis, then the animal is a giraffe.

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2-3 Reteaching (continued)

Biconditionals and Defi nitions

A good defi nition:

• is reversible; the statement works both ways

• can be written as a true biconditional

• avoids using vague, imprecise, or diffi cult words

Problem

Is the following a good defi nition for a square?

Defi nition: A square is a rectangle with four congruent sides.

Is the defi nition reversible? Yes.

A rectangle with four congruent sides is a square.

Can the defi nition be rewritten as a biconditional? Yes.

A fi gure is a square if and only if it is a rectangle with four congruent sides.

Is the defi nition clear and understandable? Yes.

Th e defi nition is a good defi nition for a square.

Exercises

State whether each statement is a good defi nition. Explain your answer.

7. A parallelogram is a quadrilateral with two pairs of parallel sides.

8. A triangle is a three-sided fi gure whose angle measures sum to 1808.

9. A juice drink is a beverage that contains less than 100% juice.

10. In basketball, the top scorer in a game is the player who scores the most points

in the game.

11. A tree is a large, green, leafy plant.

The defi nition is good. The statement is reversible. It can be written as a true biconditional and is clear and understandable.

The defi nition is good. The statement is reversible. It can be written as a true biconditional and is clear and understandable.

The defi nition is not good. The statement is not reversible because a beverage that contains 0% juice, such as water, should not be called a juice drink.

The defi nition is good. The statement is reversible and can be written as a true biconditional and is clear and understandable.

The defi nition is not good because the term “large” is vague. There are large, green, leafy plants that are not trees.

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2-4 ReteachingDeductive Reasoning

Th e Law of Detachment states that if a conditional statement is true, then any

time the conditions for the hypothesis exist, the conclusion is true.

If the conditional statement is not true, or the conditions of the hypothesis do not

exist, then you cannot make a valid conclusion.

Problem

What can you conclude from the following series of statements?

If an animal has feathers and can fl y, then it is a bird.

A crow has feathers and can fl y.

Is the conditional statement true? Yes.

Do the conditions of the hypothesis exist? Yes.

Th erefore, you can conclude that a crow is a bird.

If an animal has feathers and can fl y, then it is a bird.

A bat can fl y.

Is the conditional statement true? Yes.

Do the conditions of the hypothesis exist? No; a bat does not have feathers.

Th erefore, no conclusions can be made with the given information.

Exercises

Use the Law of Detachment to make a valid conclusion based on each conditional. Assume the conditional statement is true.

1. If it is Monday, then Jim has tae kwon do practice.

Th e date is Monday, August 25.

2. If the animal is a whale, then the animal lives in the ocean.

Daphne sees a beluga whale.

3. If you live in the city of Miami, then you live in the state of Florida.

Jani lives in Florida.

4. If a triangle has an angle with a measure greater than 90,

then the triangle is obtuse.

In nGHI , m/HGI 5 110.

5. A parallelogram is a rectangle if its diagonals are congruent.

Lincoln draws a parallelogram on a piece of paper.

Conclusion: Jim has tae kwon do practice.

Conclusion: The beluga whale lives in the ocean.

Conclusion: No conclusion is possible.

Conclusion: kGHI is obtuse.

Conclusion: No conclusion is possible.

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2-4 Reteaching (continued)

Deductive Reasoning

You can use the Law of Syllogism to string together

two or more conditionals and draw a conclusion

based on the conditionals.

Notice that the conclusion of each conditional

becomes the hypothesis of the next conditional.

Problem

What can you conclude from the following two conditionals?

If a polygon is a hexagon, then the sum of its angle measures is 720.

If a polygon’s angle measures sum to 720, then the polygon has six sides.

Th e conclusion of the fi rst conditional matches the hypothesis of the second

conditional: the sum of the angle measures of a polygon is 720.

You can conclude that if a polygon is a hexagon, then it has six sides.

Exercises

If possible, use the Law of Syllogism to make a conclusion. If it is not possible to make a conclusion, tell why.

6. If you are climbing Pikes Peak, then you are in Colorado. If you are in

Colorado, then you are in the United States.

7. If the leaves are falling from the trees, then it is fall.

If it is September 30, then it is fall.

8. If it is spring, then the leaves are coming back on the trees.

If it is April 14, then it is spring.

9. If plogs plunder, then fl egs fret.

If fl egs fret, then gops groan.

10. If you make a 95 on the next test, you will pass the course.

If you pass the course, you will not have to take summer school.

11. Use the Law of Detachment and the Law of Syllogism to make a valid

conclusion from the following series of true statements. Explain your

reasoning and which statements you used in your reasoning.

I. If an animal is an insect, then it has a head, thorax, and abdomen.

II. If an animal is a spider, then it has a head and abdomen.

III. If an animal has a head, thorax, and abdomen, then it has six legs.

IV. Humberto saw an insect called a grasshopper.

Therefore: p s

p q

q r

r s

3

Conclusion: If you are climbing Pikes Peak, then you are in the United States.

No valid conclusion is possible. The two statements stand alone.

Conclusion: If it is April 14, then the leaves are coming back on the trees.Conclusion: If plogs plunder, then gops groan.

Conclusion: If you make a 95 on the next test, then you will not have to take summer school.

Sample: Conclusion: The grasshopper has a head, a thorax, an abdomen, and six legs. I used the Law of Detachment, the Law of Syllogism, and statements I, III, and IV.

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2-5 ReteachingReasoning in Algebra and Geometry

When you solve equations you use the Properties of Equality.

Exercises

Support each conclusion with a reason.

1. Given: 6x 1 2 5 12 2. Given: m/1 1 m/2 5 90

Conclusion: 6x 5 10 Conclusion: m/1 5 90 2 m/2

3. Given: x 5 m/C 4. Given: q 2 x 5 r

Conclusion: 2x 5 m/C 1 x Conclusion: 4(q 2 x) 5 4r

5. Given: m/Q 2 m/R 5 90, 6. Given: CD 5 AF 2 2CD

m/Q 5 4m/R Conclusion: 3CD 5 AF

Conclusion: 4m/R 2 m/R 5 90

7. Given: 5(y 2 x) 5 20 8. Given: m/AOX 5 2m/XOB

Conclusion: 5y 2 5x 5 20 2m/XOB 5 140

Conclusion: m/AOX 5 140

9. Order the steps below to complete the proof.

Given: m/P 1 m/Q 5 90, m/Q 5 5m/P

Prove: m/Q 5 75

a) 6m/P 5 90 by the Distributive Property

b) m/Q 5 5 ? 15 5 75 by the Substitution and Multiplication Properties

c) m/P 5 15 by the Division Property

d) m/P 1 5m/P 5 90 by the Substitution Property

Property

Multiplication Property

Division Property

Substitution Property

Words

You can multiply by the same number on each side of an equation.

You can divide each side of an equation by the same number.

You can exchange a part of an expression with an equivalent value.

Example

Addition Property You can add the same number to each side of an equation. If x 2, then x 2 4.

Subtraction Property You can subtract the same numberfrom each side of an equation. If y 8, then y 3 5.

If z 2, then 5z 10.

If 6m 12, then m 2.

If 3x 5 3 and x 2, then 3(2) 5 3.

Subtraction Property

Substitution Property

Distributive Prop.

Substitution Prop.

Addition Property

d, a, c, b

Subtraction Property

Addition Property Multiplication Property

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2-5 Reteaching (continued)

Reasoning in Algebra and Geometry

Several other important properties are also needed to write proofs.

Exercises

Match the property to the appropriate statement.

10. RT > RT a) Refl exive Property of Equality

11. If /YER > /IOP

and /IOP > /WXZ b) Refl exive Property of Congruence

then /YER > /WXZ

12. If PQ > MN c) Symmetric Property of Equality

then MN > PQ

13. If XT 5 YZ d) Symmetric Property of Congruence

and YZ 5 UP

then XT 5 UP e) Transitive Property of Equality

14. m/1 5 m/1

15. If m/RQS 5 m/TEF f ) Transitive Property of Congruence

then m/TEF 5 m/RQS

16. Writing Write six new mathematical statements that represent each of the

properties given above.

Example Words

Reflexive Property of EqualityAB AB Any value is equal to itself.

Reflexive Property of CongruenceZ Z Any geometric object is congruent to itself.

Symmetric Property of Equality If XY ZA, then ZA XY. You can change the order of an equality.

Symmetric Property of CongruenceIf Q R, then R Q.

You can change the order of a congruencestatement.

Transitive Property of Equality If KL MN and MN TR, then KL TR.

If two values are equal to a third value, then they are equal to each other.

Transitive Property of Congruence If 1 2 and 2 3, then 1 3.

If two values are congruent to a third value,then they are congruent to each other.

b

f

d

e

a

Check students’ work.

c

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2-6 ReteachingProving Angles Congruent

A theorem is a conjecture or statement that you prove true using deductive reasoning. You prove each step using any of the following: given information, defi nitions, properties, postulates, and previously proven theorems.

Th e proof is a chain of logic. Each step is justifi ed, and then the Laws of Detachment and Syllogism connect the steps to prove the theorem.

Vertical angles are angles on opposite sides of two intersecting lines. In the fi gure at the right, /1 and /3 are vertical angles. /2 and /4 are also vertical angles. Th e Vertical Angles Th eorem states that vertical angles are always congruent. Th e symbol > means is congruent to.

Problem

Given: m/BOF 5 m/FOD

Prove: 2m/BOF 5 m/AOE

Statements Reasons

1) m/BOF 5 m/FOD 1) Given

2) m/BOF 1 m/FOD 5 m/BOD 2) Angle Addition Postulate

3) m/BOF 1 m/BOF 5 m/BOD 3) Substitution Property

4) 2(m/BOF) 5 m/BOD 4) Combine like terms.

5) /AOE > /BOD 5) Vertical Angles are >.

6) m/AOE 5 m/BOD 6) Defi nition of Congruence

7) 2m/BOF 5 m/AOE 7) Substitution Property

ExercisesWrite a paragraph proof.

1. Given: /AOB and /XOZ are vertical angles.

m/AOB 5 80

m/XOZ 5 6x 1 5

Prove: x 5 12.5

12

34

B

F

DE

A

O

Because it is given that lAOB and lXOZ are vertical angles, and all vertical angles are O , it follows that mlAOB 5 mlXOZ. Because it is also given that mlAOB 5 80 and mlXOZ 5 6x 1 5, the use of substitution yields the equation 80 5 6x 1 5. Using the Subtraction Property of Equality, 75 5 6x , and using the Division Property of Equality, it follows that x 5 12.5.

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2-6 Reteaching (continued)

Proving Angles Congruent

Find the value of each variable.

2. 3. 4.

You can use numbers to help understand theorems that may seem confusing.

Congruent Supplements Th eorem: If two angles are supplements of the same angle (or of congruent angles), then the two angles are congruent.

1

2

3

If /2 and /3 are both supplementary to /1, then /2 > /3.

Th ink about it: Suppose m/1 5 50. Any angle supplementary to /1 must have a measure of 130. So, supplements of /1 must be congruent.

Congruent Complements Th eorem: If two angles are complements of the same angle (or of congruent angles), then the two angles are congruent.

If /4 and /5 are both complementary to /6, then /4 > /5.

Th ink about it: Suppose m/6 5 30. Any complement of /6 has a measure of 60. So, all complements of /6 must be congruent.

Exercises

Name a pair of congruent angles in each fi gure. Justify your answer.

5. Given: /2 is complementary to /3. 6. Given: /AYZ > /BYW

7. Reasoning Explain why the following statement is true. Use numbers in your explanation. “If /1 is supplementary to /2, /2 is supplementary to /3, /3 is supplementary to /4, and /4 is supplementary to /5, then /1 > /5.”

(6y 16) (4y 6)

(7x 10)

(3x 50)

(7g 5)(82 4g)

12

3A B

YW Z

15711

l1 O l3; Congruent Complements Theorem

lAYW O lBYZ ; Congruent Supplements Theorem

Suppose ml1 5 50; therefore ml2 5 130, ml3 5 50, ml4 5 130, and ml5 5 50. Because ml1 5 ml5, l1 O l5.

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3-1 Reteaching Lines and Angles

Not all lines and planes intersect.

• Planes that do not intersect are parallel planes.

• Lines that are in the same plane and do not intersect are parallel.

• Th e symbol 6 shows that lines or planes are parallel: *

AB)

6 *

CD)

means

“Line AB is parallel to line CD.”

• Lines that are not in the same plane and do not intersect are skew.

Parallel planes: plane ABDC 6 plane EFHG

plane BFHD 6 plane AEGC

plane CDHG 6 plane ABFE

Examples of parallel lines: *

CD)

6 *

AB)

6 *

EF)

6 *

GH)

Examples of skew lines: *

CD)

is skew to *

BF)

, *

AE)

, *

EG)

, and *

FH)

.

Exercises

In Exercises 1–3, use the fi gure at the right.

1. Shade one set of parallel planes.

2. Trace one set of parallel lines with a solid line.

3. Trace one set of skew lines with a dashed line.

In Exercises 4–7, use the diagram to name each of the following.

4. a line that is parallel to *

RS)

5. a line that is skew to *

QU)

6. a plane that is parallel to NRTP

7. three lines that are parallel to *

OQ)

In Exercises 8–11, describe the statement as true or false. If false, explain.

8. plane HIKJ 6 plane IEGK 9. *

DH)

6 *

GK)

10. *

HJ)

and *

HD)

are skew lines. 11. *

FG)

6 *

KI)

A B

C D

E F

G H

N O

PQ

R S

UTD

GF

E

HI

J K

Answers may vary. Sample:

Answers may vary. Sample: *

NO)

Answers may vary. Sample: *

RS)

plane OSUQ

*

NP)

, *

RT)

, and *

SU)

False; these planes intersect.

False; the lines intersect.False; the lines are in

different planes, and since they do not intersect they are skew.

true

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3-1 Reteaching (continued)

Lines and Angles

Th e diagram shows lines a and b intersected by line x.

Line x is a transversal. A transversal is a line that intersects

two or more lines found in the same plane.

Th e angles formed are either interior angles or

exterior angles.

Interior Angles Exterior Angles

between the lines cut by the transversal

/3, /4, /5, and /6 in diagram above

outside the lines cut by the transversal

/1, /2, /7, and /8 in diagram above

Four types of special angle pairs are also formed.

Exercises

Use the diagram at the right to answer Exercises 12–15.

12. Name all pairs of corresponding angles.

13. Name all pairs of alternate interior angles.

14. Name all pairs of same-side interior angles.

15. Name all pairs of alternate exterior angles.

Use the diagram at the right to answer Exercises 16 and 17. Decide whether the angles are alternate interior angles, same-side interior angles, corresponding, or alternate exterior angles.

16. /1 and /5 17. /4 and /6

1 23 4

a

b

x

5 6

7 8

Angle Pair Definition Examples

alternate interior 3 and 64 and 5

1 and 82 and 7

3 and 54 and 6

1 and 53 and 72 and 64 and 8

alternate exterior

same-side interior

corresponding

inside angles on opposite sides of thetransversal, not a linear pair

outside angles on opposite sides of thetransversal, not a linear pair

inside angles on the same side of thetransversal

in matching positions above or belowthe transversal, but on the same sideof the transversal

34

5 687

1 2s

t

c

1

56 7

8

4 23

l1 and l5; l2 and l6; l3 and l8; l4 and l7

alternate exterior angles same-side interior angles

l4 and l5; l3 and l6

l1 and l8; l2 and l7

l4 and l6; l3 and l5

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3-2 Reteaching Properties of Parallel Lines

When a transversal intersects parallel lines, special congruent and supplementary

angle pairs are formed.

Congruent angles formed by a transversal intersecting parallel lines:• corresponding angles (Postulate 3-1)

/1 > /5 /2 > /6

/4 > /7 /3 > /8

• alternate interior angles (Th eorem 3-1)

/4 > /6 /3 > /5

• alternate exterior angles (Th eorem 3-3)

/1 > /8 /2 > /7

Supplementary angles formed by a transversal intersecting parallel lines:• same-side interior angles (Th eorem 3-2)

m/4 1 m/5 5 180 m/3 1 m/6 5 180

Identify all the numbered angles congruent to the given angle. Explain.

1. 2. 3.

4. Supply the missing reasons in the two-column proof.

Given: g 6 h, i 6 j

Prove: /1 is supplementary to /16.

Statements Reasons

1) /1 > /3 1) 9

2) g 6 h; i 6 j 2) Given

3) /3 > /11 3) 9

4) /11 and /16 are supplementary. 4) 9

5) /1 and /16 are supplementary. 5) 9

1 234

5 67 8

,

m

q

12

3

45

67136

a bx

1 23

4 567

c

d

f

721

23

45

95

ij

e

ij

g

h1 2

34

5 678

9 10

12 11

13 141516

Vertical angles are congruent.

Corresponding angles are congruent.

Same-side interior angles are supplementary.

Substitution property

l6; vert. ' are O; l2; corresp. ' areO; l4; alt.

ext. ' are O.

l2; vert. ' are O; l5; alt. int. ' are O; l7; corresp. ' are O.

l2; vert. ' areO; l4; alt. ext. ' are O.

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3-2 Reteaching (continued)

Properties of Parallel Lines

You can use the special angle pairs formed by parallel lines and a transversal to

fi nd missing angle measures.

Problem

If m/1 5 100, what are the measures of /2 through /8?

Supplementary angles: m/2 5 180 2 100 m/2 5 80

m/4 5 180 2 100 m/4 5 80

Vertical angles: m/1 5 m/3 m/3 5 100

Alternate exterior angles: m/1 5 m/7 m/7 5 100

Alternate interior angles: m/3 5 m/5 m/5 5 100

Corresponding angles: m/2 5 m/6 m/6 5 80

Same-side interior angles: m/3 1 m/8 5 180 m/8 5 80

Problem

What are the measures of the angles in the fi gure?

(2x 1 10) 1 (3x 2 5) 5 180 Same-Side Interior Angles Theorem

5x 1 5 5 180 Combine like terms.

5x 5 175 Subtract 5 from each side.

x 5 35 Divide each side by 5.

Find the measure of these angles by substitution.

2x 1 10 5 2(35) 1 10 5 80 3x 2 5 5 3(35) 2 5 5 100

2x 2 20 5 2(35) 2 20 5 50

To fi nd m/1, use the Same-Side Interior Angles Th eorem:

50 1 m/1 5 180, so m/1 5 130

ExercisesFind the value of x. Th en fi nd the measure of each labeled angle.

5. 6. 7.

234

5 678

100

1 (2x 10)

(3x 5)

(2x 20)

x

(2x 10)

(4x 10)

(8x 10)

2x(3x 20)

(2x 15)50; 50; 9015; 130; 50

40; 100; 80; 65

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3-3 Reteaching

Proving Lines Parallel

Special angle pairs result when a set of parallel lines is intersected

by a transversal. Th e converses of the theorems and postulates

in Lesson 3-2 can be used to prove that lines are parallel.

Postulate 3-2: Converse of Corresponding Angles Postulate

If /1 > /5, then a 6 b.

Th eorem 3-4: Converse of the Alternate Interior Angles Th eorem

If /3 > /6, then a 6 b.

Th eorem 3-5: Converse of the Same-Side Interior Angles Th eorem

If /3 is supplementary to /5, then a 6 b.

Th eorem 3-6: Converse of the Alternate Exterior Angles Th eorem

If /2 > /7, then a 6 b.

Problem

For what value of x is b 6 c?

Th e given angles are alternate exterior angles. If they are

congruent, then b 6 c.

2x 2 22 5 118

2x 5 140

x 5 70

ExercisesWhich lines or line segments are parallel? Justify your answers.

1. 2. 3.

Find the value of x for which g n h. Th en fi nd the measure of each labeled angle.

4. 5. 6.

a

b

1 23 4

5 6

7 8

b

c

(2x 22)

118

A

BC

D

W X

Y Z Q

N

O

P

126 g

h(2x 16)

g

h

(6x)(20x 2)

g

h(3x 6)

(4x 18)

WX n YZ because the O

' are alt. int. '.

OP n QN because the O

angles are alt. int. '.

7; 42; 13824; 78; 78

55; 126

AB n CD because the O '

are alt. ext. angles.

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A fl ow proof is a way of writing a proof and a type of graphic organizer. Statements

appear in boxes with the reasons written below. Arrows show the logical

connection between the statements.

Problem

Write a fl ow proof for Th eorem 3-1: If a transversal intersects

two parallel lines, then alternate interior angles are congruent.

Given: / 6 m

Prove: /2 > /3

ExercisesComplete a fl ow proof for each.

7. Complete the fl ow proof for Th eorem 3-2 using the

following steps. Th en write the reasons for each step.

a. /2 and /3 are supplementary. b. /1 > /3

c. / 6 m d. /1 and /2 are supplementary.

Th eorem 3-2: If a transversal intersects two parallel lines, then

same side interior angles are supplementary.

Given: / 6 m

Prove: /2 and /3 are supplementary.

8. Write a fl ow proof for the following:

Given: /2 > /3

Prove: a 6 b

t

,

m

12

3

Vertical angles are .

If lines, then corresponding are .

1 3

1 2 Transitive Property of

2 3Given

, m

t

,

m3

21

a b

31

2

3-3 Reteaching (continued)

Proving Lines Parallel

Given

Vertical are .

Substitution If corresponding, then lines

are parallel.

2 3

3 1 a b

2 1

, m 1 3

Given

linear pair

Substitution

2 and 3 aresupplementary.

1 and 2 aresupplementary.

corr.

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3-4 Reteaching Parallel and Perpendicular Lines

You can use angle pairs to prove that lines are parallel. Th e postulates and

theorems you learned are the basis for other theorems about parallel and

perpendicular lines.

Th eorem 3-7: Transitive Property of Parallel Lines

If two lines are parallel to the same line, then they

are parallel to each other.

If a 6 b and b 6 c, then a 6 c. Lines a, b, and c can be in

diff erent planes.

Th eorem 3-8: If two lines are perpendicular to the same

line, then those two lines are parallel to each other.

Th is is only true if all the lines are in the same plane.

If a ' d and b ' d, then a 6 b.

Th eorem 3-9: Perpendicular Transversal Th eorem

If a line is perpendicular to one of two parallel lines,

then it is also perpendicular to the other line.

Th is is only true if all the lines are in the same plane.

If a 6 b and c, and a ' d, then b ' d, and c ' d.

Exercises

1. Complete this paragraph proof of Th eorem 3-7.

Given: d 6 e, e 6 f

Prove: d 6 f

Proof: Because it is given that d 6 e, then /1 is supplementary

to /2 by the Th eorem. Because

it is given that e 6 f , then /2 > /3 by the

Postulate. So, by substitution, /1 is supplementary to / . By the

Th eorem, d 6 f .

2. Write a paragraph proof of Th eorem 3-8.

Given: t ' n, t ' o

Prove: n 6 o

a

b

c

a

b

d

a

b

c

d

d

e

fh

12

3

n

o

t

1

2

Same-Side Int. Angles

Corresponding Angles

3

Converse of the Same-Side Int. Angles

Given that t ' n and t ' o, ml1 5 90 and ml2 5 90, by def. of perpendicular lines. Thus l1 O l2. So, n n o because of the Converse of the Corr. ' Post.

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Problem

A carpenter is building a cabinet. A decorative door will be set

into an outer frame.

a. If the lines on the door are perpendicular to the top

of the outer frame, what must be true about the lines?

b. Th e outer frame is made of four separate pieces of

molding. Each piece has angled corners as shown.

When the pieces are fi tted together, will each set of

sides be parallel? Explain.

c. According to Th eorem 3-8, lines that are perpendicular

to the same line are parallel to each other. So, since each

line is perpendicular to the top of the outer frame, all the

lines are parallel.

Draw the pieces as fi tted together to determine the measures of the new angles formed. Use this to decide if each set of sides will be parallel.

Ddae

Determine whether each set of sides will be parallel.

Dew

The angles for the top and bottom pieces are 358. The angles for the sides are 558.

Th e new angle is the sum of the angles that come together. Since

35 1 55 5 90, the pieces form right angles. Two lines that are

perpendicular to the same line are parallel. So, each set of sides is parallel.

Exercises

3. An artist is building a mosaic. Th e mosaic consists of the

repeating pattern shown at the right. What must be true of

a and b to ensure that the sides of the mosaic are parallel?

4. Error Analysis A student says that according to Th eorem 3-9,

if *

AD)

6 *

CF)

and *

AD)

'*

AB)

, then *

CF)

'*

AB)

. Explain the

student’s error.

35

55

35

55

40 40a

b

65 65

a

b

A

BC

D

EF

3-4 Reteaching (continued)

Parallel and Perpendicular Lines

a 5 50 and b 5 25

*

AB)

and *

CF)

are in different planes.

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3-5 Reteaching Parallel Lines and Triangles

Triangle Angle-Sum Theorem:

Th e measures of the angles in a triangle add up to 180.

Problem

In the diagram at the right, nACD is a right triangle.

What are m/1 and m/2?

Step 1

m/1 1 m/DAB 5 90 Angle Addition Postulate

m/1 1 30 5 90 Substitution Property

m/1 5 60 Subtraction Property of Equality

Step 2

m/1 1 m/2 1 m/ABC 5 180 Triangle Angle-Sum Theorem

60 1 m/2 1 60 5 180 Substitution Property

m/2 1 120 5 180 Addition Property of Equality

m/2 5 60 Subtraction Property of Equality

Exercises

Find ml1.

1. 2. 3.

4. 5. 6.

Algebra Find the value of each variable.

7. 8. 9.

A

B

C

D

1 2

6030

1

34 1

351

75 54

59 1

50 50

1

33 28

1

6

11

127X

Y

Z ZYX

27 9536

ZYX

3218

72

56 55 51

31 80 119

42; 138; 36

117; 63; 22

90; 90; 58

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In the diagram at the right, /1 is an exterior angle of the triangle.

An exterior angle is an angle formed by one side of a polygon and an

extension of an adjacent side.

For each exterior angle of a triangle, the two interior angles that are not next to

it are its remote interior angles. In the diagram, /2 and /3 are remote interior

angles to /1.

Th e Exterior Angle Th eorem states that the measure of an exterior angle is equal to

the sum of its remote interior angles. So, m/1 5 m/2 1 m/3.

Problem

What are the measures of the unknown angles?

m/ ABD 1 m/BDA 1 m/BAD 5 180 Triangle Angle-Sum Theorem

45 1 m/1 1 31 5 180 Substitution Property

m/1 5 104 Subtraction Property of Equality

m/ABD 1 m/BAD 5 m/2 Exterior Angle Theorem

45 1 31 5 m/2 Substitution Property

76 5 m/2 Subtraction Property of Equality

Exercises

What are the exterior angle and the remote interior angles for each triangle?

10. 11. 12.

exterior: exterior: exterior:

interior: interior: interior:

Find the measure of the exterior angle.

13. 14. 15.

1

2

3

A

B D45 1 2

31

1

2

3 4

M N

O

P

J

K

ML

X45

70

T

U

VW

90

53E

F

GH

4671

3-5 Reteaching (continued)

Parallel Lines and Triangles

l1, l2 lOMN, lMNO lJKL, lLJK

l4 lNOP lJLM

115 143 117

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3-6 Reteaching Constructing Parallel and Perpendicular Lines

Parallel Postulate Th rough a point not on a line, there is exactly one line parallel to the given line.

Problem

Given: Point D not on *

BC)

Construct: *

DJ)

parallel to *

BC)

Step 1 Draw *

BD)

.

Step 2 With the compass tip on B, draw an arc that intersects

*

BD)

between B and D. Label this intersection point F. Continue the arc to intersect *

BC)

at point G.

Step 3 Without changing the compass setting, place the compass tip on D and draw an arc that intersects *

BD)

above B and D. Label this intersection point H.

Step 4 Place the compass tip on F and open or close the compass so it reaches G. Draw a short arc at G.

Step 5 Without changing the compass setting, place the compass tip on H and draw an arc that intersects the fi rst arc drawn from H. Label this intersection point J.

Step 6 Draw *

DJ)

, which is the required line parallel to *

BC)

.

CB

D

CB

DF

G

H

CB

DF

G

H

CB

DF

G

J

H

CB

DF

G

J

H

CB

D

F

G

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3-6 Reteaching (continued)

Constructing Parallel and Perpendicular Lines

Exercises

Construct a line parallel to line m and through point Y.

1. 2. 3.

Perpendicular Postulate Th rough a point not on a line, there is exactly one line perpendicular to the given line.

Problem

Given: Point D not on *

BC)

Construct: a line perpendicular to *

BC)

through D

Step 1 Construct an arc centered at D that intersects *

BC)

at two points. Label those points G and H.

Step 2 Construct two arcs of equal length centered at points G and H.

Step 3 Construct the line through point D and the intersection of the arcs from Step 2.

Step 1 Step 2 Step 3

Construct a line perpendicular to line n and through point X.

4. 5. 6.

m

Y

Y

m Y

m

B C

D

B

G H

C

D

BG H

C

D

BG H

C

D

X

nX

n X

n

Z

Z

Z

Sample answers shown.

Sample answers shown.

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3-7 Reteaching Equations of Lines in the Coordinate Plane

To fi nd the slope m of a line, divide the change in the y values by the change in the

x values from one point to another. Slope is rise

run or

change in y

change in x.

Problem

What is the slope of the line through the points (25, 3) and (4, 9)?

Slope 5change in ychange in x 5

9 2 34 2 (25)

569 5

23

Exercises

Find the slope of the line passing through the given points.

1. (25, 2), (1, 8) 2. (1, 8), (2, 4) 3. (22, 23), (2, 24)

If you know two points on a line, or if you know one point and the slope of a line,

then you can fi nd the equation of the line.

Problem

Write an equation of the line that contains the points J(4, 25)

and K(22, 1). Graph the line.

If you know two points on a line, fi rst fi nd the slope using

m 5y2 2 y1

x2 2 x1.

m 51 2 (25)

22 2 4 5626 5 21

Now you know two points and the slope of the line. Select one

of the points to substitute for (x1, y1). Th en write the equation

using the point-slope form y 2 y1 5 m(x 2 x1).

y 2 1 5 21(x 2 (22)) Substitute.

y 2 1 5 21(x 1 2) Simplify within parentheses. You may leave your equation in this form or further simplify to fi nd the slope-intercept form.

y 2 1 5 2x 2 2

y 5 2x 2 1

Answer: Either y 2 1 5 21(x 1 2) or y 5 2x 2 1 is acceptable.

424 O

2

4

2

4

x

y

2

K

J

y x 1

1 –4 214

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Exercises

Graph each line.

4. y 5 12 x 2 4 5. y 2 4 5

13 (x 1 3) 6. y 2 3 5 26(x 2 3)

If you know two points on a line, or if you know one point and the slope of a line,

then you can fi nd the equation of the line using the formula y 2 y1 5 m(x 2 x1).

Use the given information to write an equation for each line.

7. slope 21, y-intercept 6 8. slope 4

5, y-intercept 23

9. 10.

11. passes through (7, 24) and (2, 22) 12. passes through (3, 5) and (26, 1)

Graph each line.

13. y 5 4 14. x 5 24 15. y 5 22

Write each equation in slope-intercept form.

16. y 2 7 5 22(x 2 1) 17. y 1 2 513 (x 1 5) 18. y 1 5 5 2

32 (x 2 3)

x

y

4

4

2

2

4

4

O 2

2x

y

4

4

2

2

4

4

O 2

2x

y

4

4

2

2

4

4

O 2

2

424 O

2

4

2

4

x

y

2 42 O

2

4

2

4

x

y

2

3-7 Reteaching (continued)

Equations of Lines in the Coordinate Plane

y 5 2x 1 6 y 5 45 x 2 3

y 5 23x 2 8

y 5 214x 2 13

4

y 1 4 5 225 (x 2 7) y 2 5 5 4

9 (x 2 3)

horizontal line through (0, 4) vertical line through (–4, 0)

horizontal line through (0, –2)

y 5 22x 1 9 y 5 13 x 21

3y 5 23

2 x 212

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3-8 Reteaching Slopes of Parallel and Perpendicular Lines

Remember that parallel lines are lines that are in the same plane that do not

intersect, and perpendicular lines intersect at right angles.

Problem

Write an equation for the line that contains G(4, 23)

and is parallel to *

EF)

: 212x 1 2y 5 6. Write another

equation for the line that contains G and is

perpendicular to *

EF)

. Graph the three lines.

Step 1 Rewrite in slope-intercept form: y 5 14x 1 3.

Step 2 Use point-slope form to write an equation

for each line.

Parallel line: m 514 Perpendicular line: m 5 24

y 2 (23) 514(x 2 4) y 2 (23) 5 24(x 2 4)

y 5 14x 2 4 y 5 24x 1 13

Exercises

In Exercises 1 and 2, are lines m1 and m2 parallel? Explain.

1. 2.

Find the slope of a line (a) parallel to and (b) perpendicular to each line.

3. y 5 3x 1 4 4. y 5 15x 1 3

45 5. y 2 3 5 24(x 1 1) 6. 4x 2 2y 5 8

Write an equation for the line parallel to the given line that contains point C.

7. y 5 12x 1 4; C(24, 23) 8. y 5 2

12x 2 3; C(3, 1)

9. y 5 13x 2 2; C(24, 3) 10. y 5 22x 2 4; C(3, 3)

4 6 824 O

2

4

6

8

10

12

2

x

y

2

E

x 314y

G x 414y

y 4x 13

F

2 4 6246 O

2

4

6

2

4

6

x

y

m2

m1

2 626 O

2

4

6

2

4

6

x

y

(5, 3)

( 5, 2)

( 4, 1)

(3, 3)

m2m1

Yes; both lines have a slope of 0. Yes; both lines have a slope of 3.

n : 3; ': 2 13

n : 2; ':2 12

n : 24; ': 14

n : 15; ': 25

y 5 12x 2 1

y 5 13x 1 41

3

y 5 212x 1 21

2

y 5 22x 1 9

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3-8 Reteaching (continued)

Slopes of Parallel and Perpendicular Lines

Write an equation for the line perpendicular to the given line that contains P.

11. P(5, 3); y 5 4x 12. P(2, 5); 3x 1 4y 5 1

13. P(2, 6); 2x 2 y 5 3 14. P(2, 0); 2x 2 3y 5 29

Problem

Given points J(21, 4), K(2, 3), L(5, 4), and M(0, 23), are *

JK)

and *

LM)

parallel,

perpendicular, or neither?

2

13 2

75 Their slopes are not equal, so they are not parallel.

13 ?

75 2 21 The product of their slopes is not –1, so they are not perpendicular.

Exercises

Tell whether *

JK)

and *

LM)

are parallel, perpendicular, or neither.

15. J(2, 0), K(21, 3), L(0, 4), M(21, 5) 16. J(24, 25), K(5, 1), L(6, 0), M(4, 3)

17. *

JK)

: 6y 1 x 5 7 18. *

JK)

: 3x 1 2y 5 5 19. *

JK)

: 2x 2 y 5 1

*

LM)

: 16 5 25y 2 x *

LM)

: 4x 1 5y 5 222 *

LM)

: x 1 2y 5 21

20. *

JK)

: y 5 15x 1 2 21.

*

JK)

: 2y 1 12x 5 22 22.

*

JK)

: y 5 21

*

LM)

: y 5 5x 2 12

*

LM)

: 2x 1 8y 5 8 *

LM)

: x 5 0

23. Right Triangle Verify that nABC is a right triangle for

A(0, 24), B(3, 22), and C(21, 4). Graph the triangle

and explain your reasoning.

neither

y 5 214x 1 41

4 y 5 43x 1 21

3

y 5 212x 1 7 y 5 23

2x 1 3

mJK 5 21; mLM 5 21; parallel mJK 523; mLM 5 23

2; perpendicular

mJK 515; mLM 5 5;

neither

mJK 5 216; mLM 5 21

5; neither

mJK 5 232; mLM 5 24

5; neither

mJK 5 2; mLM 5 212;;

perpendicular

mJK 5 214; mLM 5 21

4;

parallelmJK 5 0; mLM 5 undefined; perpendicular

kABC is a right triangle because AB ' BC. The slope of

*

AB)

is 23. The slope of *

BC)

is 232. 424 O

2

4

2

4

x

y

2

A

B

C

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Given ABCD > QRST , fi nd corresponding parts using the names. Order matters.

For example, ABCD Th is shows that /A corresponds to /Q.

QRST Th erefore, /A > /Q.

For example, ABCD Th is shows that BC corresponds to RS.

QRST Th erefore, BC > RS.

Exercises

Find corresponding parts using the order of the letters in the names.

1. Identify the remaining three pairs of corresponding angles and sides between ABCD and QRST using the circle technique shown above.

Angles: ABCD ABCD ABCD Sides: ABCD ABCD ABCD

QRST QRST QRST QRST QRST QRST

2. Which pair of corresponding sides is hardest to identify using this technique?

Find corresponding parts by redrawing fi gures.

3. Th e two congruent fi gures below at the left have been redrawn at the right. Why are the corresponding parts easier to identify in the drawing at the right?

4. Redraw the congruent polygons at the right in the same orientation. Identify all pairs of corresponding sides and angles.

5. MNOP > QRST . Identify all pairs of congruent sides and angles.

A

B

C

D

F

GH

K

A

B C

D F

G H

K

A

B

C

DE

T

P

Q

S R

M N

OP

RS

TQ

4-1 ReteachingCongruent Figures

lB O lR, lC O lS, lD O lT , AB O QR, CD O ST , and DA O TQ

Answers may vary. Sample: AD and QT

Answers may vary. Sample: The drawing at the right shows fi gures in same orientation.

Check students’ work. lA and lP, lB and lQ, lC and lR, lD and lS, lE and lT , AB and PQ, BC and QR, CD and RS, DE and ST , and EA and TP all correspond.

lM O lQ, lN O lR, lO O lS, lP O lT , MN O QR, NO O RS, OP O ST , and PM O TQ

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Problem

Given nABC > nDEF , m/A 5 30, and m/E 5 65, what is m/C?

How might you solve this problem? Sketch both triangles, and put all the information on both diagrams.

m/A 5 30; therefore, m/D 5 30. How do you know? Because /A and /D are corresponding parts of congruent triangles.

Exercises

Work through the exercises below to solve the problem above.

6. What angle in nABC has the same measure as /E? What is the measure of that angle? Add the information to your sketch of nABC.

7. You know the measures of two angles in nABC. How can you fi nd the measure of the third angle?

8. What is m/C? How did you fi nd your answer?

Before writing a proof, add the information implied by each given statement to your sketch. Th en use your sketch to help you with Exercises 9–12.

Add the information implied by each given statement.

9. Given: /A and /C are right angles.

10. Given: AB > CD and AD > CB.

11. Given: /ADB > /CBD.

12. Can you conclude that /ABD > /CDB using the given information above? If so, how?

13. How can you conclude that the third side of both triangles is congruent?

A

B

C F

E

D30

65

A B

CD

4-1 Reteaching (continued)

Congruent Figures

lB; 65

Answers may vary. Sample: Use Triangle Angle-Sum Thm. Set sum of all three angles equal to 180.

85; answers may vary. Sample: mlC 5 180 2 (60 1 35)

mlA 5 mlC 5 90, DA ' AB and DC ' BC

ABCD is a parallelogram because it has opposite sides that are congruent.

Yes; use the Third Angles Thm.

The third side is shared by both triangles and is congruent by the Refl . Prop. of Congruence.

AD n BC

65

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4-2 Reteaching Triangle Congruence by SSS and SAS

You can prove that triangles are congruent using the two postulates below.

Postulate 4-1: Side-Side-Side (SSS) PostulateIf all three sides of a triangle are congruent to all three sides of another

triangle, then those two triangles are congruent.

If JK > XY , KL > YZ, and JL > XZ, then nJKL > nXYZ.

In a triangle, the angle formed by any two sides is called the included angle for those sides.

Postulate 4-2: Side-Angle-Side (SAS) PostulateIf two sides and the included angle of a triangle are congruent

to two sides and the included angle of another triangle, then

those two triangles are congruent.

If PQ > DE, PR > DF , and /P > /D, then nPQR > nDEF .

/P is included by QP and PR. /D is included by ED and DF .

Exercises

1. What other information do you need to prove

nTRF > nDFR by SAS? Explain.

2. What other information do you need to prove

nABC > nDEF by SAS? Explain.

3. Developing Proof Copy and complete the fl ow proof.

Given: DA > MA, AJ > AZ

Prove: nJDA > nZMA

L

J KZ

XY

P

R

Q

E

F

D

R

DF

T

A

BC

EF

D

J

A

ZD

M

Given

Vertical are .

JDA ZMAAJ AZ

DF O TR; by the Refl exive Property of Congruence, RF O FR. It is given that lTRF O lDFR. These are the included angles for the corresponding congruent sides.

lB O lE; These are the included angles between the corresponding congruent sides.

DA MA

JAD ZAM

SASGiven

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Would you use SSS or SAS to prove the triangles congruent? If there is not enough information to prove the triangles congruent by SSS or SAS, write not enough information. Explain your answer.

4. 5. 6.

7. Given: PO > SO, O is the 8. Given: HI > HG, FH ' GI

midpoint of NT . Prove: nFHI > nFHG

Prove: nNOP > nTOS

9. A carpenter is building a support for a bird feeder. He wants

the triangles on either side of the vertical post to be congruent.

He measures and fi nds that AB > DE and that AC > DF . What would he need to measure to prove that the triangles are

congruent using SAS? What would he need to measure to prove

that they are congruent using SSS?

10. An artist is drawing two triangles. She draws each so that two sides are 4 in. and 5 in.

long and an angle is 558. Are her triangles congruent? Explain.

P

QM

CJ Y D

G C M

A

B

X

ZY

C

P

N

O

S

T

GH

I

F

A B E D

C F

4-2 Reteaching (continued)

Triangle Congruence by SSS and SAS

Answers may vary. Sample: Maybe; if both the 558 angles are between the 4-in. and 5-in. sides, then the triangles are congruent by SAS.

For SAS, he would need to determine if lBAC O lEDF; for SSS, he would need to determine if BC O EF.

Statements: 1) PO O SO; 2) O is the midpoint of NT ; 3) NO O TO; 4) lNOP O lTOS; 5) kNOP O kTOS; Reasons: 1) Given; 2) Given; 3) Def. of midpoint; 4) Vert. ' are O; 5) SAS

Not enough information; two pairs of corresponding sides are congruent, but the congruent angles are not the included angles.

Not enough information; you need to know if GC O DY.

Not enough information; only two pairs of corresponding sides are congruent. You need to know if AB O XY or lZ O lC.

Statements: 1) FH O FH; 2) HI O HG, FH ' GI; 3) lFHG and lFHI are rt. '; 4) lFHG O lFHI; 5) kFHI O kFHG; Reasons: 1) Refl . Prop.; 2) Given; 3) Def. of perpendicular; 4) All rt. ' are O; 5) SAS

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4-3 Reteaching Triangle Congruence by ASA and AAS

Problem

Can the ASA Postulate or the AAS Th eorem be applied directly to prove the

triangles congruent?

a. Because /RDE and /ADE are right

angles, they are congruent. ED > ED by

the Refl exive Property of >, and it is given

that /R > /A. Th erefore, nRDE>nADE by the AAS Th eorem.

b. It is given that CH > FH and /F > /C. Because /CHE and /FHB are vertical

angles, they are congruent. Th erefore,

nCHE > nFHB by the ASA Postulate.

Exercises

Indicate congruences.

1. Copy the top fi gure at the right. Mark the fi gure with the angle

congruence and side congruence symbols that you would need

to prove the triangles congruent by the ASA Postulate.

2. Copy the second fi gure shown. Mark the fi gure with the angle

congruence and side congruence symbols that you would need

to prove the triangles congruent by the AAS Th eorem.

3. Draw and mark two triangles that are congruent by either the ASA Postulate or

the AAS Th eorem.

What additional information would you need to prove each pair of triangles congruent by the stated postulate or theorem?

4. ASA Postulate 5. AAS Th eorem 6. ASA Postulate

7. AAS Th eorem 8. AAS Th eorem 9. ASA Postulate

E

R D A

C

E

F

B

H

A B C

D

J K

M L

X Y

U V

Z

DOY

G

P

R

A

T

LC

Y A

Check students’ work.

1.

2.

lABD O lCBD lZXY O lZVUlJMK O lLKM, lJKM O lLMK , lJMK O lLMK, or lJKM O lLKM

lY O lO lP O lA lCYL O lALY

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10. Provide the reason for each step in the two-column proof.

Given: TX 6 VW , TU > VU , /XTU > /WVU , /UWV is a right angle.

Prove: nTUX > nVUW

Statements Reasons

1) /UWV is a right angle.

2) VW ' XW

3) TX 6 VW

4) TX ' XW

5) /UXT is a right angle.

6) /UWV > /UXT

7) TU > VU

8) /XTU > /WVU

9) nTUX > nVUW

1) 9

2) 9

3) 9

4) 9

5) 9

6) 9

7) 9

8) 9

9) 9

11. Write a paragraph proof.

Given: WX 6 ZY ; WZ 6 XY

Prove: nWXY > nYZW

12. Developing Proof Complete the proof by fi lling in the blanks.

Given: /A > /C, /1 > /2

Prove: nABD > nCDB

Proof: /A > /C and /1 > /2 are given. DB > BD by 9.

So, nABD > nCDB by 9.

13. Write a paragraph proof.

Given: /1 > /6, /3 > /4, LP > OP

Prove: nLMP > nONP

T

X U

V

W

W

X Y

Z

A B

CD

1

24

3

P

L M N O

1

2

34

56

4-3 Reteaching (continued)

Triangle Congruence by ASA and AAS

Refl . Prop. of Congruence

l3 O l4 is given. Therefore, ml3 5 ml4, by def. of O ls. Because l2 and l3 are linear pairs, and l4 and l5 are linear pairs, the pairs of angles are suppl. Therefore, l2 O l5 by the Congruent Suppl. Thm. l1 O l6 and LP O OP are given, so kLMP O kONP, by the AAS Thm.

It is given that WX n ZY and WZ n XY, so lXWY O lZYW and lXYW O lZWY, by the Alternate Interior ' Thm. WY O YW by the Refl exive Property of O. So, by ASA Post. kWXY O kYZW.

AAS

Given

Defi nition of perpendicular lines

Given

Perpendicular Transversal Theorem

Defi nition of perpendicular lines

All right angles are congruent.

Given

Given

AAS Theorem

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If you can show that two triangles are congruent, then you can show that all the

corresponding angles and sides of the triangles are congruent.

Problem

Given: AB 6 DC, /B > /D

Prove: BC > DA

In this case you know that AB 6 DC. AC forms a transversal and creates a pair of

alternate interior angles, /BAC and /DCA.

You have two pairs of congruent angles, /BAC > /DCA and /B > /D. Because

you know that the shared side is congruent to itself, you can use AAS to show

that the triangles are congruent. Th en use the fact that corresponding parts are

congruent to show that BC > DA. Here is the proof:

Statements Reasons

1) AB 6 DC 1) Given

2) /BAC > /DCA 2) Alternate Interior Angles Th eorem

3) /B > /D 3) Given

4) AC > CA 4) Refl exive Property of Congruence

5) nABC > nCDA 5) AAS

6) BC > DA 6) CPCTC

Exercises

1. Write a two-column proof.

Given: MN > MP, NO > PO

Prove: /N > /P

Statements Reasons

1) 9 1) Given

2) MO > MO 2) 9

3) 9 3) 9

4) /N > /P 4) 9

A B

CD

M

P N

O

4-4 ReteachingUsing Corresponding Parts of Congruent Triangles

MN O MP, NO O PO

kMNO O kMPO

Refl exive Property of O

SSS

CPCTC

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2. Write a two-column proof.

Given: PT is a median and an altitude of nPRS.

Prove: PT bisects /RPS.

Statements Reasons

1) PT is a median of nPRS.

2) 9

3) 9

4) PT is an altitude of nPRS.

5) PT ' RS

6) /PTS and /PTR are right angles.

7) 9

8) 9

9) 9

10) /TPS > /TPR

11) 9

1) 9

2) Defi nition of median

3) Defi nition of midpoint

4) 9

5) 9

6) 9

7) All right angles are congruent.

8) Refl exive Property of Congruence

9) SAS

10) 9

11) 9

3. Write a two-column proof.

Given: QK > QA; QB bisects /KQA.

Prove: KB > AB

4. Write a two-column proof.

Given: ON bisects /JOH , /J > /H

Prove: JN > HN

K Q

A

B

O

HN

J

4-4 Reteaching (continued)

Using Corresponding Parts of Congruent Triangles

T is the midpoint of RS.

RT O ST

lPTS O lPTR

PT O PT

kPTS O kPTR

PT bisects lRPS.

Given

Given

Defi nition of altitude

Defi nition of perpendicular

CPCTC

Defi nition of angle bisector

Statements Reasons

1) QK O QA; QB bisects lKQA. 1) Given

2) lKQB O lAQB 2) Def. of l bis.

3) BQ O BQ 3) Refl . Prop. of Congruence

4) kKBQ O kABQ 4) SAS

5) KB O AB 5) CPCTC

Statements Reasons

1) ON bisects lJOH, lJ O lH 1) Given

2) lJON O lHON 2) Def. of l bis.

3) ON O ON 3) Refl . Prop. of Congruence

4) kJON O kHON 4) AAS

5) JN O HN 5) CPCTC

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4-5 ReteachingIsosceles and Equilateral Triangles

Two special types of triangles are isosceles triangles and equilateral triangles.

An isosceles triangle is a triangle with two congruent sides.

Th e base angles of an isosceles triangle are also congruent.

An altitude drawn from the shorter base splits an isosceles

triangle into two congruent right triangles.

An equilateral triangle is a triangle that has

three congruent sides and three congruent

angles. Each angle measures 608.

You can use the special properties of isosceles and equilateral triangles to fi nd or

prove diff erent information about a given fi gure.

Look at the fi gure at the right.

You should be able to see that one of the triangles is

equilateral and one is isosceles.

Problem

What is m/A?

nABC is isosceles because it has two base angles that are congruent. Because the

sum of the measures of the angles of a triangle is 180,

and m/B 5 40, you can solve to fi nd m/A.

m/A 1 m/B 1 m/BEA 5 180 There are 1808 in a triangle.

m/A 1 40 1 m/A 5 180 Substitution Property

2 m/A 1 40 5 180 Combine like terms.

2 m/A 5 140 Subtraction Property of Equality

m/A 5 70 Division Property of Equality

Problem

What is FC?

nCFG is equilateral because it has three congruent angles.

CG 5 (2 1 2) 5 4, and CG 5 FG 5 FC. So, FC 5 4.

A

B

C

E

D F

40

A

B

C

D

E

F

G

y

2 2

x

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Problem

What is the value of x?

Because x is the measure of an angle in an equilateral

triangle, x 5 60.

Problem

What is the value of y?

m/DCE 1 m/DEC 1 m/EDC 5 180 There are 1808 in a triangle.

60 1 70 1 y 5 180 Substitution Property

y 5 50 Subtraction Property of Equality

Exercises

Complete each statement. Explain why it is true.

1. /EAB > 9

2. /BCD > 9 > /DBC

3. FG > 9 > DF

Determine the measure of the indicated angle.

4. /ACB

5. /DCE

6. /BCD

Algebra Find the value of x and y.

7. 8.

9. Reasoning An exterior angle of an isosceles triangle has a measure 140.

Find two possible sets of measures for the angles of the triangle.

40

A

B

C

D

E

F

G

y

2 2

x

A B C

D

E F G

40

A

B

C

D

E FG

50

y x

110

x

y

115

4-5 Reteaching (continued)

Isosceles and Equilateral Triangles

lEBA; base angles of an isosceles triangle are congruent.

lCDB; the angles of an equilateral triangle are congruent.

GD; the sides of an equilateral triangle are congruent.

60

65

55

35; 35 65; 50

40, 40, 100; 40, 70, 70

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4-6 Reteaching Congruence in Right Triangles

Two right triangles are congruent if they have congruent hypotenuses and if they

have one pair of congruent legs. Th is is the Hypotenuse-Leg (HL) Th eorem.

A P

Q RC B

nABC > nPQR because they are both right triangles, their hypotenuses are

congruent (AC > PR) , and one pair of legs is congruent (BC > QR) .

Problem

How can you prove that two right triangles that have one pair of congruent legs

and congruent hypotenuses are congruent (Th e Hypotenuse-Leg Th eorem)?

Both of the triangles are right triangles.

/B and /E are right angles.

AB > DE and AC > DF .

How can you prove that nABC > nDEF ?

Look at nDEF . Draw a ray starting at F that passes through E. Mark a

point X so that EX 5 BC. Th en draw DX to create nDEX .

See that EX > BC. (You drew this.) /DEX > /ABC. (All right angles

are congruent.) DE > AB. (Th is was given.) So, by SAS,

nABC > nDEX .

DX > AC (by CPCTC) and AC > DF . (Th is was given.). So, by the Transitive

Property of Congruence, DX > DF . Th en, /DEX > /DEF . (All right angles are

congruent.) By the Isosceles Th eorem, /X > /F . So, by AAS, nDEX > nDEF .

Th erefore, by the Transitive Property of Congruence, nABC > nDEF .

Problem

Are the given triangles congruent by the Hypotenuse-Leg Th eorem?

If so, write the triangle congruence statement.

/F and /H are both right angles, so the triangles are both right.

GI > IG by the Refl exive Property and FI > HG is given.

So, nFIG > nHGI .

A

B C

D

E F

D

E FX

F I

HG

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Exercises

Determine if the given triangles are congruent by the Hypotenuse-Leg Th eorem. If so, write the triangle congruence statement.

1. 2.

3. 4.

Measure the hypotenuse and length of the legs of the given triangles with a ruler to determine if the triangles are congruent. If so, write the triangle congruence statement.

5. 6.

7. Explain why nLMN > nOMN. Use the Hypotenuse-Leg Th eorem.

8. Visualize nABC and nDEF , where AB 5 EF and CA 5 FD. What else must

be true about these two triangles to prove that the triangles are congruent

using the Hypotenuse-Leg Th eorem? Write a congruence statement.

T

U V

L

MN

TR S

Z

L

NM

S

V R

O

P Q

X Y

Z

A B

CM

E F

G H

J

I

N

L M O

4-6 Reteaching (continued)

Congruence in Right Triangles

not congruent

lB and lE are right angles, or lC and lD are right angles. kABC O kDEF or kABC O kFED.

kRSZ O kTSZ

kOPQ O kZYX

kEFG O kHIJ

kLMN O kRVS

kABC O kCMA

Because lNML and lNMO are right angles, both triangles are right triangles. It is given that their hypotenuses are congruent. Because they share a leg, one pair of corresponding legs is congruent. All criteria are met for the triangles to be congruent by the Hypotenuse-Leg Theorem.

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4-7 Reteaching Congruence in Overlapping Triangles

Sometimes you can prove one pair of triangles congruent and then use

corresponding parts of those triangles to prove another pair congruent.

Often the triangles overlap.

Problem

Given: AB > CB,

AE > CD, /AED > /CDE

Prove: nABE > nCBD

Th ink about a plan for the proof. Examine the triangles you are trying to prove

congruent. Two pairs of sides are congruent. If the included angles, /A and /C, were congruent, then the triangles would be congruent by SAS.

If the overlapping triangles nAED and nCDE were congruent, then the angles

would be congruent by corresponding parts. When triangles overlap, sometimes it

is easier to visualize if you redraw the triangles separately.

Now use the plan to write a proof.

Given: AB > CB, AE > CD, /AED > /CDE

Prove: nABE > nCBD

Statements Reasons

1) AE > CD, /AED > /CDE

2) ED > ED

3) nAED > nCDE

4) /A > /C

5) AB > CB

6) nABE > nCBD

1) Given

2) Refl exive Property of >

3) SAS

4) CPCTC

5) Given

6) SAS

A

E

B

D

C

A

E D

B

E

B

D

C

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Separate and redraw the overlapping triangles. Identify the vertices.

1. nGLJ and nHJL 2. nMRP and nNQS 3. nFED and nCDE

Fill in the blanks for the two-column proof.

4. Given: /AEG > /AFD, AE > AF , GE > FD

Prove: nAFG > nAED

Statements Reasons

1) /AEG > /AFD, AE > AF , GE > FD 1) 9

2) 9 2) SAS

3) AG > AD, /G > /D 3) 9

4) 9 4) Given

5) GE 5 FD 5) 9

6) GF 1 FE 5 GE, FE 1 ED 5 FD 6) 9

7) GF 1 FE 5 FE 1 ED 7) 9

8) 9 8) Subtr. Prop. of Equality

9) 9 9) 9

Use the plan to write a two-column proof.

5. Given: /PSR and /PQR are

right angles, /QPR > /SRP.

Prove: nSTR > nQTP

Plan for Proof:

Prove nQPR > nSRP by AAS. Th en use CPCTC and

vertical angles to prove nSTR > nQTP by AAS.

G

J

H

K

L PQRS

M NO

T U C

DE

F

B

A

DEFG

P

T

Q

RS

4-7 Reteaching (continued)

Congruence in Overlapping Triangles

G

L J L J

H N

QS R P

MF C

E ED D

kAEG O kAFD

Given

CPCTC

Def. of O

Seg. Addition Post.

Substitution Property

SAS

GE O FD

GF 5 ED

kAFG O kAED

Statements: 1) lPSR and lPQR are rt. '; lQPR O lSRP; 2) lPSR O lRQP; 3) PR O RP; 4) kQPR O kSRP; 5) lSTR O lQTP; 6) PQ O RS; 7) kSTR O kQTP; Reasons: 1) Given; 2) Rt. ' are congruent; 3) Refl . Prop. of O; 4) AAS; 5) Vert. ' are O; 6) CPCTC; 7) AAS

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5-1 Reteaching Midsegments of Triangles

Connecting the midpoints of two sides of a triangle creates a segment called

a midsegment of the triangle.

Point X is the midpoint of AB.

Point Y is the midpoint of BC.

So, XY is a midsegment of nABC.

Th ere is a special relationship between a midsegment and the

side of the triangle that is not connected to the midsegment.

Triangle Midsegment Th eorem

• Th e midsegment is parallel to the third side

of the triangle.

• Th e length of the midsegment is half the length

of the third side.

XY 6 AC and XY 5 12 AC.

Connecting each pair of midpoints, you can see that

a triangle has three midsegments.

XY , YZ, and ZX are all midsegments of nABC.

Because Z is the midpoint of AC, XY 5 AZ 5 ZC 5 12 AC.

Problem

QR is a midsegment of nMNO.

What is the length of MO?

Start by writing an equation using the Triangle

Midsegment Th eorem.

12 MO 5 QR

MO 5 2QR

5 2(20)

5 40

So, MO 5 40.

B

A C

YX

B

A C

YX

B

A C

YX

Z

N

R

O?M

Q 20

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Problem

AB is a midsegment of nGEF . What is the value of x?

2AB 5 GF

2(2x) 5 20

4x 5 20

x 5 5

Exercises

Find the length of the indicated segment.

1. AC 2. TU 3. SU

4. MO 5. GH 6. JK

Algebra In each triangle, AB is a midsegment. Find the value of x.

7. 8. 9.

10. 11. 12.

E

BF

20

2x

A

G

B

E

CA

D15

R

26

STQ

U

R ST

UV

10.6

L MN

OP

8.8G

H

F

E

I15

K

O

LN

J4.5

M

B

N5x 7

O

A3x

R

A2x 5

SB

3x 15

T

L

B

M

A

K15x 6.5

10x

LB

J

3x 11

H

A

2x E

FAD

Bx 7

2x 17 Q

BA

P 27 R

x

5-1 Reteaching (continued)

Midsegments of Triangles

30

4.4

7

11

5

10

1.3

13.5

30 9

13 5.3

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5-2 Reteaching Perpendicular and Angle Bisectors

Perpendicular BisectorsTh ere are two useful theorems to remember about perpendicular bisectors.

Perpendicular Bisector Th eorem

If a point is on the perpendicular

bisector of a segment, then it is

equidistant from the endpoints

of the segment.

X is on the perpendicular

bisector, so it is

equidistant from the

endpoints A and B.

Converse of the Perpendicular Bisector Th eorem

If a point is equidistant from the

endpoints of a segment, then it is

on the perpendicular bisector of

the segment.

Because X is equidistant

from the endpoints C

and D, it is on the

perpendicular bisector

of the segment.

Problem

What is the value of x?

Since A is equidistant from the endpoints of the segment, it is on the

perpendicular bisector of EG. So, EF 5 GF and x 5 4.

Exercises

Find the value of x.

1. 2. 3.

4. 5. 6.

A B

X

33

C D

X33

A

E F G

7 7

x 4

A B

M

P

x8

L MP

N55

42x

F GJ

H2x 3 x 2

3 3

G

2xx 6

D F

E

S

2x12 xP R

QR S

U

T10x 3 5x 2

8

6

2

4

5

1

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Angle BisectorsTh ere are two useful theorems to remember about angle bisectors.

Angle Bisector Th eorem

If a point is on the bisector

of an angle, then the point is

equidistant from the sides of

the angle.

X is on the angle

bisector and is therefore

equidistant from the

sides of the angle.

Converse of the Angle Bisector Th eorem

If a point in the interior of an

angle is equidistant from the

sides of an angle, then the point

is on the angle bisector.

Because X is on the

interior of the angle

and is equidistant

from the sides, X is

on the angle bisector.

Problem

What is the value of x?

Because point A is in the interior of the angle and it is equidistant

from the sides of the angle, it is on the bisector of the angle.

/BCA > /ECA

x 5 40

Exercises

Find the value of x.

7. 8. 9.

A

B

C

X4

4

D

E

F

X

A

B

C

E

x40

8

8

70x

38 38

3x 12 5

5

(4x)

(2x 20)

5-2 Reteaching (continued)

Perpendicular and Angle Bisectors

4 1070

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5-3 Reteaching Bisectors in Triangles

The Circumcenter of a TriangleIf you construct the perpendicular bisectors of all three sides of

a triangle, the constructed segments will all intersect at one point.

Th is point of concurrency is known as the circumcenter of the triangle.

It is important to note that the circumcenter of a triangle can lie inside,

on, or outside the triangle.

Th e circumcenter is equidistant from the three

vertices. Because of this, you can construct a

circle centered on the circumcenter that passes

through the triangle’s vertices. Th is is called a

circumscribed circle.

Problem

Find the circumcenter of right nABC.

First construct perpendicular bisectors of the two legs,

AB and AC. Th ese intersect at (2, 2), the circumcenter.

Notice that for a right triangle, the circumcenter is

on the hypotenuse.

Exercises

Coordinate Geometry Find the circumcenter of each right triangle.

1. 2. 3.

Coordinate Geometry Find the circumcenter of kABC.

4. A(0, 0) 5. A(27, 3) 6. A(25, 2)

B(0, 8) B(9, 3) B(3, 2)

C(10, 8) C(27, 27) C(3, 6)

Circumcenter

2

4

6

2 4 622

A

B

C

D

y

x

2

4

6

2 4 622

Q

R

S

y

x

4

2 44 O

3L

M

y

x

N

4

O 2 424

2

y

xH K

J

(5, 4)

(3, 2) (1, 1) (0, 0)

(1, 22) (21, 4)

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5-3 Reteaching (continued)

Bisectors in Triangles

The Incenter of a TriangleIf you construct angle bisectors at the three vertices of a triangle, the

segments will intersect at one point. Th is point of concurrency where

the angle bisectors intersect is known as the incenter of the triangle.

It is important to note that the incenter of a triangle will always lie

inside the triangle.

Th e incenter is equidistant from the sides

of the triangle. You can draw a circle centered

on the incenter that just touches the three

sides of the triangle. Th is is called an

inscribed circle.

Problem

Find the value of x.

Th e angle bisectors intersect at P. Th e incenter P is equidistant

from the sides, so SP 5 PT . Th erefore, x 5 9.

Note that PV , the continuation of the angle bisector, is not the

correct segment to use for the shortest distance from P to AC.

Exercises

Find the value of x.

7. 8. 9.

10. 11. 12.

Incenter

B

S

A

TV

C

P9

10

x

14

x

2x

5x 6

x 6

2x 6

2x 7

4x2x 12

4 x

4x 8

5x

14

3.5 16 8

2 12

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5-4 Reteaching Medians and Altitudes

A median of a triangle is a segment that runs from one vertex of the triangle

to the midpoint of the opposite side. Th e point of concurrency of the

medians is called the centroid.

Th e medians of nABC are AM , CX , and BL.

Th e centroid is point D.

An altitude of a triangle is a segment that runs from one vertex perpendicular

to the line that contains the opposite side. Th e orthocenter is the point of

concurrency for the altitudes. An altitude may be inside or outside the triangle,

or a side of the triangle.

Th e altitudes of nQRS are QT , RU , and SN .

Th e orthocenter is point V.

Determine whether AB is a median, an altitude, or neither.

1. 2.

3. 4.

5. Name the centroid. 6. Name the orthocenter.

M

A

BC

L XD

Q

RS

U

N

T

V

AO

X

B3

3.2

A

ZT B

A

B DC

A

F

B

E

S

U

TB

W Z

Q

P R

T

S

altitude

altitude

median

neither

Z P

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Th e medians of a triangle intersect at a point two-thirds of the distance from a

vertex to the opposite side. Th is is the Concurrency of Medians Th eorem.

CJ and AH are medians of nABC

and point F is the centroid.

CF 5 23 CJ

Problem

Point F is the centroid of nABC. If CF 5 30, what is CJ?

CF 5 23 CJ Concurrency of Medians Theorem

30 523 3 CJ Fill in known information.

32 3 30 5 CJ Multiply each side by 32.

45 5 CJ Solve for CJ.

Exercises

In kVYX , the centroid is Z. Use the diagram to solve the problems.

7. If XR 5 24, fi nd XZ and ZR.

8. If XZ 5 44, fi nd XR and ZR.

9. If VZ 5 14, fi nd VP and ZP.

10. If VP 5 51, fi nd VZ and ZP.

11. If ZO 5 10, fi nd YZ and YO.

12. If YO 5 18, fi nd YZ and ZO.

In Exercises 13–16, name each segment.

13. a median in nDEF

14. an altitude in nDEF

15. a median in nEHF

16. an altitude in nHEK

A

B

C

H

JF

V

R

YPX

O Z

F E

IJHG

D

K

L

5-4 Reteaching (continued)

Medians and Altitudes

16; 8

66; 22

21; 7

34; 17

20; 30

12; 6

DL

FK

HL

HK or KE

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5-5 Reteaching Indirect Proof

In an indirect proof, you prove a statement or conclusion to be true by proving the

opposite of the statement to be false.

Th ere are three steps to writing an indirect proof.

Step 1: State as a temporary assumption the opposite (negation) of what you

want to prove.

Step 2: Show that this temporary assumption leads to a contradiction.

Step 3: Conclude that the temporary assumption is false and that what you want

to prove must be true.

Problem

Given: Th ere are 13 dogs in a show; some are long-haired and the rest are

short-haired. Th ere are more long-haired than short-haired dogs.

Prove: Th ere are at least seven long-haired dogs in the show.

Step 1: Assume that fewer than seven long-haired dogs are in the show.

Step 2: Let / be the number of long-haired dogs and s be the number of short-

haired dogs. Because / 1 s 5 13, s 5 13 2 /. If / is less than 7, s is

greater than or equal to 7. Th erefore, s is greater than /. Th is contradicts

the statement that there are more long-haired than short-haired dogs.

Step 3: Th erefore, there are at least seven long-haired dogs.

Exercises

Write the temporary assumption you would make as a fi rst step in writing an indirect proof.

1. Given: an integer q; Prove: q is a factor of 34.

2. Given: nXYZ; Prove: XY 1 XZ . YZ.

3. Given: rectangle GHIJ; Prove: m/G 5 90

4. Given: XY and XM ; Prove: XY 5 XM

Write a statement that contradicts the given statement.

5. Whitney lives in an apartment.

6. Marc does not have three sisters.

7. /1 is a right angle.

8. Lines m and h intersect.

Assume q is not a factor of 34.

Assume XY 1 XZ K YZ.

Assume mlG u 90.

Assume XY u XM.

l1 is an acute angle.

Lines m and h do not intersect.

Whitney does not live in an apartment.

Marc has three sisters.

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Problem

Given: /A and /B are not complementary.

Prove: /C is not a right angle.

Step 1: Assume that/C is a right angle.

Step 2: If /C is a right angle, then by the Triangle Angle-Sum Th eorem,

m/A 1 m/B 1 90 5 180. So m/A 1 m/B 5 90. Th erefore, /A

and /B are complementary. But /A and /B are not complementary.

Step 3: Th erefore, /C is not a right angle.

Exercises

Complete the proofs.

9. Arrange the statements given at the right to complete the steps of

the indirect proof.

Given: XY R YZ

Prove: /1 R /4

Step 1: 9 A. But XY R YZ.

Step 2: 9 B. Assume /1 > /4.

Step 3: 9 C. Th erefore, /1 R /4.

Step 4: 9 D. /1 and /2 are supplementary, and

/3 and /4 are supplementary.

Step 5: 9 E. According to the Converse of the Isosceles

Triangle Th eorem, XY 5 YZ or XY > YZ.

Step 6: 9 F. If /1 > /4, then by the Congruent

Supplements Th eorem, /2 > /3.

10. Complete the steps below to write a convincing argument using indirect

reasoning.

Given: nDEF with /D R /F

Prove: EF R DE

Step 1: 9

Step 2: 9

Step 3: 9

Step 4: 9

A

C B

X Z

Y

1 2 3 4

D F

E

5-5 Reteaching (continued)

Indirect Proof

D

B

F

E

A

C

Assume EF O DE.

If EF O DE, then by the Isosceles Triangle Theorem, lD O lF.

But lD j lF.

Therefore, EF j DE.

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5-6 Reteaching Inequalities in One Triangle

For any triangle, if two sides are not congruent, then the larger angle is opposite

the longer side (Th eorem 5-10). Conversely, if two angles are not congruent, then

the longer side is opposite the larger angle (Th eorem 5-11).

Problem

Use the triangle inequality theorems to answer the questions.

a. Which is the largest angle of nABC?

AB is the longest side of nABC. /C lies opposite AB.

/C is the largest angle of nABC.

b. What is m/E? Which is the shortest side of nDEF ?

m/D 1 m/E 1 m/F 5 180 Triangle Angle-Sum Theorem

30 1 m/E 1 90 5 180 Substitution

120 1 m/E 5 180 Addition

m/E 5 60 Subtraction Property of Equality

/D is the smallest angle of nDEF . Because FE lies opposite /D,

FE is the shortest side of nDEF .

Exercises

1. Draw three triangles, one obtuse, one acute, and one right. Label the vertices.

Exchange your triangles with a partner.

a. Identify the longest and shortest sides of each triangle.

b. Identify the largest and smallest angles of each triangle.

c. Describe the relationship between the longest and shortest sides and

the largest and smallest angles for each of your partner’s triangles.

Which are the largest and smallest angles of each triangle?

2. 3. 4.

Which are the longest and shortest sides of each triangle?

5. 6. 7.

A B

C

6 8

9

30D E

F

D E

F

7 6

4P Q

R8

6

3

A

C B

5

7

10

15

D E

F

130

P Q

R

3570

V

35

S

T100

Check students’ work. The longest side will be opposite the largest angle. The shortest side will be opposite the smallest angle.

largest: lDEF; smallest: lDFE

largest: lPQR; smallest: lPRQ

largest: lACB; smallest: lCBA

longest: DF; shortest: FE longest: PQ; shortest: RQ longest: SV ; shortest: ST

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5-6 Reteaching (continued)

Inequalities in One Triangle

For any triangle, the sum of the lengths of any two sides is greater than the length

of the third side. Th is is the Triangle Inequality Th eorem.

AB 1 BC . AC

AC 1 BC . AB

AB 1 AC . BC

Problem

A. Can a triangle have side lengths 22, 33, and 25?

Compare the sum of two side lengths with the third side length.

22 1 33 . 25 22 1 25 . 33 25 1 33 . 22

A triangle can have these side lengths.

B. Can a triangle have side lengths 3, 7, and 11?

Compare the sum of two side lengths with the third side length.

3 1 7 , 11 3 1 11 . 7 11 1 7 . 3

A triangle cannot have these side lengths.

C. Two sides of a triangle are 11 and 12 ft long. What could be the length of the

third side?

Set up inequalities using x to represent the length of the third side.

x 1 11 . 12 x 1 12 . 11 11 1 12 . x

x . 1 x . 21 23 . x

Th e side length can be any value between 1 and 23 ft long.

Exercises

8. Can a triangle have side lengths 2, 3, and 7?

9. Can a triangle have side lengths 12, 13, and 7?

10. Can a triangle have side lengths 6, 8, and 9?

11. Two sides of a triangle are 5 cm and 3 cm. What could be the length of the

third side?

12. Two sides of a triangle are 15 ft and 12 ft. What could be the length of the third

side?

A

BC

no

yes

yes

less than 8 cm and greater than 2 cm

less than 27 ft and greater than 3 ft

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5-7 Reteaching Inequalities in Two Triangles

Consider nABC and nXYZ. If AB > XY , BC > YZ, and m/Y . m/B, then XZ . AC. Th is is the Hinge Th eorem (SAS Inequality Th eorem).

Problem

Which length is greater, GI or MN?

Identify congruent sides: MO > GH and NO > HI .

Compare included angles: m/H . m/O.

By the Hinge Th eorem, the side opposite the larger included angle is longer.

So, GI . MN .

Problem

At which time is the distance between the tip of a clock’s hour hand and the tip of its minute hand greater, 3:00 or 3:10?

Th ink of the hour hand and the minute hand as two sides of a triangle whose lengths never change, and the distance between the tips of the hands as the third side. 3:00 and 3:10 can then be represented as triangles with two pairs of congruent sides. Th e distance between the tips of the hands is the side of the triangle opposite the included angle.

At 3:00, the measure of the angle formed by the hour hand and minute hand is 908. At 3:10, the measure of the angle is less than 908.

So, the distance between the tip of the hour hand and the tip of the minute hand is greater at 3:00.

Exercises

1. What is the inequality relationship between LP and XA in the fi gure at the right?

2. At which time is the distance between the tip of a clock’s hour hand and the tip of its minute hand greater, 5:00 or 5:15?

A

CB

X

YZ

80

M

ON

85

G

HI

L

PM

X

A

Y 93

5:00

XA S LP

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Consider nLMN and nPQR. If LM > PQ, MN > QR, and PR . LN , then m/Q . m/M . Th is is the Converse of the Hinge Th eorem (SSS Inequality Th eorem).

Problem

TR . ZX. What is the range of possible values for x?

Th e triangles have two pairs of congruent sides, because RS 5 XY and TS 5 ZY . So, by the Converse of the Hinge Th eorem, m/S . m/Y .

Write an inequality:

72 . 5x 1 2 Converse of the Hinge Theorem

70 . 5x Subtract 2 from each side.

14 . x Divide each side by 5.

Write another inequality:

m/Y . 0 The measure of an angle of a triangle is greater than 0.

5x 1 2 . 0 Substitute.

5x . 22 Subtract 2 from each side.

x . 2 25 Divide each side by 5.

So, 2 25 , x , 14.

Exercises

Find the range of possible values for each variable.

3. 4.

5. Reasoning An equilateral triangle has sides of length 5, and an isosceles triangle has side lengths of 5, 5, and 4. Write an inequality for x, the measure of the vertex angle of the isosceles triangle.

L

NM

P

QR

67

R

72

12

S

T

X

Z

Y

12

(5x 2)

58E G

HD

(3x 2)

12

C11

F

86

C 14

A

D

(2x 4)

B

8

5-7 Reteaching (continued)

Inequalities in Two Triangles

23 R x R 20

60 S x S 0

2 R x R 45

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6-1 Reteaching The Polygon Angle-Sum Theorems

Interior Angles of a Polygon

Th e angles on the inside of a polygon are called interior angles.

Polygon Angle-Sum Th eorem:

Th e sum of the measures of the angles of an n-gon is (n 2 2)180.

You can write this as a formula. Th is formula works for regular and irregular polygons.

Sum of angle measures 5 (n 2 2)180

Problem

What is the sum of the measures of the angles in a hexagon?

Th ere are six sides, so n 5 6.

Sum of angle measures 5 (n 2 2)180

5 (6 2 2)180 Substitute 6 for n.

5 4(180) Subtract.

5 720 Multiply.

Th e sum of the measures of the angles in a hexagon is 720.

You can use the formula to fi nd the measure of one interior angle of a regular polygon if you know the number of sides.

Problem

What is the measure of each angle in a regular pentagon?

Sum of angle measures 5 (n 2 2)180

5 (5 2 2)180 Substitute 5 for n.

5 3(180) Subtract.

5 540 Multiply.

Divide by the number of angles:

Measure of each angle 5 540 4 5

5 108 Divide.

Each angle of a regular pentagon measures 108.

Interior angle

A pentagon has 5 interior angles.

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6-1 Reteaching (continued) The Polygon Angle-Sum Theorems

Exercises

Find the sum of the interior angles of each polygon.

1. quadrilateral 2. octagon 3. 18-gon

4. decagon 5. 12-gon 6. 28-gon

Find the measure of an interior angle of each regular polygon. Round to the nearest tenth if necessary.

7. decagon 8. 12-gon 9. 16-gon

10. 24-gon 11. 32-gon 12. 90-gon

Exterior Angles of a PolygonTh e exterior angles of a polygon are those formed by extending sides. Th ere is one exterior angle at each vertex.

Polygon Exterior Angle-Sum Th eorem:

Th e sum of the measures of the exterior angles of a polygon is 360.

A pentagon has fi ve exterior angles. Th e sum of the measures of the exterior angles is always 360, so each exterior angle of a regular pentagon measures 72.

Exercises

Find the measure of an exterior angle for each regular polygon. Round to the nearest tenth if necessary.

13. octagon 14. 24-gon 15. 34-gon

16. decagon 17. heptagon 18. hexagon

19. 30-gon 20. 28-gon 21. 36-gon

22. Draw a Diagram A triangle has two congruent angles, and an exterior angle that measures 140. Find two possible sets of angle measures for the triangle. Draw a diagram for each.

Exterior angle

360

144

45 15 10.6

36 51.4 60

1012.912

40, 40, 100; 40, 70, 70

150

165

157.5

176168.8

1440

1080

1800

2880

4680

140

14010040

70

40

70

40

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6-2 Reteaching Properties of Parallelograms

ParallelogramsRemember, a parallelogram is a quadrilateral with both pairs of opposite sides

parallel. Here are some attributes of a parallelogram:

Th e opposite sides are congruent.

Th e consecutive angles are supplementary.

Th e opposite angles are congruent.

Th e diagonals bisect each other.

You can use these attributes to solve problems about parallelograms.

Problem

Find the value of x.

Because the consecutive angles are supplementary,

x 1 60 5 180

x 5 120

Problem

Find the value of x.

Because opposite sides are congruent,

x 1 7 5 15

x 5 8

Problem

Find the value of x and y.

Because the diagonals bisect each other, y 5 3x and 4x 5 y 1 3.

4x 5 y 1 3

4x 5 3x 1 3 Substitute for y.

x 5 3 Subtraction Property of 5

y 5 3x Given

y 5 3(3) Substitute for x.

y 5 9 Simplify.

A B

CD

A B

CD

60

x

x 7

15

3x 4x

y 3 y

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6-2 Reteaching (continued)

Properties of Parallelograms

Exercises

Find the value of x in each parallelogram.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. Writing Write a statement about the consecutive angles of a parallelogram.

14. Writing Write a statement about the opposite angles of a parallelogram.

15. Reasoning One angle of a parallelogram is 47. What are the measures of the

other three angles in the parallelogram?

x

130

5x 2

6x

35

x193x 2

2x 3x

y 4 y

3x 1

y

x 3 y

2

3x 2

2x 4

2x 6

x 4

(x 35) (2x 25)

(3x 15) (2x 5)

110 (3x 7)

10x

5x 10

(3x)

(x 60)

3x 4

4x 1

50

35 7

4

8 40

21

30

Consecutive angles of a parallelogram are supplementary.

Opposite angles of a parallelogram are congruent.

47, 133, and 133

5

2

3

2

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6-3 Reteaching Proving That a Quadrilateral Is a Parallelogram

Is a quadrilateral a parallelogram?

Th ere are fi ve ways that you can confi rm that a quadrilateral is a parallelogram.

If both pairs of opposite sides are parallel,

then the quadrilateral is a parallelogram.

If both pairs of opposite sides are congruent,

then the quadrilateral is a parallelogram.

If both pairs of opposite angles are congruent,

then the quadrilateral is a parallelogram.

If the diagonals bisect each other, then the

quadrilateral is a parallelogram.

If one pair of sides is both congruent and parallel,

then the quadrilateral is a parallelogram.

Exercises

Can you prove that the quadrilateral is a parallelogram based on the given information? Explain.

1. 2. 3.

4. 5. 6.

yes; opposite sides O no; not enough info yes; 1 pair of sides O and n

yes; opposite sides nyes; opposite ' O

Yes; diagonals bisect each other.

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6-3 Reteaching (continued) Proving That a Quadrilateral Is a Parallelogram

Determine whether the given information is suffi cient to prove that quadrilateral WXYZ is a parallelogram.

7. WY bisects ZX 8. WX 6 ZY ; WZ > XY

9. VZ > VX ; WX > YZ 10. /VWZ > /VYX ; WZ > XY

You can also use the requirements for a parallelogram to solve problems.

Problem

For what value of x and y must fi gure ABCD be a parallelogram?

In a parallelogram, the two pairs of opposite angles are congruent.

So, in ABCD, you know that x 5 2y and 5y 1 54 5 4x. You can

use these two expressions to solve for x and y.

Step 1: Solve for y. 5y 1 54 5 4x

5y 1 54 5 4(2y) Substitute 2y for x.

5y 1 54 5 8y Simplify.

54 5 3y Subtract 5y from each side.

18 5 y Divide each side by 3.

Step 2: Solve for x. x 5 2y Opposite angles of a parallelogram are congruent.

x 5 2(18) Substitute 18 for y.

x 5 36 Simplify.

For ABCD to be a parallelogram, x must be 36 and y must be 18.

Exercises

For what value of x must the quadrilateral be a parallelogram?

11. 12. 13.

14. 15. 16.

W X

YZ

V

2y

(5y 54) x

4x

A B

D C

3(x 6)

4x

x 32x

x 9

3x 3

3x 3 2x

x 3

2x 55x

(3x 28) 4x

x 16.5

18 3

8 14 5.5

3

no no

no yes

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Rhombuses, rectangles, and squares share some characteristics. But they also have

some unique features.

A rhombus is a parallelogram with four congruent sides.

A rectangle is a parallelogram with four congruent angles.

Th ese angles are all right angles.

A square is a parallelogram with four congruent sides and four

congruent angles. A square is both a rectangle and a rhombus.

A square is the only type of rectangle that can also be a rhombus.

Here is a Venn diagram to help you see the relationships.

Th ere are some special features for each type of fi gure.

Rhombus: Th e diagonals are perpendicular.

Th e diagonals bisect a pair of opposite angles.

Rectangles: Th e diagonals are congruent.

Squares: Th e diagonals are perpendicular.

Th e diagonals bisect a pair of opposite angles (forming two 458

angles at each vertex).

Th e diagonals are congruent.

Exercises

Decide whether the parallelogram is a rhombus, a rectangle, or a square.

1. 2. 3. 4.

Rhombus Square Rectangle

F G

I H

6-4 Reteaching Properties of Rhombuses, Rectangles, and Squares

rectangle rhombus rhombus square

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6-4 Reteaching (continued)

Properties of Rhombuses, Rectangles, and Squares

List the quadrilaterals that have the given property. Choose among parallelogram, rhombus, rectangle, and square.

5. Opposite angles are supplementary. 6. Consecutive sides are >.

7. Consecutive sides are '. 8. Consecutive angles are >.

You can use the properties of rhombuses, rectangles, and squares to

solve problems.

Problem

Determine the measure of the numbered angles in rhombus DEFG.

/1 is part of a bisected angle. m/DFG 5 48, so m/1 5 48.

Consecutive angles of a parallelogram are supplementary.

m/EFG 5 48 1 48 5 96, so m/DGF 5 180 2 96 5 84.

Th e diagonals bisect the vertex angle, so m/2 5 84 4 2 5 42.

Exercises

Determine the measure of the numbered angles in each rhombus.

9. 10.

Determine the measure of the numbered angles in each fi gure.

11. rectangle ABCD 12. square LMNO

Algebra TUVW is a rectangle. Find the value of x and the length of each diagonal.

13. TV 5 3x and UW 5 5x 2 10 14. TV 5 2x 2 4 and UW 5 x 1 10

15. TV 5 6x 1 4 and UW 5 4x 1 8 16. TV 5 7x 1 6 and UW 5 9x 2 18

17. TV 5 8x 2 2 and UW 5 5x 1 7 18. TV 5 10x 2 4 and UW 5 3x 1 24

482 1

E

G F

D

3521

122

1

30

60

1A

2

34B

D C 21

M N

L O

3

35; 55

78; 90

90; 90; 4560; 30; 60; 30

5; 15; 15 14; 24; 24

12; 90; 90

4; 36; 36

2; 16; 16

3; 22; 22

rectangle, square rhombus, square

rectangle, square rectangle, square

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6-5 Reteaching Conditions for Rhombuses, Rectangles, and Squares

A parallelogram is a rhombus if either of these conditions is met:

A B

CD

1) Th e diagonals of the parallelogram are perpendicular. (Th eorem 6-16)

D E

FG

2) A diagonal of the parallelogram bisects a pair of opposite angles. (Th eorem 6-17)

A parallelogram is a rectangle if the diagonals of the parallelogram are congruent.

WY > XZ

Exercises

Classify each of the following parallelograms as a rhombus, a rectangle, or a square. For each, explain.

1. MO > PN 2. 3. AC > BD

Use the properties of rhombuses and rectangles to solve problems.

Problem

For what value of x is ~DEFG a rhombus?

In a rhombus, diagonals bisect opposite angles.

So, m/GDF 5 m/EDF .

(4x 1 10) 5 (5x 1 6) Set angle measures equal to each other.

10 5 x 1 6 Subtract 4x from each side.

4 5 x Subtract 6 from each side.

W X

YZ

T U

VW

M N

OP

A B

CD

D E

FG

(4x 10)

(5x 6)

Rectangle; the diagonals are O.

Rhombus; the diagonals bisect opposite angles.

Square; the diagonals are O and '.

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6-5 Reteaching (continued)

Conditions for Rhombuses, Rectangles, and Squares

Exercises

4. For what value of x is ~WXYZ a rhombus?

W X

YZ

222x

5. SQ 5 14. For what value of x is ~PQRS a rectangle?

Solve for PT. Solve for PR.

P Q

RS

T

3x 111 x

3

3

6. For what value of x is ~RSTU a rhombus? What is m/SRT ? What is m/URS?

R S

TU

(2x 46)

(x 2)

7. LN 5 54. For what value of x is ~LMNO a rectangle?

L M

NO

P

4x

21

2x

3

3

3

8. Given: ~ABCD, AC ' BD at E.

Prove: ABCD is a rhombus.

Statements Reasons

1) AE > CE 1) 9

2) AC ' BD at E 2) 9

3) 9 3) Defi nition of perpendicular lines

4) 9 4) 9

5) 9 5) Refl exive Property of Congruence

6) nAED > nCED 6) 9

7) AD > CD 7) 9

8) 9 8) Opposite sides of a ~ are >.

9) 9 9) 9

10) ABCD is a rhombus. 10) 9

lAED and lCED are right angles.

Diagonals of a ~ bisect each other.

Given

11

12

6; 7; 14

48; 50; 100

All right angles are congruent.

SAS Postulate

CPCTC

Transitive Property of Congruence

Defi nition of rhombus

lAED O lCED

DE O DE

AB O CD, AD O BC

AB O BC O CD O DA

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6-6 Reteaching Trapezoids and Kites

A trapezoid is a quadrilateral with exactly one pair of parallel sides.

Th e two parallel sides are called bases. Th e two nonparallel sides

are called legs.

A pair of base angles share a common base.

/1 and /2 are one pair of base angles.

/3 and /4 are a second pair of base angles.

In any trapezoid, the midsegment is parallel to the bases. Th e length

of the midsegment is half the sum of the lengths of the bases.

MN 512(QR 1 ZX)

An isosceles trapezoid is a trapezoid in which the legs are congruent.

An isosceles trapezoid has some special properties:

Each pair of base angles is congruent. Th e diagonals are congruent.

Exercises

1. In trapezoid LMNO, what is the measure of /OLM ?

What is the measure of /LMN ?

2. WXYZ is an isosceles trapezoid and WY 5 12. What is XZ?

3. XZ is the midsegment of trapezoid EFGH. If FG 5 8 and EH 5 12,

what is XZ?

1 2

3 4

base

leg

base

leg

Q

M N

Z X

R

D C

A B A B

CD

AC BD

70

O

L

M

N

W X

YZ

XE

G

F

H Z

70110

12

10

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6-6 Reteaching (continued)

Trapezoids and Kites

A kite is a quadrilateral in which two pairs of consecutive sides are

congruent and no opposite sides are congruent.

In a kite, the diagonals are perpendicular. Th e diagonals look like the

crossbars in the frame of a typical kite that you fl y.

Notice that the sides of a kite are the hypotenuses of four right triangles

whose legs are formed by the diagonals.

Problem

Write a two-column proof to identify three pairs of congruent triangles

in kite FGHJ.

Statements Reasons

1) m/FKG 5 m/GKH 5 m/HKJ 5 m/JKF 5 90 1) Th eorem 6-22

2) FG > FJ 2) Given

3) FK > FK 3) Refl exive Property of Congruence

4) nFKG > nFKJ 4) HL Th eorem

5) JK > KG 5) CPCTC

6) KH > KH 6) Refl exive Property of Congruence

7) nJKH > nGKH 7) SAS Postulate

8) JH > GH 8) Given

9) FH > FH 9) Refl exive Property of Congruence

10) nFJH > nFGH 10) SSS Postulate

So nFKG > nFKJ , nJKH > nGKH , and nFJH > nFGH.

ExercisesIn kite FGHJ in the problem, mlJFK 5 38 and mlKGH 5 63. Find the following angle and side measures.

4. m/FKJ 5. m/FJK 6. m/FKG

7. m/KFG 8. m/FGK 9. m/GKH

10. m/KHG 11. m/KJH 12. m/JHK

13. If FG 5 4.25, what is JF?

14. If HG 5 5, what is JH?

15. If JK 5 8.5, what is GJ?

J G

F

H

K

F

JK

G

H

90

38

27

4.25

52

63

90

27

5

17

52 90

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6-7 Reteaching Polygons in the Coordinate Plane

Below are some formulas that can help you classify fi gures on a coordinate plane.

To determine if line segments that form sides or diagonals are congruent, use the Distance Formula:

d 5 "(x2 2 x1)2 1 (y2 2 y1)2.

In the fi gure at the right, the length of AB is

"(5 2 1)2 1 (4 2 1)2 5 "42 1 32 5 5.

In the fi gure above right the length of BC is

"(5 2 0)2 1 (4 2 4)2 5 "52 1 02 5 5.

So, AB > BC. Th e fi gure is an isosceles triangle.

To fi nd the midpoint of a side or diagonal, use the Midpoint Formula.

M 5 ax1 1 x2

2 , y1 1 y2

2 bIn the fi gure above, the midpoint of AB is a1 1 5

2 , 1 1 4

2 b 5 a62,

52b 5 (3, 2.5)

To determine whether line segments that form sides or diagonals are parallel or perpendicular, use the Slope Formula.

m 5y2 2 y1x2 2 x1

In the fi gure at the right, the slope of AB is (4 2 1)

(5 2 1)5

34

.

Th e slope of TS is (6 2 3)

(5 2 1)5

34

. Th e line segments

are parallel.

Lines with equal slopes are parallel.

Lines with slopes that have a product of 21 are perpendicular.

Exercises

1. How could you use the formulas to determine if a polygon on a coordinate

plane is a rhombus?

2. How could you use the formulas to determine if a trapezoid on a coordinate

plane is isosceles?

3. How could you use the formulas to determine if a quadrilateral on a

coordinate plane is a kite?

y

x

C(0, 4)B(5, 4)

A(1, 1)

O

y

x

T(1, 3)

S(5, 6)

B(5, 4)

A(1, 1)

Answers may vary. Sample: Use the Distance Formula to prove that the sides are all the same length.

Answers may vary. Sample: Use the Distance Formula to prove that the non-parallel pair of sides are congruent.

Answers may vary. Sample: Use the Distance Formula to show that there are two pairs of adjacent congruent sides and no side is congruent to the side opposite it.

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6-7 Reteaching (continued)

Polygons in the Coordinate Plane

Problem

Is nABC scalene, isosceles, or equilateral?

Find the lengths of the sides using the Distance Formula.

BA 5 "(6)2 1 (1)2 5 "36 1 1 5 "37

BC 5 "(2)2 1 (4)2 5 "20

CA 5 "(4)2 1 (23)2 5 "16 1 9 5 "25 5 5

Th e sides are all diff erent lengths. So, nABC is scalene.

Problem

Is quadrilateral GHIJ a parallelogram?

Find the slopes of the opposite sides.

slope of GH 54 2 3

0 2 (23)5

13

; slope of JI 521 2 (22)

4 2 1 513

slope of HI 5 21 2 44 2 0

5254

; slope of GJ 5 22 2 31 2 (23)

5254

So, JI 6 GH and HI 6 GJ . Th erefore, GHIJ is a parallelogram.

Exercises

kJKL has vertices at J(22, 4), K(1, 6), and L(4, 4).

4. Determine whether nJKL is scalene, isosceles, or equilateral. Explain.

5. Determine whether nJKL is a right triangle. Explain.

6. Trapezoid ABCD has vertices at A(2, 1), B(12, 1), C(9, 4), and D(5, 4). Which

formula would help you fi nd out if this trapezoid is isosceles? Is this an

isosceles trapezoid? Explain.

y

x

A

C

B

2

2 6

4

O

y

x

G

I

J

H

2

22

Distance Formula; yes; the distances AD and BC are both "18 or 3"2. Because these distances represent the legs, ABCD is isosceles.

Isosceles; use the Distance Formula to fi nd that JK 5 KL 5"13, and JL 5 6.

No; use the Slope Formula. The slope of JK 5 6 2 41 2 (22)

5 23, the slope of

KL 5 6 2 41 2 4 5 22

3, and the slope of JL 5 4 2 422 2 4 5 0, so no sides are perpendicular.

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6-8 Reteaching Applying Coordinate Geometry

You can use variables instead of integers to name the coordinates of a polygon in the coordinate plane.

Problem

Use the properties of each fi gure to fi nd the missing coordinates.

rhombus MNPQ

M is at the origin (0, 0). Because diagonals of a rhombus bisect each other, N has x-coordinate a2. Because the x-axis is a horizontal line of symmetry for the rhombus, Q has coordinates (a

2, 2b).

square ABCD

Because all sides are congruent, D has coordinate (0, x). Because all angles are right, C has coordinates (x, x).

Exercises

Use the properties of each fi gure to fi nd the missing coordinates.

1. parallelogram OPQR 2. rhombus XYZW 3. square QRST

4. A quadrilateral has vertices at (a, 0), (2a, 0), (0, a), and (0, 2a). Show that it is a square.

5. A quadrilateral has vertices at (a, 0), (0, a 1 1), (2a, 0), and (0, 2a 2 1). Show that it is a rhombus.

6. Isosceles trapezoid ABCD has vertices A(0, 0), B(x, 0), and D(k, m). Find the coordinates of C in terms of x, k, and m. Assume AB 6 CD.

bN(?, b)

P(a, 0)

Q(?, ?)

M(?, ?) A(0, 0)

B(?, ?) C(?, ?)

D(x, 0)

P(k, m)Q(?, ?)

R (x, 0)O (0, 0)

Y(0, b)

Z(a, 0)

W(?, ?)

X(?, ?) R(a, 0)

S(?, ?)T(?, ?)

Q(0, 0)

Q(x 1 k, m) X(2a, 0); W(0, 2b) S(a, 2a), T(0, 2a)

Each side has a length of "2a2 1 2a 1 1. Therefore, the fi gure is a rhombus.

C(x 2 k, m)

Sample: Each side has a length of a"2, making the fi gure a rhombus. One pair of opposite sides has a slope of 1, and the other pair has a slope of 21. Because the product of the slopes is 21, the sides are perpendicular and the rhombus is a square.

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6-8 Reteaching (continued)

Applying Coordinate Geometry

You can use a coordinate proof to prove geometry theorems. You can use the Distance Formula, the Slope Formula, and the Midpoint Formula when writing coordinate proofs. With the Midpoint Formula, using multiples of two to name coordinates makes computation easier.

Problem

Plan a coordinate proof to show that the diagonals of a square are congruent.

Draw and label a square on a coordinate grid. In square ABCD, AB 5BC 5CD 5DA. Draw in the diagonals, AC and BD.

Prove that AC 5BD. Use the Distance Formula.

CA 5 "(0 2 a)2 1 (a 2 0)2 5 "a2 1 a2 5 "2a2

BD 5 "(a 2 0)2 1 (a 2 0)2 5 "a2 1 a2 5 "2a2

So, CA 5 BD. Th e diagonals of the square are congruent.

Exercises

7. How would you use a coordinate proof to prove that the diagonals of a square are perpendicular?

8. How would you use a coordinate proof to prove that the diagonals of a rectangle are congruent?

9. How would you use a coordinate proof to prove that if the midpoints of the sides of a trapezoid are connected they will form a parallelogram?

10. How would you use a coordinate proof to prove that the diagonals of a parallelogram bisect one another?

11. Classify quadrilateral ABCD with vertices A(0, 0), B(a, 2b), C(c, 2b), D(a 1 c, 0) as precisely as possible. Explain.

12. Classify quadrilateral FGHJ with vertices F(a, 0), G(a, 2c), H(b, 2c), and J(b, c) as precisely as possible. Explain.

A(0, a)

x

y

D(a, a)

C(a, 0)B(0, 0)

Answers may vary. Sample: Use the Slope Formula to prove that the product of the slopes of the diagonals is 21.

Answers may vary. Sample: Use the Distance Formula to prove that the lengths of the diagonals are equal.

Answers may vary. Sample: Use the Midpoint Formula to fi nd the midpoints, then use the Slope Formula to show that opposite sides in the new fi gure have equal slopes.

Answers may vary. Sample: Use the Midpoint Formula to show that the midpoints of the diagonals are the same point.

isosceles trapezoid; one pair of sides parallel, other opposite pair of sides O, and diagonals O

Trapezoid; one pair of sides is parallel, the other pair of sides is not parallel.

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6-9 Reteaching Proofs Using Coordinate Geometry

A coordinate proof can be used to prove geometric relationships. A coordinate proof uses variables to name coordinates of a fi gure on a coordinate plane.

Problem

Use coordinate geometry to prove that the diagonals of a rectangle are congruent.

AC 5 "(k 2 0)2 1 (m 2 0)2

5 "k2 1 m2

BD 5 "(0 2 k)2 1 (m 2 0)2

5 "(2k)2 1 m2

5 "k2 1 m2

AC > BD

Exercises

Use coordinate geometry to prove each statement.

1. Diagonals of an isosceles 2. Th e line containing 3. Th e segmentstrapezoid are congruent. the midpoints of two joining the midpoints sides of a triangle is of a rectangle form parallel to the a rhombus. third side.

C(k, m)

B(k, 0)

D(0, m)

A(0, 0)

(b, c)

( a, 0)

( b, c)

(a, 0)(a, b)

(0, 0) (a, 0)

(0, b)(b, c)

(0, 0) (a, 0)Use the Distance Formula to fi nd the lengths of the diagonals. Each has a length of "c2 1 (b 1 a)2. So, the diagonals are congruent.

You need to use the Midpoint Formula to fi nd the midpoints of two sides, and the Slope Formula to show that the line connecting the midpoints is parallel to the third side. The midpoints are (b

2, c

2)

and (b 1 a2

, c2). The line

connecting the midpoints has slope 0 and is therefore parallel to the third side.

You need to fi nd the midpoints using the Midpoint Formula and lengths of segments connecting the midpoints with the Distance Formula. The midpoints are (a2, 0), (a, b2), (a2, b), and (0, b2). The segments joining these points are congruent, as they each have a length of 12"a2 1 b2.

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Th e example used the Distance Formula to prove two line segments congruent. When planning a coordinate proof, write down the formulas that you will need to use, and write what you can prove using those formulas.

Exercises

State whether you can reach each conclusion below using coordinate methods. Give a reason for each answer.

4. AB 5 12CD.

5. nABC is equilateral.

6. Quadrilateral ABCD is a square.

7. Th e diagonals of a quadrilateral form right angles.

8. Quadrilateral ABCD is a trapezoid.

9. nABC is a right triangle.

10. Quadrilateral ABCD is a kite.

11. Th e diagonals of a quadrilateral form angles that measure 30 and 150.

12. m/D 5 33

13. nABC is scalene.

14. Th e segments joining midpoints of an equilateral triangle form an equilateral triangle.

15. Quadrilateral KLMN is an isosceles trapezoid.

6-9 Reteaching (continued) Proofs Using Coordinate Geometry

Yes; use the Distance Formula to show that AB 5 12CD.

Yes; use the Distance Formula to show all sides are equal.

Yes; use the Distance Formula to show all sides are equal and the Slope Formula to show all sides are perpendicular if the product of the slopes of any two adjacent sides is 21.

Yes; use the Slope Formula to show the diagonals are ' if the product of their slopes is 21.

Yes; use the Slope Formula to show that one pair of sides is parallel because the slopes are equal.

Yes; use the Slope Formula to show that the product of the slopes of two sides is 21.

No; I do not have coordinate methods to measure angles.

No; I do not have coordinate methods to measure angles.

Yes; use the Midpoint Formula to fi nd the midpoints of the sides and the distance formula to show that the measure of each side is equal in the new triangle.

Yes; use the Distance Formula to show that all sides have a different length.

Yes; use the Slope Formula to show that one pair of sides is parallel (have equal slopes), and use the Distance Formula to show that the legs are congruent.

Yes; use the Distance Formula to check that there are two pairs of adjacent sides of the same length, and that all four sides are not the same length.

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7-1 ReteachingRatios and Proportions

Problem

About 15 of every 1000 light bulbs assembled at the Brite Lite Company are

defective. If the Brite Lite Company assembles approximately 13,000 light bulbs

each day, about how many are defective?

Set up a proportion to solve the problem. Let x represent the number of defective

light bulbs per day.

15

1000 5x

13,000

15(13,000) 5 1000x Cross Products Property

195,000 5 1000x Simplify.

195,000

1000 5 x Divide each side by 1000.

195 5 x Solve for the variable.

About 195 of the 13,000 light bulbs assembled each day are defective.

Exercises

Use a proportion to solve each problem.

1. About 45 of every 300 apples picked at the Newbury Apple Orchard are rotten.

If 3560 apples were picked one week, about how many apples were rotten?

2. A grocer orders 800 gal of milk each week. He throws out about 64 gal of

spoiled milk each week. Of the 9600 gal of milk he ordered over three months,

about how many gallons of spoiled milk were thrown out?

3. Seven of every 20 employees at V & B Bank Company are between the ages of

20 and 30. If there are 13,220 employees at V & B Bank Company, how many

are between the ages of 20 and 30?

4. About 56 of every 700 picture frames put together on an assembly line have

broken pieces of glass. If 60,000 picture frames are assembled each month,

about how many will have broken pieces of glass?

Algebra Solve each proportion.

5. 3001600 5

x4800 6. 40

140 5700

x 7. x

2000 517

400

8. 35x 5

1502400 9.

x1040 5

2905200 10.

x42,000 5

87500

11. x

380 5180

5700 12. 120090,000 5

270x 13. 325

x 57306

56,200

534

768

4627

4800

900 2450

7308

12 20,250 2500

85

560 58

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7-1 Reteaching (continued)

Ratios and Proportions

In a proportion, the products of terms that are diagonally across the equal sign

from each other are the same. Th is is called the Cross Products Property because

the products cross at the equal sign.

a

b d

c b c

a db c a d

Proportions have other properties:

Property (1) ab 5

cd is equivalent to

ba 5

dc . Use reciprocals of the ratios.

Property (2) ab 5

cd is equivalent to

ac 5

bd . Switch b and c in the proportion.

Property (3) ab 5

cd is equivalent to

a 1 bb 5

c 1 dd . Add the denominator to the numerator.

Problem

How can you use the Cross Products Property to verify Property (3)?

ab 5

cd is equivalent to ad 5 bc.

a 1 b

b 5c 1 d

d is equivalent to (a 1 b)d 5 b(c 1 d) . Cross Products Property

ad 1 bd 5 bc 1 bd Distributive Property

ad 5 bc Subtraction Property of Equality

So, ab 5

cd is equivalent to

a 1 bb 5

c 1 dd .

Exercises

Use the proportion x

10 52z . Complete each statement. Justify your answer.

14. x2 5uu

15. 10x 5uu

16. x 1 1010 5

uu

17. Th e ratio of width to length of a rectangle is 7 i 10. Th e width of the rectangle is

91 cm. Write and solve a proportion to fi nd the length.

18. Th e ratio of the two acute angles in a right triangle is 5 i 13. What is the

measure of each angle in the right triangle?

10z , Prop. of Proportions (2)

710 5

91x ; 130 cm

25, 65, 90

z2, Prop. of Proportions (1) 2 1 z

z , Prop. of Proportions (3)

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7-2 Reteaching Similar Polygons

Similar polygons have corresponding angles that are congruent and

corresponding sides that are proportional. An extended proportion can be

written for the ratios of corresponding sides of similar polygons.

Problem

Are the quadrilaterals at the right similar? If so, write a similarity

statement and an extended proportion.

Compare angles: /A > /X , /B > /Y .

/C > /Z , /D > /W

Compare ratios of sides: ABXY 5

63 5 2

CDZW 5

94.5 5 2

BCYZ 5

84 5 2

DAWX 5

42 5 2

Because corresponding sides are proportional and corresponding angles are

congruent, ABCD , XYZW .

Th e extended proportion for the ratios of corresponding sides is:

ABXY 5

BCYZ 5

CDZW 5

DAWX

Exercises

If the polygons are similar, write a similarity statement and the extended proportion for the ratios of corresponding sides. If the polygons are not similar, write not similar.

1. 2.

3. 4.

8

4

C

9

D4A

6

B

Y

3

X 2 W

4.5

Z

L

R

Q

S

M

K

60

40

36

18 24

30

B

A

Z X

Y

C40

20

21

28

14

30

P 6 Q

6

S

6

R6

T 8 U

4 4

8W V77

7

7

7 7

7

7

M L R Q

PNJ K

60

KML M QSR, KMQS 5

MLSR 5

LKRQ BCA M YZX, BC

YZ 5CAZX 5

ABXY

not similarnot similar

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7-2 Reteaching (continued)

Similar Polygons

Problem

nRST , nUVW. What is the scale factor?

What is the value of x?

Identify corresponding sides: RT corresponds to UW , TS corresponds to WV , and

SR corresponds to VU .

RT

UW 5TS

WV Compare corresponding sides.

42 5

7x Substitute.

4x 5 14 Cross Products Property

x 5 3.5 Divide each side by 4.

Th e scale factor is 42 5

73.5 5 2. Th e value of x is 3.5.

Exercises

Give the scale factor of the polygons. Find the value of x. Round answers to the nearest tenth when necessary.

5. ABCD , NMPO 6. nXYZ , nEFD

7. LMNO , RQTS 8. OPQRST , GHIJKL

4 7

6

2

T

R S

W

U Vx

D C

A B4

5

6

M

P

N

Ox

3

2.4

F E

D

10

x

X Y

Z

15

14

O N

L M

11.5

14

Q R

ST x

9.8

20

20

9

15 15

Q

RxS

T

O

K

JI

H

G

L

P

5 i3; 3.6

10 i7; 8.1

3 i2; 9.3

4 i3; 12

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7-3 Reteaching Proving Triangles Similar

Problem

Are the triangles similar? How do you know? Write a similarity statement.

Given: DC 6 BA

Because DC 6 BA, /A and /D are alternate Compare the ratios of the

interior angles and are therefore >. Th e same lengths of sides:

is true for /B and /C. So, by AA , Postulate, ABXY 5

BCYZ 5

CAZX 5

32

nABX , nDCX . So, by SSS , Th eorem,

nABC , nXYZ.

Exercises

Determine whether the triangles are similar. If so, write a similarity statement and name the postulate or theorem you used. If not, explain.

1. 2. 3.

4. 5. 6.

7. Are all equilateral triangles similar? Explain.

8. Are all isosceles triangles similar? Explain.

9. Are all congruent triangles similar? Are all similar triangles

congruent? Explain.

D

C

B

AX

B

A C

6 7.5

9 X Z

Y

4 5

6

A B

C

1230

X

Y Z830 R

Q S

8 10

13

U

T V6

54E I

Q US O

A C

BS

TR

502020

110

N

M P70

70

4

K

J L2

3

6

AQ

RXB C

40

40

kABC M kZYX by the AA M Postulate. kQEU M kSIO by the

SSS M Theorem.

kABC M kRST by the AA M Postulate.

kBAC M kXQR by the SAS M Postulate.

Not similar; not all corresponding sides are proportional.

Not similar; the congruent angles are not corresponding.

No; the vertex angles may differ in measure.

All congruent triangles are similar, with ratio 1 i1. Similar triangles are not necessarily congruent.

Yes; by the SSS M Theorem or by the AA M Postulate

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7-3 Reteaching (continued)

Proving Triangles Similar

10. Provide the reason for each step in the two-column proof.

Given: LM ' MO

PN ' MO

Prove: nLMO , nPNO

Statements Reasons

1) LM ' MO, PN ' MO 1) 9

2) /PNO and /LMO are

right '. 2) 9

3) /PNO > /LMO 3) 9

4) /O > /O 4) 9

5) nLMO , nPNO 5) 9

11. Developing Proof Complete the proof by fi lling in

the blanks.

Given: AB 6 EF , AC 6 DF

Prove: nABC , nFED

Proof: AB 6 EF and AC 6 DF are given. EB is a

transversal by 9.

/E > /B by 9.

Similarly, /EDF > /BCA by 9.

So, nABC , nFED by 9.

12. Write a paragraph proof.

Given: AD and EC intersect at B.

Prove: nABE , nDBC

O

P

N M

L

A B

C

D

E F

12

15

10

5

4 3

E

A

CB

D

13

Given

Defi nition of a perpendicular line

Defi nition of a transversal

Answers may vary. Sample: BDAB 5

412 5

13 5

CBEB 5

515 5

13 5

CDEA 5

313

10. So,

BDAB 5

CBEB 5

CDAE . Therefore, the sides of kABE and

kDBC are proportional and kABE M kDBC,

by SSS M Theorem.

Alt. Ext. ' Thm.

All right ' are O.

Refl exive Property of O

AA M Postulate

AA M Postulate

Alt. Int. ' Thm.

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7-4 Reteaching Similarity in Right Triangles

Theorem 7-3

If you draw an altitude from the right angle to the hypotenuse of a right triangle, you create three similar triangles. Th is is Th eorem 7-3.

nFGH is a right triangle with right /FGH and the altitude of the hypotenuse JG. Th e two triangles formed by the altitude are similar to each other and similar to the original triangle. So, nFGH , nFJG , nGJH .

Two corollaries to Th eorem 7-3 relate the parts of the triangles formed by the altitude of the hypotenuse to each other by their geometric mean.

Th e geometric mean, x, of any two positive numbers a and b can be found with the proportion ax 5

xb.

Problem

What is the geometric mean of 8 and 12?

8x 5

x12

x2 5 96

x 5 Á96 5 Á16 ? 6 5 4Á6

Th e geometric mean of 8 and 12 is 4Á6.

Corollary 1 to Theorem 7-3

Th e altitude of the hypotenuse of a right triangle divides the hypotenuse into two segments. Th e length of the altitude is the geometric mean of these segments.

Since CD is the altitude of right nABC, it is the geometric mean of the segments of the hypotenuse AD and DB:

ADCD 5

CDDB .

FJ

HG

A

C B

D

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Corollary 2 to Theorem 7-3

Th e altitude of the hypotenuse of a right triangle divides the hypotenuse into two segments. Th e length of each leg of the original right triangle is the geometric mean of the length of the entire hypotenuse and the segment of the hypotenuse adjacent to the leg. To fi nd the value of x, you can write a proportion.

segment of hypotenuseadjacent leg 5

adjacent leghypotenuse 4

8 58

4 1 x Corollary 2

4(4 1 x) 5 64 Cross Products Property

16 1 4x 5 64 Simplify.

4x 5 48 Subtract 16 from each side.

x 5 12 Divide each side by 4.

Exercises

Write a similarity statement relating the three triangles in the diagram.

1. 2.

Algebra Find the geometric mean of each pair of numbers.

3. 2 and 8 4. 4 and 6 5. 8 and 10 6. 25 and 4

Use the fi gure to complete each proportion.

7. i

u5

fk 8. i

j 5j

u 9.

uf 5

f

u

10. Error Analysis A classmate writes the proportion 35 55

(3 1 b) to fi nd b. Explain why the proportion is incorrect and provide the right answer.

8

4x

P

N

T O

MF

H G

g

hi

j

k

f

3 5

b

7-4 Reteaching (continued)

Similarity in Right Triangles

kNOP M kTNP M kTONkFHG M kHMG M kFMH

f

4 102"6 4"5

The altitude is the geometric mean for the two segments of the hypotenuse, not for one segment and the entire hypotenuse. 35 5

5b

h

i

k

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7-5 ReteachingProportions in Triangles

Th e Side-Splitter Th eorem states the proportional relationship in a triangle in which a line is parallel to one side while intersecting the other two sides.

Theorem 7-4: Side-Splitter TheoremIn nABC, GH 6 AB. GH intersects BC and AC. Th e

segments of BC and AC are proportional: AGGC 5

BHHC

Th e corollary to the Side-Splitter Th eorem extends the proportion to three parallel lines intercepted by two transversals.

If AB 6 CD 6 EF , you can fi nd x using the proportion:

27 5

3x

2x 5 21 Cross Products Property

x 5 10.5 Solve for x.

Theorem 7-5: Triangle-Angle-Bisector TheoremWhen a ray bisects the angle of a triangle, it divides the opposite side into two segments that are proportional to the other two sides of the triangle.

In nDEF , EG bisects /E. Th e lengths of DG and GF are

proportional to their adjacent sides DE and EF : DGDE 5

GFEF .

To fi nd the value of x, use the proportion 36 5x8.

6x 5 24

x 5 4

Exercises

Use the fi gure at the right to complete each proportion.

1. uMN 5

SRLM 2. NO

u5

LMSR

3. MNRQ 5uQP 4.

SQLN 5

RP

uAlgebra Solve for x.

5. 6. 7.

A

B

CG

H

A

C

E F

D

B32

7 x

E D

G

F

6

3

x8

L M N O

PQ

RS

x3

46 12 x

68 x

53

2.5

RQ

QPNO

MO

2 9 1.5

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7-5 Reteaching (continued)

Proportions in Triangles

Algebra Solve for x.

8. 9. 10.

11. 12. 13.

In kABC, AB 5 6, BC 5 8, and AC 5 9.

14. Th e bisector of /A meets BC at point N. Find BN and CN.

15. XY 6 CA. Point X lies on BC such that BX 5 2, and Y is on BA. Find BY.

16. Error Analysis A classmate says you can use the Corollary to the Side-Splitter Th eorem to fi nd the value of x. Explain what iswrong with your classmate’s statement.

17. An angle bisector of a triangle divides the opposite side of the triangle into segments 6 and 4 in. long. Th e side of the triangle adjacent to the 6-in. segment is 9 in. long. How long is the third side of the triangle?

18. Draw a Diagram nGHI has angle bisector GM , and M is a point on HI . GH 5 4, HM 5 2, GI 5 9. Solve for MI. Use a drawing to help you fi nd the answer.

19. Th e lengths of the sides of a triangle are 7 mm, 24 mm, and 25 mm. Find the lengths to the nearest tenth of the segments into which the bisector of each angle divides the opposite side.

x3

1 1.5

x

32

4

x

64

4.8

2x

15

12.5x 2

2x 2

x

8

10.5x 1

x 2

129

B

AC

x15

12

3

G

HIM

4 9

2

4.5

5.6 mm and 19.4 mm; 3.4 mm and 3.6 mm; 5.3 mm and 18.8 mm

BN 5 3 15, CN 5 4

45

1.5

2 23

7.24.5

6 in.

3 56

The corollary states that the segments on the transversal, not the segments on the parallel lines, are proportional.

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8-1 Reteaching The Pythagorean Theorem and Its Converse

Th e Pythagorean Th eorem can be used to fi nd the length of a side of a right triangle.

Pythagorean Th eorem: a2 1 b2 5 c2, where a and b are the legs of a right triangle, and c is the hypotenuse.

Problem

What is the value of g? Leave your answer in simplest radical form.

Using the Pythagorean Th eorem, substitute g and 9 for the legs and 13 for the hypotenuse.

a2 1 b2 5 c2

g2 1 92 5 132 Substitute.

g2 1 81 5 169 Simplify.

g2 5 88 Subtract 81 from each side.

g 5 "88 Take the square root.

g 5 "4(22) Simplify.

g 5 2"22

Th e length of the leg, g , is 2"22.

Exercises

Identify the values of a, b, and c. Write ? for unknown values. Th en, fi nd the missing side lengths. Leave your answers in simplest radical form.

1. 2. 3. 4.

5. A square has side length 9 in. What is the length of the longest line segment that can be drawn between any two points of the square?

6. Right nABC has legs of lengths 4 ft and 7 ft. What is the length of the triangle’s hypotenuse?

7. Televisions are sold by the length of the diagonal across the screen. If a new 48-in. television screen is 42 in. wide, how tall is the screen to the nearest inch?

13

g9

8

47

710

6 345

9"2 in.

"65 ft

23 in.

a 5 4, b 5 8,c 5 ?; 4"5 a 5 7, b 5 7, c 5 ?; 7"2

a 5 3, b 5 ?,c 5"45 ; 6a 5 6, b 5 ?,

c 5 10; 8

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Use Th eorems 8-3 and 8-4 to determine whether a triangle is acute or obtuse.

Let a and b represent the shorter sides of a triangle and c represent the longest side.

If a2 1 b2 . c2, then the triangle is acute

If a2 1 b2 , c2, then the triangle is obtuse.

Problem

A triangle has side lengths 6, 8, and 11. Is the triangle acute, obtuse, or right?

a 5 6, b 5 8, c 5 11 Identify a, b, and c.

a2 1 b2 5 62 1 82 Substitute to fi nd a2 1 b2.

a2 1 b2 5 36 1 64

a2 1 b2 5 100

c2 5 112 Substitute to fi nd c2.

c2 5 121

100 , 121 Compare a2 1 b2 and c2.

a2 1 b2 , c2, so the triangle is obtuse.

Exercises

Th e lengths of the sides of a triangle are given. Classify each triangle as acute, right, or obtuse.

8. 7, 9, 10 9. 18, 16, 24 10. 3, 5, 5"2

11. 10, 10, 10"2 12. 8, 6, 10.5 13. 7, 7"3, 14

14. 22, 13, 23 15. 17, 19, 26 16. 21, 28, 35

17. A local park in the shape of a triangle is being redesigned. Th e fencing around the park is made of three sections. Th e lengths of the sections of fence are 27 m, 36 m, and 46 m. Th e designer of the park says that this triangle is a right triangle. Is he correct? Explain.

18. Your neighbor’s yard is in the shape of a triangle, with dimensions 120 ft, 84 ft, and 85 ft. Is the yard an acute, obtuse, or a right triangle? Explain.

8-1 Reteaching (continued)

The Pythagorean Theorem and Its Converse

acute

obtuse

obtuse

obtuse

right right

rightacute

acute

No; the triangle is obtuse; 272 1 362 5 2025 and 462 5 2116. So, by the Pythagorean Inequality Theorems, this triangle is obtuse, because a2 1 b2

R c2.

Obtuse; 842 1 852R 1202

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8-2 Reteaching Special Right Triangles

In a 458-458-908 triangle, the legs are the same length.

hypotenuse 5 "2 3 leg

Problem

What is the value of the variable, s?

10 5 s"2 In a 458-458-908 triangle, the hypotenuse is "2 times the length of the leg.

s 5 10

"2 Divide both sides by "2.

10

"2?"2

"25

10"22 5 5"2 Rationalize the denominator.

s 5 5"2

In a 458-458-908 triangle the length of the leg is "2

2 3 hypotenuse.

Exercises

Complete each exercise.

1. Draw a horizontal line segment on centimeter grid paper so that

the endpoints are at the intersections of grid lines.

2. Use a protractor and a straightedge to construct a 458-458-908 triangle.

3. Use the 458-458-908 Triangle Th eorem to calculate the lengths of the legs.

Round to the nearest tenth.

4. Measure the lengths of the legs to the nearest tenth of a centimeter.

Compare your calculated results and your measured results.

Use the diagrams below each exercise to complete Exercises 5–7.

5. Find the length of the leg

of the triangle.

24

4545

6. Find the length of the

hypotenuse of the triangle.

7

45

45

7. Find the length of the leg

of the triangle.

16

45

45

10

s

4545

12"27"2

8"2

Sample at the right. Check students’ work.

Sample: hypotenuse 5 9 cm; leg 5 6.4 cm

They are the same.

45 45

Check students’ work.

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In a 308-608-908 triangle, the longer leg is opposite the 608 angle and the shorter

leg is opposite the 308 angle.

longer leg 5 "3 3 shorter leg

hypotenuse 5 2 3 shorter leg

Problem

Find the value of each variable.

5 5 "3 s In a 308-608-908 triangle the length of the longer leg is "3 times the length of the shorter leg.

5

"35 s Divide both side by "3 .

s 5 5

"3?"3

"35

5"3

3 Rationalize the denominator.

Th e length of the hypotenuse is twice the length of the shorter leg.

t 5 2a5"33 b 5 10"3

3

Exercises

Complete each exercise.

8. Draw a horizontal line segment on centimeter grid paper so that the

endpoints are at the intersections of grid lines.

9. Use a protractor and a straightedge to construct a 308-608-908 triangle with

your segment as one of its sides.

10. Use the 308-608-908 Triangle Th eorem to calculate the lengths of the other

two sides. Round to the nearest tenth.

11. Measure the lengths of the sides to the nearest tenth of a centimeter.

12. Compare your calculated results with your measured results.

13. Repeat the activity with a diff erent segment.

For Exercises 14–17, fi nd the value of each variable.

14. 15. 16. 17.

5

t60

30s

2d60

30e 4

g60

30f

6

x

60

30y

12

v

60 60

u

8-2 Reteaching (continued)

Special Right Triangles

Sample:

Check students’ work.

Sample: 3.5 and 7

Sample: 3.5 and 7

Sample: They are approximately equal.

Check students’ work.

d 5 2"3; e 5 4 f 5 4"33 , g 5 8"3

3x 5 3"3; y 5 3 u 5 6"3; v 5 12

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8-3 Reteaching Trigonometry

Use trigonometric ratios to fi nd the length of a side of a right triangle.

sin A 5opposite

hypotenuse

cos A 5adjacent

hypotenuse

tan A 5opposite

adjacent

Problem

What is the value of x to the nearest tenth?

First, identify the information given.

Th e angle measure is 29. Th e length of the side opposite the angle is x. Th e length of the hypotenuse is 13.

sin 298 5opposite

hypotenuse Use the sine ratio.

sin 298 5 x13 Substitute.

13(sin 298) 5 x Multiply by 13.

6.3 < x Solve for x using a calculator.

Exercises

Find the value of t to the nearest tenth.

1. 2.

Find the missing lengths in each right triangle. Round your answers to the nearest tenth.

3. 4. 5.

hyp

B

C A

opp

adj

29

13x

44

22

t

4610.5

t

8D

FE 205

X Z

Y60

50ST

10

R

15.8 10.9

2.7; 7.5 8.7; 10 8.4; 13.1

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When you know the length of one or more sides in a right triangle and are looking for the angle measures of the triangle, you should use inverse trigonometric ratios.

sin21(x) is the measure of the angle where opposite

hypotenuse5 x.

Similarly, cos21(x) is the measure of the angle where adjacent

hypotenuse5 x, and

tan21(x) is the measure of the angle where opposite

adjacent5 x.

Problem

Find the measure of /T to the nearest degree.

First, identify the information given. Th e length of the side adjacent to the angle is 33. Th e length of the hypotenuse is 55.

cos T 5adjacent

hypotenuse Use the cosine ratio.

cos T 5 3355 5 0.6 Fill in known information.

T 5 cos21(0.6) Use the inverse of the cosine ratio.

T < 538 Use a calculator to solve.

Th e measure of /T is about 53.

Exercises

Find mlM to the nearest degree.

6. 7. 8.

9. 10. 11.

55

33

X

T Z

14

4N M

O

52

M L

K V

M S

60 48

X

M

Y50

37.5MV

P140

63

M

C

G

125

41.25

8-3 Reteaching (continued)

Trigonometry

73 24 53

41 27 18

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8-4 Reteaching Angles of Elevation and Depression

Angle of ElevationSuppose you are looking up at an airplane. Th e angle formed by a

horizontal line and your line of sight to the airplane is called the

angle of elevation.

Angle of DepressionNow suppose you are standing on a cliff and looking down at a river

below. Th e line stretches horizontally from your point of view on the

cliff . Your angle of sight to the river below forms an angle of depression with the horizontal line.

You can use your knowledge of trigonometric ratios to determine

distances and lengths using angles of elevation and depression.

Using the Angle of Elevation

Problem

Suppose you are looking up at the top of a building. Th e angle

formed by your line of sight and a horizontal line is 358. You are

standing 80 ft from the building and your eyes are 4 ft above the

ground. How tall is the building, to the nearest foot?

Look at the diagram and think about what you know. You can see

that a right triangle is formed by a horizontal line, your line of

sight, and the building. You know an angle and one length.

Remember: tan A 5opposite length

adjacent length

Let the opposite length be x.

tan 35 5x

80

80 tan 35 5 x

x < 56 ft

Your eyes are 4 ft above the ground, so add 4 to the value of x to fi nd the total

height of the building: 56 ft 1 4 ft 5 60 ft.

horizon line

line of sight

horizon line

Line of sight

35x

80 ft4 ft 4

#

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Using the Angle of Depression

Problem

Suppose you are a lifeguard looking down at a swimmer in

a swimming pool. Your line of sight forms a 558 angle with

a horizontal line. You are 10 ft up in your seat. How far is

the swimmer from the base of the lifeguard stand?

Look at the diagram and think about what you know.

You can see that a right triangle is formed by the

horizon line, your line of sight, and a vertical distance

that is the same as your height in the seat. You know an

angle and one length.

Remember: tan A 5opposite length

adjacent length

Let the unknown side length be x.

tan 55 510x

x 5 10tan 55

x < 7 ft

Th e swimmer is 7 ft from the base of the lifeguard stand.

Exercises

Find the value of x. Round the lengths to the nearest tenth of a unit.

1. 2.

3. 4.

5510 ft

x

45

30 ft5 ft

x

#

8

40

900 ft

x

42

11 m

x 38

500 ydx

8-4 Reteaching (continued)

Angles of Elevation and Depression

35 ft 1072.6 ft

12.2 m 307.8 yd

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8-5 Reteaching Vectors

A vector is a quantity with size (magnitude) and direction.

For example, you could represent a car traveling due north

at 30 mi/h for 3 h with a vector on a coordinate grid.

Th e y-axis is the north-south direction. Draw the vector with

its tail at the origin and its head at the point that represents the

completion of the 3-h trip. Th e vector has a direction, due north,

and a magnitude, 90 mi.

Because vectors have magnitude and direction, you can

determine both of these values for a vector.

Direction of a Vector You can use a compass arrangement on the coordinate grid

to describe a vector’s direction.

EW

N

S

30

Th is vector is 308

south of east.

EW

N

S

40

Th is vector is 408

east of north.

EW

N

S

20

Th is vector is 208

north of west.

Exercises

Sketch a vector with the given direction.

1. 408 north of east 2. 308 east of south 3. 508 north of west

Use compass directions to describe the direction of each vector.

4. 5. 6.

EW

N

S

90

E

N

W

S20

E

N

W

S

38 E

N

W

S29

E

N

W

S

40 E

N

W

S30

E

N

W

S

50

208 south of west or 708 west of south

388 east of north or 528 north of east

298 south of east or 618 east of south

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8-5 Reteaching (continued)

Vectors

Magnitude of a VectorTh e magnitude of a vector is its length. You can use the distance formula to

determine the length, or magnitude, of a vector.

Problem

What is the magnitude of each vector?

4 2 2 4

2

2

4

4

x

y(4, 5)

Th e vector is from the origin to (4, 5).

"(4 2 0)2 1 (5 2 0)2 5 "41

< 6.4

Th e magnitude is 6.4.

4 2 2 4

2

4

4

x

y

2

( 3, 4)

Th e vector is from the origin to (23, 24).

"(23 2 0)2 1 (24 2 0)2 5 "25

5 5

Th e magnitude is 5.

You can indicate a vector by using an ordered pair. For example, k5, 7l is a vector

with its tail at the origin and its head at (5, 7).

Exercises

Find the magnitude of the given vector to the nearest tenth.

7. k2, 3l 8. k28, 2l 9. k23, 3l

10. 11. 12.

Find the magnitude and direction of each vector. Round to the nearest tenth.

13. k212, 5l 14. k15, 28l 15. k27, 224l

E

N

W

S

O4 2 2 4

2

2

4

4

4 2 2 4

2

4

4

E

N

2W

S

O

(1, 4)

4 2 2 4

2

4

E

N

S

O

( 2, 4) 4

2W

3.6 8.2 4.2

4.54.15.7

13; 22.68 north of west or 67.48 west of north

17; 28.18 south of east or 61.98 east of south

25; 16.38 west of south or 73.78 south of west

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9-1 Reteaching Translations

A translation is a type of transformation. In a translation, a geometric fi gure changes position, but does not change shape or size. Th e original fi gure is called the preimage and the fi gure following transformation is the image.

Th e diagram at the right shows a translation in the coordinate plane. Th e preimage is nABC. Th e image is nArBrCr.

Each point of nABC has moved 5 units left and 2 units up. Moving left is in the negative x direction, and moving up is in the positive y direction. So, the rule for the translation is (x, y) u (x 2 5, y 1 2).

All translations are isometries because the image and the preimage are congruent. In this case, nABC > nArBrCr.

Problem

What are the images of the vertices of WXYZ for the translation (x, y) u (x 1 5, y 2 1)? Graph the image of WXYZ.

W(24, 1) S (24 1 5, 1 2 1), or Wr(1, 0)

X(24, 4) S (24 1 5, 4 2 1), or Xr(1, 3)

Y(21, 4) S (21 1 5, 4 2 1), or Yr(4, 3)

Z(21, 1) S (21 1 5, 1 2 1), or Zr(4, 0)

Exercises

Use the rule to fi nd the images of the vertices for the translation.

1. nMNO (x, y) u (x 1 2, y 2 3) 2. square JKLM (x, y) u (x 2 1, y)

A

CB

A

CB

2

24

2

4

y

x

X Y

W Z

X Y

W Z

2

24

y

M

O

N

3 1

2

2

y

x

J K

LM3 3

3

3y

x

M9(0, 1), N9(21, 22), O9(1, 23) J9(22, 2), K9(1, 2), L9(1, 21), M9(22, 21)

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9-1 Reteaching (continued)

Translations

Problem

What rule describes the translation of ABCD to ArBrCrDr?

To get from A to Ar (or from B to Br or C to Cr or D to Dr), you move 8 units left and 7 units down. Th e rule that describes this translation is (x, y) u (x 2 8, y 2 7).

Exercises• On graph paper, draw the x- and y-axes, and label

Quadrants I–IV.

• Draw a quadrilateral in Quadrant I. Make sure that the vertices are on the intersection of grid lines.

• Trace the quadrilateral, and cut out the copy.

• Use the cutout to translate the fi gure to each of the other three quadrants.

Name the rule that describes each translation of your quadrilateral.

3. from Quadrant I to Quadrant II

4. from Quadrant I to Quadrant III

5. from Quadrant I to Quadrant IV

6. from Quadrant II to Quadrant III

7. from Quadrant III to Quadrant I

Refer to ABCD in the problem above.

8. Give the image of the vertices of ABCD under the translation (x, y) u (x 2 2, y 2 5).

9. Give the image of the vertices of ABCD under the translation (x, y) u (x 1 2, y 2 4).

10. Give the image of the vertices of ABCD under the translation (x, y) u (x 1 1, y 1 3).

2

2

2

4

4

6

246 4 6

B

C

AD

A

B

C

D

y

xO

x

y

Check students’ work.

Check students’ work.

Check students’ work.

Check students’ work.

A9(0, 23), B9(1, 1), C9(4, 21), D9(5, 24)

A9(4, 22), B9(5, 2), C9(8, 0), D9(9, 23)

A9(3, 5), B9(4, 9), C9(7, 7), D9(8, 4)

Check students’ work.

Quadrant II Quadrant I

Quadrant III Quadrant IV

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9-2 Reteaching Refl ections

A refl ection is a type of transformation in which a geometric fi gure is fl ipped over a

line of refl ection.

In a refl ection, a preimage and an image have opposite orientations, but are the

same shape and size. Because the preimage and image are congruent, a refl ection

is an isometry.

Problem

What are the refl ection images of nMNO across

x- and y-axes? Give the coordinates of the vertices

of the images.

From the graph you can see that the refl ection image of nMNO across the x-axis

has vertices at (2, 23), (3, 27), and (5, 24). Th e refl ection image of nMNO

across the y-axis has vertices at (22, 3), (23, 7), and (25, 4).

Exercises

Use a sheet of graph paper to complete Exercises 1–5.

1. Draw and label the x- and y-axes on a sheet of graph paper.

2. Draw a scalene triangle in one of the four quadrants. Make sure that the

vertices are on the intersection of grid lines.

3. Fold the paper along the axes.

4. Cut out the triangle, and unfold the paper.

5. Label the coordinates of the vertices of the refl ection images across the x- and y-axes.

2

4

6

2 4 6x

O

y

M

N

O

fold

fold

Fold the paper along

the x-axis and y-axis.

Cut out the

triangle. Unfold the paper.

Copy the fi gure onto

a piece of paper.

Check students’ work.

Check students’ work.

Check students’ work.

Check students’ work.

Check students’ work.

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9-2 Reteaching (continued)

Refl ections

To graph a refl ection image on a coordinate plane, graph the images of each

vertex. Each vertex in the image must be the same distance from the line of

refl ection as the corresponding vertex in the preimage.

Refl ection across the x-axis:

(x, y) u (x, 2y)

Th e x-coordinate does not change.

Th e y-coordinate tells the distance

from the x-axis.

Refl ection across the y-axis:

(x, y) u (2x, y)

Th e y-coordinate does not

change.

Th e x-coordinate tells the

distance from the y-axis.

Problem

nABC has vertices at A(2, 4), B(6, 4), and C(3, 1). What is the

image of nABC refl ected over the x-axis?

Step 1: Graph Ar, the image of A. It is the same distance from

the x-axis as A. Th e distance from the y-axis has not

changed. Th e coordinates for Ar are (2, 24).

Step 2: Graph Br. It is the same distance from the x-axis as B. Th e distance from the y-axis has not changed. Th e

coordinates for Br are (6, 24).

Step 3: Graph Cr. It is the same distance from the x-axis as C. Th e coordinates for Cr are (3, 21).

Each fi gure is refl ected across the line indicated. Find the coordinates of the vertices for each image.

6. nFGH with vertices F(21, 3), G(25, 1), and H(23, 5) reflected across x–axis

7. nCDE with vertices C(2, 4), D(5, 2), and E(6, 3) reflected across x–axis

8. nJKL with vertices J(21, 25), K(22, 23), and L(24, 26) reflected

across y–axis

9. Quadrilateral WXYZ with vertices W(23, 4), X(24, 6), Y(22, 6), and Z(21, 4)

reflected across y–axis

B

4 62

2

2

4

4

x

yA

C

B

62

2

2

4

4

x

y

A

C

F9(21, 23) , G9(25, 21) , H9(23, 25)

C9(2, 24) , D9(5, 22) , E9(6, 23)

J9(1, 25) , K9(2, 23) , L9(4, 26)

W9(3, 4) , X9(4, 6) , Y9(2, 6) , Z9(1, 4)

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9-3 ReteachingRotations

A turning of a geometric fi gure about a point is a rotation. Th e center of rotation is

the point about which the fi gure is turned. Th e number of degrees the fi gure turns

is the angle of rotation. (In this chapter, rotations are counterclockwise unless

otherwise noted.)

A rotation is an isometry. Th e image and preimage are congruent.

ABCD is rotated about Z. ABCD is the preimage and ArBrCrDr is the image. Th e center of rotation is point Z. Th e angle of

rotation is 828.

A composition of rotations is two or more rotations in

combination. If the center of rotation is the same, the measures

for the angles of rotation can be added to fi nd the total rotation

of the combination.

Exercises

Complete the following steps to draw the image of kXYZ under an 808 rotation about point T .

1. Draw /XTXr so that m/XTXr 5 80 and TX > TXr.

2. Draw /ZTZr so that m/ZTZr 5 80 and TZ > TZr.

3. Draw /YTYr so that m/YTYr 5 80 and TY > TYr.

4. Draw XrZr, XrYr, and YrZr to complete nXrYrZr.

Copy kXYZ to complete Exercises 5–7.

5. Draw the image of nXYZ under a 1208 rotation about T .

6. Draw a point S inside nXYZ. Draw the image of nXYZ

under a 1358 rotation about S.

7. Draw the image of nXYZ under a 908 rotation about Y .

A

BC

D

A B

CD

Z82

Y

X

T

ZY

X ZX

YZ

T

Y

X Z

XY

Z

T

Y

X Z

X

Z

Y

SY

X Z

X

Z

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In a regular polygon, the center is the same distance from every vertex. Regular

polygons can be divided into a number of congruent triangles. Th e number of

triangles is the same as the number of sides of the polygon. Th e measure of each

central angle (formed by one vertex, the center, and an adjacent vertex)

is 360

number of congruent triangles.

In regular hexagon MNOPQR, the center and the vertices can be

used to divide the hexagon into six congruent triangles. Th e

measure of each central angle is 360

6 , or 60.

Problem

In regular pentagon QRSTU, what is the image of point Q for a

rotation of 1448 about point Z?

First fi nd the measure of the central angle of a regular pentagon.

m/RZS 5 3605 5 72

When Q rotates 728, it moves one vertex counterclockwise. When

Q rotates 1448, it moves two vertices counterclockwise. So, for a

rotation of 1448, the image of Q is point T .

Exercises

Point Z is the center of regular pentagon QRSTU. Find the image of the given point or segment for the given rotation.

8. 2168 rotation of S about Z 9. 1448 rotation of TU about Z

10. 3608 rotation of Q about Z 11. 2888 rotation of R about Z

12. What is the measure of the angle of rotation that maps T onto U?

(Hint: How many vertices away from T is U, counterclockwise?)

13. What is the measure of the angle of rotation for the regular hexagon ABCDEF

that maps A onto C?

14. What is the measure of the angle of rotation for the regular octagon DEFGHIJK

that maps F onto K?

M N

O

PQ

R

60

S

ZR

Q U

T

9-3 Reteaching (continued)

Rotations

U

Q

RS

S

2888

1208 or 2408

1358 or 2258

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9-4 Reteaching Symmetry

A fi gure with symmetry can be refl ected onto itself or rotated onto itself.

Line Symmetry Line symmetry is also called refl ectional symmetry. If a fi gure has line symmetry,

it has a line of symmetry that divides the fi gure into two congruent halves.

A fi gure may have one or more than one line of symmetry.

Th e heart has line symmetry. Th e dashed line is its one line

of symmetry.

Rotational and Point Symmetry If a fi gure has rotational symmetry, there is a rotation of

one-half turn (1808) or less for which the fi gure is its own image.

Th e star at the right has rotational symmetry. Th e smallest angle

needed for the star to rotate onto itself is 458. Th is is the angle

of rotation.

A fi gure has point symmetry if a 1808 rotation maps the fi gure

onto itself. Th e letter S has point symmetry.

Three-Dimensional SymmetryTh e regular hexagonal prism has refl ectional symmetry in

a plane; the plane divides the prism into two congruent halves.

Th e regular hexagonal prism also has rotational symmetry.

Any rotation that is a multiple of 608 around the dashed line

will map the prism onto itself.

45

S

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Consider the following types of symmetry: rotational, point, and (line) refl ectional.

Problem

What types of symmetry does the fl ag have?

Th e fl ag has four lines of symmetry shown by the dotted lines.

It has 908 rotational symmetry and point symmetry about its center.

Exercises

Describe the symmetries in each fl ag.

1. 2.

3. 4.

5. 6.

Tell whether each three-dimensional object has refl ectional symmetry about a plane, rotational symmetry about a line, or both.

7. a pen 8. a juice box

9. a candle 10. a sofa

Switzerland

IsraelSouth Africa

Canada United Kingdom

Honduras

Somalia

9-4 Reteaching (continued)

Symmetry

two lines of symmetry (vertical and horizontal), 1808 rotational symmetry (point symmetry)

one line of symmetry (vertical)

one line of symmetry (vertical)

both

both

refl ectional symmetry in a plane

refl ectional symmetry in a plane

one line of symmetry (horizontal)

two lines of symmetry (vertical and horizontal), 1808 rotational symmetry (point symmetry)

one line of symmetry (vertical)

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9-5 Reteaching Dilations

A dilation is a transformation in which a fi gure changes size. Th e preimage and image of a dilation are similar. Th e scale factor of the dilation is the same as the scale factor of these similar fi gures.

To fi nd the scale factor, use the ratio of lengths of corresponding sides. If the scale factor of a dilation is greater than 1, the dilation is an enlargement. If it is less than 1, the dilation is a reduction.

ExamplenXrYrZr is the dilation image of nXYZ. Th e center of dilation is X. Th e image of the center is itself, so Xr = X.

Th e scale factor, n, is the ratio of the lengths of corresponding sides.

n 5 XrZrXZ 5

3012 5

52

Th is dilation is an enlargement with a scale factor of 52.

Exercises

For each of the dilations below, A is the center of dilation. Tell whether the dilation is a reduction or an enlargement. Th en fi nd the scale factor of the dilation.

1. 2.

3. AB 5 2; ArBr 5 3 4. DE 5 3; DrEr 5 6

5. Th e image of a mosquito under a magnifying glass is six times the mosquito’s actual size and has a length of 3 cm. Find the actual length of the mosquito.

X X

12

18

Z

Y Y

Z

10

BB

CC

8

A A

10

A A

BB

CC

8

A A

B

B

D

C

D

C

B

D

A

B

CE

D

CE

enlargement; 54enlargement; 32 enlargement; 2

reduction; 45

0.5 cm

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9-5 Reteaching (continued)

Dilations

Problem

Quadrilateral ABCD has vertices A(22, 0), B(0, 2), C(2, 0), and D(0, 22).Find the image of ABCD under the dilation centered at the origin with scale factor 2. Th en graph ABCD and its image.

To fi nd the image of the vertices of ABCD, multiply the x-coordinates and y-coordinates by 2.

A(22, 0)u Ar(24, 0) B(0, 2)u Br(0, 4)

C(2, 0)u Cr(4, 0) D(0, 22)uDr(0, 24)

Exercises

Use graph paper to complete Exercise 6.

6. a. Draw a quadrilateral in the coordinate plane.

b. Draw the image of the quadrilateral under dilations centered at the origin with scale factors 12, 2, and 4.

Graph the image of each fi gure under a dilation centered at the origin with the given scale factor.

7. 8.

9. 10.

4 2 2

2

4

x

y

OA A

B

C

D

B

C

D

4 4

4

4

x

y

O

scale factor 2 scale factor 12

4 2 2 4

2

2

4

4

x

y

O

4 2 2 4

4

4

x

y

O

scale factor 23 scale factor 3

2

4 2 2 4

4

4

2

x

y

O

Check students’ work.

Check students’ graphs.

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9-6 Reteaching Compositions of Refl ections

Two congruent fi gures in a plane can be mapped onto one another by a single

refl ection, compositions of refl ections, or glide refl ections.

Compositions of two refl ections may be either translations or rotations.

If a fi gure is refl ected across two parallel lines, it is a translation.

If a fi gure is refl ected across intersecting lines, it is a rotation.

For both translations and rotations, the preimage and the image have the

same orientation.

Th e arrow is refl ected fi rst across line / and then across

line m. Lines / and m are parallel. Th ese two refl ections

are equivalent to translation of the arrow downward.

Th e triangle is refl ected fi rst across line / and

then across line m. Lines / and m intersect at

point X. Th ese two refl ections are equivalent

to a rotation. Th e center of rotation is X and

the angle of rotation is twice the angle of

intersection, in this case 2 3 908, or 1808.

A composition of a translation and a refl ection is a glide refl ection. For both

refl ections and glide refl ections, the image and the preimage have opposite

orientations.

nNrOrPr is the image of nNOP, for a glide refl ection

where the translation is (x, y) S (x 1 4, y 2 1) and

the line of refl ection is y 5 21.

m

<

BA

C

A

CB

m

X<

Reflection

Glide

2

42

4

2N N

O

O

P

P

x

y

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9-6 Reteaching (continued)

Compositions of Refl ections

Problem

What transformation maps the fi gure ABCD onto the

fi gure EFGH shown at the right?

Th e transformation is a glide refl ection. It involves a translation, or glide, followed

by a refl ection in a line parallel to the translation vector.

Exercises• Draw two pairs of parallel lines that intersect as shown at the right.

• Draw a nonregular quadrilateral in the center of the four lines.

• Use paper folding and tracing to refl ect the fi gure and its

images so that there is a fi gure in each of the nine sections.

• Label the fi gures 1 through 9 as shown.

Describe a transformation that maps each of the following.

1. fi gure 4 onto fi gure 6 2. fi gure 1 onto fi gure 2

3. fi gure 7 onto fi gure 5 4. fi gure 2 onto fi gure 9

5. fi gure 1 onto fi gure 5 6. fi gure 6 onto fi gure 7

7. fi gure 8 onto fi gure 9 8. fi gure 2 onto fi gure 8

P(2, 3) m P9 by a glide refl ection with the given translation and line of refl ection. What are the coordinates of P9? (Hint: it may help to graph the transformations.)

9. (x, y) S (x 1 3, y 2 2); y 5 0 10. (x, y) S (x 2 4, y 1 2); x 5 0

11. (x, y) S (x, y 2 3); y 5 x 12. (x, y) S (x 2 2, y 2 3); y 5 4

A

B

CD

E

F

GHA

B

CD

E

F

GH

12 3

45

6

7 89

translation

translation

refl ection

refl ection

rotation

rotation

glide refl ection

glide refl ection

(5, 21) (2, 5)

(0, 2) (0, 8)

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9-7 Reteaching Tessellations

Tessellations are repeating patterns of fi gures that completely cover a plane

without overlaps or gaps. A tessellation can be made by congruent copies of

one fi gure or by congruent copies of multiple shapes.

If the fi gures in a tessellation are polygons, the sum of the measures of the angles

around any vertex is 3608. To determine if a regular polygon will tessellate,

fi nd the measure of each angle of the polygon. If 360 is a multiple of this measure,

the fi gure will tessellate.

A tessellation is shown at the right. Th e repeating

fi gure in the tessellation is shown below.

Problem

Will a regular 24-gon tessellate a plane?

a 5 the measure of one angle of a 24-gon (all angles of a regular polygon

are congruent)

a 5180(n 2 2)

n Polygon Angle-Sum Theorem

a 5180(24 2 2)

24 Substitute 24 for n.

a 5 165 Solve.

No multiple of 165 is equal to 360. A regular 24-gon will not tessellate.

Exercises

Find the repeating fi gure for each tessellation.

1. 2.

3. Will a regular 22-gon tessellate a plane? Explain.No; a regular 22-gon has angles that measure about 163.6°, which is not a factor of 360.

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Tessellations have symmetry. Th e types of possible symmetry are refl ectional

(or line) symmetry, rotational symmetry, translational symmetry, and glide

refl ectional symmetry. A tessellation may have more than one type of symmetry.

Problem

What are the symmetries of the tessellation below?

Th e tessellation has line symmetry as shown by the

dotted lines. It has rotational symmetry about the

points shown. It has translational symmetry and

glide refl ection symmetry.

Exercises

Copy the fi gure at the right onto stiff paper or cardboard. Th en cut it out.

4. Use the cutout to make a tessellation.

5. List the symmetries of the tessellation.

List the symmetries of each tessellation.

6. 7. 8.

9-7 Reteaching (continued)

Tessellations

line symmetry, rotational symmetry, translational symmetry, glide refl ectional symmetry

line symmetry, rotational symmetry, translational symmetry, glide refl ectional symmetry

rotational symmetry, translational symmetry

line symmetry, rotational symmetry, translational symmetry, glide refl ectional symmetry

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10-1 Reteaching

Areas of Parallelograms and Triangles

Th e area of a parallelogram is base 3 height. Th e base can be any side of the parallelogram. Th e height is the length of the corresponding altitude.

Problem

What is the area of ~ABCD? AB is the correct base to use for the given altitude.

A 5 bh Substitute and simplify.

A 5 8(16) 5 128 in.2

Problem

What is the value of x?

Step 1: Find the area of the parallelogram using the altitude perpendicular to LM .

A 5 bh Substitute and simplify.

A 5 9(3) 5 27 m2

Step 2: Use the area of the parallelogram to fi nd the value of x.

A 5 bh Substitute.

27 5 4.5x Simplify.

x 5 6 m

Exercises

Find the area of each parallelogram.

1. 2. 3.

Find the value of h for each parallelogram.

4. 5. 6.

A B

CD

8 in.

16 in.

10 in.

L M

N

x

3 m 4.5 m

9 m

P

9 ft

7 ft 8 ft

3.5 mm

12 mm13 mm

5.5 m

10 m

4 m

h

12 yd

7.5 yd

15 yd

16 m

h

9 m

4 m

h

14 mi

5 mi

9.8 mi

63 ft2

6 yd2.25 m 3.5 mi

42 mm2

40 m2

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Th e area of a triangle is 12 3 base 3 height. Th e base is any side of the triangle. Th e height is the length of the corresponding altitude.

Problem

A triangle has an area of 18 in.2. Th e length of its base is 6 in. What is the corresponding height?

Draw a sketch. Th en substitute into the area formula, and solve for h.

A 5 12bh Substitute.

18 5 12(6)h 5 3h Simplify.

h 5 6 in.

Exercises

7. Use graph paper. Draw an obtuse, an acute, and a right triangle, each with an area of 12 square units. Label the base and height of each triangle.

8. Draw a diff erent obtuse, acute, and right triangle, each with an area of 12 square units. Label the base and height of each triangle.

9. A triangle has height 5 cm and base length 8 cm. Find its area.

10. A triangle has height 11 in. and base length 10 in. Find its area.

11. A triangle has area 24 m2 and base length 8 m. Find its height.

12. A triangle has area 16 ft2 and height 4 ft. Find its base.

Find the area of each triangle.

13. 14. 15.

16. Th e fi gure at the right consists of a parallelogram and a triangle. What is the area of the fi gure?

6 in.

h

12 cm

15 cm12.2 in.9.2 in.

6 in. 10 in.

7 in.14 ft

11 ft

2.5 ft

18 m

50 m

40 m

22 m

10-1 Reteaching (continued)

Areas of Parallelograms and Triangles

90 cm2 56 in.2 13.75 ft2

1170 m2

20 cm2

Sample:

Samples:

55 in.2

6 m

8 ft

4

6

4

6

4

6

4

6

3

8

38

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10-2 Reteaching

Areas of Trapezoids, Rhombuses, and Kites

Th e area of a trapezoid is 12h(b1 1 b2), where h is the length of the height and

b1 and b2 are the lengths of the two parallel bases.

Problem

What is the area of trapezoid WXYZ?

Draw an altitude to divide the trapezoid into a rectangle and

a 308-608-908 triangle. In a 308-608-908 triangle, the length of the

longer leg is "3

2 times the length of the hypotenuse.

h 5 "32 (8) 5 4"3 in.

Use the formula for the area of a trapezoid.

A 5 12h(b1 1 b2) Substitute.

512 (4"3 )(12 1 16) Simplify.

5 56"3 in.2

Th e area of trapezoid WXYZ is 56"3 in.2

Find the area of each trapezoid. If necessary, leave your answer in simplest radical form.

1. 2. 3.

4. 5. 6.

7. Find the area of a trapezoid with bases 9 m and 12 m and height 6 m.

8. Find the area of a trapezoid with bases 7 in. and 11 in. and height 3 in.

9. Find the area of a trapezoid with bases 14 in. and 5 in. and height 11 in.

W X

YZ

12 in.

16 in.

8 in.

60

W X

YZ

12 in.

16 in.

8 in.60

4 3 in.

15 in.

17 in.

7 in.

12 ft

16 ft

4.5 ft5 in.

3 in.8 in.

45

10 m

5 m

7 m

4 m

60

30

24 cm

D C

20 cm

112 in.2

75 m2

63 ft2

16 "3 m2

19.5 in.2

(240 2 50"3) cm2

63 m2

27 in.2

104.5 in.2

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Th e area of a rhombus or a kite is 12d1d2, where d1 and d2 are the lengths of the

diagonals. Th e diagonals bisect each other and intersect at right angles.

Problem

What is the area of rhombus ABCD?

First fi nd the length of each diagonal.

d1 5 7 1 7 5 14 Diagonals of a rhombus bisect each other.

82 5 72 1 x2 Pythagorean Theorem

64 5 49 1 x2 Simplify.

x 5 "15

d2 5 2x 5 2"15 cm

Use the formula for the area of a rhombus.

A 5 12 d1d2 Substitute.

512(14)(2"15) Simplify.

5 14"15 cm2

Th e area of rhombus ABCD is 14"15 cm2.

Find the area of each rhombus. Leave your answer in simplest radical form.

10. 11. 12.

Find the area of each kite. Leave your answer in simplest radical form.

13. 14. 15.

16. Find the area of a kite with diagonals 13 cm and 14 cm.

17. Find the area of a rhombus with diagonals 16 ft and 21 ft.

D

A B

C

x

7 cm

8 cm

13.5 ft 15 ft3 cm

60

7 yd 30

4 ft

9 ft9 ft

12 ft

7 yd

2.5 yd

30 12 cm

45

45 60

30

10-2 Reteaching (continued)

Areas of Trapezoids, Rhombuses, and Kites

405 ft2

168 ft2

144 ft2

18 "3 cm2

91 cm2

(36 1 36"3) cm2

98 "3 yd2

23.75 "3 yd2

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10-3 ReteachingAreas of Regular Polygons

If you circumscribe a circle around a regular polygon:

Th e radius of the circle is also the radius of the polygon, that is, the distance from the center to any vertex of the polygon.

Th e apothem is the perpendicular distance from the center to any side of the polygon.

Problem

What are m/1, m/2, and m/3?

To fi nd m/1, divide the number of degrees in a circle by the number of sides.

m/1 5 3608 5 45

Th e apothem bisects the vertex angle, so you can fi nd m/2 using m/1.

m/2 5 12 m/1 5 1

2 (45) 5 22.5

Find /3 using the fact that /3 and /2 are complementary angles.

m/3 5 90 2 22.5 5 67.5

For any regular polygon, the area is A 5 12ap. You can use the properties of special

triangles to help you fi nd the area of regular polygons.

Problem

What is the area of a regular quadrilateral (square) inscribed in a circle with radius 4 cm?

Draw one apothem to the base to form a 458-458-908 triangle. Using the 458-458-908 Triangle Th eorem, fi nd the length of the apothem.

4 5 "2a The hypotenuse 5 "2 ? leg in a 458-458-908 triangle.

a 5 4

"2 Simplify.

a 5 4

"2?"2

"25 2"2 cm Rationalize the denominator.

Th e apothem has the same length as the other leg, which is half as long as a side. To fi nd the square’s area, use the formula for the area of a regular polygon.

A 5 12 ap

5 12 (2"2)(16"2) Substitute p 5 4(4"2) 5 16"2.

5 32 cm2

Th e area of a square inscribed in a circle with radius 4 cm is 32 cm2.

radius

apothem

12

3

4 cm

45

45

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Exercises

Each regular polygon has radii and apothem as shown. Find the measure of each numbered angle.

1. 2. 3.

Find the area of each regular polygon with the given apothem a and side length s.

4. pentagon, a 5 4.1 m, s 5 6 m 5. hexagon, a 5 10.4 in., s 5 12 in.

6. 7-gon, a 5 8.1 cm, s 5 7.8 cm 7. octagon, a 5 11.1 ft, s 5 9.2 ft

8. nonagon, a 5 13.2 in., s 5 9.6 in. 9. decagon, a 5 8.6 m, s 5 5.6 m

Find the area of each regular polygon. Round your answer to the nearest tenth.

10. 11. 12.

13. Reasoning How does the area of an equilateral triangle with sides 24 ft compare to the area of a regular hexagon with sides 24 ft? Explain.

Find the area of each regular polygon with the given radius or apothem. If necessary, leave your answer in simplest radical form.

14. 15. 16.

1

2

3

1

23

1

23

20 m

12 in. 14 ft

5√2 in.

6 cm

6 m

10-3 Reteaching (continued)

Areas of Regular Polygons

The hexagon can be divided into 6 equilateral triangles with sides 24 ft, so it has 6 times the area of the equilateral triangle.

90; 45; 4560; 30; 60 30; 15; 75

374.4 in.2

62.4 in.2 509.2 ft.2

570.24 in.2

100 in.2

61.5 m2

1039.2 m2

72 cm2 108"3 m2

240.8 m2

221.13 cm2 408.48 ft2

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10-4 Reteaching Perimeters and Areas of Similar Figures

Corresponding sides of similar fi gures are in proportion. Th e relationship between the lengths of corresponding sides in the two fi gures is called the scale factor. Th e perimeters and areas are related by the scale factor.

Scale Factor

Ratio of Perimeters

Ratio of Areas

ab

ab

a2

b2

Problem

Th e hexagons at the right are similar. What is the ratio (smaller to larger) of their perimeters and their areas?

Th e ratio of the corresponding sides is 612.

PsmallerPlarger

56

12 512 Simplify.

Th e ratio of the areas is the square of the ratio of the corresponding sides.

AsmallerAlarger

512

22 514

Problem

Th e rectangles at the right are similar. Th e area of the smaller rectangle is 72 in.2. What is the area of the larger rectangle?

Th e ratio of corresponding sides is ab 569 5

23.

Set up a proportion and solve:

AsmallerAlarger

5a2

b2

72Alarger

522

32 Substitute.

Alarger 5 72 (94) 5 162 in.2 Cross Products Property

Exercises

Th e fi gures in each pair are similar. Compare the fi rst fi gure to the second. Give the ratio of the perimeters and the ratio of the areas.

1. 2. 3.

12 cm 6 cm

6 in. 9 in.

3 in.6 in.

4 cm 7 cm15 ft 6 ft1 i2; 1 i4 4 i7; 16 i49

5 i2; 25 i4

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Th e fi gures in each pair are similar. Th e area of one fi gure is given. Find the area of the other fi gure to the nearest whole number.

4. 5.

Area of smaller pentagon 5 112 m2 Area of smaller rectangle 5 78 in.2

6. 7.

Area of larger triangle 5 75 cm2 Area of smaller octagon 5 288 ft2

Th e scale factor of two similar polygons is given. Find the ratio of their perimeters and the ratio of their areas.

8. 4 i3 9. 5 i8 10. 37 11. 9

2

12. Th e area of a regular nonagon is 34 m2. What is the area of a regular nonagon with sides fi ve times the sides of the smaller nonagon?

13. A town is installing a sandbox in the park. Th e sandbox will be in the shape of a regular hexagon. On the plans for the sandbox, the sides are 4 in. and the

area is about 42 in.2. If the actual area of the sandbox will be 168 ft2, what will be the length of one side of the sandbox?

14. Th e longer sides of a parallelogram are 6 ft. Th e longer sides of a similar parallelogram are 15 ft. Th e area of the smaller parallelogram is 27 ft2. What is the area of the larger parallelogram?

15. Th e shortest side of a pentagon is 4 cm. Th e shortest side of a similar pentagon is 9 cm. Th e area of the larger pentagon is 243 cm2. What is the area of the smaller pentagon?

16. Th e scale factor of two similar fl oors is 5 i6. It costs $340 to tile the smaller fl oor. At that rate, how much would it cost to tile the larger fl oor?

8 m 12 m

4 in. 8 in.

5 cm 2 cm

8 ft10 ft

10-4 Reteaching (continued)

Perimeters and Areas of Similar Figures

252 m2

12 cm2 450 ft2

850 m2

8 ft

168.75 ft2

48 cm2

$489.60

312 in.2

4 i3; 16 i9 5 i8; 25 i64 37; 9

4992; 81

4

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10-5 Reteaching Trigonometry and Area

You can use the trigonometric ratios of

sine, cosine, and tangent to help you fi nd

the area of regular fi gures.

Problem

What is the area of a regular hexagon with side 12 cm?

Area 512ap, where a 5 apothem

p 5 perimeter

p 5 6 ? 12 5 72 cm, because the fi gure is 6-sided.

Area 512 a(72) 5 36a cm2

To fi nd a, examine nAOB above. Th e apothem is measured along OM ,

which divides nAOB into congruent triangles.

AM 512 AB 5 6

m/AOM 512 m/AOB

512 a360

6 b5 30

Divide 360 by 6 because there are six congruent central angles.

So, by trigonometry, tan 308 5AM

a .

tan 308 56a

a 5 6tan 308

Finally, area 512 ap 5 1

2 a 6tan 308

b(72) < 374.1 cm2.

Exercises

Find the area of each regular polygon. Round your answers to the nearest tenth.

1. octagon with side 2 in. 2. decagon with side 4 cm

3. pentagon with side 10 in. 4. 20-gon with side 40 in.

A

BC

hypotenuse

opposite side

adjacentside

A M

a

O

B

sin /A 5opposite side

hypotenuse5

CBAB

cos /A 5adjacent side

hypotenuse5

ACAB

tan /A 5opposite side

adjacent side5

CBAC

19.3 in.2 123.1 cm2

172.0 in.2 50,510.0 in.2

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10-5 Reteaching (continued)

Trigonometry and Area

Th eorem 10-8 Area of a Triangle Given SAS

A 5 12 bc(sin A)

Problem

What is the area of the triangle? Round your answer

to the nearest tenth.

A 5 12 bc(sin A)

512 (8)(11)(sin 628) Substitute.

< 38.84969 cm2 Use a calculator.

< 38.8 cm2 Round to the nearest tenth.

Th e area of the triangle is about 38.8 cm2.

Exercises

Find the area of each triangle. Round your answers to the nearest tenth.

5. 6. 7.

8. 9. 10.

11. ABCDEF is a regular hexagon with center H and side 12 m.

Find each measure. If necessary, round your answers to

the nearest tenth.

a. m/BHC b. m/BHG

c. BG d. HG

e. perimeter of ABCDEF f. area of ABCDEF

B

CA

c

b

Included angle

62

8 cm

11 cm

8 m

7 m

11542

16 in.

10 in.

88

18 cm

12 cm

20 mm

22 mm38

25 ft30 ft 124

8.5 m

15 m

60

A

B

C

D

E

F H

G

12 m

25.4 m2

135.4 mm2

60 30

6.0 m 10.4 m

72.0 m 374.1 m2

53.5 in.2

310.9 ft2

107.9 cm2

55.2 m2

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10-6 Reteaching Circles and Arcs

Th e circumference is the measure of the outside edge

of a circle. Sections of the circumference are called arcs.

Th ere are three types of arcs.

Th e circumference of a circle is C 5 2pr, or C 5 pd.

Problem

A circle has a 4.5-ft radius. What is the circumference of the circle?

C 5 2pr

C 5 2(3.14)(4.5) Substitute 3.14 for p and 4.5 for r.

C 5 28.26 ft Simplify.

Th e measure of an arc is in degrees. Th e arc’s length depends on the size

of the circle because it represents a fraction of the circumference.

Length of AB0

5m AB0

360 ? 2pr

Problem

What is the length of the darkened arc? Leave your answer in terms of p.

Length of AB0

5m AB0

360 ? 2pr

5300360 ? 2p(15) Substitute 300 for m AB

0

and 15 for r.

5 25p m Simplify.

Th e length of the darkened arc is 25p m.

A

B

T

C

Major arcsMinor arcs Semicircles

name

degrees

example

measure

less than 180 exactly 180

named by two endpoints:

measure measure ofcorresponding central angle

named by three endpoints:

measure 180

more than 180

named by three endpoints:

measure 360 measureof related minor arc

ABC is a semicircle. CAB is a major arc.AB and CB are minor arcs.

D

4.5 ft

60

15 m

A

B

C

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Exercises

Name the following in (P.

1. the minor arcs

2. the major arcs

3. the semicircles

Find the measure of each arc in (A.

4. WX0

5. XY0

6. WY0

7. WXZ1

8. YZW1

9. XZY1

10. ZWX1

11. WZ0

12. WYX1

Find the circumference of each circle. Leave your answers in terms of π.

13. 14. 15.

Find the length of each arc. Leave your answers in terms of π.

16. SV0

17. UV0

18. SUT1

19. UTV1

20. UT0

21. VT0

22. Th e wheel of a car is shown at the right. How far does the

hubcap of the tire travel in one complete rotation? How far

does the tire itself travel in one complete rotation?

23. How far does the tip of a minute hand on a clock travel in 20 minutes if the

distance from the center to the tip is 9 cm? Leave your answer in terms of p.

Th en, round your answer to the nearest tenth.

P

ST

U

QR

103

A

WX

YZ

3 ft9 in.

3.5 m

V S

U

T

R5 cm

30

3 in.

8 in.

10-6 Reteaching (continued)

Circles and Arcs

Answers may vary. Sample: RQ0

; RU0

; RS0

; QU0

; QT0

; UT0

; US0

; TS0

Answers may vary. Sample: RUS1

; RSU1

; RUQ1

; QSU1

; QST1

; UQS1

; UQT1

; TQS1

RUT1

; RST1

; SUQ1

; SRQ1

6π ft

90 77 167

270 193 283

180 90 270

9π in. 7π m

18.8 in.; 69.1 in.

6π cm; 18.8 cm

10π3 cm5π

6 cm 5π cm

20π3 cm 5π

2 cm 25π6 cm

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10-7 ReteachingAreas of Circles and Sectors

Finding the Area of a CircleTh e area of a circle is A 5 pr2. So, to fi nd the area of a circle, you need to know the radius, r. Sometimes you are given the radius directly. Sometimes you are given the diameter and have to divide by 2 to fi nd the radius.

Problem

What is the area of (S?

r 5 62 5 3 Find the radius.

A 5 p(3)2 Substitute.

A 5 9p Simplify.

A < 28.3 Approximate.

Th e area of (S is about 28.3 units2.

Finding the Area of a SectorTo fi nd the area of a sector, fi nd the area of the circle, then multiply by the fraction of the circumference covered by the arc of the sector.

Problem

(B has a radius of 2 and mAC0

5 60. What is the area of sector ABC?

First, fi nd the area of (B : A 5 (2)2 ? p 5 4p

Th en, fi nd the fraction of the circumference covered by the arc.

mAC0

360 560

360 56

36 516

Last, multiply by the fraction.

Area of sector ABC 5 16 ? 4p 5 4

6p 523p < 2.1

Exercises

Find the areas of circles and sectors described below. Write your answer in terms of π and round to the nearest tenth.

1. circle with radius 5 2. circle with diameter 16

3. sector ABC in (B of radius 4 and mAC0

5 90

4. sector RST in (S of diameter 12 and mRT0

5 45

6S

602

A

B

C

25π N 78.5

4π N 12.6

92 π N 14.1

64π N 201.1

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Finding the Area of a SegmentTo fi nd the area of a segment, subtract the area of the triangle from the area of the sector. So, use your knowledge of how to fi nd the area of a sector.

Problem

What is the area of the shaded segment?

a. Find the area of sector ABC.Circle: A 5 (9)2 ? p 5 81p

Fraction: mAC0

360 5120360 5

13

Multiply: Area of sector ABC 5 13 ? 81p 5 27p

b. Find the area of nABC.

Triangle: A 5 12(AB)(BC) sin /ABC

A 5 12(9)2 sin 120 5 81"3

4

c. Subtract to fi nd the area of the segment.

Segment: A 5 27p 2 81"34 < 49.7

Th e area of the shaded segment is about 49.7 units2.

Exercises

Find the area of each shaded segment.

5. 6. 7.

Find the area of the shaded region. Leave your answer in terms of π and in simplest radical form.

8. 9. 10.

1209

C

A B

604

S

RT

8A B

C

1406

S

12 45

36

15032

45

10-7 Reteaching (continued)

Areas of Circles and Sectors

83

π 2 4"3 N 1.4

126π 1 36"2

16π 2 32 N 18.3

189π 1 81

14π 2 18sin 140 N 32.4

160π 1 64"2

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10-8 Reteaching Geometric Probability

Problem

If a dart lands at random on the poster at the right, what is the probability that the dart will land inside one the polygons?

Find the sum of the areas of the polygons.

area of polygons 5 area of parallelogram 1 area of triangle

5 (12)(10) 1 12 (10)(16)

5 120 1 80

5 200 in.2

Find the total area of the poster.

A 5 (24)(36) 5 864 in.2

Calculate the probability.

P(polygon) 5area of polygons

total area

5200864

< 23%

Exercises

Complete each exercise.

1. Use a compass to draw a circle with radius 1 in. on an index card.

2. Calculate the theoretical probability that if a tack is dropped on the card, its tip will land in the circle.

3. Lift a tack 12 in. above the index card and drop it. Repeat this 25 times. Record how many times the tip of the tack lands in the circle. (Ignore the times that the tack bounces off the card.) Calculate the experimental probability:

P 5number of times tip landed in circle

25

4. How do the probabilities you found in Exercises 2 and 3 compare?

5. If you repeated the experiment 100 times, what would you expect the results to be?

6. If a dart lands at random on the poster at the right, what is the probability that the dart will land in a circle?

36 in.

24 in.

10 in.

16 in.12 in.

10 in.

9 in.

6 in.

3 ft

2 ft

Check students’ work. Sample shown:

N 21%

Check students’ work.

Sample: The ratio would tend to be close to 21%.

Sample: They are approximately equal.

N 43%

1 in.

5 in.

3 in.

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For Exercises 7–10 give your answer as a ratio and as a percent. For Exercises 7 and 8 use square ABCD at the right.

7. Point P in square ABCD is chosen at random.

Find the probability that P is in square AXYZ.

8. Find the probability that P is not in square AXYZ.

For Exercises 9 and 10 use rectangle ABCD at the right.

9. Point P in rectangle ABCD is chosen at random.

Find the probability that P is in square QRST.

10. Find the probability that P is not in square QRST.

Give your answer in terms of π, then as a percent.

11. Point P in square ABCD is chosen at random.

Find the probability that P is in not in (S.

12. Point P in (S is chosen at random.

Find the probability that P is in not in square ABCD.

Point P in (S is chosen at random. Find the probability that P is in sector ABC. Give your answer in terms of a ratio, then as a percent.

13. 14. 15.

16. Th e cycle of the light on George Street at the intersection of George Street and Main Street is 10 seconds green, 5 seconds yellow, and 60 seconds red. If you reach the intersection at a random time, what is the probability that the light is red?

A B

CD

X

YZ

2

24

4

A B

CD

5

Q R

ST

6

3

3

S

A B

CD

10

10A B

CD

S12

8

A

B

C

660 A

120

B

C

3A

B

C

4

10-8 Reteaching (continued)

Geometric Probability

416 5

14 5 25%

34 5 75%

310 5 30%

710 5 70%

12π4 N 21.5%

12 2π N 36.3%

80%

1

6? 36π

36π5 1

6N 16.7%

1

3? 9π

9π5 1

3N 33.3%

1

4? 16π

16π5 1

45 25%

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11-1 Reteaching

Space Figures and Cross Sections

A polyhedron is a three-dimensional fi gure with faces that are polygons. Faces intersect at edges, and edges meet at vertices.

Faces, vertices, and edges are related by Euler’s Formula: F 1 V 5 E 1 2.

For two dimensions, such as a representation of a polyhedron by a net, Euler’s Formula is F 1 V 5 E 1 1. (F is the number of regions formed by V vertices linked by E segments.)

Problem

What does a net for the doorstop at the right look like? Label the net with its appropriate dimensions.

Exercises

Complete the following to verify Euler’s Formula.

1. On graph paper, draw three other nets for the polyhedron shown above. Let each unit of length represent 14 in.

2. Cut out each net, and use tape to form the solid fi gure.

3. Count the number of vertices, faces, and edges of one of the fi gures.

4. Verify that Euler’s Formula, F 1 V 5 E 1 2, is true for this polyhedron.

Draw a net for each three-dimensional fi gure.

5. 6.

3 in.

4 in.

8 in.

5 in.

Doorstop

3 in.

4 in.4 in.

8 in.5 in.5 in.

5 in.

5 in.8 in.

3 in.

5 in.4 in.

3 in.

4 in.

Sample:

Samples:

Check students’ work.

6 vertices, 5 faces, 9 edges

F 1 V 5 E 1 2, 5 1 6 5 9 1 2, 11 5 11

3 in.

8 in.

5 in.4 in.

3 in.

4 in.

3 in.

4 in.

3 in. 3 in.

8 in.

5 in.

4 in.

3 in.

5 in. 5 in.

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A cross section is the intersection of a solid and a plane. Cross sections can be many diff erent shapes, including polygons and circles.

Th e cross section of this solid and this plane is a rectangle. Th is cross section is a horizontal plane.

To draw a cross section, visualize a plane intersecting one face at a time in parallel segments. Draw the parallel segments, then join their endpoints and shade the cross section.

Exercises

Draw and describe the cross section formed by intersecting the rectangular prism with the plane described.

7. a plane that contains the vertical line of symmetry

8. a plane that contains the horizontal line of symmetry

9. a plane that passes through the midpoint of the top left edge, the midpoint of the top front edge, and the midpoint of the left front edge

10. What is the cross section formed by a plane that contains a vertical line of symmetry for the fi gure at the right?

11. Visualization What is the cross section formed by a plane that is tilted and intersects the front, bottom, and right faces of a cube?

11-1 Reteaching (continued)

Space Figures and Cross Sections

Answers may vary. a rectangle or a square

a rectangle

an isosceles triangle

Answers may vary. a hexagon or a rectangle

a triangle

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11-2 Reteaching Surface Areas of Cylinders and Prisms

A prism is a polyhedron with two congruent parallel faces called bases. Th e

non-base faces of a prism are lateral faces. Th e dimensions of a right prism

can be used to calculate its lateral area and surface area.

Th e lateral area of a right prism is the product of the perimeter of the base and

the height of the prism.

L.A. 5 ph

Th e surface area of a prism is the sum of the lateral area and the areas of the bases

of the prism.

S.A. 5 L.A. 1 2B

Problem

What is the lateral area of the regular hexagonal prism?

L.A. 5 ph

p 5 6(4 in.) 5 24 in. Calculate the perimeter.

L.A. 5 24 in. 3 13 in. Substitute.

L.A. 5 312 in.2 Multiply.

Th e lateral area is 312 in.2.

Problem

What is the surface area of the prism?

S.A. 5 L.A. 1 2B

p 5 2(7 m 1 8 m) Calculate the perimeter.

p 5 30 m Simplify.

L.A. 5 ph

L.A. 5 30 m 3 30 m Substitute.

L.A. 5 900 m2 Multiply.

B 5 8 m 3 7 m Find base area.

B 5 56 m2 Multiply.

S.A. 5 L.A. 1 2B

S.A. 5 900 m2 1 2 3 56 m2 Substitute.

S.A. 5 1012 m2 Simplify.

Th e surface area of the prism is 1012 m2.

4 in.

13 in.

8 m7 m

30 m

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A cylinder is like a prism, but with circular bases. For a right cylinder, the radius of

the base and the height of the cylinder can be used to calculate its lateral area and

surface area.

Lateral area is the product of the circumference of the base (2pr) and the height

of the cylinder. Surface area is the sum of the lateral area and the areas of the

bases (2pr2).

L.A. 5 2prh or pdh S.A. 5 2prh 1 2pr2

Problem

Th e diagram at the right shows a right cylinder. What are the lateral

area and surface area of the cylinder?

L.A. 5 2prh or pdh

L.A. 5 2p 3 4 in. 3 9 in. Substitute for r and h.

L.A. 5 72p in.2 Multiply.

Th e lateral area is 72p in.2.

S.A. 5 2prh 1 2pr2

S.A. 5 2p 3 4 in. 3 9 in. 1 2p 3 (4 in.)2 Substitute for r and h.

S.A. 5 72p in.2 1 32p in.2 Multiply.

S.A. 5 104p in.2 Add.

Th e surface area is 104p in.2.

Exercises

Find the lateral area and surface area of each fi gure. Round your answers to the nearest tenth, if necessary.

1. 2. 3.

4. A cylindrical carton of raisins with radius 4 cm is 25 cm tall. If all surfaces

except the top are made of cardboard, how much cardboard is used to make

the raisin carton? Round your answer to the nearest square centimeter.

r 4 in.

9 in.

18 cm

5 cm

12 in.

12 in.

12 in.

3 m

5 m10 m

11-2 Reteaching (continued)

Surface Areas of Cylinders and Prisms

565.5 cm2; 722.6 cm2

679 cm2

576 in.2; 864 in.2

138.3 m2; 153.3 m2

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11-3 ReteachingSurface Areas of Pyramids and Cones

A pyramid is a polyhedron in which the base is any polygon and the lateral faces are triangles that meet at the vertex. In a regular pyramid, the base is

a regular polygon. Th e height is the measure of the altitude of a pyramid,

and the slant height is the measure of the altitude of a lateral face. Th e

dimensions of a regular pyramid can be used to calculate its lateral

area (L.A.) and surface area (S.A.).

L.A. 5 12p/, where p is the perimeter of the base and / is slant height of

the pyramid.

S.A. 5 L.A. 1 B, where B is the area of the base.

Problem

What is the surface area of the square pyramid to the nearest tenth?

S.A. 5 L.A. 1 B

L.A. 512 p/ Find lateral area fi rst.

p 5 4(4 m) 5 16 m Find the perimeter.

/2 5 "22 1 102 5 "104 Use the Pythagorean Theorem.

/ < 10.2

L.A. 512 (16m)(10.2) 5 81.6 m2 Substitute to fi nd L.A.

B 5 (4 m)(4 m) 5 16 m2 Find area of the base.

S.A. 5 81.6 m2 1 16 m2 5 97.6 m2 Substitute to fi nd S.A.

Th e surface area of the square pyramid is about 97.6 m2.

ExercisesUse graph paper, scissors, and tape to complete the following.

1. Draw a net of a square pyramid on graph paper.

2. Cut it out, and tape it together.

3. Measure its base length and slant height.

4. Find the surface area of the pyramid.

In Exercises 5 and 6, round your answers to the nearest tenth, if necessary.

5. Find the surface area of a square pyramid with base length 16 cm and slant

height 20 cm.

6. Find the surface area of a square pyramid with base length 10 in. and

height 15 in.

s

,

4 m

10 m

Sample:

Sample: base 5 3 cm, slant height 5 4 cm

Check students’ work.

Sample: 33 cm2

896 cm2

416.2 in.2

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A cone is like a pyramid, except that the base of a cone is a circle.

Th e radius of the base and cylinder height can be used to calculate

the lateral area and surface area of a right cone.

L.A. 5 pr /, where r is the radius of the base and / is slant height

of the cone.

S.A. 5 L.A. 1 B, where B is the area of the base (B 5 pr2).

Problem

What is the surface area of a cone with slant height 18 cm and height 12 cm?

Begin by drawing a sketch.

Use the Pythagorean Th eorem to fi nd r, the radius of the base

of the cone.

r2 1 122 5 182

r2 1 144 5 324

r2 5 180

r < 13.4

Now substitute into the formula for the surface area of a cone.

S.A. 5 L.A. 1 B

5 pr/ 1 pr2

5 p(13.4)(18) 1 180p

< 1323.2

Th e surface area of the cone is about 1323.2 cm2.

In Exercises 7–10, round your answers to the nearest tenth, if necessary.

7. Find the surface area of a cone with radius 5 m and slant height 15 m.

8. Find the surface area of a cone with radius 6 ft and height 11 ft.

9. Find the surface area of a cone with radius 16 cm and slant height 20 cm.

10. Find the surface area of a cone with radius 10 in. and height 15 in.

,

r

18 cm12 cm

r

11-3 Reteaching (continued)

Surface Areas of Pyramids and Cones

314.2 m2

349.3 ft2

1809.6 cm2

880.5 in.2

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11-4 Reteaching Volumes of Prisms and Cylinders

Problem

Which is greater: the volume of the cylinder or the volume

of the prism?

Volume of the cylinder: V 5 Bh

5 pr2 ? h

5 p(3)2 ? 12

< 339.3 in.3

Volume of the prism: V 5 Bh

5 s2 ? h

5 62 ? 12

5 432 in.3

Th e volume of the prism is greater.

Exercises

Find the volume of each object.

1. the rectangular prism part of the milk container

2. the cylindrical part of the measuring cup

Find the volume of each of the following. Round your answers to the nearest tenth, if necessary.

3. a square prism with base length 7 m and height 15 m

4. a cylinder with radius 9 in. and height 10 in.

5. a triangular prism with height 14 ft and a right triangle base with legs

measuring 9 ft and 12 ft

6. a cylinder with diameter 24 cm and height 5 cm

12 in.

6 in.6 in.

6 in.

12 in.

7 cm7 cm

10 cm

6.3 cm7 cm

490 cm3

735 m3

2261.9 cm3

2544.7 in.3

756 ft3

77.175π or about 242.5 cm3

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Problem

What is the volume of the triangular prism?

Sometimes the height of a triangular base in a triangular prism is not given.

Use what you know about right triangles to fi nd the missing value.

Th en calculate the volume as usual.

hypotenuse 5 18 cm Given

short leg 5 9 cm 308-608-908 triangle theorem

long leg 5 9"3 cm 308-608-908 triangle theorem

Volume of prism: V 5 Bh

V 5 a12b (9)(9"3)(12)

V < 841.8 cm3

Th e volume of the triangular prism is about 841.8 cm3.

Exercises

Find the volume of each prism. Round to the nearest tenth.

7. 8. 9.

10. 11. 12.

Find the volume of each composite fi gure to the nearest tenth.

13. 14. 15.

12 cm

18 cm

30

60

45

455 cm

3 2 cm

25 in.

27 in.

9 in. 10 mm

16 mm

22 mm

3 in.

1 in.

3 m

2 m

7 m

6 in.

17 in.

6 in.6 in.

6 ft

3 ft2 ft

4 ft8 ft

10 in.

3 in.3 in.

2 in.

4 in.5 in.

11-4 Reteaching (continued)

Volumes of Prisms and Cylinders

22.5 cm3

7.1 in.3

76 ft3 96.8 in.3 111.4 in.3

265.0 in.342 m3

2833.8 in.3

1208.0 mm3

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11-5 Reteaching Volumes of Pyramids and Cones

Problem

What is the volume of the square pyramid?

Sometimes the height of a triangular face in a square pyramid is

not given. Here the slant height and the lengths of the sides of

the base are given. Use what you know about right triangles

to fi nd the missing value. Th en calculate the volume as usual.

72 1 x2 5 252 Use the Pythagorean Theorem.

49 1 x2 5 625 Substitute.

x2 5 625 2 49 Isolate the variable.

x2 5 576 Simplify.

x 5 24 cm Find the square root of each side.

Volume of the pyramid:

V 5 13 Bh Use the formula for volume of a pyramid.

513(14 3 14)(24) Substitute.

5 1568 cm3 Simplify.

Th e volume of the square pyramid is 1568 cm3.

Exercises

Find the volume of each pyramid. Round to the nearest whole number.

1. 2. 3.

4. 5. 6.

25 cm

14 cm

14 cm

45 cm

54 cm

54 cm 32 in.32 in.

34 in.

150 m2

3 m

13 in.

10 in.10 in.

36 yd

400 yd2

18 cm

8 cm2

34,992 cm3 10,240 in.3 150 m3

400 in.3 4800 yd3 48 cm3

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11-5 Reteaching (continued)

Volumes of Pyramids and Cones

Problem

What is the volume of the cone?

Find the height of the cone.

132 5 h2 1 52 Use the Pythagorean Theorem.

169 5 h2 1 25 Substitute.

h2 5 144 Simplify.

h 5 12 Take the square root of each side.

Find the volume of the cone.

V 5 13pr

2h Use the formula for the volume of a cone.

513p(5)2 ? 12 Substitute.

5 100p Simplify.

< 314.2

Th e volume of the cone is about 314.2 cm2.

Exercises

7. From the fi gures shown below, choose the pyramid with volume closest

to the volume of the cone at the right.

A. B. C.

Find the volume of each fi gure. Round your answers to the nearest tenth.

8. 9. 10. 11.

h 13 cm

5 cm

2.5 in.

7 in.

5 in.

5 in.

7 in.

3 in.

15 in.

3 in.

2 in.8 in.

8 in.

10 cm

6 cm

h

17 m15 cm

2 ft

6 ft

12.7 mh

4.1 m

1005.3 m3

18.8 ft3

B

301.6 cm3211.6 m3

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11-6 Reteaching Surface Areas and Volumes of Spheres

Problem

What are the surface area and volume of the sphere?

Substitute r 5 5 into each formula, and simplify.

S.A. 5 4pr 2 V 5 4

3pr 3

5 4p(5)2 543p(5)3

5 100p 5500p

3

< 314.2 < 523.6

Th e surface area of the sphere is about 314.2 in.2. Th e volume of the sphere

is about 523.6 in.3.

Exercises

Use the fi gures at the right to guide you in completing the following.

1. Use a compass to draw two circles, each with radius 3 in.

Cut out each circle.

2. Fold one circle in half three successive times. Number the

central angles 1 through 8.

3. Cut out the sectors, and tape them together as shown.

4. Take the other circle, fold it in half, and tape it to the

rearranged circle so that they form a quadrant of a sphere.

5. Th e area of one circle has covered one quadrant of a sphere. How many circles

would cover the entire sphere?

6. How is the radius of the sphere related to the radius of the circle?

Find the volume and surface area of a sphere with the given radius or diameter. Round your answers to the nearest tenth.

7. 8. 9.

5 in.

6

2

84

3

57

1

6

8

2

435

7

1

13

57 8

6

42

10 in. 3 cm 24 m

V 5 523.6 in.3; S.A. 5 314.2 in.2

V 5 113.1 cm3; S.A. 5 113.1 cm2

V 5 7,238.2 m3; S.A. 5 1809.6 m2

Check students’ work.

Check students’ work.

Check students’ work.

Check students’ work.

four circles

They are the same.

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11-6 Reteaching (continued)

Surface Areas and Volumes of Spheres

Find the volume and surface area of the sphere. Round to the nearest tenth.

10. 11. 12.

13. 14. 15.

A sphere has the volume given. Find its surface area to the nearest whole number.

16. 1436.8 mi3 17. 808 cm3 18. 72 m3

Find the volume of each sphere with the given surface area. Round to the nearest whole number.

19. 435 yd2 20. 907 cm2 21. 28 m2

22. Visualization Th e region enclosed by the

semicircle at the right is revolved completely

about the x-axis.

a. Describe the solid of revolution that is formed.

b. Find its volume in terms of p.

c. Find its surface area in terms of p.

23. Th e sphere at the right fi ts snugly inside a cube

with 18 cm edges. What is the volume of the

sphere? What is the surface area of the sphere?

Leave your answers in terms of p.

14 in. 700 m

2 cm

10 m

2 ft

7 m

24 2 4

2

4

y

xO

18 cm

616 mi2 420 cm2 84 m2

853 yd3 2569 cm3 14 m3

a sphere

85

13 π units3

64π units2

972π cm3; 324π cm2

S.A. 5 2463.0 in.2; V 5 11,494.0 in.3

S.A. 5 1256.6 m2;V 5 4188.8 m3

S.A. 5 6,157,521.6 m2; V 5 1,436,755,040.2 m3

S.A. 5 50.3 ft2;V 5 33.5 ft3

S.A. 5 12.6 cm2; V 5 4.2 cm3

S.A. 5 153.9 m2;V 5 179.6 m3

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11-7 Reteaching Areas and Volumes of Similar Solids

When two solids are similar, their corresponding

dimensions are proportional.

Rectangular prisms A and B are similar because the

ratio of their corresponding dimensions is 23.

height: 8 m

12 m 523 length:

2 m3 m 5

23 width:

4 m6 m 5

23

Th e ratio of the corresponding dimensions of similar solids is called the scale

factor. All the linear dimensions (length, width, and height) of a solid must

have the same scale factor for the solids to be similar.

Areas and Volumes of Similar SolidsArea Volume• Th e ratio of corresponding areas of similar

solids is the square of the scale factor.

• Th e ratio of the volumes of similar

solids is the cube of the scale factor.

• Th e ratio of the areas of prisms A and

B is 22

32, or 49.

• Th e ratio of the volumes of prisms

A and B is 23

33, or 8

27.

Problem

Th e pyramids shown are similar, and they have volumes of 216 in.3 and

125 in.3 Th e larger pyramid has surface area 250 in.2

What is the ratio of their surface areas?

What is the surface area of the smaller pyramid?

By Th eorem 11-12, if similar solids have similarity ratio a ib, then the

ratio of their volumes is a3ib3.

So,

a3

b3 5216125

ab 5

65 Take the cube root of both sides to get a ib.

a2

b2 53625 Square both sides to get a2

ib2.

Ratio of surface areas 5 36 i25

If the larger pyramid has surface area 250 in.2, let the smaller pyramid have

surface area x. Th en,

250

x 53625

36x 5 6250

x < 173.6 in.2

Th e surface area of the smaller pyramid is about 173.6 in2.

3 m6 m

12 m

2 m4 m

8 m

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11-7 Reteaching (continued)

Areas and Volumes of Similar Solids

ExercisesFind the scale factors.

1. Similar cylinders have volumes of 200p in.3 and 25p in.3

2. Similar cylinders have surface areas of 45p in.2 and 20p in.2

Are the two fi gures similar? If so, give the scale factor.

3. 4.

Each pair of fi gures is similar. Use the given information to fi nd the scale factor of the smaller fi gure to the larger fi gure.

5. 6.

Find the ratio of volumes.

7. Two cubes have sides of length 4 cm and 5 cm.

8. Two cubes have surface areas of 64 in.2 and 49 in.2

Th e surface areas of two similar fi gures are given. Th e volume of the larger fi gure is given. Find the volume of the smaller fi gure.

9. S.A. 5 16 cm2 10. S.A. 5 6 ft2 11. S.A. 5 45 m2

S.A. 5 100 cm2 S.A. 5 294 ft2 S.A. 5 80 m2

V 5 500 cm3 V 5 3430 ft3 V 5 320 m3

Th e volumes of two similar fi gures are given. Th e surface area of the smaller fi gure is given. Find the surface area of the larger fi gure.

12. V 5 12 in.3 13. V 5 6 cm3 14. V 5 40 ft3

V 5 96 in.3 V 5 384 cm3 V 5 135 ft3

S.A. 5 12 in.2 S.A. 5 6 cm2 S.A. 5 20 ft2

18 m

8 m9 m

4 m

10 in.

4 in.5 in.

2 in.

6 in.

3 in.

V 135 in.3 V 320 in.3 S.A. 32 cm2 S.A. 162 cm2

yes; 1 i2 no

3 i4 4 i9

64 i125

512 i343

32 cm3 10 ft3 135 m3

48 in.2 96 cm2 45 ft2

2 i 1

3 i 2

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12-1 Reteaching Tangent Lines

A tangent is a line that touches a circle at exactly one point.

In the diagram, AB is tangent to (Q. You can apply theorems

about tangents to solve problems.

Theorem 12-1If a line is tangent to a circle, then that line forms a right angle

with the radius at the point where the line touches the circle.

Theorem 12-2If a line in the same plane as a circle is perpendicular to a radius at its endpoint on

the circle, then the line is tangent to the circle.

Problem

Use the diagram at the right to solve the problems below. GH is tangent to (K .

What is the measure of /G?

Because GH is tangent to (K , it

forms a right angle with the radius.

Th e sum of the angles of a triangle

is always 180. Write an equation

to fi nd m/G.

m/G 1 m/H 1 m/K 5 180

m/G 1 90 1 68 5 180

m/G 1 158 5 180

m/G 5 22

What is the length of the radius?

You can use the Pythagorean

Th eorem to fi nd missing lengths.

HK2 1 HG2 5 GK2

r2 1 122 5 (9 1 r)2

r2 1 144 5 (9 1 r)(9 1 r)

r2 1 144 5 81 1 18r 1 r2

63 5 18r

3.5 5 r

So, the measure of /G is 22 and the length of the radius is 3.5 units.

Exercises

In each circle, what is the value of x?

1. 2. 3.

A

B

R

Q

68 9

12G

K

H

rr

x

28

O O59x

O33x

62 31 57

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12-1 Reteaching (continued)

Tangent Lines

In each circle, what is the value of r?

4. 5. 6.

Theorem 12-3If two segments are tangent to a circle from the same point

outside the circle, then the two segments are equal in length.

In the diagram, AB and BC are both tangent to (D. So, they

are also congruent.

When circles are drawn inside a polygon so that the sides

of the polygon are tangents, the circle is inscribed in the

fi gure. You can apply Th eorem 12-3 to fi nd the

perimeter, or distance around the polygon.

Problem

(M is inscribed in quadrilateral ABCD. What is the perimeter of ABCD?

ZA 5 AW 5 9 WB 5 BX 5 5

CY 5 XC 5 2 YD 5 DZ 5 3

Now add to fi nd the length of each side:

AB 5 AW 1 WB 5 9 1 5 5 14 BC 5 BX 1 CX 5 5 1 2 5 7

CD 5 CY 1 YD 5 2 1 3 5 5 DA 5 DZ 1 ZA 5 3 1 9 5 12

14 1 7 1 5 1 12 5 38

Th e perimeter is 38 in.

Exercises

Each polygon circumscribes a circle. What is the perimeter of each polygon?

7. 8. 9.

r

r

3

1 O rr

9 15

O r

r

8 14

O

A B

C

D

MW

9 in.

5 in.

2 in.

3 in.

A

Z

XD Y

C

B

6 cm

8 cm

3 cm

4 ft7.5 ft

6 ft

7 cm

3.5 cm

5.25 cm 8.75 cm

4

34 cm 35 ft 49 cm

8 8.25

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12-2 Reteaching Chords and Arcs

Several relationships between chords, arcs, and the central angles of a circle are listed below. Th e converses of these theorems are also true.

Th eorem 12-4 Congruent central angles have congruent arcs.

Th eorem 12-5 Congruent central angles have congruent chords.

Th eorem 12-6 Congruent chords have congruent arcs.

Th eorem 12-7 Chords equidistant from the center are congruent.

Problem

What is the value of x?

EF 5 FG 5 3.2 Given

AB > DC Chords equidistant from the center of a circle are congruent.

DC 5 DG 1 GC Segment Addition Postulate

AB 5 x 1 GC Substitution

DG 5 GC 5 3.5 Given

x 5 3.5 1 3.5 5 7 Substitution

Th e values of x is 7.

Exercises

In Exercises 1 and 2, the circles are congruent. What can you conclude?

1. 2.

Find the value of x.

3. 4. 5.

x

3.5

E F

G

A

B

D

C

3.2

G H I

LK J

W X

OZ M

Y

12

9

9

x 8

3.2

x6.3

10

x

GI O L J and GI0

O LJ0

24

Answers may vary. Samples below:

3.2 5

WX O ZY and WX0

O ZY0

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Useful relationships between diameters, chords, and arcs are listed below. To bisect a fi gure means to divide it exactly in half.

Th eorem 12-8 In a circle, if a diameter is perpendicular to a chord, it bisects that chord and its arc.

Th eorem 12-9 In a circle, if a diameter bisects a chord that is not a diameter of the circle, it is perpendicular to that chord.

Th eorem 12-10 If a point is an equal distance from the endpoints of a line segment, then that point lies on the perpendicular bisector of the segment.

Problem

What is the value of x to the nearest tenth?

In this problem, x is the radius. To fi nd its value draw radius BD, which becomes the hypotenuse of right nBED. Th en use the Pythagorean Th eorem to solve.

ED 5 CE 5 3 A diameter perpendicular to a chord bisects the chord.

x2 5 32 1 42 Use the Pythagorean Theorem.

x2 5 9 1 16 5 25 Solve for x2.

x 5 5 Find the positive square root of each side.

Th e value of x is 5.

Exercises

Find the value of x to the nearest tenth.

6. 7. 8.

Find the measure of each segment to the nearest tenth.

9. Find c when r 5 6 cm and d 5 1 cm.

10. Find c when r 5 9 cm and d 5 8 cm.

11. Find d when r 5 10 in. and c 5 10 in.

12. Find d when r 5 8 in. and c 5 15 in.

64

x x

A

B

C

D

E

14

20

x

x

9

12

9

x

1.9

c

dr

12-2 Reteaching (continued)

Chords and Arcs

9.8

11.8 cm

8.2 cm

8.7 in.

2.8 in.

4.915.9

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12-3 Reteaching Inscribed Angles

Two chords with a shared endpoint at the vertex of an angle

form an inscribed angle. Th e intercepted arc is formed where

the other ends of the chords intersect the circle.

In the diagram at the right, chords AB and BC form inscribed

/ABC. Th ey also create intercepted arc AC0

.

Th e following theorems and corollaries relate to inscribed

angles and their intercepted arcs.

Th eorem 12-11: Th e measure of an inscribed angle is half

the measure of its intercepted arc.

• Corollary 1: If two inscribed angles intercept the same

arc, the angles are congruent. So, m/A > m/B.

• Corollary 2: An angle that is inscribed in a semicircle

is always a right angle. So, m/W 5 m/Y 5 90.

• Corollary 3: When a quadrilateral is inscribed in a

circle, the opposite angles are supplementary. So,

m/X 1 m/Z 5 180.

Th eorem 12-12: Th e measure of an angle formed by a tangent

and a chord is half the measure of its intercepted arc.

Problem

Quadrilateral ABCD is inscribed in (J .

m/ADC 5 68; CE)

is tangent to (J

What is m/ABC? What is mCB0

? What is m/DCE?

m/ABC 1 m/ADC 5 180 Corollary 3 of Theorem 12-11

m/ABC 1 68 5 180 Substitution

m/ABC 5 112 Subtraction Property

mDB0

5 mDC0

1 mCB0

Arc Addition Postulate

180 5 110 1 mCB0

Substitution

70 5 mCB0

Simplify.

mCD0

5 110 Given

m/DCE 5 12 mCD0

Theorem 12-12

m/DCE 5 12 (110) Substitution

m/DCE 5 55 Simplify.

So, m/ABC 5 112, mCB0

5 70, and m/DCE 5 55.

A

B

C

AB

W

X

Y

Z

A

BD

C110

J

E

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Exercises

In Exercises 1–9, fi nd the value of each variable.

1. 2. 3.

4. 5. 6.

7. 8. 9.

Find the value of each variable. Lines that appear to be tangent are tangent.

10. 11. 12.

Points A, B, and D lie on (C. mlACB 5 40. mAB0

R mAD0

. Find each measure.

13. mAB0

14. m/ADB 15. m/BAC

16. A student inscribes a triangle inside a circle. Th e triangle divides the circle into

arcs with the following measures: 468, 1028, and 2128. What are the measures

of the angles of the triangle?

17. A student inscribes NOPQ inside (Y . Th e measure of m/N 5 68 and

m/O 5 94. Find the measures of the other angles of the quadrilateral.

a24

a

88a 9690

56

b

c94

a

ab

c98

34

76a

200

40

b

a

80140

a

b

c

74

40

A

C B

a

b

106

a

112a

b

108

60

ab

c

12-3 Reteaching (continued)

Inscribed Angles

48

28; 47; 105

60; 70

212

40

56; 124

20

66; 54; 120

70

23; 51; 106

mlP 5 112; mlQ 5 86

90; 53; 100 120; 60

82; 130; 118

44 87

40

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12-4 Reteaching Angle Measures and Segment Lengths

Problem

In the circle shown, m BC0

5 15 and mDE0

5 35.

What are m/A and m/BFC?

Because *

AD)

and *

AE)

are secants, m/A can be found using Th eorem 12-14.

m/A 5 12(m DE

0

2 m BC0

)

512(35 2 15)

5 10

Because BE and CD are chords, m/BFC can be found using Th eorem 12-13.

m/BFC 5 12(m DE

0

1 m BC0

)

512(35 1 15)

5 25

So, m/A 5 10 and m/BFC 5 25.

Exercises

Algebra Find the value of each variable.

1. 2. 3.

4. 5. 6.

7. 8. 9.

AB

C

D

E

F

70116x

x

20

28

40

139

x

x20

90x 240

x

120

65

x

24

z

y 96

x

y

z 232 xy

z

18046

93

35

36; 60; 48 64; 64; 52 46; 90; 44

156

60

42

55

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12-4 Reteaching (continued)

Angle Measures and Segment Lengths

Segment LengthsHere are some examples of diff erent cases of Th eorem 12-15.

A. Chords intersecting inside a circle:

part ? part 5 part ? part

6x 5 18

x 5 186 5 3

B. Secants intersecting outside a circle:

outside ? whole 5 outside ? whole

x(x 1 6) 5 2(18 1 2)

x2 1 6x 5 40

x2 1 6x 2 40 5 0

(x 1 10)(x 2 4) 5 0

x 5 210 or x 5 4

C. Tangent and secant intersecting outside a circle:

tangent ? tangent 5 outside ? whole

x(x) 5 4(4 1 5)

x2 5 4(9)

x2 5 36

x 5 26 or x 5 6

Exercises

Algebra Find the value of each missing variable.

10. 11. 12.

13. 14. 15.

9 6

2x

x

2

6

18

x

45

2

y

5

15

34

8

z

7

y

5

4

5

zz

4

10

y

21

4z

8

6

3Í2

6

4

11

12

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12-5 Reteaching Circles in the Coordinate Plane

Writing the Equation of a Circle from a DescriptionTh e standard equation for a circle with center (h, k) and radius r is (x 2 h)2 1 (y 2 k)2 5 r2. Th e opposite of the coordinates of the center appear in the equation. Th e radius is squared in the equation.

Problem

What is the standard equation of a circle with center (22, 3) that passes through the point (22, 6)?

Step 1 Graph the points.

Step 2 Find the radius using both given points. Th e radius is the distance from the center to a point on the circle, so r 5 3.

Step 3 Use the radius and the coordinates of the center to write the equation.

(x 2 h)2 1 (y 2 k)2 5 r2

(x 2 (22))2 1 (y 2 3)2 5 32

(x 1 2)2 1 (y 2 3)2 5 9

Step 4 To check the equation, graph the circle. Check several points on the circle.

For (1, 3): (1 1 2)2 1 (3 2 3)2 5 32 1 02 5 9

For (25, 3): (25 1 2)2 1 (3 2 3)2 5 (23)2 1 02 5 9

For (22, 0): (22 1 2)2 1 (0 2 3)2 5 02 1 (23)2 5 9

Th e standard equation of this circle is (x 1 2)2 1 (y 2 3)2 5 9.

Exercises

Write the standard equation of the circle with the given center that passes through the given point. Check the point using your equation.

1. center (2, 24); point (6, 24) 2. center (0, 2); point (3, 22)

3. center (21, 3); point (7, 23) 4. center (1, 0); point (0, 5)

5. center (24, 1); point (2, 22) 6. center (8, 22); point (1, 4)

2

4

6

2 246

( 2, 6)

Center( 2, 3)

y

x

2

4

6

2 24

( 2, 6)

( 5, 3) (1, 3)

( 2, 0)6

y

x

(x 2 2)2 1 (y 1 4)2 5 16;(6 2 2)2 1 (24 1 4)2 5 16

x2 1 (y 2 2)2 5 25; (3 2 0)2 1 (22 2 2)2 5 25

(x 1 1)2 1 (y 2 3)2 5 100; (7 1 1)2 1 (23 2 3)2 5 100

(x 1 4)2 1 (y 2 1)2 5 45; (2 1 4)2 1 (22 2 1)2 5 45

(x 2 1)2 1 y2 5 26; (0 2 1)2 1 52 5 26

(x 2 8)2 1 (y 1 2)2 5 85; (1 2 8)2 1 (4 1 2)2 5 85

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Writing the Equation of a Circle from a GraphYou can inspect a graph to fi nd the coordinates of the circle’s center. Use the center and a point on the circle to fi nd the radius. It is easier if you use a horizontal or vertical radius.

Problem

What is the standard equation of the circle in the diagram at the right?

Step 1 Write the coordinates of the center. Th e center is at C(25, 3).

Step 2 Find the radius. Choose a vertical radius: CZ. Th e length is 6, so the radius is 6.

Step 3 Write the equation using the radius and the coordinates of the center.

(x 2 h)2 1 (y 2 k)2 5 r2

(x 2 (25))2 1 (y 2 3)2 5 62

(x 1 5)2 1 (y 2 3)2 5 36

Step 4 Check two points on the circle.

For (1, 3): (1 1 5)2 1 (3 2 3)2 5 62 1 02 5 36

For (211, 3): (211 1 5)2 1 (3 2 3)2 5 62 1 02 5 36

Th e standard equation of this circle is (x 1 5)2 1 (y 2 3)2 5 36.

Exercises

Write the standard equation of each circle. Check two points using your equation.

7. 8.

2

4

2C

6

2 24

( 2, 6)

( 11, 3) (1, 3)

( 5, 3)

68

y

x

Z

8

(0, 4) (7, 4)

8x

O

6

4

6

10

(3, 4)

8 4x

y

O

12-5 Reteaching (continued)

Circles in the Coordinate Plane

x2 1 (y 2 4)2 5 49 Sample: 02 1 (23 2 4)2 5 49 72 1 (4 2 4)2 5 49

(x 1 2)2 1 (y 1 4)2 5 25 Sample: (3 1 2)2 1 (24 1 4)2 5 25 (27 1 2)2 1 (24 1 4)2 5 25

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12-6 Reteaching Locus: A Set of Points

A locus is a set of points that all meet a condition or conditions. Finding a locus is a strategy that can be used to solve a word problem.

Problem

A family on vacation wants to hike on Oak Mountain and fi sh at North Pond and along the White River. Where on the river should they fi sh to be equidistant from North Pond and Oak Mountain?

Draw a line segment joining North Pond and Oak Mountain.

Construct the perpendicular bisector of that segment.

Th e family should fi sh where the perpendicular bisector meets the White River.

Exercises

Describe each of the following, and then compare your answers with those of a partner.

1. the locus of points equidistant from your desk and your partner’s desk

2. the locus of points on the fl oor equidistant from the two side walls of your classroom

3. the locus of points equidistant from a window and the door of your classroom

4. the locus of points equidistant from the front and back walls of your classroom

5. the locus of points equidistant from the fl oor and the ceiling of your classroom

Use points A and B to complete the following.

6. Describe the locus of points in a plane equidistant from A and B.

7. How many points are equidistant from A and B and also lie on *

AB)

? Explain your reasoning.

8. Describe the locus of points in space equidistant from A and B.

9. Draw AB. Describe the locus of points in space 3 mm from AB.

A

B

North Pond

Oak Mountain

White River

Check students’ work.

the line perpendicular to AB at its midpoint

one, the midpoint of AB

the plane perpendicular to AB at its midpoint

Check students’ drawings; a cylinder with radius 3 mm and central axis AB and two hemispheres centered at A and B of radius 3 mm at the bases of the cylinder.

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12-6 Reteaching (continued)

Locus: A Set of Points

10. Two students meet every Saturday afternoon to go running. Describe how they could use the map to fi nd a variety of locations to meet that are equidistant from their homes.

Use what you know about geometric fi gures to answer the following questions.

11. Sam tells Tony to meet him in the northeast section of town, 1 mi from the town’s center. Tony looks at his map of the town and picks up his cell phone to call Sam for more information. Why?

12. How can city planners place the water sprinklers at the park so they are always an equal distance from the two main paths of the park?

13. An old pirate scratches the following note into a piece of wood: “Th e treasure is 50 ft from a cedar tree and 75 ft from an oak.” Under what conditions would this give you one point to dig? two? none?

14. A ski resort has cut a wide path through mountain trees. Skiers will be coming down the hill, but the resort also needs to install the chairlift in the same space. What design allows skiers to ski down the hill with the maximum amount of space between them and the trees and the huge poles that support the chairlift?

15. A telecommunications company is building a new cell phone tower and wants to cover three diff erent villages. What location allows all three villages to get equal reception from the new tower

Path a Path b

Green Park

They could draw a line segment with their homes as the endpoints, and then fi nd its perpendicular bisector through the town. Everywhere the line passes through a street is a possible meeting place.

The sprinklers can be placed along two lines that bisect the angles formed by the two paths.

when the trees are 125 ft apart; when the trees are less than 125 ft apart; when the trees are more than 125 ft apart

The locus of points Sam describes is roughly a quarter of a circle, with a radius of 1 mi.

The resort should install the chairlift to the far left or right of the open area so that skiers can ski down the center line, equidistant from the line of poles and the line of trees.

The tower could be placed at the circumcenter of a triangle formed by the three villages.