Geo_Book_Answers

Answers to Exercises 

0 

CHAPTER 0 • CHAPTER CHAPTER 0 • CHAPTER 

LESSON 0.1 

1. possible answers: snowflakes and crystals; 

flowers and starfish 

2. Answers might include shapes and patterns, 

perspective, proportions, or optical illusions. 

3. Answers will vary. Bilateral symmetry. 

4. A, B, C, and F 

5. A, B, D, and E 

6. 4 of diamonds; none 

7. The line of reflection is a line along the 

surface of the lake. 

8. One line of symmetry is a vertical line through 

the middle of the Taj Mahal, and the other is a 

horizontal line between the Taj Mahal and the 

reflecting pool. The pool reflects the building, 

giving the scene more reflectional symmetry than 

the building itself has. 

9. Designs will vary, but all should have 2-fold 

rotational symmetry. 

2 lines of reflectional 

symmetry 

It is impossible to have 

2 lines of reflectional 

symmetry without 

rotational symmetry. 

10. Answers will vary. 


ANSWERS TO EXERCISES 1 

Answers to Exercises


LESSON 0.2 

1. compass and straightedge 

2. Answer will be the Astrid or 8-Pointed Star, 

possibly with variations. 

3. Answers will be a design from among the three 

choices. 

4. The first design has 4-fold symmetry and four 

lines of reflectional symmetry. The triangle has 

3-fold rotational symmetry and three lines of 

reflectional symmetry. The third design has 6-fold 

rotational symmetry if you ignore color and 2-fold 

rotational symmetry if you don’t. 

5. possible answer: 

2 lines of 

reflectional 

symmetry 

2 ANSWERS TO EXERCISES 


7. 3 lines of reflectional 

symmetry 

3-fold rotational 

symmetry: rotated 

120, 240, 360

LESSON 0.3 

1. Design with reflectional symmetry is drawn on 

this background. 

2. Design should have rotational symmetry. 

Possible answer: 



5. Drawing should resemble the hexagons in the 

given figure, except that each radius, and the side 

of each hexagon, should measure 1 inch. 




LESSON 0.4 

1. Possible answer: The designs appear to go in 

and out of the page (they appear 3-D). The squares 

appear to spiral (although there are no curves in 

the drawing). The spiral appears to go down as if 

into a hole. Lines where curves meet look wavy. 

Bigger squares appear to bulge out. 

2. Possible answer: When zebras group together, 

their stripes make it hard for predators to see 

individual zebras. 


3. Designs will vary. 


5. Possible answers: 2-D: rectangle, triangle, 

trapezoid; 3-D: cylinder, cone, prism. The palace 

facade has one line of reflectional symmetry 

(bilateral symmetry).

LESSON 0.5 

1. possible answers: Scotland, Nigeria 




Cut the middle ring. 

5. 

6. Answers will be a lusona from among the three 

choices. 

7. It means to solve a problem boldly and 

decisively or in a creative way not considered by 

others. 

8. The square knot has reflectional symmetry 

across a horizontal line. The less secure granny 

knot has 2-fold rotational symmetry. 

9. The result is a regular pentagon. 




LESSON 0.6 

1. possible answers: Morocco, Iran, Spain, 

Malaysia 

2. Alhambra 

3. 2.1 cm 

4. in terms of the original square tile: 4-fold in the 

center (center of orange) or corner (center of 

white); 2-fold in the midpoint of the edge of the 

tile (where two orange shapes meet) 



7. Designs will vary, but should contain a regular 

hexagon. 

8. Tessellations will var y.

CHAPTER 0 REVIEW 

1. possible answer: Islamic, Hindu, Celtic 

2. Sometimes the large hexagons appear as blocks 

with a corner removed, and sometimes they appear 

as corners with small cubes nestled in them. 

3. Compass: A geometry tool used to construct 

circles. 

Straightedge: A geometry tool used to construct 

straight lines. 


5. 

6. possible answers: hexagon: honeycomb, 

snowflake; pentagon: starfish, flower 

7. Answers will vary. Should be some form of an 

interweaving design. 

8. Wheel A has four lines of reflectional symmetry; 

WheelChasfivelinesofreflectionalsymmetry. 

Wheels B and D do not have reflectional symmetry. 

9. Wheels B and D have only rotational symmetry. 

Wheels A and B have 4-fold, Wheel C has 5-fold, 

and Wheel D has 3-fold rotational symmetry. 

10, 11. Drawing should contain concentric circles 

and symmetry in some of the rings. 

12. The mandala should contain all the required 

elements. 

13a. The flag of Puerto Rico is not symmetric 

because of the star and the colors. 

13b. The flag of Kenya does not have rotational 

symmetry because of the spearheads. 

13c. possible answers: 

Japan 

Nigeria 






LESSON 1.1 

1. sample answers: point: balls, where lines cross; 

segment: lines of the parachute, court markings; 

collinear: points along a line of the court structure; 

coplanar: each boy’s hand, navel, and kneecap, 

points along two strings 

2. PT ,TP 

3. any two of the following: AR ,RA ,AT ,TA ,RT ,TR 

4. any two of the following: MA ,MS ,AS ,AM ,SA ,SM 

5. A 

B 

6. 7. 

y 

8. AC or CA 9. PQ or QP 

10. TR or RT, RI or IR, and TI or IT 

11. 12. y 

13. mAB 14.3 cm 14. mCD 6.7 cm 

15–17. Check each length to see if it is correct. 

Refer to text for measurements. 

18. R is the midpoint of PQ. X is the midpoint of 

WY . Y is the midpoint of XZ.No midpoints are 

shown in ABC. 

19. possible answers: 

20. 

A 

S 

L 

P 

M 

or 


K 

B 

T 

1 

A C E D 

B 

AC CE 

BC CD 

Q 

D 

–3 

S 

11 

R 

3 

–3 

E 

x 

3 

x 

21. AB ,AC 22. PM,PN 

23. XY ,XZ 24. 

25. 26. 

27. 

28. 10 

29–31. 

32. Yes, P(6, 2), Q(5, 2), R(4, 6); the slope 

between any two of the points is 4 

1 . 

P 

Q 


Q 

N 

34. 

R 

P 

Y 

T 

35. Answers will vary. Possible answers: 

8 cm 

y 

A 

y 

B 

–5 

A 

C 

X 

R´ 

G 

11 cm 

C 

D 

Q´ 

x 

5 

O 

–5 

P´ 

y 

D 

x 

R 

B 

M 

Y 

8 cm 

A 

A (4, 0) 

N 

x 

11 cm 

B 

M

USING YOUR ALGEBRA SKILLS 1 

1. (3, 4) 

2. (9, 1.5) 

3. (5.5, 5.5) 

4. (6, 44) 

5. Yes. The coordinates of the midpoint of a 

segment with endpoints (a, b) and (c, d) are found 

a c 

by taking the average of the x-coordinates, , 2 

b d 

and the average of the y-coordinates, .Thus 2 the 

a c b d 

midpoint is , . 2 2 

6. (3, 2) and (6, 4). To get the first point of 

trisection, sum the coordinates of points A and B 

to get (9, 6), then multiply those coordinates by 1 

 

3 to 

get (3, 2). To get the second point of trisection, sum 

the coordinates of points A and B to get (9, 6), then 

multiply those coordinates by 2 

3 to get (6, 4). This 

works because the coordinates of the first point 

are (0, 0). 

7. Find the midpoint, then find the midpoint of 

each half. 

8a. Midpoints for Figure 1 are (5.5, 6.5); for 

Figure 2, (16, 6.75); and for Figure 3, (29.75, 5.5). 

8b. For these figures the midpoints of the two 

diagonals are the same point. 




LESSON 1.2 

1. TEN, NET, E; FOU, UOF, 1; 

ROU, UOR, 2 

2. 

N 

3. 

4. 

5. S, P, R, Q; none in the second figure 


A 

D 

7. 90° 

8. 120° 

9. 45° 

10. 135° 

11. 45° 

12. 135° 

13. 30° 

14. 90° 

15. Yes; mXQA mXQY 45° 90° 135°, 

which equals mAQY. 

16. 69° 

17. 110° 

18. 40° 

19. 125° 

20. 55° 

21. SML has the greater measure because 

mSML 30° and mBIG 20°. 

22. 

23. 

A T 

G 

M 

B 

A 44 

I 

C 

B 

S 

90 

B 


L 

24. 

25. 

26. no 

C 

A 

27. 

28. 

29. 

30. One possibility is 4:00. 

31. 

H 

32. 

C 

A 

T 

R 

135 

D E 

22 

22 

Y 

90 

6 8 

33. G T 

I 

D 

T H 

A 

67.5 67.5 

A 

N 

D 

N 

O 

S

34. 

B 

35. MY; CK; mI 36. SEU; EUO; MO 

37. x 54° 38. y 102° 

39. z 32° 40. 288° 

41. 242° 42. They add to 360°. 

43. no 

44. You will miss the target because the incoming 

angle is too big. 

Laser light 

source 

W 

Target 

O 

I 

Mirror 

T 

45. 180 km. The towns can be represented by 

three collinear points, P, S, and G.Because S is 

between P and G, PS SG PG. 

46. 

12 cm 

47. 

48. MS DG means that the distance between 

M and S equals the distance between D and G. 

MS DG means that segment MS is congruent 

to segment DG. The first statement equates two 

numbers. The second statement concerns the 

congruence between two geometric figures. 

However, they convey the same information and 

are marked the same way on a diagram. 

49. Answers will vary. Possible answer. 

40 70 

A 4 cm B 

6 cm 6 cm 

4.36 cm 

2.18 cm 2.18 cm 

C 

40 

70 

D 6 cm 

E 

F 




1–3. possible answers: 

1. 

D 

2. 

3. 

4. 

5. 

6. 

7. 

8. 

45 

O G 

R 

B 

A 

D 

T E 

M 

A 

C 

40 

I G 

A 

P E 

R 

B 

B 

50 

140 

C 

LESSON 1.3 

9. B is a Zoid. A Zoid is a creature that has in its 

interior a small triangle with a large black dot 

taking up most of its center. 

10. A good definition places an object in a class 

and also differentiates it from other objects in that 

class. A good definition has no counterexamples. 


E 

G 

S 

D 

D 

E 

40 

11. The measures of complementary angles sum 

to 90°, whereas the measures of supplementary 

angles sum to 180°. 

12. No, supplementary angles can be unconnected, 

while a linear pair must share a vertex and 

a common side. 

13a. an angle; measures less than 90° 

13b. angles; have measures that add to 90° 

13c. a point; divides a segment into two congruent 

segments 

13d. a geometry tool; is used to measure the sizes 

of angles in degrees 

14. 

If AB and CD intersect at point P so that P is 

between A and B, and P is between C and D, then 

APC and BPD are a pair of vertical angles. 

15. true 

16. true 

17. true 

18. false 

19. false 

A 

B 

P 

C 

D

20. false 

21. true 

22. false; Though the converse is true, a 

counterexample is 

C T 

23. false 

D 

24. false 

C T 

25. possible answers: (3, 0), (0, 2), or (6, 6) 

26. possible answers: (2, 3) or (5, 1) 

27. The reflected ray and the ray that passes 

through (called the refracted ray) are mirror 

images of each other. Or they form congruent 

angles with the mirror. 

B 

50 80 

A 

40 

140 

C 

A 

Mirror 

A 

50 

Reflected 

segment 

28. One possible answer is A(8, 8), B(4.5, 6.5), 

C(11, 1). 

C 

29. 12 cm 

30. 36° 


32. 

33. 

A 

–12 –6 

B 

1 pt. 

34a. 120° 

 

360° 

1 

, 2 

3 3 

left 

60° 

34b. 360° 

1 

, 1 

6 6 

missing 

34c. 360° 

 

9 

40° 

S 

T 

3 pt. 

6 

y 

R 

1 pt. 

3 pt. 

4 pt. 

2 pt. 

5 pt. 6 pt. 

x 

Infinitely 

many points 

0 pt. 

2 pt. 




1. 


3. 

LESSON 1.4 

4. octagon 

5. hexagon 

6. heptagon 

7. pentagon 

8–10. One possible answer for each is shown. 

8. pentagon FIVER 

9. quadrilateral FOUR 

10. equilateral quadrilateral BLOC 

11a. a polygon; has eight sides 

11b. a polygon; has at least one diagonal outside 

of the polygon 

11c. a polygon; has 20 sides 

11d. a polygon; has all sides of equal length 

12. One possibility is C and Y are consecutive 

angles; CY and YN are consecutive sides. 

13. 9; possible answer: 

14. AC, AD , BD, BE, CE 

15. TIN 

16. WEN 

17a. a 44, b 58, c 34 

17b. mT 87° and mI 165° 


18. PA FI and IVE ANC 



21. 84 in. 

22. 5.25 cm 

23. AB 14 m, CD 25 m 

24. complementary angles: AOS and SOC; 

vertical angles: OCT and ECR or TCE and 

RCO 

25. 


5 

130 130 

5 cm 

20 

130 

E 

A B 

C D 

F 

27. All are possible except two points. 

0 pts. 

3 pts. 

5 pts. 

1 pt. 

4 pts. 

6 pts.

LESSON 1.5 

1. D 2. A 

3. C 4. B 

5. C 

6. C 

7. 


2a 


A 


A 


A 

12. possible answers: (1, 1), (1, 0), or (4, 3) 

13. possible answers: (3, 3), (3, 4), (11, 4), 

(11, 3), (0.5, 0.5), or (7.5, 0.5) 

14. possible answers: (3, 1), (1, 9), (2, 2), 

(4, 8), (0, 1), or (1, 6) 


A 

4 cm 

A T 

A 

R 

b 

12 cm 

2a 

C 

4 cm 

120 

P 

45 

9 cm 

3a 

6 cm 6 cm 

B A 

40 40 

C 

80 

4 cm 

C 

3a 

b – 2a 

4 cm 

B 

B 

C 

Z 

C 

A 

7 cm 

45 

C 

9 cm 

S M L 

B 

B 


A 

A 

17. true 

18. true 

19. False, a diagonal connects nonconsecutive 

vertices. 

20. False, an angle bisector divides an angle into 

two congruent angles. 

21. true 

(–3, 5) 

(–4, 1) 

40 

40 

22. (4,1) → (3,2) (1,1) → (2, 2) (2, 4) → 

(3, 1) (3, 5) → (2, 2) Yes, the quadrilaterals are 

congruent. 

23. Find the midpoint of each rod. All the 

midpoints lie on the same line; place the edge 

of a ruler under this line. 

24–26. Sample answers for 24–26 

24. P 

25. 

P 

E 

14 cm 

10 cm 

y 

F 

(2, 4) 

(1, 1) 

N T 

26. Q U 

C 

10 cm 

A 

10 cm 

D A 

x 

B 

E 

E 

N T 


A 



LESSON 1.6 

1. first figure: quadrilateral ABCD with two 

congruent sides and one right angle; second 

figure kite EFGH; third figure: trapezoid IJKL 

with two right angles; fourth figure parallelogram 

MNPQ 

2. B 3. D 

4. F 

6. A, D, F 

5. C 

7. D I 

8. 

9. 

10. 

11. 

Z O 

E 

T 

R 

B 

N 

L 

A 

E U 

Q 

12. square 

13. possible answers: (2, 6), (2, 4); (6, 2), 

(2, 4); (3, 1), (1, 3) 

14. 90 cm 

15. (5, 6) 

16. S(9, 0), I(4, 2) 

17. S(3, 0), I(1, 5) 

18. A(7, 6), N(5, 9) or A(5, 2), N(7, 1) 

19a. Possible answer: Flip one triangle and align it 

so that a congruent pair of sides forms a diagonal 

of a kite. 


I 

F 

H 

G 

19b. Possible answer: Rotate one triangle so that 

a congruent pair of sides forms a diagonal of a 

parallelogram. 

20. There are three possibilities: a rhombus, a 

concave kite, or a parallelogram. 

21a. right triangles 

21b. isosceles right triangles 

22. 

23. 

24. 

25. no 

y 

5 cm 

72 72 

72 72 

72 

C (0, 5) 

B (4, 4) 

x 

A (5, 0) 

5 cm

LESSON 1.7 

1. Answers will vary. Sample answers:The green 

areaintheirrigationphotoisacircle,thewaterisa 

radius,and a path on the far side of the circle appears 

to be tangent to the circle.The wood bridge is an arc of 

a circle,and the railings are arcs of concentric circles. 

The horizontal support beam under the bridge is 

achord. 

2. three of the following: AB, BD, EC, EF 

3. EC 

4. AP, EP, FP,BP, CP 

5. five of the following: EF ,AE ,AB ,BC ,CD ,DF , 

EB ,ED ,FC ,AC ,DB ,AF ,AD ,BF 

6. EDC 

or EFC 

,EBC 

or EAC 

 

7. two of the following: ECD 

,EDF 

,FEC 

,DEC 

,... 

8. FG , HB 

9. either F or B 

10. possible answers: cars, trains, motorcycles; 

washing machines, dishwashers, vacuum cleaners; 

compact disc players, record players, car racing, 

Ferris wheel 

11. mPQ 110°; mPRQ 

250° 

12. 

13. possible answers: concentric rings on cross 

sections of trees (annual rings),bull’s-eye or target, 

ripples from a rock falling into a pond. 

14. 

15. 

65 

215 

P Q 

16. Equilateral quadrilateral. (The figure is 

actually a rhombus, but students have not yet 

learned the properties needed to conclude that the 

sides are parallel.) 

17. equilateral; 3 to 1 

18. 

19. yes; yes 

20. yes; no 

21. no; no 

8 

–5 

–5 

y 

P 

s 

5 

–5 

5 

–5 

B 

y 

y 

A 

Q 

5 

5 

x 

x 

8 16 

x 




22. 80°. The slices of pizza do not overlap, so 

60° x 140°, where x is the measure in 

degrees of the angle of the second slice. 

23. 

24. not possible 

25. 

26. 

27. about 0.986° 

28. 15° 

29. 

30. 

C A 

C B 

A 100 

F 

A 

P A 

T R 


B 

T 

31. 

32. not possible 

33. R 

34. 

35. 

36. 

37. 

T 

E 

K 

60 60 

120 120 60 120 

4a + 2b 

50 70 50 70 

55 

2p 

I 2p 

U 

2p 2p 

E 

120 

I 

Q 

G 

T 

120 

60

LESSON 1.8 

1. 2. 

3. 4. 

5. 6. 

7. 8. 

9. 60 boxes 10. 

5 

3 

4 

11. 12. 

13. B, D 14. B 

15. C 16. D 

17. A 

18. 19. 

20. true 21. true 

2 

3 

4 

22. False. The two lines are not necessarily in the 

same plane, so they might be skew. 


25. true 

26. False. They divide space into seven or eight 

parts. 

27. true 28. (3, 1) 

29. perimeter 20.5 cm; m(largest angle) 100° 

30. 

13 cm 

120 

8 cm 




LESSON 1.9 

1. Sample answer: Furniture movers might 

visualize how to rotate a couch to get it up a 

narrow staircase. 

2. yes 

3. W ; O M E N Nadine is ahead. 

4. 28 posts 

5. 28 days 

6. 0 ft (The poles must be touching!) 

7. 

8. 

9. 

10. 

A B 

A 

3 cm 

A 

A 

Quadrilaterals 

Trapezoids 

11. no 

12. C(2, 3), A(0, 0), B(0, 5), D(2, 1) 

13. Y(4, 1), C(3, 1), N(0, 3) 

14. (x, y) → (x 3, y 2) 

15. AB CP , EF GH ; i k, j k 

16. perimeter 34 cm 

17. Left photo: Three points determine a plane. 

Middle photo: Two intersecting lines determine a 

plane. Right photo: A line and a point not on the 

line determine a plane. 


B 

B 

A 

Polygons 

3 cm 

3 cm 

Triangles 

B 

Obtuse Isosceles 

18. pyramid with hexagonal base 

19. prism with hexagonal base 

20. pyramid with square base 

21. x 15, y 27 

22. x 12, y 4 

23. 

24. 

25. point, line, plane 

26. AB 27. AB 

28. vertex 29. AB 

30. AB CD 31. protractor 

32. ABC 

34. congruent to 

33. AB CD 

35. The distance is two times the radius. 

P 

36. They bisect each other and are perpendicular. 

P 

r 

PQ = 2r 

A 

B 

r 

Q 

Q


1. true 

2. False; it is written as QP . 

3. true 

4. False; the vertex is point D. 

5. true 

6. true 

7. False; its measure is less than 90°. 

8. false; two possible counterexamples: 

C 

9. true 

10. true 

11. False; they are supplementary. 

12. true 

13. true 

14. true 

15. False; it has five diagonals. 

16. true 

17. F 

18. G 

19. L 

20. J 

21. C 

22. I 

23. no match 

24. A 

25. no match 

26. T 

27. 

28. 

A 

T 

N 

P 

E Y 

K 

D 

B 

APD and APC 

are a linear pair. 

A 

G 

5 

7 

R 

I E 

C 

A 

O 

D 

S 

P 

APD and APC 

are the same angle. 

3 

P 

29. 

30. 

31. 

32. 

33. 

34. 

35. 

36. 

37. 

Y 

C 

2 in. 

R T 

A P 

40 

P 

5 in. 

A 

B 

125 

A 

T 

D 

E 

3 in. 




38. 114° 

39. x 2, y 1 

40. x 12, y 4 

41. x 4, y 2.5 

42. x 10, y 8 

43. AB 16 cm 

44. 96° 

45. 105° 

46. 30° 

47. 

48. 66 inches 

49. 3 feet 

50. The path taken by the midpoint of the ladder is 

anarcofacircleoraquarter-circleiftheladder 

slides all the way from the vertical to the horizontal. 

Shed 

Quadrilateral 

Rectangle Square Rhombus 

Path of flashlight 

Trapezoid 


51. He will get home at 5:46 (assuming he goes 

inside before he gets blown back again). 

52. (2, 3) 

53. 

54. 

55. 

56. 

57.


CHAPTER 2 • CHAPTER 2 

CHAPTER 2 • CHAPTER 

LESSON 2.1 

1. “All rocks sink.” Stony needs to find one rock 

that will not sink. 

2. If two angles are formed by drawing a ray from a 

line, then their measures add up to 180°. 

3. 10,000, 100,000, ....Each term is 10 times the 

previous term. 

4. 5 ,1,....(Written 6 with the common 

denominator 6, the pattern becomes 

1 , 2 , 3 , 4 , 5 , 6 

6 6 6 6 6 6 ,....) 

5. 17, 21 6. 28, 36 

7. 21, 34 8. 49, 64 

9. 10, 24 10. 64, 128 

11. 12. 

13. 14. 

15. 16. 

17. 1, 4, 7, 10, 13 18. 15, 21, 28, 36, 45 


20. sample answers:3,6,12,24,48,...and 4,8,12, 

16, 20, . . . 


22. 7th term: 56; 10th term: 110; 25th term: 650 

23. Conjecture is false; 14 2 196 but 41 2 1681. 

24. 11,111 11,111 123,454,321. For the sixth 

line, if 111,111 is multiplied by itself, then the 

product will be 12,345,654,321. But 1,111,111,111 

1,111,111,111 1,234,567,900,987,654,321. 

25. 26. 

27. 28. 


30. sample answer: 

31. collinear 32. isosceles 

33. dodecagon 34. parallel 

35. protractor 36. radius 

37. diagonal 38. regular 

39. 90° 40. perpendicular 

41. sample answer: 42. sample answer: 

A G 

N T 

I 

43. sample answer: 44. possible answer: 

45. Methods will vary. It is not possible to draw a 

second triangle with the same angle measures and 

side length that is not congruent to the first. 

C 

40 60 

A 9 cm B 

A C Clearly AC 

does not 

bisect A 

or C. 




1. See table below. 



1. (Lesson 2.2) 

LESSON 2.2 

n 1 2 3 4 5 6 … n … 20 

f (n) 3 9 15 21 27 33 … 6n 3 … 117 

2. (Lesson 2.2) 

n 1 2 3 4 5 6 … n … 20 

f (n) 1 2 5 8 11 14 … … 56 

3. (Lesson 2.2) 

n 1 2 3 4 5 6 … n … 20 

f (n) 4 4 12 20 28 36 … … 148 

4. (Lesson 2.2) 

Number of sides 3 4 5 6 … n … 35 

Number of triangles formed 1 2 3 4 … n 2 … 33 

5. (Lesson 2.2) 

Figure number 1 2 3 4 5 6 … n … 200 

Number of tiles 8 16 24 32 40 48 … 8n … 1600 

6. (Lesson 2.2) 


Number of tiles 1 5 9 13 17 21 … 4n 3 … 797 

7. (Lesson 2.2) 


Number of matchsticks 5 9 13 17 21 25 … 4n 1 … 801 

Number of matchsticks 

in perimeter of figure 

5 8 11 14 17 20 … 3n 2 … 602 






3n 4 

8n 12

8. 8n is steeper; the coefficient of n. 

f (n) 

50 

40 

30 

20 

10 

2 

9. C n H 2n2 

H C C C C C C C C 

10. y 3 

x 

2 

3 

11. 

L 

12. O 

H H H H H H H H 

H H H H H H H H 

E 

4 

Octane (C 8 H 18 ) 

T 

4n 1 

4n 3 

3n 2 

6 8 

L 

8n 

M 

n 

Q 

Y 

H 

13. 

14. 

15. Márisol should respond by noticing that all 

the triangles José drew were isosceles, but it is 

possible to draw a triangle with no two sides 

congruent. In other words, she should show him a 

counterexample. 

16. Methods will vary. It is not possible to draw 

another triangle with the given side lengths and 

angle measure that is not congruent to the first. 

9 cm 

R E 

T C 

2 

2 

C 

45 

A 8 cm B 

2 

1 1 

1 1 

3 3 

1 1 

1 1 2 

17. She could try eating one food at a time; 

inductive. 

18. 2600 




1. 

See table below. 

2. n 1, 36 

LESSON 2.3 

3. You can draw a diagonal from one vertex to all 

the other vertices except three: the two adjacent 

vertices and the vertex itself. 

4. n(n 3) 

, 

2 

560 

5. n(n 1) 

,595 

2 

6. n(n 1) 

,595 

2 

7. Answers will include relationships between 

points in Exercises 5 and 6 and vertices of a polygon. 

The total number of segments connecting n random 

points is the number of diagonals of the n-sided 

polygon, n(n 3) 

, 2 plus the number of sides, n. 

Thus, the total number of segments connecting n 

random points is n(n 3) 2n 

n(n 3) 2n 

 

2 2 

n2 3n 

2n n 

 

2 

2 n 

 

2 n(n 1) 

. 2 

1. (Lesson 2.3) 


 

2 

8. Use n(n 1) 

2 

to get 780 direct lines. Use a central 

hub with a line to each house to get 40 lines. The 

art shows the direct-line solution and the practical 

solution for six houses. 

9. Use points for the 10 teams and segments 

connecting them to represent four games played 

between them. So, use 4 n(n 1) 

 

2 to get 180 games 

played. 

10. If n(n 

1) 

 

2 66, then n(n 1) 132. What 

two consecutive numbers multiply to equal 132? 

12 times 11. Thus, there were 12 people at the 

party. 

11. true 

12. true 

13. False; an isosceles right triangle has two sides 

congruent. 

14. false 

A 

E 

B C 

D 

AED and BED are 

not a linear pair. 

15. False; they are parallel. 

16. true 

17. False; a rectangle is a parallelogram with all of 

its angles congruent. 

18. False; a diagonal is a segment in a polygon 

connecting any two nonconsecutive vertices. 

19. true 

20. 5049 

Lines 1 2 3 4 5 … n … 35 

Regions 2 4 6 8 10 … 2n … 70

LESSON 2.4 

1. inductive; deductive 

2. mB 65°; deductive 

3. inductive 

4. DG 258 cm; deductive 

5. 180°; 180°; The sum of the five angles is 180°; 

inductive. 

64 

27 

20 

37 

25 

32 

36 

6. LNDA is a parallelogram; deductive. 

7. Possible answer: AB CD, so AB BC 

CD BC.Because AB BC AC and CD 

BC BD, AC BD, so AC BD. 

8. 3; 10; 7 

9. Just over 45°; if m 90°, then 1 m 2 45°. 

10. 48°; 17°; 62°; mCPB 

11. Possible answers: Because mAPC 

mBPD, add the same measure to both sides to get 

mAPC mCPD mBPD mCPD.By 

angle addition, mAPC mCPD mAPD 

and mBPD mCPD mCPB.Therefore, 

mAPD mCPB. 


13. The pattern cannot be generalized because 

once the river is straight, it cannot get any shorter. 

14. 900, 1080 

15. 75, 91 

16. 4 

5 ,12 

17. 

18. 

19. 

20. 

21. Sample answer: There were three sunny days 

in a row, so I assumed it would be sunny the fourth 

day, but it rained. 

22. L 23. M 

24. A 25. B 

26. E 27. C 

28. G 29. D 

30. H 31. I 

32. 

W 

33. 

34. 

35. 

I 

B 

D 

L 

O 

K 

140 

G 

C 

D 

N 

D 

T 

E 

O 




LESSON 2.5 

1. a 60°, b 120°, c 120° 

2. a 90°, b 90°, c 50° 

3. a 77°, b 52°, c 77°, d 51° 

4. a 60°, b c 120°, d f 115°, e 65°, 

g i 125°, h 55° 

5. a 90°, b 163°, c 17°, d 110°, e 70° 

6. The measures of the linear pair of angles add up 

to 170°, not 180°. 

7. The angles at which he should cut measure 45°. 

8. Greatest: 120°. Smallest: 60°. One possible 

explanation: The tree is perpendicular to the 

horizontal. The angle of the hill measures 30°. 

The smaller angle and the angle between the hill 

and the horizontal form a pair of complementary 

angles, so the smaller angle equals 90° 30° 60°. 

The smaller angle and larger angle form a linear 

pair, so the larger angle equals 180° 60° 120°. 

9. The converse is not true. ; sample counterexample: 

A 

55 

125 

B 

10. each must be a right angle 

11. Let the measures of the congruent angles be x. 

They are supplementary, so x x 180°, 2x 180°, 

x 90°. Thus each angle is a right angle. 

12. The ratio is 1. The ratio does not change as 

long as the lines don’t coincide. Because the 

demonstration does not explain why, it is not 

a proof. 

13. P 

14. 

3 

A 

5 


T 

22. (Lesson 2.5) 

15. 

16. 

17. 

6 

18. Possible answer: All the cards look exactly as 

they did, so it must be the 4 of diamonds, because 

it has rotational symmetry while the others do not. 

19. 22.5° 

20. 

21. C n H 2n 

H H 

H C C 

H H 


23. handshake problem: n(n 1) 

2 

pieces of string 

24. 3160 intersections 

25. n(n 2) 

 

2 yields 760 handshakes. 

26. 21; 252 

27. n(n 3) 

2 

3 4 

H 

C 

H 

H 

C 

H 

H 

C 

H 

80( 79) 

; 

2 

560; there are 35 vertices. 

28. M is the midpoint of AY;deductive. 

Rectangle 1 2 3 4 5 6 … n … 200 

Perimeter of rectangle 10 14 18 22 26 30 … 4n 6 … 806 

O 

A 

T 

H 

C 

H 

C 

H 

H 

C 

H 

3160 

Number of squares 6 12 20 30 42 56 … … 40,602 

(n 1)(n 2)

LESSON 2.6 

1. 63° 2. 90° 

3. no 4. 57° 

5. yes 6. 113° 

7. a d 64°, b c 116°, e g i j 

k 108°, f h s 72°, m 105°, n 79°, 

p 90°, q 116°, t 119°; Possible explanation: 

Using the Vertical Angles Conjecture, n 79°. 

Using the Linear Pair Conjecture, p 90°. Using the 

Corresponding Angles Conjecture and b 116°, 

q 116°. 

8. Possible answer: In the 

diagram, lines and m are parallel 

and intersected by transversal k. 

Using the Corresponding Angles 

Conjecture,1 2. Using the 

Vertical Angles Conjecture, 

2 3. Because 1and3arebothcongruent 

to 2,theymustbecongruenttoeachother.So 

1 3. Therefore,if two parallel lines are cut by a 

transversal, then alternate exterior angles are 

congruent. 

9. 56° 114° 170° 180°. Thus, the lines 

marked as parallel cannot really be parallel. 

10. Alternate interior angles measure 55°, but 

55° 45° 180°. 

11. The incoming and 

outgoing angles measure 

45°. Possible explanation: 

Yes, the alternate interior 

angles are congruent, and 

thus, by the Converse of the 

Parallel Lines Conjecture, 

the mirrors are parallel. 

12. Explanations will vary. 

Sample explanation:“I used the 

protractor to make corresponding 

angles congruent when I 

drew line PQ.” 

26. (Lesson 2.6) 

A 

45 

90 

45 

90 

45 

P 

45 

1 

k 

2 

3 

Q 

B 

 

m 

13. No, tomorrow could be a holiday. Converse:“If 

tomorrow is a school day, then yesterday was part 

of the weekend;” false. 

14. 42° 15. 20° 

16. No, the lines are not parallel. 

17. isosceles triangle 

18. a parallelogram that is not also a rectangle or a 

rhombus 

19. 18 cm 20. 39° 

21. 3486 

22. 30 squares (one 4-by-4, four 3-by-3, nine 

2-by-2, and sixteen 1-by-1) 

23. The triangle moved to the left 1 unit.Yes, 

congruent to original. 

24. The quadrilateral was reflected across both 

axes or rotated 180° about the origin.Yes, congruent 

to original. 

25. The pentagon was reflected across the line 

x y. Yes, congruent to original. 

L 

M' 

E 

E' 

y 

4 

5 

N 

M 

y 

O' 

L' 

O 

6 

N' 



Number of yellow squares 2 3 4 5 6 7 … n 1 … 36 

Number of blue squares 3 5 7 9 11 13 … 2n 1 … 71 

Total number of squares 5 8 11 14 17 20 … 3n 2 … 107 

y 

5 

5 

4 

x 

x 

x 





1. 2 

2. 12 

 

13 

97 

3. 46 

2.1 

4. y 23 

5. x 4 

6. Any point of the form (3p, 4p).Possible answer: 

(6,8),(9,12),and (12,16).To find another 

point go right 3 and down 4. 


7. 66.7 mi/h 

8. At 6 m/s, Skater 1 is 4 m/s faster than Skater 2 at 

2m/s. 

9. A 100% grade has a slope of 1. It has an 

inclination of 45°, so you probably could not 

drive up it. You might be able to walk up it. Grades 

higher than 100% are possible, but the angle of 

inclination would be greater than 45°. 

10. The slope of the adobe house flat roof is 

approximately 0. The Connecticut roof is steeper, 

with a slope of about 2, and will shed the snow.


1. poor inductive reasoning, but Diana was 

probably just being funny 



4. 19, 30 

5. S, 36 

6. 2, 5, 10, 17, 26, 37 

7. 1, 2, 4, 8, 16, 32 

8. 9. 

10. 900 

11. 930 



14. n2 ,302 900 

15. n(n 

1) 100( 101) 

, 2 2 

5050 

16. n(n 1) 

 

2 

741; therefore, n 39 

17. n(n 1) 

 

2 

2926; therefore, n 77 

18. n 2, 54 2 52 

19a. possible answers: EFC and AFG, AFE 

and CFG, FGD and BGH, or BGF and 

HGD 

19b. possible answers: AFE and EFC, AFG 

and CFG, BGF and FGD, or BGH and 

HGD (there are four other pairs) 

19c. possible answers: EFC and FGD, AFE 

and BGF, CFG and DGH, or AFG and 

BGH 

12. (Chapter 2 Review) 

19d. possible answers: AFG and FGD or 

CFG and BGF 

20. possible answers: GFC by the Vertical 

Angles Conjecture, BGF by the Corresponding 

Angles Conjecture, and DGH, using the 

Alternate Exterior Angles Conjecture 

21. True. Converse: “If two polygons have the 

same number of sides, then the two polygons are 

congruent”; false. Counterexamples may vary; 

students might draw a concave quadrilateral and a 

convex quadrilateral. 

22. 

4.5 cm 

56 

56 

124 

7 cm 

7 cm 

56 

124 56 

4.5 cm 

23. PV RX and SU VX ; explanations will vary. 

24. The bisected angle measures 50° because of 

AIA. So each half measures 25°. The bisector is a 

transversal, so the measure of the other acute angle 

in the triangle is also 25° by AIA. However, this 

angle forms a linearpairwiththeanglemeasuring 

165°,and 25° 165° 180°, which contradicts the 

Linear Pair Conjecture. 

25. a 38°, b 38°, c 142°, d 38°, e 50°, 

f 65°, g 106°, h 74°; The angle with measure e 

forms a linear pair with an angle with measure 130° 

because of the CorrespondingAngles Conjecture.So 

e measures 50° because of the Linear Pair Conjecture. 

The angle with measure f is half of the angle with 

measure 130°, so f65°.The angle with measure g is 

congruent to the angle with measure 106° by the 

CorrespondingAngles Conjecture,so g106°. 

n 1 2 3 4 5 6 … n … 20 

f(n) 2 1 4 7 10 13 … … 55 


3n 5 

n 1 2 3 4 5 6 … n … 20 

f(n) 1 3 6 10 15 21 … … 210 

n(n 1) 

 

2 





CHAPTER 3 • CHAPTER 3 

CHAPTER 3 • CHAPTER 

1. 

2. 

3. 

4. 


LESSON 3.1 

6. m3 m1 m2; possible answer: 


A B 

8. 

A 

C 

E 

L 

G 

1 2 3 

C 

AB 

AB 

AB CD 

AB 2EF CD 

copy 

AC BC 

AB 


F 

D 

E 

B 

EF 

CD 

CD 

EF 

2 

1 

9. For Exercise 7, trace the triangle. For Exercise 8, 

trace the segment onto three separate pieces of 

patty paper. Lay them on top of each other, and 

slide them around until the segments join at the 

endpoints and form a triangle. 

10. One method: Draw DU . Copy Q and 

construct COY QUD.Duplicate DUA 

at point O.Construct OYP UDA. 

Q 

U 

11. One construction method is to create congruent 

circles that pass through each other’s center. One 

side of the triangle is between the centers of the 

circles; the other sides meet where the circles 

intersect. 

12. a 50°, b 130°, c 50°, d 130°, e 50°, 

f 50°, g 130°, h 130°, k 155°, l 90°, 

m 115°, n 65° 

13. west 

14. An isosceles triangle is a triangle that has at 

least one line of reflectional symmetry.Yes, all 

equilateral triangles are isosceles. 

15. 

16. new coordinates: A(0, 0), Y(4, 0), D(0, 2) 

17. Methods will vary. It isn’t possible to draw a 

second triangle with the same side lengths that is 

not congruent to the first. 

11 cm 8 cm 

10 cm 

6 

Y 

y 

Y' 

–6 D A' A 6 

–6 

A 

D 

D' 

C 

O 

x 

P 

Y

1. 

2. 

3. 

4. 

5. 

A B 

Q D 

C D 

1 

CD 

2 

Edge of the paper 

Original segment 

1 

CD 

2 

2AB – 

1 

CD 

2 

M N 

AB CD 

LESSON 3.2 

AB AB 

1 

CD 

2 

MN = 

1 

(AB + CD) 

2 

6. Exercises 1–5 with patty paper: 

Exercise 1 This is the same as Investigation 1. 

Exercise 2 

Step 1 Draw a segment on patty paper. Label it QD . 

Step 2 Fold your patty paper so that endpoints Q 

and D coincide. Crease along the fold. 

Step 3 Unfold and draw a line in the crease. 

Step 4 Label the point of intersection A. 

Step 5 Fold your patty paper so that endpoints Q 

and A coincide. Crease along the fold. 


Step 7 Label the point of intersection B. 

Step 8 Fold your patty paper so that endpoints A 

and D coincide. Crease along the fold. 


Step 10 Label the point of intersection C. 

Exercise 3 This is the same as Investigation 1. 

Exercise 4 

Step 1 Do Investigation 1 to get 1 

2 CD. 

Step 2 On a second piece of patty paper, trace AB 

two times so that the two segments form a segment 

of length 2AB. 

Step 3 Lay the first piece of patty paper on top of 

the second so that the endpoints coincide and the 

shorter segment is on top of the longer segment. 

Step 4 Trace the rest of the longer segment with a 

different colored writing utensil. That will be the 

answer. 

Exercise 5 

Step 1 Trace segments AB and CD so that the two 

segments form a segment of length AB CD. 

Step 2 Fold your patty paper so that points A and 

D coincide. Crease along the fold. 


7. The perpendicular bisectors all intersect in one 

point. 

8. The medians all intersect in one point. 

C 

A 

M 

I 

9. GH appears to be parallel to EF, and its length 

is half the length of EF. 

G 

B 

N 

D H 

E 

L 

F 

L 

A 




10. The quadrilateral appears to be a rhombus. 

11. 

Any point on the perpendicular bisector of the 

segment connecting the two offices would be 

equidistant from the two post offices. Therefore, 

any point on one side of the perpendicular bisector 

would be closer to the post office on that side. 

12. It is a parallelogram. 

T 

E 

R 

D 

I 

V 

Ness Station 

F L 

A 

O 

C 

S 

Umsar 

Station 

13. The triangles are not necessarily congruent, 

but their areas are equal. A cardboard triangle 

would balance on its median. 


14. One way to balance it is along the median. The 

two halves weigh the same. 

sample figure: 

C 

D 

A 

Area CDB = 158 in 2 

Area DAB = 158 in 2 

15. F 16. E 

17. B 18. A 

19. D 20. C 

21. B, C, D, E, H, I, O, X (K in some fonts, though 

not this one) 

22. Methods will vary. 

10 cm 

40 70 

A B 

C 

B 

Ruler 

It is not possible to draw a second triangle with the 

same angle and side measures that is not congruent 

to the first.

1. 

LESSON 3.3 

The answer depends on the angle drawn and where 

P is placed. 

2. C 

3. 

Two altitudes fall outside the triangle, and one falls 

inside. 

4. From the point, swing arcs on the line to 

construct a segment whose midpoint is that point. 

Then construct the perpendicular bisector of the 

segment. 

5. Construct perpendiculars from point Q and 

point R.Mark offQS and RE congruent to QR . 

Connect points S and E. 

Q 

S 

A 

B 

U 

B 

D 

I 

O T 

R 

E 

P 

B 

G 

6. The two folds are parallel. 

7. Fold the patty paper through the point so that 

two perpendiculars coincide to see the side closest 

to the point. Fold again using the perpendicular 

of the side closest to the point and the third 

perpendicular; compare those sides. 

8. Draw a line. Mark two points on it, and label 

them A and C. Construct a perpendicular at C. 

Mark off CB congruent to CA. The altitude CD is 

also the angle bisector, median, and perpendicular 

bisector. 

A 

C 

9. 

10. 

11. 

12. 

Q 

T 

P 

D 

B 

O B 

U 

E L 

A B 

Complement 

of A 

A 





14. 

15. 

16. 

17. 

F 

A 

C D 

B 

Y 

I T 

E 

F 

13. (Lesson 3.3) 


18. not congruent 



5 cm 

6 cm 

40 

5 cm 

8 cm 

9 cm 

9 cm 

Rectangular pattern with triangles 

Rectangle 1 2 3 4 5 6 … n … 35 

Number of shaded 

triangles 

2 9 20 35 54 77 … … 2484 

40 

8 cm 

6 cm 

(2n 1)(n 1)

1. D 

2. F 

3. A 

4. C 

5. E 

6. 

7. 

8. 

9a, b. 

9c. 

10. 

R 

M 

M 

z 

M 

Altitude 

M O S 

135 

z 

z 

E 

45 

90 45 

45 

E 

S 

LESSON 3.4 

A 

P 

R A 

R P 

A 

U 

B 

M 

Angle 

bisector 

Median 

P 

11. 

12. The angle bisectors are perpendicular. The 

sum of the measures of the linear pair is 180°. The 

sum of half of each must be 90°. 

13. If a point is equally distant from the sides of 

an angle, then it is on the bisector of an angle. This 

is true for points in the same plane as the angle. 

Mosaic answers: Square pattern constructions: 

perpendiculars, equal segments, and midpoints; 

The triangles are not identical, as the downward 

ones have longer bases. 

14. y 110° 

15. mR 46° 

16. Angle A is the largest; mA 66°, 

mB 64°, mC 50°. 

17. STOP 

18. 

A 

N 

L 

G 

T 

O 

I 

S 




19. 

20. 

B 

5.6 cm 

A 

A 

130 

3.5 cm C 

B 

6.5 cm 

C 

21. No, the triangles don’t look the same. 

40 60 40 

8 cm 

60 

8 cm 


22a. A web of lines fills most of the plane, except 

a U-shaped region and a V-shaped region. (The 

U-shaped region is actually bounded by a section 

of a parabola and straight lines. If AB were 

extended to AB , the U would be a complete 

parabola.) 

B 

C 

D 

Parabola 

A 

22b. a line segment parallel to AB and half the 

length (The segment is actually the midsegment 

of ABD.)

LESSON 3.5 

1. 2. 

3. Construct a segment with length z. Bisect the 

z 

segment to get length . 2 Bisect again to get a 

z . Construct a square with 

segment with length 

4 

z 

each side of length . 4 

4. 

5. sample construction: 

T 

6. Draw a line and construct ML perpendicular to 

it. Swing an arc from point M to point G so that 

MG RA. From point G, swing an arc to construct 

RG. Finish the parallelogram by swinging an arc of 

length RA from R and swinging an arc of length GR 

from M. There is only one possible parallelogram. 

G 

A 

1 z 

4 

P 

A 

1 z 

2 

L 

M A 

7. 1 S, 2 U 

x 

x 

A 

R 

R 

z 

x 

x 

x 

8. 1 S and 2 U by of the Alternate 

Interior Angles Conjecture 

9. The ratios appear to be the same. 

10. 1 3 and 2 4 by the 

Corresponding Angles Conjecture 

11. A parallelogram 

12. Using the Converse of the Parallel Lines 

Conjecture, the angle bisectors are parallel: 

DAB ABC, so AD BC . 

13. Construct the perpendicular bisector of each 

of the three segments connecting the fire stations. 

Eliminate the rays beyond where the bisectors 

intersect. A point within any region will be closest 

to the fire station in that region. 

14. Z O 

15. B 

M 

16. 

D 

R 

D 

B 

A C 

T 

R E 

W 

K C 

RC = KE = 8 cm 

I 

60 

R O 

17. a 72°, b 108°, c 108°, d 108°, e 72°, 

f 108°, g 108°, h 72°, j 90°, k 18°, l 90°, 

m 54°, n 62°, p 62°, q 59°, r 118°; 

Explanations will vary. Sample explanation: 

c is 108° because of the Corresponding Angles 

Conjecture. Using the Vertical Angles Conjecture, 

2m 108°, so m 54°. p n because of the 

Corresponding Angles Conjecture. Using the 

Linear Pair Conjecture, n 62°, so p 62°. 

Using the Linear Pair Conjecture, r p 180°. 

Because p 62°, r 118°. 





1. perpendicular 

2. neither 


4. parallel 

5. possible answer: (2, 5) and (7, 11) 

6. possible answer: (1, 5) and (2, 12) 

7. Ordinary; no two slopes are the same, so no 

sides are parallel, although TE EM because their 

slopes are opposite reciprocals. 

8. Ordinary, for the same reason as in Exercise 7— 

none of the sides are quite parallel. 

9. trapezoid: KC RO 

10a. Slope HA slope ND 1 

6 ; 

slope HD slope NA 6. Quadrilateral HAND 

is a rectangle because opposite sides are parallel 

and adjacent sides are perpendicular. 


10b. Midpoint HN midpoint AD 1 

diagonals of a rectangle bisect each other. 

11a. Yes, the diagonals are perpendicular. 

Slope OE 1; slope VR 1. 

11b. Midpoint VR midpoint OE (2, 4). 

The diagonals of OVER bisect each other. 

11c. OVER appears to be a rhombus. Slope 

2 ,3.The 

OV slope RE 1 

5 and slope OR slope 

VE 5, so opposite sides are parallel. Also, all of 

the sides appear to have the same length. 

12a. Both slopes equal 1 

2 . 

12b. The segments are not parallel because they 

are coincident. 

12c. distinct 

13. (3, 6)

LESSON 3.6 

1. Sample description: Construct one of the 

segments, and mark arcs of the correct length from 

the endpoints. Draw sides to where those arcs meet. 

M 

A 

M 

2. 

O 

Sample description: Construct O. Mark off 

distances OD and OT on the sides of the angle. 

Connect D and T. 

3. 

I 

I 

D 

O 

M A 

O D 

O 

Y 

I Y 

A 

Y 

S 

T 

T 

Sample description: Construct IY. Construct I at 

I and Y at Y. Label the intersection of the rays 

point G. 

4. Yes, students’ constructions must be either 

larger or smaller than the triangle in the book. 

Sample description: Draw one side with a different 

length than the lengths in the book. Duplicate an 

S 

S 

G 

angle at each end of that segment congruent to one 

of the angles in the book. Where they meet is the 

third vertex of the triangle. 

5. 

A 

A 

C 

Sample description: Construct A and mark off 

the distance AB. From B swing an arc of length BC 

to intersect the other side of A at two points. 

Each gives a different triangle. 

6. 

C 

y x 

____ 

2 

A 

y 

x 

C 

T 

B 

B 

y x 

____ 

2 

x 

Sample description: Mark the distance y, mark 

back the distance x, and bisect the remaining 

length of y x. Using an arc of that length, mark 

arcs on the ends of segment x. The point where 

they intersect is the vertex angle of the triangle. 

7. 

Sample description: Draw an angle. Mark off equal 

segments on the sides of the angle. Use a different 

compass setting to draw intersecting arcs from the 

ends of those segments. 

8. Sample description: Draw an angle and mark 

off unequal distances on the sides. At the endpoint 

of the longer segment (not the angle vertex), swing 

an arc with the length of the shorter segment. From 

the endpoint of the shorter segment, swing an arc 

the length of the longer segment. Connect the 

endpoints of the segments to the intersection 

points of the arcs to form a quadrilateral. 




9. 

Sample description: Draw a segment and draw an 

angle at one end of the segment. Mark off a 

distance equal to that segment on the other side of 

the angle. Draw an angle at that point and mark off 

the same distance. Connect that point to the other 

end of the original segment. 

10. 

Sample description: Draw an angle and mark off 

equal lengths on the two sides. Use that length to 

determine another point that distance from the 

points on the sides. Connect that point with the 

two points on the side of the angle. 

11. Answers will vary. The angle bisector lies 

between the median and the altitude. The order of 

the points is either M, R, S or S, R, M. One possible 

conjecture: In a scalene obtuse triangle the angle 

bisector is always between the median and the 

altitude. 

mABC = 111 

B 

R Median 

A S M 

Altitude Angle 

bisector 


C 

12. new coordinates: E(4, 6), A(7, 0), T(1, 2) 

13. 

–5 

14. half a cylinder 

15. 503 

16. 

R 

–5 

–5 

y 

T' 

T 

E 

E' 

110 

E 

3.2 cm 

110 110 

A 5.5 cm C 

A' 

A 

Reflectional Rotational 

Figure symmetries symmetries 

Trapezoid 0 0 

Kite 1 0 

Parallelogram 0 2 

Rhombus 2 2 

Rectangle 2 2 

x

1. incenter 

LESSON 3.7 

Because the station needs to be equidistant from 

the paths, it will need to be on each of the angle 

bisectors. 

2. circumcenter 

3. incenter 

The center of the circular sink must be equidistant 

from the three counter edges, that is, the incenter of 

the triangle. 


To find the point equidistant from three points, 

find the circumcenter of the triangle with those 

points as vertices. 

5. Circumcenter. Find the perpendicular bisectors 

of two of the sides of the triangle formed by the 

classes. Locate the pie table where these two lines 

intersect. 

6. 

7. 

Stove 

Fridge Sink 

8. Yes, any circle with a larger radius would not 

fit within the triangle. To get a circle with a larger 

radius tangent to two of the sides would force the 

circle to pass through the third side twice. 

9. No, on an obtuse triangle the circle with the 

largest side of the triangle as the diameter of the 

circle creates the smallest circular region that 

contains the triangle. The circumscribed circle of 

an acute triangle does create the smallest circular 

region that contains the triangle. 

10. For an acute triangle, the circumcenter is 

inside the triangle; for an obtuse triangle, the 

circumcenter is outside the triangle. The 

circumcenter of a right triangle lies on the 

midpoint of the hypotenuse. 

11. For an acute triangle, the orthocenter is inside 

the triangle; for an obtuse triangle, the orthocenter 

is outside the triangle. The orthocenter of a right 

triangle lies on the vertex of the right angle. 

12. The midsegment appears parallel to side MA 

and half the length. 

13. The base angles of the isosceles trapezoid 

appear congruent. 

T 

A O 

M 

A 

M H T 

S 




14. The measure of A is 90°. The angle inscribed 

in a semicircle appears to be a right angle. 

15. The two diagonals appear to be perpendicular 

bisectors of each other. 

T 

M 

16. 

17. 

A 

9 

y 

Construct the incenter by bisecting the two angles 

shown. Any other point on the angle bisector of the 

third angle must be equidistant from the two 

unfinished sides. From the incenter, make congruent 

arcs that intersect the unfinished sides. The 

intersection points are equidistant from the incenter. 

Use two congruent arcs to find another point that 

is equidistant from the two points you just 

constructed. The line that connects this point and 

the incenter is the angle bisector of the third angle. 

18. Answers should describe the process of 

discovering that the midpoints of the altitudes are 

collinear for an isosceles right triangle. 

19. a triangle 

20. 

M 

6.0 cm 6.0 cm 

60 6.0 cm 60 

R 

O 

60 60 

6.0 cm 

A 

x + y = 9 

H 

9 

T 

6.0 cm 


x 

21. 

K 

4.8 cm 

Y 

40 40 

E 

6.4 cm 

22. construction of an angle bisector 

23. construction of a perpendicular line through a 

point on a line 

24. construction of a line parallel to a given line 

through a point not on the line 

25. construction of an equilateral triangle 

26. construction of a perpendicular bisector 

T

LESSON 3.8 

1. The center of gravity is the centroid. She needs 

to locate the incenter to create the largest circle 

within the triangle. 

2. AM 20; SM 7; TM 14; UM 8 

3. BG 24; IG 12 4. RH 42; TE 45 

5. The points of concurrency are the same point 

for equilateral triangles because the segments are 

the same. 

6. 

7. ortho-/in-/centroid/circum-; the order changes 

when the angle becomes larger than 60°. The 

points become one when the triangle is equilateral. 

8. Start by constructing a quadrilateral, then make 

a copy of it. Draw a diagonal in one, and draw a 

different diagonal in the second. Find the centroid 

of each of the four triangles. Construct a segment 

connecting the two centroids in each quadrilateral. 

Place the two quadrilaterals on top of each other 

matching the congruent segments and angles.Where 

the two segments connecting centroids intersect is 

the centroid of the quadrilateral. 

A 

D 

Circumcenter 

M 1 

Centroid 

Incenter 

Orthocenter 

Orthocenter 

Incenter 

Circumcenter 

M 2 

C 

Centroid 

B 

D 

M4 A 

M 3 

C 

B 


10. The shortest chord through P is a segment 

perpendicular to the diameter through P, which is 

the longest chord through P. 

11. 

12. rule: 2n 2, possible answer: 

13. 

O 

P 

H H H 

H C C C C 

H H H 

A 

C 

14. a 128°, b 52°, c 128°, d 128°, 

e 52°, f 128°, g 52°, h 38°, 

k 52°, m 38°, n 71°, p 38° 

15. Construct altitudes from the two accessible 

vertices to construct the orthocenter. Through the 

orthocenter, construct a line perpendicular to the 

southern boundary of the property. This method 

will divide the property equally only if the southern 

boundary is the base of an isosceles triangle. 

Altitude to 

missing vertex 

16. 1580 greetings 

C 

H H H 

C 

C 

B' 

B 

C 

H H H 

H 





1. False; a geometric construction uses a straightedge 

and a compass. 

2. False; a diagonal connects two non-consecutive 

vertices. 


5. false 

6. False; the lines can’t be a given distance from a 

segment because the segment has finite length and 

the lines are infinite. 

7. false 

D 

A B 


10. False; the orthocenter does not always lie 

inside the triangle. 

11. A 12. B or K 13. I 14. H 

15. G 16. D 17. J 18. C 

19. 20. 

21. 22. 

23. Construct a 90° angle and bisect it twice. 

24. 

25. incenter 

26. Dakota Davis should locate the circumcenter 

of the triangular region formed by the three stones, 

which is the location equidistant from the stones. 

27. 

B 

A 

B 

A C 

C 

z 

Copy 


C 

28. 

29. 

30. mA mD.You must first find B. 

mB 180° 2(mA). 

B 

2y 

31. 

32. 

A 

A B 

I 

y 

R 

1_ 

2 z 

y 

1_ 

2 z 

y y x 

Segment 

5x 

P R 

3x 4x 

x 

2y 

Q 

B A 

D F 

T 

D 

33. rotational symmetry 

34. neither 35. both 

36. reflectional symmetry 

37. D 38. A 39. C 40. B 

41. False; an isosceles triangle has two congruent 

sides. 

D 

4x 

y

42. true 

43. False; any non-acute triangle is a 


44. False; possible explanation: The orthocenter is 

the point of intersection of the three altitudes. 

45. true 

46. False; any linear pair of angles is a 


47. False; each side is adjacent to one congruent 

side and one noncongruent side, so two consecutive 

sides may not be congruent. 

48. false; 

49. False; the measure of an arc is equal to the 

measure of its central angle. 

50. false; TD 2DR 

51. False; a radius is not a chord. 

52. true 

53. False; inductive reasoning is the process of 

observing data, recognizing patterns, and making 

generalizations about those patterns. 

54. paradox 

55a. 2 and 6 or 3 and 5 

55b. 1 and 5 

55c. 138° 

56. 55 


58a. yes 

58b. If the month has 31 days, then the month is 

October. 


58c. no 

59. 



62. a 38°, b 38°, c 142°, d 38°, e 50°, 

f 65°, g 106°, h 74°. 

Possible explanation: The angle with measure c is 

congruent to an angle with measure 142° because 

of the Corresponding Angles Conjecture, so 

c 142°. The angle with measure 130° is 

congruent to the bisected angle by the 

Corresponding Angles Conjecture. The angle with 

measure f has half the measure of the bisected 

angle, so f 65°. 

63. Triangles will vary. Check that the triangle is 

scalene and that at least two angle bisectors have 

been constructed. 

64. mFAD 30° so mADC 30°, but 

its vertical angle has measure 26°. This is a 

contradiction. 

65. minimum: 101 regions by 100 parallel lines; 

maximum: 5051 regions by 100 intersecting, 

noncurrent lines 

n 1 2 3 4 5 6 … n … 20 

f(n) 1 2 5 8 11 14 … … 56 


f(n) 3n 4 

n 1 2 3 4 5 6 … n … 20 

f(n) 0 3 8 15 24 35 … … 399 

f(n) n 2 1 

Q 

Q' 






LESSON 4.1 

1. The angle measures change, but the sum 

remains 180°. 

2. 73° 

3. 60° 

4. 110° 

5. 24° 

6. 3 360° 180° 900° 

7. 3 180° 180° 360° 

8. 69°; 47°; 116°; 93°; 86° 

9. 30°; 50°; 82°; 28°; 32°; 78°; 118°; 50° 

10. 

11. 

12. First construct E, using the method used in 

Exercise 10. 

A 

13. 

E 

M 

R 

A 

R 

M A 

G 

L 

14. From the Triangle Sum Conjecture 

mA mS mM 180°. Because M is a 

right angle, mM 90°. By substitution, 

mA mS 90° 180°. By subtraction, 

mA mS 90°. So two wrongs make a right! 

15. Answers will vary. See the proof on page 202. 

To prove the Triangle Sum Conjecture, the Linear 

Pair Conjecture and the Alternate Interior Angles 

Conjecture must be accepted as true. 

16. It is easier to draw PDQ if the Triangle Sum 

Conjecture is used to find that the measure of 

D is 85°. Then PD can be drawn to be 7 cm, and 


R 

Fold 

4 

G 

L 

A 

E 

R 

angles P and D can be drawn at each endpoint 

using the protractor. 

40 

85 

P 7 cm D 

17. The third angles of the triangles also have the 

same measures; are equal in measure 

18. You know from the Triangle Sum Conjecture 

that mA mB mC 180°, and mD 

mE mF 180°. By the transitive property, 

mA mB mC mD mE mF. 

You also know that mA mD, and mB 

mE. You can substitute for mD and mE in the 

longer equation to get mA mB mC 

mA mB mF. Subtracting equal terms 

from both sides, you are left with mC mF. 

19. For any triangle, the sum of the angle measures 

is 180°, by the Triangle Sum Conjecture. Since the 

triangle is equiangular, each angle has the same 

measure, say x. So x x x 180°, and x 60°. 

20. false 

21. false 

22. false 

23. false 

55 

24. true 

25. eight; 100 

Q

LESSON 4.2 

1. 79° 

2. 54° 

3. 107.5° 

4. 44°; 35 cm 

5. 76°; 3.5 cm 

6. 72°; 36°; 8.6 cm 

7. 78°; 93 cm 

8. 75 m; 81° 

9. 160 in.; 126° 

10. a 124°, b 56°, c 56°, d 38°, e 38°, 

f 76°, g 66°, h 104°, k 76°, n 86°, 

p 38°; Possible explanation: The angles with 

measures 66° and d form a triangle with the angle 

with measure e and its adjacent angle. Because d, 

e, and the adjacent angle are all congruent, 

3d 66° 180°. Solve to get d 38°. This is 

also the measure of one of the base angles of the 

isosceles triangle with vertex angle measure h. 

Using the Isosceles Triangle Conjecture, the other 

base angle measures d, so 2d h 180°, or 

76° h 180°. Therefore, h 104°. 

11. a 36°, b 36°, c 72°, d 108°, e 36°; 

none 

12a. Yes. Two sides are radii of a circle. Radii must 

be congruent; therefore, each triangle must be 

isosceles. 

12b. 60° 

13. NCA 

14. IEC 

15. 

16. 


Fold 2 

D 

A 

M N 

G 

Fold 1 

K 

105 60 

45 

C 

Fold 3 

Fold 4 


19. parallel 

20. parallel 

21. neither 

22. parallelogram 

23. 40 

0 8 16 24 32 40 

P 

F 

B 

H 

E 

24. New: (6, 3), (2, 5), (3, 0). Triangles are 

congruent. 

25. New: (3, 3), (3, 1), (1, 5). Triangles 

are congruent. 





1. false 2. true 

3. not a solution 4. solution 

5. not a solution 6. x 7 

7. y 4 8. x 8 

9. x 4.2 10. n 1 

2 

11. x 2 12. t 18 

13. n 2 

5 14a. x 9 

 

4 

9 

14b. x ; 4 The two methods produce identical 

results. Multiplying by the lowest common 

denominator (which is comprised of the factors of 

both denominators) and then reducing common 

factors (which clears the denominators on either 

side) is the same as simply multiplying each numerator 

by the opposite denominator (or cross multiplying). 

Algebraically you could show that the two methods 

are equivalent as follows: 

a c 

b 

d 

bda b bd c d 

abd 

 

b 

bcd 

 

d 

ad bc 

The method of “clearing fractions” results in the 

method of “cross multiplying.” 


15. You get an equation that is always false, so 

there is no solution to the equation. 

16. Camella is not correct. Because the equation 

0 0 is always true, the truth of the equation does 

not depend on the value of x. Therefore, x can be 

any real number. Camella’s answer, x 0, is only 

one of infinitely many solutions. 

17. 

x 

2x 

2x 

If x equals the measure of the vertex angle, then 

the base angles each measure 2x. Applying the 

Triangle Sum Conjecture results in the following 

equation: 

x 2x 2x 180° 

5x 180° 

x 36° 

The measure of the vertex angle is 36° and the 

measure of each base angle is 72°.

1. yes 

2. no 

3. no 

4 5 

9 

5 6 

12 

LESSON 4.3 

4. yes 

5. a, b, c 

6. c, b, a 

7. b, a, c 

8. a, c, b 

9. a, b, c 

10. v, z, y, w, x 

11. 6 length 102 

12. By the Triangle Inequality Conjecture, the 

sum of 11 cm and 25 cm should be greater than 

48 cm. 

13. b 55°, but 55° 130° 180°, which is 

impossible by the Triangle Sum Conjecture. 

14. 135° 

15. 72° 

16. 72° 

17. a b c 180° and x c 180°. Subtract c 

from both sides of both equations to get x 180 c 

and a b 180 c. Substitute a b for 180 c 

in the first equation to get x a b. 

18. 45° 

19. a 52°, b 38°, c 110°, d 35° 

20. a 90°, b 68°, c 112°, d 112°, e 68°, 

f 56°, g 124°, h 124° 

21. By the Triangle Sum Conjecture, the third 

angle must measure 36° in the small triangle, but it 

measures 32° in the large triangle. These are the 

same angle, so they can’t have different measures. 

22. ABE 

23. FNK 

24. cannot be determined 




LESSON 4.4 

1. Answers will vary. Possible answer: If three 

sides of one triangle are congruent to three sides of 

another triangle, then the triangles are congruent 

(all corresponding angles are also congruent). 

2. Answers will vary. Possible answer: The picture 

statement means that if two sides of one triangle 

are congruent to two sides of another triangle, and 

the angles between those sides are also congruent, 

then the two triangles are congruent. 

If you know this: then you also know this: 

3. Answers will vary. Possible answer: 

4. SAS 

5. SSS 


7. SSS 

8. SAS 

9. SSS (and the Converse of the Isosceles Triangle 

Conjecture) 

10. yes, ABC ADE by SAS 

11. Possible answer: Boards nailed diagonally in 

the corners of the gate form triangles in those 

corners. Triangles are rigid, so the triangles in the 

gate’s corners will increase the stability of those 

corners and keep them from changing shape. 

12. FLE by SSS 


25. (Lesson 4.4) 

Side length 1 2 3 4 5 … n … 20 

Elbows 4 4 4 4 4 4 4 

T’s 0 4 8 12 16 76 

13. Cannot be determined. SSA is not a congruence 

conjecture. 

14. AIN by SSS or SAS 

15. Cannot be determined. Parts do not correspond. 

16. SAO by SAS 

17. Cannot be determined. Parts do not correspond. 

18. RAY by SAS 

19. The midpoint of SD and PR is (0, 0). 

Therefore, DRO SPO by SAS. 

20. Because the LEV is marking out two triangles 

that are congruent by SAS, measuring the length 

of the segment leading to the finish will also 

approximate the distance across the crater. 

21. 22. 

Crosses 0 1 4 9 16 361 

23. a 37°, b 143°, c 37°, d 58°, 

e 37°, f 53°, g 48°, h 84°, k 96°, 

m 26°, p 69°, r 111°, s 69°; Possible 

explanation: The angle with measure h is the vertex 

angle of an isosceles triangle with base angles 

measuring 48°, so h 2(48) 180°, and h 84°. 

The angle with measure s and the angle with 

measure p are corresponding angles formed by 

parallel lines, so s p 69°. 

24. 3 cm third side 19 cm 


26a. y 6 b. y 13 

c. y 

3 

3 

x 

4 

2 

27. (5, 3) 

4n 4 (n 1) 2

LESSON 4.5 

1. If two angles and the included side of one triangle 

are congruent to the corresponding side and angles 

of another triangle, then the triangles are congruent. 

2. If two angles and a non-included side of one 

triangle are congruent to the corresponding side 

and angles of another triangle, then the triangles 

are congruent. 

If you know this: then you also know this: 

3. Answers will vary. Possible answer: 

4. ASA 


6. SAA 


8. ASA 


10. FED by SSS 

11. WTA by ASA or SAA 

12. SAT by SAS 

13. PRN by ASA or SAS; SRE by ASA 

14. Cannot be determined. Parts do not 

correspond. 

15. MRA by SAS 

16. Cannot be determined.AAA does not guarantee 

congruence. 

17. WKL by ASA 

18. Yes, ABC ADE by SAA or ASA. 

19. Slope AB slope CD 3 and slope BC 

slope DA 1 

3 ,soAB BC,CD DA , and 

BC DA . ABC CDA by SAA. 

20. 

21. The construction is the same as the 

construction using ASA once you find the third 

angle, which is used here. (Finding the third angle 

is not shown.) 

22. Construction will show a similar but larger 

(or smaller) triangle constructed from a drawn 

triangle by duplicating two angles on either end of a 

new side that is not congruent to the corresponding 

side. 

23. Draw a line segment. Construct a perpendicular. 

Bisect the right angle. Construct a triangle with 

two congruent sides and with a vertex that 

measures 135°. 

24. 125 

25. False. One possible counterexample is a kite. 

26. None. One triangle is determined by SAS. 

27. 

28a. about 100 km southeast of San Francisco 

28b. Yes. No, two towns would narrow it down 

to two locations. The third circle narrows it down 

to one. 

Eureka 

L 

K 

Reno 

Sacramento 

San 

Francisco 

CALIFORNIA 

Los Angeles 

Miles 

0 50 100 200 

0 100 

Kilometers 

400 

O R EGO N 

Elko 

N E VA D A 

Las 

Vegas 

M 

Boise 

I DAHO 

U TAH 

ARIZO NA 




LESSON 4.6 

1. Yes. BD BD (same segment), A C 

(given), and ABD CBD (given), so DBA 

DBC by SAA. AB CB by CPCTC. 

2. Yes. CN WN and C W (given), and 

RNC ONW (vertical angles), CNR 

WON by ASA. RN ON by CPCTC. 

3. Cannot be determined. The congruent parts 

lead to the ambiguous case SSA. 

4. Yes. S I, G A (given), and TS IT 

(definition of midpoint), so TIA TSG by 

SAA. SG IA by CPCTC. 

5. Yes. FO FR and UO UR (given), and UF 

UF (same segment), so FOU FRU by SSS. 

O R by CPCTC. 

6. Yes. MN MA and ME MR (given), and 

M M (same angle), so EMA RMN by 

SAS. E R by CPCTC. 

7. Yes. BT EU and BU ET (given), and 

UT UT (same segment), so TUB UTE by 

SSS. B E by CPCTC. 

8. Cannot be determined. HLF LHA by 

ASA, but HA and HF are not corresponding sides. 

9. Cannot be determined. AAA does not guarantee 

congruence. 

10. Yes. The triangles are congruent by SAS. 

11. Yes. The triangles are congruent by SAS, and 

the angles are congruent by CPCTC. 

12. Draw AC and DF to form ABC and DEF. 

AB CB DE FE because all were drawn with 

the same radius. AC DF for the same reason. 

ABC DEF by SSS. Therefore, B E by 

CPCTC. 

21. (Lesson 4.6) 



14. KEI by ASA 

15. UTE by SAS 

16. 

17. 

18. a 112°, b 68°, c 44°, d 44°, e 136°, 

f 68°, g 68°, h 56°, k 68°, l 56°, m 124°; 

Possible explanation: f and g are measures of base 

angles of an isosceles triangle, so f g.The vertex 

angle measure is 44°, so subtract 44° from 180° and 

divide by 2 to get f 68°. The angle with measure 

m is the exterior angle of a triangle. Add the remote 

interior angle measures 56° and 68° to get 

m 124°. 

19. ASA. The “long segment in the sand” is a 

shared side of both triangles 

20. (4, 1) 


22. Value C is always decreasing. 

23. x 3, y 10 

Number of sides 3 4 5 6 7 … 12 … n 

Number of struts needed 

to make polygon rigid 

0 1 2 3 4 … 9 … n 3

1. See flowchart below. 


1. (Lesson 4.7) 

LESSON 4.7 



 

 

 

 

 

1 

SE SU 

? Given 

2 

E U 

4 

ESM USO 

? ? 

5 

MS SO 

? Given ASA Congruence 

Conjecture 

? CPCTC 

3 

1 2 

1 

2 

? 

 

Vertical Angles Conjecture 

2. (Lesson 4.7) 

I is midpoint of CM 

Given 

I is midpoint of BL 

? 

 

1 NS is a median 

3 

4 

CI IM 

Definition 

of midpoint 

IL IB 

? 

 

5 1 2 

? 

 

3 NS NS 

Given Same segment 

6 

? ? 

 

 

 

2 

S is a midpoint 

4 

WS SE 

6 

ESN 

WSN ? 

7 

Definition 

of median 

1 NS is an 

angle bisector 

Given 

Given 

3. (Lesson 4.7) 

4. (Lesson 4.7) 

2 

5 

Definition 

of midpoint 

WN NE 

? 

 

Definition of ? 

 


3 W E 

4 

1 ? 

 

? 

 

NS NS 

? 

 

Definition of 

midpoint 

? 

 

Vertical Angles Conjecture 

Given 

2 

? 

 

5 

? ? 

 

 

 

CIL MIB 

by SAS 

7 

? 

 

CPCTC 

W ? 

 

SSS ? CPCTC 

Given ? SAA 

? CPCTC 

Same segment 

WNS ENS 

6 

WN NE 

 

CL MB 

 

7 NEW is 

isosceles 

? 

 

E 

Definition of 

isosceles triangle 





6. Given: ABC with BA BC, CD AD 

Show: BD is the angle bisector of ABC. 

B 

1 

2 

A 

C 

BA BC CD AD BD BD 

Given Given 

Same segment 

ABD CBD 

SSS 

1 2 

CPCTC 

→ 

BD bisects ABC 

Definition of angle 

bisector 

7. The angle bisector does not go to the midpoint 

of the opposite side in every triangle, only in an 

isosceles triangle. 

8. NE, because it is across from the smallest angle 

in NAE. It is shorter than AE, which is across 

from the smallest angle in LAE. 

9. The triangles are congruent by SSS, so the two 

central angles cannot have different measures. 

10. PRN by ASA; SRE by ASA 

11. Cannot be determined. Parts do not 

correspond. 


D 

5. (Lesson 4.7) 

1 

2 

SA NE 

? 

? 

Given 

SE NA 

Given 

3 

4 

5 

3 4 

AIA Conjecture 

SN SN 

12. ACK by SSS 

13. a 72°, b 36°, c 144°, d 36°, e 144°, 

f 18°, g 162°, h 144°, j 36°, k 54°, 

m 126° 

14. The circumcenter is equidistant from all three 

vertices because it is on the perpendicular bisector 

of each side. Every point on the perpendicular 

bisector of a segment is equidistant from the 

endpoints. Similarly, the incenter is equidistant 

from all three sides because it is on the angle 

bisector of each angle, and every point on an angle 

bisector is equidistant from the sides of the angle. 

15. ASA. The fishing pole forms the side. 

“Perpendicular to the ground” forms one angle. 

“Same angle on her line of sight” forms the other 

angle. 

16. 2 

7 

17. 

18. y 

4 

Y 

X 

B O 

B' Y'4 

? 

? 6 

 

? 

? 7 

? 

? 

? 

1 2 

ESN ANS 

SA NE 

AIA Conjecture ASA 

CPCTC 

Same segment 

O' 

This proof shows that in a parallelogram, 

opposite sides are congruent. 

X' 

x

1. 6 

2. 90°; 18° 

3. 45° 



6. 1 

Isosceles ABC 

with AC BC 

and CD bisects 

C 

Given 

3 

AD BD 

CPCTC 

4. (Lesson 4.8) 

LESSON 4.8 

2 

ADC BDC 

Conjecture A 

(Exercise 4) 

4 

CD is a median 

Def. of median 

 

1 

CD is the bisector of C 

2 ? 1 2 

Given Definition of 


5. (Lesson 4.8) 

1 ABC is isosceles with 

AC BC, and CD is 

the bisector of C 

Given 

3 

4 

ABC is isosceles 

with AC BC 

Given 

CD CD 

Same segment 

2 ADC BDC 

Conjecture A 

3 

1 and 2 form 

a linear pair 

Definition of 

linear pair 

7. 

1 

AC BC 

2 

CD CD 

3 

D is 

Given Same segment 

midpoint 

of AB 

Given 

5 

ADC BDC 

4 

AD BD 

SSS 

Def. of midpoint 

6 ACD DCB 

CPCTC 

8 

CD is altitude 

of ABC 

Conjecture B 

(Exercise 5) 

7 CD is angle 

bisector of ACB 

Def. of angle bisector 

9 

CD AB 

Def. of altitude 

8. Yes. First show that the three exterior triangles 

are congruent by SAS. 

4 1 2 

? 

 

CPCTC 

5 ADC BDC 

? 

 

5 1 and 2 are 

supplementary 

Linear Pair 

Conjecture 

SAS 

7 

Definition of 

perpendicular 

CD is an altitude 

6 1 and 2 are 

right angles 

8 

Congruent 

supplementary 

angles are 90 

CD AB 

? 

 

? 

 

Definition of 

altitude 




9. 

C 

A D B 

Drawing the vertex angle bisector as an auxiliary 

segment, we have two triangles. We can show them 

to be congruent by SAS, as we did in Exercise 4. 

Then, A B, by CPCTC. Therefore, base angles 

of an isosceles triangle are congruent. 

10. The proof is similar to the one on page 245, 

but in reverse, and using the Converse of the 

Isosceles Triangle Conjecture. 

11. 

30 

12. a 128°, b 128°, c 52°, d 76°, e 104°, 

f 104°, g 76°, h 52°, j 70°, k 70°, l 40°, 

m 110°, n 58° 

13. between 16 and 17 minutes 


14. y 5 

x 

3 

16 

15. 120 

16. (4, 6) or (4, 0) or any point at which the 

x-coordinate is either 1 or 7 and the y-coordinate 

does not equal 3 

17. Hugo and Duane can locate the site of the 

fireworks by creating a diagram using SSS. 

340 m/s 5 s 

= 1.7 km 

18. C n H 2n 

H 

Duane 

Fireworks 

H 

H H 

H 

C 

C C 

1.5 km 

H 

H 

C C 

H 

H 

C C 

H 

H 

H H 

H 

340 m/s 3 s 

= 1.02 km 

Hugo


1. Their rigidity gives strength. 

2. The Triangle Sum Conjecture states that the 

sum of the measures of the angles in every triangle 

is 180°. Possible answers: It applies to all triangles; 

many other conjectures rely on it. 

3. The angle bisector of the vertex angle is also the 

median and the altitude. 

4. The distance between A and B is along the 

segment connecting them. The distance from A 

to C to B can’t be shorter than the distance from A 

to B. Therefore, AC CB AB. Points A, B, and C 

form a triangle. Therefore, the sum of the lengths 

of any two sides is greater than the length of the 

third side. 

5. SSS, SAS, ASA, or SAA 

6. In some cases, two different triangles can 

be constructed using the same two sides and 

non-included angle. 


8. ZAP by SAA 

9. OSU by SSS 


11. APR by SAS 

12. NGI by SAS 


14. DCE by SAA or ASA 

15. RBO or OBR by SAS 

16. AMD UMT by SAS, AD UT by 

CPCTC 



19. TRI ALS by SAA, TR AL by 

CPCTC 

20. SVE NIK by SSS, EL KV by 

overlapping segments property 



23. LAZ IAR by ASA, LRI IZL by 

ASA, and LRD IZD by ASA 

24. yes. PTS TPO by ASA or SAA 

25. ANG is isosceles, so A G. However, 

the sum of mA mN mG 188°. The 

measures of the three angles of a triangle must sum 

to 180°. 

26. ROW NOG by ASA, implying that 

OW OG . However, the two segments shown are 

not equal in measure. 

27. a g e d b f c. Thus, c is the 

longest segment, and a and g are the shortest. 

28. x 20° 

29. Yes. TRE SAE by SAA, so sides are 

congruent by CPCTC. 

30. Yes. FRM RFA by SAA. RFM 

FRA by CPCTC. Because base angles are 

congruent, FRD is isosceles. 

31. x 48° 

32. The legs form two triangles that are congruent 

by SAS. Because alternate interior angles are 

congruent by CPCTC, the seat must be parallel to 

the floor. 

33. Construct P and A to be adjacent. The 

angle that forms a linear pair with the conjunction 

of P and A is L. Construct A. Mark off the 

length AL on one ray. Construct L. Extend the 

unconnected sides of the angles until they meet. 

Label the point of intersection P. 

P 

A y L 

34. Construct P. Mark off the length PB on one 

ray. From point B, mark off the two segments that 

intersect the other ray of P at distance x. 

S 1 

A 

L P 

P z 

B 

x 

S 2 

x 





36. Given three sides, only one triangle is possible; 

therefore, the shelves on the right hold their shape. 

The shelves on the left have no triangles and move 

freely as a parallelogram. 



TMI RME 

2 

? ? 

 

 

 

37. Possible method: Construct an equilateral 

triangle and bisect one angle to obtain 30°. 

Adjacent to that angle, construct a right angle 

and bisect it to obtain 45°. 

38. d, a b, c, e, f 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Vertical angles ? 

TM ME TMI EMR 

T E 

or 

R I 

TI RE 

M is midpoint 

of TE and IR 

1 ? 

3 ? ? 

? Definition 

of midpoint 

5 

? ? 

? SAS 

6 

? ? 

? CPCTC 

7 ? ? 

? Converse of 


? Given 

4 ? ? 

IM MR 

? Definition of midpoint


5 


LESSON 5.1 



3. 122° 

4. 136° 

5. 108°; 36° 

6. 108°; 106° 

7. 105°; 82° 

8. 120°; 38° 

9. The sum of the interior angle measures of the 

quadrilateral is 358°. It should be 360°. 

10. The measures of the interior angles shown sum 

to 554°. However, the figure is a pentagon, so the 

measures of its interior angles should sum to 540°. 

11. 18 

12. a 116°, b 64°, c 90°, d 82°, e 99°, 

f 88°, g 150°, h 56°, j 106°, k 74°, 

m 136°, n 118°, p 99°; Possible explanation: 

The sum of the angles of a quadrilateral is 360°, so 

a b 98° d 360°. Substituting 116° for a 

and 64° for b gives d 82°. Using the larger 

quadrilateral, e p 64° 98° 360°. 

Substituting e for p, the equation simplifies to 

2e 198°, so e 99°. The sum of the angles of a 

pentagon is 540°, so e p f 138° 116° 

540°. Substituting 99° for e and p gives f 88°. 

1. (Lesson 5.1) 

13. 17 

14. 15 

15. the twelfth century 

16. The angles of the trapezoid measure 67.5° and 

112.5°; 67.5° is half the value of each angle of a regular 

octagon, and 112.5° is half the value of 360° 135°. 

135 

67.5 

17. Answers will vary; see the answer for 

Developing Proof on page 259. Using the Triangle 

Sum Conjecture, a b j c d k e f 

l g h i 4(180°), or 720°. The four angles in 

the center sum to 360°, so j k l i 360°. 

Subtract to get a b c d e f g 

h 360°. 

18. x 120° 

19. The segments joining the opposite midpoints 

of a quadrilateral always bisect each other. 

20. D 

21. Counterexample: The base angles of an 

isosceles right triangle measure 45°; thus they are 

complementary. 

Number of sides of polygon 7 8 9 10 11 20 55 100 

Sum of measures of angles 900° 1080° 1260° 1440° 1620° 3240° 9540° 17640° 

2. (Lesson 5.1) 

Number of sides of 5 6 7 8 9 10 12 16 100 

equiangular polygon 

Measure of each angle 108° 120° 128 4 

° 

7 135° 140° 144° 150° 1571 ° 1762 

2 5 ° 

of equiangular polygon 




LESSON 5.2 

1. 360° 

2. 72°; 60° 

3. 15 

4. 43 

5. a 108° 

6. b 45 1 

3 ° 

7. c 51 3 

°, 

7 

d 115 5 

7 ° 

8. e 72°, f 45°, g 117°, h 126° 

9. a 30°, b 30°, c 106°, d 136° 

10. a 162°, b 83°, c 102°, d 39°, e 129°, 

f 51°, g 55°, h 97°, k 83° 


12. Yes. The maximum is three. The minimum is 

zero. A polygon might have no acute interior 

angles. 

11. (Lesson 5.2) 

? 

Linear Pair Conjecture 

? 


? 



1 a b 180 

2 c d 180 

3 e f 180 

13. Answers will vary. Possible proof using 

the diagram on the left: a b i 180°, c d 

h 180°, and e f g 180° by the Triangle 

Sum Conjecture. a b c d e f g 

h i 540° by the addition property of equality. 

Therefore, the sum of the measures of the angles of 

a pentagon is 540°. To use the other diagram, 

students must remember to subtract 360° to 

account for angle measures k through o. 

14. regular polygons: equilateral triangle and 

regular dodecagon; angle measures: 60°, 150°, 

and 150° 

15. regular polygons: square, regular hexagon, and 

regular dodecagon; angle measures: 90°, 120°, 

and 150° 

16. Yes. RAC DCA by SAS. AD CR by 

CPCTC. 

17. Yes. DAT RAT by SSS. D R by 

CPCTC. 

540° 

4 

a + b + c + d + e + f = ? 

Addition property 

of equality 

? 

? 180° 

5 

a + c + e = 

Triangle Sum Conjecture 

6 

b + d + f = ? 

360° 

Subtraction property 

of equality

LESSON 5.3 

1. 64 cm 2. 21°; 146° 3. 52°; 128° 

4. 15 cm 5. 72°; 61° 6. 99°; 38 cm 

7. w 120°, x 45°, y 30° 

8. w 1.6 cm, x 48°, y 42° 


10. Answers may vary. This proof uses the Kite 

Angle Bisector Conjecture. 

Y 

2 

B N 

1 X 

E 

Given: Kite BENY with vertex angles B and N 

Show: Diagonal BN is the perpendicular bisector 

of diagonal YE. 

From the definition of kite, BE BY.From the 

Kite Angle Bisector Conjecture, 1 2. BX 

BX because they are the same segment. By SAS, 

BXY BXE. So by CPCTC, XY XE. 

Because YXB and EXB form a linear pair, they 

are supplementary, so mYXB mEXB 

180°. By CPCTC, YXB EXB, or mYXB 

mEXB, so by substitution, 2mYXB 180°, or 

mYXB 90°. So mYXB mEXB 90°. 

Because XY XE and YXB and EXB are right 

angles, BN is the perpendicular bisector of YE. 

11. possible answer: E I 

I 


Z I 

Q U 

The other base is ZI . Q and U are a pair of base 

angles. Z and I are a pair of base angles. 

9. (Lesson 5.3) 

1 BE BY 

Given 

2 EN YN 

3 

Given 

BN BN 

 

Same segment 

? 

? 

T 

E 

K 


OW is the other base. S and H are a pair of 

base angles. O and W are a pair of base angles. 

SW HO . 

14. Only one kite is possible 

F 

because three sides determine 

a triangle. 

N 

B 

15. 

E 

16. infinitely many, possible construction: 

E 

W O 

S H 

I 

B O 

S H 

N 

W 

17. 80°, 80°, 100°, 100° 

18. Because ABCD is an isosceles trapezoid, A 

B. AGF BHE by SAA. Thus, AG BH 

by CPCTC. 

19. a 80°, b 20°, c 160°, d 20°, e 80°, 

f 80°, g 110°, h 70°, m 110°, n 100°; 

Possible explanation: Because d forms a linear pair 

with e and its congruent adjacent angle,d 2e 

180°.Substituting d 20° gives 2e 160°,so e 80°. 

Using theVerticalAngles Conjecture and d 20°, the 

unlabeled angle in the small right triangle measures 

20°, which means h 70°. Because g and h are a 

linear pair, they are supplementary, so g 110°. 

? 

5 1 and 

3 

6 

and 

? 

? 4 

Congruence 

shortcut 

 

Definition of 


? 

BEN ? 

? 

? 

BYN 2 BN bisects B, BN bisects N 

SSS 

4 

CPCTC 




LESSON 5.4 

1. three; one 2. 28 

3. 60°; 140° 4. 65° 

5. 23 6. 129°; 73°; 42 cm 

7. 35 8. See flowchart below. 

9. Parallelogram. Draw a diagonal of the original 

quadrilateral. The diagonal forms two tri-angles. 

Each of the two midsegments is parallel to the 

diagonal, and thus the midsegments are parallel to 

each other. Now draw the other diagonal of the 

original quadrilateral. By the same reasoning, the 

second pair of midsegments is parallel. Therefore, 

the quadrilateral formed by joining the midpoints is 

a parallelogram. 

10. The length of the edge of the top base 

measures 30 m. We know this by the Trapezoid 

Midsegment Conjecture. 

11. Ladie drives a stake into the ground to create a 

triangle for which the trees are the other two 

vertices. She finds the midpoint from the stake to 

each tree. The distance between these midpoints is 

half the distance between the trees. 

12. Explanations will vary. 

80 

Cabin 

40 60 

60 cm 

8. (Lesson 5.4) 


1 

2 

FOA with 

midsegment LN 

Given 

IOA with 

midsegment RD 

Given 

3 

4 

LN OA 

? 

 

Triangle Midsegment Conjecture 

OA RD 

? 

 

13. If a quadrilateral is a kite, then exactly one 

diagonal bisects a pair of opposite angles. Both the 

original and converse statements are true. 

14. a 54°, b 72°, c 108°, d 72°, e 162°, 

f 18°, g 81°, h 49.5°, i 130.5°, k 49.5°, 

m 162°, n 99°; Possible explanation: The third 

angle of the triangle containing f and g measures 

81°, so using the Vertical Angles Conjecture, the 

vertex angle of the triangle containing h also 

measures 81°. Subtract 81° from 180° and divide by 

2 to get h 49.5°. The other base angle must also 

measure 49.5°. By the Corresponding Angles 

Conjecture, k 49.5°. 

15. (3, 8) 

16. (0, 8) 

17. coordinates: E(2, 3.5), Z(6, 5); the slope of 

EZ 3 

8 ,and the slope ofYT 3 

8 

18. 

F 

R 

K 

N 

There is only one kite, but more than one way to 

construct it. 

Triangle Midsegment 

Conjecture 

5 

LN RD 

? 

 

Two lines parallel to the 

same line are parallel

1. 34 cm; 27 cm 

2. 132°; 48° 

3. 16 in.; 14 in. 

4. 63 m 

5. 80 

6. 63°; 78° 

7. 

8. 

L 

D 

P 

P 

9. 

V h 

13. (Lesson 5.5) 

R 

LESSON 5.5 

T S 

A 

O 

R 

 

V w 

10. 

11. (b a, c) 


a 

b 

c 

b d 

c a 

1 

LEAN is a 

3 

AEN LNE 

parallelogram 

 


7 

2 

EA LN 

? 

? 

4 

 


? 

6 

 

ASA 

? 

8 

5 

? 

Given 

EAL NLA AET LNT 

Definition of 

parallelogram AE LN 

Opposite sides 

congruent 

 

V c 

d 

 

V b 


14. The parallelogram linkage is used for the 

sewing box so that the drawers remain parallel to 

each other (and to the ground) so that the contents 

cannot fall out. 

15. a 135°, b 90° 

16. a 120°, b 108°, c 90°, d 42°, e 69° 

ET NT 

CPCTC 

AT LT 

? 

CPCTC 

9 

EN and LA bisect 

each other 

? 

Definition of 

segment bisector 




17. x 104°, y 98°. The quadrilaterals on the 

left and right sides are kites. Nonvertex angles are 

congruent. The quadrilateral at the bottom is an 

isosceles trapezoid. Base angles are congruent, and 

consecutive angles between the bases are 

supplementary. 

18. a 84°, b 96° 

19. No. The congruent angles and side do not 

correspond. 

20. 

21. Parallelogram. Because the triangles are 

congruent by SAS, 1 2. So, the segments are 


parallel. Use a similar argument to show that the 

other pair of opposite sides is parallel. 

1 

2 

22. Kite or dart. Radii of the same circle are 

congruent. If the circles have equal radii, a rhombus 

is formed.

1. 

2. 

3. 

(0, 4) 

y 

y 

y 

(0, 6) 


(0, 1) 

(3, 8) 

(1, –1) 

(2, 9) 

x 

x 

x 

4. y x 2 

6 

5. y 13 

x 74 

 

13 

6. y x 1 

7. y 3x 5 

8. y 2 

x 

5 

8 

5 

9. y 80 4x 

10. y 3x 26 

11. y 1 

x 

4 

3 

12. y 6 

5 x 

13. y x 1 

14. y 2 3 

x 

9 

4 

9 




LESSON 5.6 

1. Sometimes true; it is true only if the 

parallelogram is a rectangle. 

2. Always true; by the definition of rectangle, all 

the angles are congruent. By the Quadrilateral Sum 

Conjecture and division, they all measure 90°, so 

any two angles in a rectangle, including consecutive 

angles, are supplementary. 

3. Always true by the Rectangle Diagonals 

Conjecture. 

4. Sometimes true; it is true only if the rectangle is 

a square. 

true 

true 

5. Always true by the Square Diagonals Conjecture. 

6. Sometimes true; it is true only if the rhombus is 

equiangular. 

true 

false 

false 

false 

7. Always true; all squares fit the definition of 

rectangle. 

8. Always true; all sides of a square are congruent 

and form right angles, so the sides become the legs 

of the isosceles right triangle and the diagonal is 

the hypotenuse. 

9. Always true by the Parallelogram Opposite 

Angles Conjecture. 

10. Sometimes true; it is true only if the 

parallelogram is a rectangle. Consecutive angles of 

a parallelogram are always supplementary, but are 

congruent only if they are right angles. 

11. 20 

12. 37° 

13. 45°, 90° 

14. DIAM is not a rhombus because it is not 

equilateral and opposite sides are not parallel. 

15. BOXY is a rectangle because its adjacent sides 

are perpendicular. 

16. Yes. TILE is a rhombus, and a rhombus is a 



17. 

18. Constructions will vary. 

B 

A 

19. one possible construction: 

I 

E V 

L O 

E 

K 

P S 

E 

20. Converse: If the diagonals of a quadrilateral 

are congruent and bisect each other, then the 

quadrilateral is a rectangle. 

Given: Quadrilateral ABCD with diagonals 

AC BD. AC and BD bisect each other 

Show: ABCD is a rectangle 

A 

8 E 

B 

D 

C 

1 2 3 

7 

6 5 

4 

B 

A 

E 

Because the diagonals are congruent and bisect 

each other, AE BE DE EC.Using the 

Vertical Angles Conjecture, AEB CED and 

BEC DEA.So AEB CED and AED 

CEB by SAS. Using the Isosceles Triangle 

Conjecture and CPCTC, 1 2 5 6, 

and 3 4 7 8. Each angle of the 

quadrilateral is the sum of two angles, one from 

each set, so for example, mDAB m1 m8. 

By the addition property of equality, m1 

m8 m2 m3 m5 m4 m6 

m7. So mDAB mABC mBCD 

mCDA. So the quadrilateral is equiangular. Using 

1 5 and the Converse of AIA, AB CD. 

Using 3 7 and the Converse of AIA, 

BC AD . Therefore ABCD is an equiangular 

parallelogram, so it is a rectangle. 

21. If the diagonals are congruent and bisect each 

other, then the room is rectangular (converse of the 

Rectangle Diagonals Conjecture). 

K

22. The platform stays parallel to the floor 

because opposite sides of a rectangle are parallel 

(a rectangle is a parallelogram). 

23. The crosswalks form a parallelogram: The 

streets are of different widths, so the crosswalks are 

of different lengths. The streets would have to meet 

at right angles for the crosswalks to form a rectangle. 

The corners would have to be right angles and the 

streets would also have to be of the same width for 

the crosswalk to form a square. 

24. Place one side of the ruler along one side of 

the angle. Draw a line with the other side of the 

ruler. Repeat with the other side of the angle. Draw 

a line from the vertex of the angle to the point 

where the two lines meet. 

25. Rotate your ruler so that each endpoint of the 

segment barely shows on each side of the ruler. 

Draw the parallel lines on each side of your ruler. 

Now rotate your ruler the other way and repeat the 

process to get a rhombus. The original segment is 

one diagonal of the rhombus. The other diagonal 

will be the perpendicular bisector of the original 

segment. 


27. Yes, it is true for rectangles. 

Given: 1 2 3 4 

Show: ABCD is a rectangle 

By the Quadrilateral Sum Conjecture, m1 

m2 m3 m4 360°. It is given that all 

four angles are congruent, so each angle measures 

26. (Lesson 5.6) 

1 

2 

3 

QU AD 

Given 

QD AU 

? 

? 

Given SSS 

DU DU 

Same segment 

4 

QUD ADU 

8 

Given 

5 

1 2 

3 4 

? CPCTC 

QU UA AD DQ 

90°. Because 4 and 5 form a linear pair, 

m4 m5 180°. Substitute 90° for m4 and 

solve to get m5 90°. By definition of congruent 

angles, 5 3, and 5 and 3 are alternate 

interior angles, so AD BC by the Converse of the 

Parallel Lines Conjecture. Similarly, 1 and 5 are 

congruent corresponding angles, so AB CD by the 

Converse of the Parallel Lines Conjecture. Thus, 

ABCD is a parallelogram by the definition of 

parallelogram. Because it is an equiangular 

parallelogram, ABCD is a rectangle. 

28. a 54°, b 36°, c 72°, d 108°, e 36°, 

f 144°, g 18°, h 48°, j 48°, k 84° 

29. possible answers: (1, 0); (0, 1); (1, 2); (2, 3) 

30. y 8 6 

x 

9 

8 

9 

or 8x 9y 86 

7 

31. y x 10 

12 

 

5 

or 7x 10y 24 

32. velocity 1.8 mi/h; angle of path 106.1° 

clockwise from the north 

60 

1.5 mi/h 

2 mi/h 

6 

QU AD 

7 QUAD is a 

QD AU 

parallelogram 

Converse of the 

Parallel Lines 

Conjecture 

9 

Definition of 

rhombus 

Definition of 

parallelogram 

QUAD is a 

rhombus 

? 




LESSON 5.7 

1. See flowchart below. 2. See flowchart below. 


4. 

1 SP OA 

1 

1 

Given 

1. (Lesson 5.7) 

Given 

2 SP OA 

Given 

3 1 2 

Same segment 


4 

SOAK is a 

parallelogram 

2. (Lesson 5.7) 

Given 

? Given 


PO PO 

Parallelogram BATH 

with diagonal BT 

4 Parallelogram BATH 

with diagonal HA 

3. (Lesson 5.7) 

? 

? 

? 

Same segment 

2 

3 

5 

SOP APO 

SAS 

6 

3 4 

CPCTC 

7 PA SO 

Converse of 


8 SOAP is a parallelogram 

SO KA 

Definition of parallelogram 

Definition of 

parallelogram 

SK 

OA ? 

 

4 

 

? 

? 

? 

? ? 

? 1 

 

? 

? WA RT 

Given 

WR AT WRT TAW 

2 

 

? 

Given 

WT WT 

SSS 

3 

? 

2 

Definition of 

parallelogram 

4 

5 

6 

BAT THB 


6. sample flowchart proof: 

1 2 

IY GO 

YOG OYI 

Opposite sides of 

rectangle are congruent 

3 4 

AIA 

1 2 

? 

SA 

 

? SA 

? 

Conjecture proved in CPCTC 

Exercise 1 

5 

BAH = 

6 

HBA 

? 

? 

? 

? 

THA 

ATH 

Conjecture proved 

in Exercise 1 

CPCTC 

5 

7 

AIA 

Same segment 

7 

3 BAT THB 

Definition of rectangle 

4 

YOG OYI 

5 

SAS 

YG IO 

CPCTC 

SOA 

? 

? 

ASA 

AKS 

3 

YO OY 

Same segment 

1 

? 6 

 

9 WATR is a 

parallelogram 

4 

? 8 

? 

? 

? 

 

? 

? 

? 

? 

? 

? 

2 RT WA by Converse of the Parallel 

Lines Conjecture 

CPCTC 

3 

RW TA 

CPCTC 

Definition of 

parallelogram 


Parallel Lines Conjecture

7. 

2 

RE AB 

3 

BR EA 

4 

AR EB 

Given Parallelogram Opposite 

Sides Conjecture 

5 

6 

1 BEAR is a parallelogram 

Given 

EBR ARB RAE BEA 

Parallelogram Opposite 

Sides Conjecture 

8. Because AR is parallel to ZT, corresponding 

3 and 2 are congruent. Opposite sides of 

parallelogram ZART are congruent so AR TZ. 

Because the trapezoid is isosceles, AR PT, and 

substituting gives ZT PT making PTZ 

isosceles and 1 and 2 congruent. By 

substitution, 1 and 3 are congruent. 

SSS 

EBR ARB RAE BEA 

CPCTC 

7 BEAR is a rectangle 

Definition of rectangle 

9. (Lesson 5.7) 

1 

Isosceles trapezoid 

GTHR 

Given 

2 

GR TH 

Given 

9. See sample flowchart below. 

10. Ifthefabricispulledalongthewarportheweft, 

nothing happens. However, if the fabric is pulled 

along the bias, it can be stretched because the 

rectangles are pulled into parallelograms. 

11. 30° angles in 4-pointed star, 30° angles in 

6-pointed star; yes 

12. Heshouldmeasurethealternateinterior 

angles to see whether they’re congruent.If they are, 

theedgesareparallel. 

13. ES : y 2x 3; QI : y 1 x 2 2 

14. (12, 7) 

15. 1 

3 

16. 

9 

6 inches 

12 3 

3 

GT GT 

 

5 

RGT HTG 

6 

GH TR 

Same segment 

SAS CPCTC 

4 

RGT HTG 

Isosceles Trapezoid 

Conjecture 

6 





1. 360° divided by the number of sides 

2. Sample answers: Using an interior angle, set the 

interior angle measure formula equal to the angle 

and solve for n. Using an exterior angle, divide into 

360°. Or find the interior angle measure and go 

from there. 

3. Trace both sides of the ruler as shown at right. 

4. Make a rhombus using the double-edged 

straightedge, and draw a diagonal connecting the 

angle vertex to the opposite vertex. 

5. Sample answer: Measure the diagonals with 

string to see if they are congruent and bisect each 

other. 

6. Sample answer: Draw a third point and connect 

it with each of the two points to form two sides of a 

triangle. Find the midpoints of the two sides and 

connect them to construct the midsegment. The 



are parallel 


are congruent 

Opposite angles 

are congruent 


distance between the two points is twice the length 

of the midsegment. 

7. x 10°, y 40° 8. x 60 cm 

9. a 116°, c 64° 10. 100 

11. x 38 cm 

12. y 34 cm, z 51 cm 


14. a 72°, b 108° 

15. a 120°, b 60°, c 60°, d 120°, e 60°, 

f 30°, g 108°, m 24°, p 84°; Possible 

explanation: Because c 60°, the angle that forms 

a linear pair with e and its congruent adjacent 

angle measures 60°. So 60° 2e 180°, and 

e 60°. The triangle containing f has a 60° angle. 

The other angle is a right angle because it forms a 

linear pair with a right angle. So f 30° by the 

Triangle Sum Conjecture. Because g is an interior 

angle in an equiangular pentagon, divide 540° by 5 

to get g 108°. 

16. 15 stones 

17. (1, 0) 

18. When the swing is motionless, the seat, the bar 

at the top, and the chains form a rectangle. When 

you swing left to right, the rectangle changes to a 

parallelogram. The opposite sides stay equal in 

length,so they stay parallel.The seat and the bar 

at the top are also parallel to the ground. 

19. a = 60°, b = 120° 

Isosceles 

Kite trapezoid Parallelogram Rhombus Rectangle Square 

No No Yes Yes Yes Yes 



Diagonals bisect 

each other No No Yes Yes Yes Yes 

Diagonals are 

perpendicular Yes No No Yes No Yes 

Diagonals are 

congruent No Yes No No Yes Yes 

Exactly one line 

of symmetry 

Yes Yes No No No No 

Exactly two lines 

of symmetry No No No Yes Yes No

20. 

Speed: 901.4 km/h. Direction: slightly west of 

north. Figure is approximate. 

21. 

E 


F 

23. 

Resultant 

vector 

S 

y 

N E 

x 

R 

50 km/h 

Q 

Y 

L 

1 

x 

2 

L z 

1 

900 km/h 

F 

R 

x 

P 

ABCD is a parallelogram 

Given 

R 

Y 

L 

1 

x 

2 

2 

AD CB 

Definition of 

parallelogram 

3 AB CD 

Definition of 

parallelogram 

4 1 3 

AIA 


F 

5 

BD BD 

Same segment 

6 2 4 

AIA 

F 

x 

25. 20 sides 

26. 12 cm 



A B 

4 

3 

1 

2 

D C 

x 

R 

Given: Parallelogram ABCD 

Show: AB CD and AD CB 

See flowchart below. 

2 

DE 

? 

1 

DENI is 

a rhombus 

3 

NE 

? 

4 

DN 

? 

5 

 

? 

? 

6 

1 

? 

3 

? 

7 

DN bisects 

IDE and INE 

? 

? 

? 

? 

? 

? 


DI 

Definition of rhombus 

NI DEN DIN 2, 4 

Given 

Definition of 

rhombus 

DN 

SSS 

CPCTC Definition of 



? 

Same segment 

ASA 

R 

Y 

7 ABD CDB 

z 

Y 

D 

z 

D 

8 AB CD 

CPCTC 

9 

AD CB 

CPCTC 






LESSON 6.1 

1. 50° 2. 55° 

3. 30° 4. 105° 

5. 76 in. 


7. Possible answer: The perpendicular to the 

tangent passes through the center of the circle. Use 

the T-square to find two diameters of the Frisbee. 

The intersection of these two lines is the center. 

8. Construct OT . Construct a line through point T 

perpendicular to OT . 

9. Construct OX ,OY , and OZ . Construct tangents 

through points X, Y, and Z. 

X 

O 

Z 

r 

10. Construct tangent circles M and N. Construct 

equilateral MNP.Construct circle P with radius s. 

s s 

M N 

P 

O 


T 

t Y 

6 

Target 

11. Construct a line and label a point T on it. On 

the line, mark off two points, L and M, each on 

the same side of T at distances r and t from T, 

respectively. Construct circle L with radius r. 

Construct circle M with radius t. 

r 

L M T 

t 

12. Start with the construction from Exercise 11. 

On line LM , mark off length TK so that TK s 

and K is on the opposite side of T from L and M. 

Construct circle K with radius s. 

r 

L M T K 

t s 

13. Sample answer: If the three points do not lie 

on the same semicircle, the tangents form a circumscribed 

triangle. If two of the three points are 

on opposite sides of a diameter, the tangent lines to 

those two points are parallel. If the points lie on the 

same semicircle, they form a triangle outside the 

circle, with one side touching (called an exscribed 

triangle). 

14. sample answer: internally tangent: wheels on a 

roller-coaster car in a loop, one bubble inside 

another; externally tangent: touching coins, a 

snowman, a computer mouse ball and its roller 

balls 

15. Constructions will vary. 

16. 360° 30° 90° 90° 150° and 

150° 

 

360° 

41.6% 

17. Angles A and B must be right angles, but this 

would make the sum of the angle measures in the 

quadrilateral shown greater than 360°. 

18a. rhombus 18b. rectangle 

18c. kite 18d. parallelogram 

19. 78° 20. 9.7 km 

21. x 55° 55° 180° and 40° y 

y 180°, so x y 70° 

22. 11 

 

21

LESSON 6.2 

1. 165°, definition of measure of an arc 

2. 84°, Chord Arcs Conj. 

3. 70°, Chord Central Angles Conj. 

4. 8 cm, Chord Distance to Center Conj. 

5. mAC 68°; mB 34° (Because OBC is 

isosceles, mB mC, mB mC 68°, 

and therefore mB 34°.) 

6. w 115°, x 115°, y 65°; Chord Arcs 

Conjecture 

7. 20 cm, Perpendicular to a Chord Conj. 

8. w 110°, x 48°, y 82°, z 120°; definition 

of arc measure 

9. x 96°, Chord Arcs Conjecture; y 96°, 

Chord Central Angles Conjecture; z 42°, 

Isosceles Triangle Conjecture and Triangle Sum 

Conjecture. 

10. x 66°, y 48°, z 66°; Corresponding 

Angles Conjecture, Isosceles Triangle Conjecture, 


11. The length of the chord is greater than the 

length of the diameter. 

12. The perpendicular bisector of the segment 

does not pass through the center of the circle. 

13. The longer chord is closer to the center; the 

longest chord, which is the diameter, passes 

through the center. 

20. (Lesson 6.2) 

1 

AB CD 

14. The central angle of the smaller circle is larger, 

because the chord is closer to the center. 

5 cm 

15. M(4, 3), N(4, 3), O(4, 3) 

16. The center of the circle is the circumcenter of 

the triangle. Possible construction: 

17. possible construction: 

O 

18. 13.8 cm 

? 

2 

AO CO 

4 

 

? 

? 

All radii of a circle 

are congruent 

3 

BO DO 

? 

? 

Given 

AOB COD 

SSS Congruence 

Conjecture 

All radii of a circle are congruent 

5 cm 

19. They can draw two chords and locate the 

intersection of their perpendicular bisectors. The 

radius is just over 5 km. 


5 

AOB 

? 

? 

COD 

CPCTC 




21. y 1 

7 x; (0, 0) is a point on this line. 

22a. true 

A 

C 

D 

X 

B 

Possible explanation: If AB is the perpendicular 

bisector of CD , then every point on AB is 

equidistant from endpoints C and D. Therefore 

AC AD and BC BD. Because CD is not the 

perpendicular bisector of AB, Cis not equidistant 

from A and B. Likewise, D is not equidistant from 

A and B. So,AC and BC are not congruent, and 

AD and BD are not congruent. Thus ACBD 

has exactly two pairs of consecutive congruent 

sides, so it is a kite. 

22b. false, isosceles trapezoid 

22c. false, rectangle 

23. 140°, 160°, 60°; 180° minus the measure of 

the intercepted arc 


24. Using the Tangent Conjecture, 2 and 4 

are right angles, so m2 m4 180°. 

According to the Quadrilateral Sum Conjecture, 

m1 m2 m3 m4 360°, so by the 

Subtraction Property of Equality, m1 

m3 180°, or m1 180° m3. The 

measure of a central angle equals the measure of 

its arc, so by substitution, m1 180° mAB . 

25. the circumcenter of the triangle formed by 

the three light switches 

26. 7 

27. Station Beta is closer. 

N 

72 

38 

Alpha 

7.2 mi 

28. D 

Downed aircraft 

8.2 mi 

4.6 mi 

48 

312 

Beta

1 OR CD 

? 

5 OC OD 

? 

All radii of 

a circle are 

congruent 

LESSON 6.3 

1. 65° 2. 30° 3. 70° 4. 50° 

5. 140°, 42° 6. 90°, 100° 7. 50° 8. 148° 

9. 44° 10. 142° 11. 120°, 60° 

12. 140°, 111° 13. 71°, 41° 14. 180° 

15. 75° 

16. The two inscribed angles intercept the same 

arc, so they should be congruent. 

17. BFE DFA (Vertical Angles Conjecture). 

BGD FHD (all right angles congruent). 

Therefore, B D (Third Angle Conjecture). 

mB 1 

2 mAC ,mD 1 

2 mEC ,AC EC 

18. Possible answer: Place the corner so that it is 

an inscribed angle. Trace the inscribed angle. Use 

the side of the paper to construct the hypotenuse 

of the right triangle (which is the diameter). Repeat 

the process. The place where the two diameters 

intersect is the center. 


It works on acute and right triangles. 

20. The camera can be placed anywhere on the 

major arc (measuring 268°) of a circle such that the 

row of students is a chord intersecting the circle to 

form a minor arc measuring 92°. This illustrates 

the conjecture that inscribed angles that intercept 

the same arc are congruent (Inscribed Angles 

Intercepting Arcs Conjecture). 

21. two congruent externally tangent circles with 

half the diameter of the original circle 

24. (Lesson 6.3) 

Given 

 

6 

OCD is isosceles 

? 

2 

ORC and ORD 

are right angles 

Definition of 

perpendicular lines 

? 

Definition of 

isosceles triangle 

22. a 108°; b 72°; c 36°; d 108°; 

e 108°; f 72°; g 108°; h 90°; l 36°; 

m 18°; n 54°; p 36° 

23. (2, 1); Possible method: Plot the 

three points. Construct the midpoint and the 

perpendicular bisector of the segments connecting 

two different pairs of points. The center is the point 

of intersection of the two lines. To check, construct 

the circle through the three given points. 


25. Start with an equilateral triangle whose vertices 

are the centers of the three congruent circles. Then 

locate the incenter/circumcenter/orthocenter/ 

centroid (all the same point because the triangle is 

equilateral) to find the center of the larger circle. To 

find the radius, construct a segment from the 

incenter of the triangle through the vertex of the 

triangle to a point on the circle. 

26. mA 60°, mB 36°, mC 90°; 

60° 36° 90° 180° 

3 mORC 90 

mORD 90 

? 

7 C ? 

? 

Definition of 

right angle 

D 

7. Isosceles Triangle 

Conjecture 

4 mORC mORD 

(ORC ORD) 

? 

8 OCR ? 

? 

9 CR ? 

? 

10 OR bisects CD 

? 

Substitution 

property 

SAA 

DR 

CPCTC 

ODR 

Definition of bisect 




LESSON 6.4 

1. Proof: mACD 1 

2 mAD mABD by the 

Inscribed Angle Conjecture. ACD ABD. 

2. Proof: By the Inscribed Angle Conjecture, 

mACB 1 

2 mAD B 1 (180°) 2 90°. ACB is a 

right angle. 

3. Proof: By the Inscribed Angle Conjecture, 

mC 1 

2 mYLI and mL 1 

2 mYCI 1 (360° 2 

mYLI 

) 180° 1 

2 mYLI 180° mC. L 

and C are supplementary. (A similar proof can be 

used to show that I and Y are supplementary.) 

4. Proof: 1 2 by AIA. mBC 2m2 

2m1 mAD by the Inscribed Angle Conjecture. 

BC AD . 

5. True. Opposite angles of a parallelogram are 

congruent. If it is inscribed in a circle, the opposite 

angles are also supplementary. So they are right 

angles, and the parallelogram is indeed 

equiangular, or a rectangle. 

6. Construct diagonals RG and DB. Diagonal RG 

is the perpendicular bisector of BD by the Kite 

Diagonal Bisector Conjecture. Because kite BRDG 

is inscribed in the circle, BD and RG are chords of 

the circle. By the Perpendicular Bisector of a Chord 

Conjecture, the perpendicular bisector of BD 

passes through the center of the circle. Because RG 

is a chord of the circle that passes through the 

center, by the definition of diameter, RG is a 

diameter. 

7. Draw in radii GR,ER,TR, and AR. Because they 

are radii of the same circle, GR ER and AR TR. 

By the Parallel Lines Intercepted Arcs Conjecture, 

GA ET . Their central angles must also be 

congruent, so GRA ERT.Thus GRA 

ERT by SAS, so GA ET by CPCTC. GATE is 

an isosceles trapezoid. 

8. x 65°, y 40°, z 148°; half the measure of 

the intercepted arc 

9. Because the radii of a circle are congruent, 

OA OB, so OAB is isosceles. By the Isosceles 

Triangle Conjecture, 2 3, so m2 m3. 

By the Triangle Sum Conjecture, m1 m2 

m3 180°, so by substitution, m1 

2(m2) 180°, or m1 180° 2(m2). 


By the Tangent Conjecture, OB BC , so 

mOBC 90°. By Angle Addition, m2 

m4 mOBC, so m2 m4 90°, 

or m2 90° m4. Substitute 90° m4 

for m2 in the earlier equation: m1 180° 

2(90° m4). Simplifying gives m1 2(m4). 

Because mAB m1, substitution gives mAB 

2(m4), or m4 1 

2 mAB . 

10a. S; An equilateral triangle is equiangular, but a 

rhombus is not equiangular. 

10b. A 

10c. N; Only one diagonal is the perpendicular 

bisector of the other. 

10d. A 

10e. S; An equilateral triangle has rotational 

symmetry and three lines of symmetry; a 

parallelogram has rotational symmetry but no line 

of symmetry. 

11. L 

12. J 

13. K 

14. D 

15. E 

16. B 

17. H 

18. G 

19. N 

20. a b b a 180°, so a b 90° 

2 

21. 21 

22. It is given that PQ RS. OP OQ OR 

OS because all radii in a circle are congruent. 

Therefore, OPQ ORS by SSS and 2 1 

by CPCTC. Now we move on to the smaller right 

triangles inside OPQ and ORS.It is given that 

OT PQ and OV RS. Therefore, OTQ and 

OVS are right angles by the definition of 

perpendicular lines and OTQ OVS because 

all right angles are congruent. Thus, OTQ 

OVS by SAA and OT OV by CPCTC.

LESSON 6.5 

1. d 5 cm 

2. C 10 cm 

3. r 12 

 

 

m 

4. C 5.5 or 11 

 

2 

5. C 12 cm 

6. d 46 m 

7. C 15.7 cm 

8. C 25.1 cm 

9. r 7.0 m 

10. C 84.8 in. 

11. 565 ft 

12. C 6 cm 

13. 16 in. 

14. Trees grow more in years with more rain; 

244 yr 

18. (Lesson 6.5) 

1 

2 

? MA MT 

Given 

3 MAST is a kite 

15. 1399 tiles 

16. g 40°, n 30°, x 70°; y 142°; z 110°; 

Conjecture: The measure of the angle formed by 

two intersecting chords is one-half the sum of the 

measures of the two intercepted arcs. In the 

diagrams, mAEN 1 

2 mAN mLG . 

17. mAGN 1 

2 mAN and mLNG 1 

2 mLG 

by the Inscribed Angle Conjecture. mAEN 

mAGN mLNG bytheTriangleExterior 

Angle Conjecture. So, mAEN 1 

2 mAN mLG 

by substitution and the distributive property. 


19. b 90°, c 42°, d 70°, e 48°, f 132°, 

g 52° 

20. 150° 

5 

 

360° 

or 12 

21. The base angles of the isosceles triangle have a 

measure of 39°. Because the corresponding angles 

are congruent, m is parallel to n. 

22. 10x 2y 

4 

? 

MS is the perpendicular bisector of AT 

? 

Given 

? 

? 

SA ST Definition of kite Kite Diagonal Bisector Conjecture 




LESSON 6.6 

1. 4398 km/h 

2. 11 m/s 

3. 37,000,000 revolutions 

4. 637 revolutions 

5. Mama; C 50 in. 

6. 168 cm 

7. d 7.6 ft. The table will fit, but the chairs may 

be a little tight in a 12-by-14 ft room. 12 chairs 

192 in., 12 spaces 96 in., C 288 in., d 91.7 in. 

7.6 ft. 

8. 0.35 ft/s 

9. a 37.5°, s 17.5°, x 20°; y 80°; z 61°; 

Conjecture: The measure of an angle formed by 

two secants that intersect outside a circle is 

one-half the difference of the larger arc measure 

and the smaller arc measure. 


10. mESA 1 

2 mEA and mSAN 1 

2 mSN by 

the Inscribed Angle Conjecture. mSAN 

mESA mECA by the Triangle Exterior Angle 

Conjecture. So, 1 

2 mSN 1 

2 mEA mECA by 

substitution. With a little more algebra, mECA 

1 

2 mSN mEA . 

11. Both triangles are isosceles, so the base angles 

in each triangle are congruent. But one of each base 

angle is part of a vertical pair. So, a b by the 

Vertical Angles Conjecture and transitivity. 

12. C 

13. 38° 

14. 48° 

15. 30 cm 

16. 400 

rev 

 

1 min 

2 . 26 

ft 

 

1 rev 

1 min 

 

60 

s 

1089 ft/s 

17. 12 cm third side 60 cm. This is based on 

the Triangle Inequality Conjecture.

1. 1 

2 ,3 


2. (3, 3) 

3. 2 5 ,3 

4. (7, 2) 

5. infinitely many solutions 

6. no solution 

7. 4. The lines intersect at the solution point, (7,2). 

5 

–5 

–10 

5. The lines are the same. There are infinitely many 

solutions. 

4 

y 

y 

y 2x 16 

5 

( 7, 2) 

y 

1 

3 

x 2 

x 

10 

y 

1 

2 

x 

3 

2 

6. The lines are parallel (the slopes are the same, 

but the y-intercepts are different). 

There is no solution. 

y 

–10 

y = –2x + 9 

8. Possible solution using the equations 

y 1 

4 

x 2 1 and y 2x 

3 1 : 3 

y 1 

x 

2 

1 

2 4 

x 

3 

1 

3 

1 

x 

2 

1 

4x 28 3x 6 

22 7x 

22 

 

7 

x 

y 1 2 

2 2 7 

1 

y 18 

 

7 

x 

–5 

5 

–5 

x 

y = –2x – 1 

The circumcenter is 22 

, 

7 

18 

. 7 Only two 

perpendicular bisectors are needed because 

the third bisector intersects at the same point. 

9a. Plan A: y 4x 20; Plan B: y 7x; 

x 6 h 40 min, y $46.67 

9b. y 

Cost ($) 

50 

40 

30 

20 

10 

(0, 20) 

The point of intersection shows when both plans 

cost the same, the answer for 9a. 

9c. Plan B; 7 1 

 

2 hours, with Plan A 

10. (3, 1), (0, 3), (3, 1) 

11a. 

11b. (6, 6) 

11c. (6, 6); (6, 6) 

11d. The diagonals intersect at their midpoints, 

which supports the conjecture that the diagonals of 

a parallelogram bisect each other. 

12. y 2 3 

x 

3 

1 

3 

13. (4, 7) 

14. 69 

, 

14 

6 

7 

y 3 

x 

4 

3 

2 

y x 12 

15. 3, 3 

2 

Plan A 

Plan B 

2 4 6 

Hours 

6 2_ , 46 

2_ 

 

3 

16. The circumcenter is the midpoint of the 

hypotenuse. For a right triangle, the perpendicular 

bisectors of the legs are two of the midsegments 

of the triangle. As with any pair of triangle 

midsegments, they intersect at the midpoint of 

the third side. 

x 

3 




LESSON 6.7 

1. 4 

 

3 

in. 2. 8 m 3. 14 cm 

4. 9 m 5. 6 ft 6. 4 m 

7. 27 in. 8. 100 cm 9. 217 m/min 

10. 4200 mi 

11. Desks are about 17 meters from the center. 

About four desks will fit because an arc with 

one-half the radius and the same central angle 

will be one-half as long as the outer arc. 

12. The measure of the central angle is 7.2° 

because of the Corresponding Angles Conjecture. 

7.2 

Therefore, 500 360 

C, so C 25,000 mi. 

13. 18°/s. No, the angular velocity is measured in 

degrees per second not in distance per second, so it 

is the same at every point on the carousel. 

14. Outer horse 2.5 m/s, inner horse 1.9 m/s. 

One horse has traveled farther in the same amount of 

time (tangential velocity), but both horses have 

rotated the same number of times (angular velocity). 

15. a 50°, b 75°, x 25°; y 45°; z 35°; 

Conjecture: The measure of the angle formed by 

an intersecting tangent and secant to a circle is 

one-half the difference of the larger intercepted arc 

measure and the smaller intercepted arc measure. 

16. Let z represent the measure of the exterior 

angle of PBA formed by AB and tangent PB . 

By the Tangent Chord Conjecture, z 1 

2 mAB .By 

the Inscribed Angle Conjecture, mBAP 1 

2 mBC . 

By the Triangle Exterior Angle Conjecture, z 

mBPA mBAP, or mBPA z mBAP. 


By substitution, mBPA 1 

2 mAB 1 

2 mBC . 

Using the distributive property, mBPA 

1 

2 (mAB mBC ). 

17. a 70°, b 110°, c 110°, d 70°, e 20°, 

f 20°, g 90°, h 70°, k 20°, m 20°, 

n 20°, p 140°, r 80°, s 100°, t 80°, 

u 120° 


19. 170° 

8 

20. y x 15 

289 

 

15 

21. 45° 

22. The sum of the lengths is 8 cm for Case 1, 

Case 2, Case 3, and Case 10. 

23. 6 

24. Yes, as long as the three points are noncollinear; 

possible answer: connect the points with 

segments, then find the point of concurrency of 

the perpendicular bisectors (same as circumcenter 

construction).



2. Draw two nonparallel chords. The intersection 

of their perpendicular bisectors is the center of the 

circle. 

Fold the paper so that two semicircles coincide. 

Repeat with two different semicircles. The center is 

the intersection of the two folds. 

Place the outside or inside corner of the L in the 

circle so that it is an inscribed right angle. Trace the 

sidesofthecorner.Drawthehypotenuseoftheright 

triangle (which is the diameter of the circle). Repeat. 

The center is the intersection of the two diameters. 

3. The velocity vector is always perpendicular to 

the radius at the point of tangency to the object’s 

circular path. 

4. Sample answer: An arc measure is between 0° and 

360°.An arc length is proportional to arc measure 

and depends on the radius of the circle. 

5. 55° 6. 65° 7. 128° 8. 118° 

9. 91° 10. 66° 11. 125.7 cm 12. 42.0 cm 

13. 15 cm 

14. 14 ft 

15. 2 57° 2 35° 180° 

16. 84° 56° 56° 158° 360° 

17. mEKL 1 

2 mEL 1 

 

2 (180° 108°) 36° 

mKLY. KE YL by Converse of the Parallel 

Lines Conjecture. 

18. mJI 360° 56° 152° 152° mMI 

. 

mJMI 1 

2 mJI 1 

 

2 mMI 

mMJI.By the 

Converse of the Isosceles Triangle Conjecture, 

JIM is isosceles. 

19. mKIM 2mKEM 140°. mKI 

 

140° 70° 70° mMI 

. mIKM 

1 

 

2 mMI 

1 

 

2 mKI 

mIMK. By the Converse of the 

Isosceles Triangle Conjecture, KIM is isosceles. 

20. Ertha can trace the incomplete circle on paper. 

She can lay the corner of the pad on the circle to 

trace an inscribed right angle. Then Ertha should 

mark the endpoints of the intercepted arc and use 

the pad to construct the hypotenuse of the right 

triangle, which is the diameter of the circle. 

21. Sample answer: Construct 

perpendicular bisectors of two 

sides of the triangle. The point 

at which they intersect (the 

circumcenter) is the center of 

the circle. The distance from 

the circumcenter to each vertex 

is the radius. 

22. Sample answer: Construct the incenter (from 

the angle bisectors) of the triangle. From the incenter, 

which is the center of the circle, construct a perpendicular 

to a side. The distance from the incenter to 

the foot of the perpendicular is the radius. 

23. Sample answer: Construct a right angle and 

label the vertex R.Mark offRE and RT with any 

lengths. From point E, swing an arc with radius RT. 

From point T, swing an arc with radius RE. Label 

the intersection of the arcs as C. Construct the 

diagonals ET and RC. Their intersection is the 

center of the circumscribed circle. The circle’s 

radius is the distance from the center to a vertex. It 

is not possible to construct an inscribed circle in a 

rectangle unless it is a square. 

24. Sample answer: Construct acute angle R.Mark 

off equal lengths RM and RH.FrompointsM and 

H, swing arcs of lengths equal to RM. Label the 

intersection of the arcs as O. Construct RHOM. The 

intersection of the diagonals is the center of the inscribed 

circle. Construct a perpendicular to a side to 

find the radius. It is not possible to construct a circumscribed 

circle unless the rhombus is a square. 

R 

E C 

R 

H 

25. 4x 3y 32, or y 4 2 

x 

3 

3 

3 

26. (3, 2) 27. d 0.318 m 

28. Melanie: 151 m/min or 9 km/h; Melody: 

94 m/min or 6 km/h. 

2(6357) 

29. 1.849 1.852 1.855 

360 60 

30. Possible 

location 

5500 ft 

Possible 

location 

M 

5280 ft 

D T 

T 

O 

7700 ft 

2(6378) 

360 60 




31. 200 

 

 

ft 63.7 ft 32. 8 m 25.1 m 

48 

33. The circumference is 360 

2(45) 12;the 

diameter is 12 cm. 

34. False. 20° 20° 140° 180°. An angle with 

measure 140° is obtuse. 

35. true 

36. false 

D 

37. true 38. true 39. true 40. true 

41. False. (7 2) 180° 900°. It could have seven 

sides. 

42. False. The sum of the measures of any triangle 

is 180°. 

43. False. The sum of the measures of one set of 

exterior angles for any polygon is 360°. The sum of 

the measures of the interior angles of a triangle is 

180° and of a quadrilateral is 360°. Neither is greater 

than 360°, so these are two counterexamples. 

44. False. The consecutive angles between the 

bases are supplementary. 

45. False. 48° 48° 132° 180° 

46. False. Inscribed angles that intercept the same 

arc are congruent. 

47. False. The measure of an inscribed angle is 

half the measure of the arc. 

48. true 

49. False. AC and BD bisect each other, but AC is 

not perpendicular to BD. 

D 

A B 

50. False. It could be isosceles. 

51. False. 100° 100° 100° 60° 360° 

52. false; AB CD 

A 

A 

90 


B 

C 

B 

C 

90 


D 

C 

53. False. The ratio of the circumference to the 

diameter is . 

54. false; 24 24 

48 48 96 

24 cm 24 cm 

55. true 56. This is a paradox. 

57. a 58°,b 61°,c 58°,d 122°,e 58°, 

f 64°,g 116°,h 52°,i 64°,k 64°, 

l 105°,m 105°,n 105°,p 75°,q 116°, 

r 90°,s 58°,t 122°,u 105°,v 75°, 

w 61°,x 29°,y 151° 

58. TAR YRA by SAS, TAE YRE 

by SAA 

59. FTO YTO by SAA, SAS, or SSS; 

FLO YLO by SAA, SAS, or SSS; 

FTL YTL by SSS, SAS, or ASA 

60. PTR ART by SSS or SAS; 

TPA RAP by SSS, SAS, SAA, or ASA; 

TLP RLA by SAA or ASA 

61. ASA 

62. mAC 84°, length of AC 11.2 35.2 in. 

63. x 63°, y 27°, w 126° 



66a. The circle with its contents has 3-fold rotational 

symmetry, the entire tile does not. 

66b. No, it does not have reflectional symmetry. 

67. 68. 9.375 cm 

69. 70. 

n 1 2 3 4 5 6 . . . n ... 20 

f(n) 5 1 3 7 11 15 ... 9 4n ... 71 

48 cm 48 cm 

150 

30



LESSON 7.1 

1. Rigid; reflected, but the size and shape do not 

change. 

2. Nonrigid; the shape changes. 

3. Nonrigid; the size changes. 

4. 5. 

6. 

 

7 

7. possible answer: a boat moving across the water 

8. possible answer: a Ferris wheel 

9a. Sample answer: Fold the paper so that the 

images coincide, and crease. 

9b. Construct a segment that connects 

two corresponding points. Construct the 

perpendicular bisector of that segment. 

10a. Extend the three horizontal segments onto 

the other side of the reflection line. Use your 

compass to measure lengths of segments and 

distances from the reflection line. 

10b. 

P 

16. (Lesson 7.1) 

11. 


13. 4-fold rotational and reflectional symmetry 


15. 7-fold symmetry: possible answers are F or J. 

9-fold symmetry: possible answers are E or H. 

Basket K has 3-fold rotational symmetry but not 

reflectional symmetry. 

16. See table below; n, n 

17. 

18. 

, or 

19. P(a, b), Q(a, b), R(a, b) 

20. possible construction: 

P 

21. 50th figure: 154 (50 shaded, 104 unshaded); 

nth figure: 3n 4 (n shaded, 2(n 2) unshaded) 

22. 46 

23. It is given that 1 2, and 2 3 

because of the Vertical Angles Conjecture, so 

1 3. Segment DC is congruent to itself. 

DCE and DCB are both right angles, so they 

are congruent. Therefore, DCB DCE by 

ASA, and BC CE by CPCTC. 

Number of sides of 

regular polygon 

3 4 5 6 7 8 . . . n 

Number of reflectional 

symmetries 

3 4 5 6 7 8 ... n 

Number of rotational 

symmetries ( 360°) 

3 4 5 6 7 8 ... n 

P 




1. 

2. reflection 

3. 

4. 

5. 

–5 

–4 

5 

5 

y 

8 

–8 

y 

rotation 

y 

translation 

5 

–5 

y 

x 

reflection 

5 

LESSON 7.2 


y 

5 

reflection 

5 

5 

7 

x 

x 

x 

x 

6. Rules that involve x or y changing signs, 

or switching places, produce reflections. 

If both x and y change signs, the rule produces a 

rotation. Rules that produce translations involve a 

constant being added to the x and/or y terms. 

5, 0 is the translation vector for Exercise 1. 

7. (x, y) → (x, y) 

8. (x, y) → (x, y) 

9. 

10. There are two possible points, one on the N 

wall and one on the W wall. 

11. 

12. by the Minimal Path Conjecture 

13. 

14. 

W 

T 

Proposed freeway 

Perry 

, 

W 

N 

W E 

S 

T 

N 

8 ball 

S 

N 

S 

H H' 

Mason 

Cue ball 

H'' 

H 

E 

E

15. possible answer: HIKED 

16. one, unless it is equilateral, in which case it 

has three 

17. two, unless it is a square, in which case it has 

four 

18. 



21. false; possible counterexample: trapezoid 

with two right angles 

22. false; possible counterexample: isosceles 

trapezoid 




LESSON 7.3 

1. 10, 10 

2. A 180° rotation. If the centers of rotation differ, 

rotate 180° and add a translation. 

3a. 20 cm 

3b. 20 cm, but in the opposite direction 

4a. 80° counterclockwise 

4b. 80° clockwise 

5. 180° 

6. 3 cm 



N 

9. 

O A 

H 

10. Two reflections across intersecting lines yield 

a rotation. The measure of the angle of rotation is 

twice the measure of the angle between the lines of 

reflection, or twice 90°, or 180°. 


O′ A′ 

H′ 

N′ 

H′′ 

O′′ A′′ 

N′′ 

Center of rotation 

11. Answers may vary. Possible answer: reflection 

across the figure’s horizontal axis and 60° 

clockwise rotation. 

12. 

13. 

, 

14. Sample answer: Draw a figure on an overhead 

transparency and then project the image onto a 

screen. 

15. possible answers: rotational: playing card, 

ceiling fan, propeller blade; reflectional: human 

body, backpack 

16. one: yes; two: no; three: yes 


A O B 

18a. 

14 11 

0 

? 

? 

5 11 

18b. 

a b 

d e 

12 

? ? 

? 

c a 2b 3c 

f 

20 

d 

9 0 

? 

? 

12 7 

d–e 

0 

13 

2a 3b 4c 

? 

? 

? 

0 d f

LESSON 7.4 

1. Answers will vary. 2. Answers will vary. 

3. 3 3 .4 2 4. 3 4 .6 

5. 3 2 .4.3.4 6. 3.4.6.43.4 2 .6 

7. 3 3 .4 2 3 2 .4.3.4 8. 3 6 3 2 .4.12 

9a. The dual of a square tessellation is a square 

tessellation. 

9b. The dual of a hexagon tessellation is a triangle 

tessellation. 

9c. If tessellation A is the dual of tessellation B, 

then tessellation B is the dual of tessellation A. 

10. The dual is a 3 4 3 8 tessellation of isosceles 

right triangles. 

11. 

12. 

13. A ring of ten pentagons fits around a decagon, 

and another decagon can fit into any two of the 

pentagons. But another ring of pentagons around 

the second decagon doesn’t leave room for a third 

decagon. 

14. 


16. y 1 

x 

2 

4 

–3 

17. possible answer: TOT 

18. 

W 

4 

–6 

y 

8-ball 

5 

x 

N 

Cue ball 

S 

E 




LESSON 7.5 


2. The dual is a 5 3 5 4 tessellation. 

3. 

4. Yes. The four angles of the quadrilateral 

will be around each point of intersection in the 

tessellation. 

5. c a c a 

b 

b 

b 

a 

c 

c b a 

a c 

b 

c b a 

a c 

b 

c 

a 

b a c b a c b 


By the Triangle Sum Conjecture, a b c 180°. 

Around each point, we have 2(a b c) 

2 180° 360°. Therefore, a triangle will fill the 

plane edge to edge without gaps or overlaps. Thus, a 

triangle can be used to create a monohedral tiling. 

6. three ways 

7. 

8. y 2x 3 

8 

–2 

y 

5 

x

LESSON 7.6 




4. regular hexagons 

5. squares or parallelograms 

6. squares or parallelograms 

7. 

8. 



11. 

A 

S 

B 

E 

12. y 2 

x 

3 

3; the slope is the opposite sign. 

–10 

5 

y 

13. 3.4.6.44.6.12 

14. 440 

rev 

 

1 min 

1 

2 28 ft min 

1290 ft/s 

1 rev 60 

s 

15. Possible explanations: 

15a. true; The kite diagonal between vertex angles 

is the perpendicular bisector of the other diagonal; 

in a square, diagonals would bisect each other 

15b. False; it could be an isosceles trapezoid. 

15c. False; it could be a rectangle. 

15d. true; Parallel lines cut off congruent arcs of a 

circle, so inscribed angles (the base angles of the 

trapezoid) are congruent. 

10 

x 




1. equilateral triangles. 

2. regular hexagons. 

3. 

4. 



7. sample design: 

LESSON 7.7 

8. False; they must bisect each other in a 



9. true 

10. true 

11. False; it could be a kite or an isosceles 

trapezoid. 

12. The path would be 1 

 

4 of Earth’s circumference, 

approximately 6280 miles, which will take 

126 hours, or around 5 1 

 

4 days. 

13a. Using the Reflection Line Conjecture, the 

line of reflection is the perpendicular bisector of 

AAand BB. Because these segments are both 

perpendicular to the reflection line, they are 

parallel to each other. Note that if AB is parallel to 

the reflection line, quadrilateral AABB will be a 

rectangle instead of a trapezoid. 

13b. Yes; it has reflectional symmetry, so legs and 

base angles are congruent. 

13c. greatest: near each of the acute vertices; 

least: at the intersection of the diagonals (where A, 

C, and B become collinear and A,C,and B 

become collinear) 

14a. 

? 

? 

3 5 6 

6 0 

1 

0 

14b. 13 30 

 

? 

2 

8 3 

12 

4 

8 

9 

0 

7 

2 

? 

? 5 

1 10 

108 9 

? 

? 

28 15 

? 

29 

50

1. parallelograms 

2. parallelograms 

3. 

4. 

LESSON 7.8 



7. Circumcenter is (3, 4); orthocenter is (10, 8). 

8. 

9. 

10. 





1. y 1 

6 x 

2. y 2x 2 

3. Centroid is 2, 2 

3 ;orthocenter is (0,5). 


4. Centroid is (4, 0); orthocenter is (3, 0). 

5. 1, 4 

3 

6. (1, 1) 

7. (5, 8)





7. False; a regular pentagon does not create a 

monohedral tessellation and a regular hexagon 

does. 


10. False; two counterexamples are given in 

Lesson 7.5. 

11. False; any hexagon with one pair of opposite 

sides parallel and congruent will create a 

monohedral tessellation. 

12. This statement can be both true and false. 

13. 6-fold rotational symmetry 

14. translational symmetry 

15. Reflectional; color arrangements will vary, but 

the white candle must be in the middle. 

16. The two towers are not the reflection (or 

even the translation) of each other. Each tower 

individually has bilateral symmetry. The center 

portion has bilateral symmetry. 



19. 3 6 3 2 .4.3.4; 2-uniform 

20. 4.8 2 ; semiregular 

21. y 1 

2 x 

y 

x 

22. 

T H 

23. Use a grid of squares. Tessellate by translation. 

24. Use a grid of equilateral triangles. Tessellate by 

rotation. 

25. Use a grid of parallelograms. Tessellate by 

glide reflection. 

26. Yes. It is a glide reflection for one pair of sides 

and midpoint rotation for the other two sides. 

27. No.Because the shape is suitable for glide 

reflection,the rows of parallelograms should 

alternatethedirectioninwhichtheylean(row1 

leans right,row 2 leans left,row 3 leans right,and 

so on). 

28. 






1. 228 m 2 

2. 41.85 cm 2 

3. 8 yd 

4. 21 cm 

5. 91 ft 2 

6. 182 m 2 

7. 96 in 2 

8. 210 cm 

9. A 42 ft 2 


4 cm 

11. 3 square units 


13. 7 1 

 

2 square units 

14. sample answers: 


16 cm 

16 cm 

12 cm 

48 cm 2 

16 cm 

64 cm2 4 cm 

4 cm 

LESSON 8.1 

8 cm 

16 cm 

8 

16 cm 

64 cm 2 

16. 23.1 m2 17. 2(4)(3) 2(5.5)(3) 57 m2 18. For a constant perimeter, area is maximized by 

a square. 100 m 4 25 m per side; A 625 m2 . 

1 

19. 

530 

 

20. 112 




8 cm 

8 cm 

6 cm 

48 cm 2 

23. 500 cm2 24a. smallest: 191.88 cm2 ; largest: 194.68 cm2 24b. Answers will vary. Sample answer: about 

193 cm2 . 

24c. Answers will vary. The smallest and largest 

area values differ at the ones place, so the digits 

after the decimal point are insignificant compared 

to the effect of the limit of precision in the 

measurements. 

25a. In one Ohio Star block, the sum of the red 

patches is 36 in2 , the sum of the blue patches is 

72 in2 ,and the yellow patch is 36in2 . 

25b. 42 

25c. About 1814 in2 of red fabric, about 3629 in2 of blue fabric, and about 1814 in2 of yellow 

fabric. The border requires 5580 in2 (if it does 

not need the extra 20%). 

26. 100; 36 64. The area of the square on the 

longer side is the same as the sum of the areas on 

the other two legs. 

27. a 76°,b 52°,c 104°,d 52°,e 76°, 

f 47°, g 90°, h 43°, k 104°, m 86°. 

Explanations will vary. 


A 

29a. 29b. 

29c. 

30° 

30° 

M

1. 20 cm 2 

2. 49.5 m 2 


4. 60 cm 2 

5. 6 cm 

6. 9 ft 

7. 30 ft 

8. 5 cm 

9. 16 m 

10. 168 cm 

11. 12 cm 

12. 3.6 ft; 10.8 ft 


12 cm 

9 cm 12 cm 


4 cm 


46 cm 35 cm 

12 cm 

45 cm 56 cm 

LESSON 8.2 

16. The length of the base of the triangle equals 

the sum of the lengths of both bases of the 

trapezoid. 

2 cm 

8 cm 

10 cm 9 cm 

12 cm 

9 cm 

7 cm 

7 cm 

3 cm 3 cm 

7 cm 9 cm 

17. 1 (To 2 see why, draw altitude PQ.) 

18. more than half, because the top card 

completely covers one corner of the bottom card 

19a. 86 in. of balsa wood and 960 in2 of Mylar 

19b. 56 in. (or less, if he tilts the kite) 

20. 3600 shingles (to cover an area of 900 ft2 ) 

21. The isosceles triangle is a right triangle 

because the angles on either side of the right 

angle are complementary. If you use the trapezoid 

area formula, the area of the trapezoid is 

(a b)(a b). If you add the areas of the 

1 

2 

three triangles, the area of the trapezoid is 

1 

2 c2 ab. 

22. 

D 

A 

h 

b 1 

b 2 

Given: trapezoid ABCD with height h. area 

of ABD 1 

2 hb1 ;area ofBCD 1 

2 hb2 ;area 

of trapezoid sum of areas of two triangles 

1 

2 hb 1 b 2 

B 

C 

23. 11 1 

 

4 square units 


25. 70 m 

26. 144 cm2 27. 828 ft2 ; 144 ft 

28. 1440 cm2 ; 220 cm 

29a. incenter 

29b. orthocenter 

29c. centroid 

30. a 34°, b 68°, c 68°, d 56°, e 56°, 

f 90°, g 34°, h 56°, m 56°, n 90°, 

p 34°. Possible explanation: Let O be the center 

of the circle. mBC 112° by the Inscribed Angle 

Conjecture, and d e mBC by the Central 

Angle Conjecture. OBA is congruent to OCA 

by SSS, so d e 56°. DEC 

is a semicircle, so 

mDE 68°. By the Inscribed Angle Conjecture, 

p 34°. Using OEC and the Triangle Sum 

Conjecture, n 90°. 

31. 32 .623.6.3.6 ANSWERS TO EXERCISES 97 



LESSON 8.3 

1a. 121,952 ft 2 

1b. 244 gal of base paint and 488 gal of finishing 

paint 

2. He should buy at least four rolls of wallpaper. 

(The area of each roll is 125 ft 2 . The total surface 

area to be papered is 480 ft 2 .) If paper cut off at the 

corners is wasted, he’ll need 5 rolls. 

3. 1552 ft 2 ; 776 ft 2 more surface area 

4. 21 

5. 336 ft 2 ; $1780 


6. $760 

7. 220 terra cotta tiles, 1107 blue tiles; $1598.15 

8. 72 cm2 9. AB 16.5 cm, BD 15.3 cm 

10. 60 cm2 by either method 

11. Because AOB is isosceles, mA 20° and 

mAOB 140°. mAB 140° and mCD 82°. 

mAC mBD because parallel lines intercept 

congruent arcs on a circle. 360°1 40° 82° 

 

2 69°. 

12. E


1. x 2 6x 5 2. 2x 2 7x 

x 1 x 

3. 6x 2 19x 10 

2x 5 

4. (3)(2x 1) 5. (x 5)(x 3) 

2x 1 

6. (2x 3)(x 4) 

x 4 

7. x 2 26x 165 8. 12x 2 13x 35 

9. x 2 8x 16 10. 4x 2 25 

x 

4 

x 

11 

1 

x 

4 

x 

x 

x 

5 

x 

x 

1 

x 2 

x 

x 2 

3 

x 

4 

4x 

4x 16 

x 5 

x 

x 2 

x 

15 

15x 

11x 165 

5 

3x 2 

x x 2 

2x 3 

x 3 

x 3 x 

4x 

5 

2x 

5 

3 

x 

2x 7 x 

7 

3x 

12x 2 

2x 

4x 2 

x 

x 5 

x 

x 2 

7 

28x 

15x 35 

5 

10x 

10x 25 

5 

11. a 2 2ab b 2 12. a 2 b 2 

a 

b 

13. (x 15)(x 4) 14. (x 12)(x 2) 

x 

4 

15. (x 5)(x 4) 

x 

4 

16. (x 3) 2 (x 3)(x 3) 

x 

3 

17. (x 6)(x 6) 

x 

6 

18. (2x 7)(2x 7) 

2x 

7 

a 

a 2 

x 

x 2 

x 

x 2 

x 

x 2 

x 

x 2 

2x 

4x 2 

b 

ab 

ab b 2 

15 

15x 

4x 60 

3 

3x 

3x 9 

19. x 4 or x 1 

20. x 10 or x 3 

21. x 3 or x 8 

22. x 1 

2 

or x 4 

23a and b. 

h 

5 

5x 

4x 20 

6 

6x 

6x 36 

7 

14x 

14x 49 

h 

x 

2 

h 4 

12 

23c. 1 

2 [h (h 4)]h 48 

23d. h 8 or h 6. The height cannot be 

negative, so the only valid solution is h 6. 

The height is 6 feet, one base is 6 feet, and the 

otherbaseis10feet. 

a 

b 

x 

x 2 

a 

a 2 

b 

ab 

ab b 2 

12x 

2x 24 




LESSON 8.4 

1. 2092 cm2 2. 74 cm 

3. 256 cm 4. 33 cm2 5. 63 cm 6. 490 cm2 7. 57.6 m 8. 25 ft 

9. 42 cm2 10. 58 cm2 11. a 1 

2 s; A 1 

2 asn 1 

2 1 

2 s s 4 s2 12. It is impossible to increase its area, because a 

regular pentagon maximizes the area. Any 

dragging of the vertices decreases the area. 

(Subsequent dragging to space them out more 

evenly can increase the area again, but never 

beyond that of the regular pentagon.) 

13. 996 cm2 14. 497 cm2 15. total surface area 13,680 in2 95 ft2 ; 

cost $8075 

16. Area is 20 square units. 

y 

y 

1_ 

2 

x 5 

(2, 6) 

y 2x 10 


x 


y 

y – 

4_ x 12 

3 

(6, 4) 

18. Conjecture: The three medians of a triangle 

divide the triangle into six triangles of equal area. 

Argument: Triangles 1 and 2 have equal area 

because they have equal bases and the same 

height. Because the centroid divides each median 

into thirds, you can show that the height of 

triangles 1 and 2 is 1 

 

3 the height of the whole 

triangle. Each has an area 1 

6 

the area of the whole 

triangle. By the same argument, the other small 

triangles also have areas 1 

 

6 the area of the whole 

triangle. 

1 

h 

y – 

1_ x 6 

3 

19. nw ny 2x 

20. 504 cm 2 

21. 840 cm 2 

2 

x

LESSON 8.5 

1. 9 in 2 

2. 49 cm 2 

3. 0.8 m 2 

4. 3 cm 

5. 3 in. 

6. 0.5 m 

7. 36 in 2 

8. 7846 m 2 

9. 25 48, or about 30.5 square units 

10. 100 128, or about 186 square units 

11. 

r 18 cm 

12. 804 m 2 

13. 11,310 km 2 

14. 154 m 2 

15. 4 times 

16. A r 2 because the 100-gon almost 

completely fills the circle. 

17. 456 cm 2 

18. 36 ft 2 

19. The triangles have equal area when the point 

is at the intersection of the two diagonals. There is 

no other location at which all four triangles have 

equal area. 

20. x mDE 2 24° 48° 

21. 90° 38° 28° 28° 180° 

22. 

6 cm 

18 cm 

12 cm 

24 cm 




1. 6 cm2 2. 64 

 

3 

cm2 3. 192 cm2 4. ( 2) cm2 5. (48 32) cm2 6. 33 cm2 7. 21 cm2 8. 105 

 

2 

cm2 9. 6 cm 

10. 7 cm 

11. 75 

12. 100 

13. 42 

14. $448 

15a. 

15b. 

15c. 

LESSON 8.6 


15d. 


17a. (144 36) cm2 ; 78.54% 

17b. (144 36) cm2 ; 78.54% 

17c. (144 36) cm2 ; 78.54% 

17d. (144 36) cm2 ; 78.54% 

18. 480 m2 19. AB 17.0 cm, AG 6.6 cm 

90 

20. True. If 24 360 

2r, then r 48 cm. 

21. True. If 360 

 

n 

24, then n 15. 

22. False. It could be a rhombus. 

23. true; Triangle Inequality Conjecture

LESSON 8.7 

1. 150 cm 2 2. 4070 cm 2 

3. 216 cm 2 4. 340 cm 2 

5. 103.7 cm 2 6. 1187.5 cm 2 

7. 1604.4 cm 2 8. 1040 cm 2 

9. 414.7 cm 2 10. 329.1 cm 2 

11. area of square 4 area of trapezoid 4 

area of triangle 

12. $1570 


14. sample tiling 

3 3 .4 2 /3 2 .4.3.4/4 4 

15. a 75°, b 75°, c 30°, d 60°, 

e 150°, f 30° 

16. About 23 days. Each sector is about 1.767 km 2 . 

17. a 50°, b 50°, c 80°, d 100°, e 80°, 

f 100°, g 80°, h 80°, k 80°, m 20°, 

n 80°. Explanations will vary. Sample 

explanation: The angle with measure d corresponds 

to the angle forming a linear pair with g.Because 

d 100°, by the Parallel Lines Conjecture, the 

angle adjacent to g measures 100°, and by the 

Linear Pair Conjecture, g 80°. The angle with 

measure f corresponds to the angle measuring 

100°, so f 100°. The angles measuring g and k 

are the base angles of an isosceles triangle, so by 

the Isosceles Triangle Conjecture, k 80°. 






1. B (parallelogram) 2. A (triangle) 

3. C (trapezoid) 4. E (kite) 

5. F (regular polygon) 6. D (circle) 

7. J (sector) 8. I (annulus) 

9. G (cylinder) 10. H (cone) 

11. 12. 

13. 

14. Sample answer: Construct an altitude from 

the vertex of an obtuse angle to the base. Cut off 

the right triangle and move it to the opposite side, 

forming a rectangle. Because the parallelogram’s 

area hasn’t changed, its area equals the area of the 

rectangle. Because the area of the rectangle is 

given by the formula A bh, the area of the 

parallelogram is also given by A bh. 

h 

15. Sample answer: Make a copy of the trapezoid 

and put the two copies together to form a 

parallelogram with base b 1 b 2 and height h. 

Thus the area of one trapezoid is given by the 

formula A 1 

b 1 

b 2 

Apothem 

b 

2 b 1 b 2h. 

16. Sample answer:Cut a circular region into a large 

number of wedges and arrange them into a shape 

that resembles a rectangle.The base length of this 

“rectangle”is r and the height is r, so its area isr 2 . 

Thustheareaof acircleisgivenbytheformulaAr 2 . 

r 

b 2 

h h 

b 1 

17. 800 cm 2 18. 5990.4 cm 2 

19. 60 cm 2 or about 188.5 cm 2 

b 

r 


h 

20. 32 cm 21. 32 cm 22. 15 cm 

23. 81 cm 2 24. 48 cm 25. 40° 

26. 153.9 cm 2 27. 72 cm 2 28. 30.9 cm 2 

29. 300 cm 2 30. 940 cm 2 31. 1356 cm 2 


y 


F (0, 0) 

A (0, 0) 

y 

D (6, 8) C (20, 8) 

R (4, 15) 

O (4, –3) 

B (14, 0) 

U (9, 5) 

34. 6 cm 35. 172.5 cm 2 

36. sample answers: 

x 

x 

37. 1250 m 2 

38. Circle. For the square, 100 4s, s 25, 

A 25 2 625 ft 2 . For the circle, 100 2r, 

r 15.9, A (15.9) 2 794 ft 2 . 

39. A round peg in a square hole is a better fit. 

The round peg fills about 78.5% of area of the 

square hole, whereas the square peg fills only about 

63.7% of the area of the round hole. 

40. giant 41. about 14 oz 

42. One-eighth of a 12-inch diameter pie; 

one-fourth of a 6-inch pie and one-eighth of a 

12-inch pie both have the same length of crust, 

which is longer than one-sixth of an 8-inch pie. 

43a. 96 ft; 40 ft 43b. 3290 ft 2 

44. $3000 45. $4160 

46. It’s a bad deal. 2r 1 44 cm. 2r 2 22 cm, 

which implies 4r 2 44 cm. Therefore r 1 2r 2 . 

The area of the large bundle is 4r 2 2 cm 2 .The 

combined area of two small bundles is 2r 2 2 cm 2 . 

Thus he is getting half as much for the same price. 

47. $2002 48. $384 (16 gal)



LESSON 9.1 

1. c 19.2 cm 2. a 12 cm 

3. b 5.3 cm 4. d 10 cm 

5. s 26 cm 6. c 8.5 cm 

7. b 24 cm 8. x 3.6 cm 

9. x 40 cm 10. s 3.5 cm 

11. r 13 cm 

12. 127 ft 

13. 512 m2 14. 11.3 cm 

15. 3, 4, 5 

16. 28 m 

17. The area of the large square is 4 area of 

triangle area of small square. 

c 2 4 1 

2 

ab (b a)2 

c 2 2ab b 2 2ab a 2 

c 2 a 2 b 2 

b a 

b 

a 

c 

9 

18. Sample answer: Yes, ABC XYZ by SSS. 

Both triangles are right triangles, so you can 

use the Pythagorean Theorem to find that 

CB ZY 3 cm. 

19. 54 cm 2 

20. 3 6 3 2 .4.3.4 

21. Mark the unnamed angles as shown in the 

figure below. By the Linear Pair Conjecture, 

p 120° 180°, so p 60°. By AIA, m q. By 

the Triangle Sum Conjecture, q p n 180°. 

Substitute m q and p 60° to get 

m 60° n 180°. m n 120°. 

m 

120° 

p 

q 

n 

Or use the Exterior Angle Conjecture to get 

q n 120°. By AIA, q m. Substituting, 

m n 120°. 

22. a 122°, b 74°, c 106°, d 16°, e 90°, 

f 74°, g 74°, h 74°, n 74°, r 32°, 

s 74°, t 74°, u 32°, v 74°. Possible 

explanation: By the Tangent Conjecture, the 

quadrilateral containing g has two right angles 

formed by radii intersecting tangent lines. Using 

c 106° and the Quadrilateral Sum Conjecture, 

g 90° 90° 106° 360°, so g 74°. 

The angle measures e and f and the measure of the 

inscribed angle that intercepts the arc with measure 

u sum to 180°. Using e 90° and f 74°, the 

inscribed angle measures 16°. Using the Inscribed 

Angle Conjecture, u 32°. 

23 . 

T1 T 2 




LESSON 9.2 

1. yes 

2. yes 

3. no 

4. no 

5. no 

6. no 

7. no 

8. No, the given lengths are not a Pythagorean 

triple. 

9. y 25 cm 

10. y 24 units 

11. y 17.3 m 

12. 6, 8, 10 

13. 60 cm 2 

14. 14.1 ft 

15. 17.9 cm 2 

16. 102 m 

17a. 1442 cm 2 

17b. 74.8 cm 

18. Sample answer: The numbers given satisfy the 

Pythagorean Theorem, so the triangle is a right 

triangle; but the right angle should be inscribed 

in an arc of 180°. Thus the triangle is not a right 

triangle. 


19. Sample answer: (BD) 2 62 32 27; 

(BC) 2 (BD) 2 92 108; then (AB) 2 (BC) 2 

(AC) 2 (36 108 144), so ABC is a right 

triangle by the Converse of the Pythagorean 

Theorem. 

20. centroid 

3 

21. 10 

22. Because mDCF 90°, mDCE (90 x)°. 

Because DCE is isosceles, mDEC (90 x)°. 

mD 180° 2 (90 x)° 2x.Because D is a 

central angle, 2x a.Therefore,x 1 

2 a. 

23. The path from C to M to T lies on a straight 

line and therefore must be shorter than the path 

from C to A to T. 

P 

C 


25. 

26. 19 

A 

U 

M 

T 

B


1. 6 2. 5 

3. 182 4. 147 

5. 8 6. 23 

7. 32 8. 210 

9. 53 10. 85 

11. 46 12. 24 

13. 125 14. 28 

15. 623 

16. Answers will vary. Possible answer: The length 

of the hypotenuse of an isosceles right triangle 

with legs of length 3 units is 18 units. This is the 

same as 2 2 2,or 32. 

2 

1 

17. The hypotenuse represents 5 units. In the 

second right triangle, the legs have lengths 4 and 2, 

so the hypotenuse has length 20.The length of 

the hypotenuse is twice the length of the hypotenuse 

of the smaller triangle, so 20 25. 

5 

2 

1 

1 

18 

3 

20 

4 

3 

2 

18. Possible answer: A right triangle with legs of 

lengths 1 and 3 units has a hypotenuse of length 

10 units. 

A right triangle with legs of lengths 6 and 2 units 

has a hypotenuse of length 40 10 

10 210. 

19. Possible answer: A right triangle with legs of 

lengths 2 and 3 units has a hypotenuse of length 

13 units. 

13 

10 

3 

2 

3 

20. x 3, y 12. Because y is twice as 

long as x, y 2x, so 12 23. 

21. 2 2 3 2 7 2 

7 

2 

6 

1 

40 

3 

2 




1. 722 cm 

2. 13 cm 

3. 10 cm, 53 cm 

4. 103 cm, 10 cm 

5. 34 cm, 17 cm 

6. 72 cm 

7. 123 cm 

8. 50 cm, 100 cm 

9. 16 cm2 1 1 

10. , 

2 2 

LESSON 9.3 

11. A 30°-60°-90° triangle must have sides whose 

lengths are multiples of 1, 2, and 3. The triangle 

shown does not reflect this rule. 

12. Possible answer: 

Use 1 2 3 2 2 2 and 4 2 48 2 8 2 . 

4 

3 


14. Possible answer: 

Use 22 12 5 2 and 62 32 45 2 . 

5 

4 

2 

3 

2 

4 

1 

1 

2 

3 

1 

4 

2 

3 

4 

45 3 

15a. CDA, AEC, AEB, BFA, BFC 

15b. MDB, MEB, MEC, MFC, MFA 


8 

1 

2 

3 

6 

2 

16. c 2 x 2 x 2 Start with the Pythagorean Theorem. 

c 2 2x 2 Combine like terms. 

c x2 Take the square root of both sides. 

17. 1693 m 2 292.7 m 2 

18. 390 m 2 

19. 

20. Construct an isosceles right triangle with legs 

of length a, construct a 30°-60°-90° triangle with 

legs of lengths a and a3,and construct a right 

triangle with legs of lengths a2 and a3. 

a 

a 

x 3 

a 

2 

21. Areas: 4.5 cm2 ,8 cm2 , 12.5 cm2 .4.5 8 12.5; that is, the sum of the areas of the 

semicircles on the two legs is equal to the area of 

the semicircle on the hypotenuse. 

22. 73 

 

6 

12.2 chih 

23. Extend the rays that form the right angle. 

m4 m5 180° by the Linear Pair 

Conjecture, and it’s given that m5 90°. 

m4 90°. m2 m3 m4 m2 

m3 90° 180°. m2 m3 90°. 

m3 m1 by AIA. m1 m2 90°. 

24. 80° 

x 

30° 

4 

3 

45° 

a 

1 

5 

a 

2 

x 

2a 

3 

2 

x 

2x 

a 

3 

1 

15° 

3 

45° 

a 

a 

2 

5

LESSON 9.4 

1. No. The space diagonal of the box is about 

33.5 in. 

2. 10 m 3. 50 km/h 

4. 8 ft 5. area: 60 m2 ; cost: $7200 

6. surface area of prism 273 180 cm2 

226.8 cm2 ; surface area of cylinder 78 cm2 

245.0 cm2 7. 

36 

cm 20.8 cm 

3 

8. 48.2 ft; 16.6 lb 

9a. 160 ft-lb 

10. about 4.6 ft 

9b. 40 lb 9c. 20 lb 

11. 

2 

2 

2 

2 

2 2 2 2 

2 

4 

2 

2 

4 

2 

2 

2 

2 

4 

2 

4 

2 

12. 

13. 12 units 

14. 182 cm 

15. Draw radii CD and CB. ABC ADC 

90°. For quadrilateral ABCD, 54° 90° mC 

90° 360°, so mC 126°. BD 126° but 

126° 226° 360°. 

16. 4, 43 

17. SAA 

18. orthocenter 

19. 115° 

20. PO 




LESSON 9.5 

1. 5 units 2. 45 units 3. 34 units 

4. 354 m 5. 52.4 units 6. isosceles 

7. rectangle 

10 

8 

6 

4 

2 

y 


9. kite 

K 

8 

J 

10. square 

M 

11a. 

2 

2 

2 

H 

D 

8 

10 

8 

4 

L 

I 2 

2 

y 

P 

y 

(1, –2) 

Circle A 

(x, y) 


4 

2 

O 

N 

4 2 2 4 

y 

A 

E 2 4 6 

G 

C 

y 

B 

4 6 8 10 

F 

x 

x 

x 

x 

x 

(0, 2) 

y 

Circle B 

(x, y) 

x 

11b. Circle A: (x 1) 2 ( y 2) 2 64; 

Circle B: x2 (y 2) 2 36 

11c. (x h) 2 (y k) 2 (r) 2 

12. (x 2) 2 y2 25 

13. Center is (0, 1), r 9. 

14. (x 3) 2 (y 1) 2 18 

15a. 14 units 

15b. 176 411 units 

15c. (x x 2 1 ) 2 y 

(y2 1 ) 2 z 

(z2 1 ) 2 

16. 86.5 units 

17. 14 units 

18. n (n 2) 3 (n 3) (n 1) 

19. 3 

, 

2 

1 

2 

20. k 2, m 6 

21. x , y 

22. 96 cm 

23. The angle of rotation is approximately 77°. 

Connect two pairs of corresponding points. 

Construct the perpendicular bisector of each 

segment. The point where the perpendicular 

bisectors meet is the center of rotation. 

24. Any long diagonal of a regular hexagon 

divides it into two congruent quadrilaterals. Each 

(6 2) 

angle of a regular hexagon is 

6 

180° 

6 12 

 

3 3 

120°, and 

the diagonal divides two of the 120° angles into 60° 

angles. Look at the diagonal as a transversal. 

The alternate interior angles are congruent, thus 

the opposite sides of a regular hexagon are parallel. 

120° 

120° 

60° 

60° 

60° 

60° 

120° 

120°

LESSON 9.6 

1. 18 cm2 56.5 cm2 2. (8 16) m2 9.1 m2 3. 456 cm2 1433 cm2 4. 120 cm2 377.0 cm2 5. 32 323 m2 45.1 m2 6. (25 48) cm2 30.5 cm2 7. 64 

 

3 

163 cm2 39.3 cm2 8. (240 18) m2 183.5 m2 9. 102 cm 

10. Possible proof: 

Given: Circle C with tangents AM and AN 

Show: AM AN 

C 

N 

M 

Add MC ,NC , and AC to the diagram as shown. 

By the Tangent Conjecture, M and N are right 

angles, so AMC and ANC are right triangles. 

Using the Pythagorean Theorem, (AM) 2 (MC) 2 

(AC) 2 and (AN) 2 (NC) 2 (AC) 2 .Solvingfor 

AM and AN, AM (AC) 2 (MC) 2 and 

AN (AC) 2 (NC) 2 .BecauseMC and NC are 

radii of the same circle, they have the same lengths. 

So, substitute MC for NC in the second equation: 

AN (AC) 2 (MC) 2 AM.Therefore, 

AM AN . 

11. 18 m 

12. 324 cm2 1018 cm2 13. 3931 cm2 A 

14. 12 63 cm 22.4 cm 

15. 77 cm 8.8 cm 

16. 76 cm 

17. Inscribed circle: 3 cm 2 . Circumscribed circle: 

12 cm 2 . The area of the circumscribed circle is four 

times as great as the area of the inscribed circle. 

18. 135° 

19. (x 3) 2 (y 3) 2 36 

20. The diameter is the transversal, and the 

chords are parallel by the Converse of the Parallel 

Lines Conjecture. The chords are congruent 

because they can be shown to be the same distance 

from the center. (Draw a perpendicular from each 

chord to the center and use AAS and CPCTC.) 

Sample construction: 

21a. Because a carpenter’s square has a right angle 

and both radii are perpendicular to the tangents, a 

square is formed. The radius is 10 in., therefore the 

diameter is 20 in. 

10 in. 

d 20 in. 

10 in. 

10 in. 

10 

in. 

21b. Possible answer: Measure the circumference 

with string and divide by . 

22. 5 

6 





1. 20 cm 

2. 10 cm 

3. obtuse 

4. 26 cm 

5. 3 

, 

2 

1 

2 

6. 1 1 

, 2 

2 

7. 2003 cm2 346.4 cm2 8. d 122 cm 17.0 cm2 9. 246 cm2 10. 72 in2 226.2 in2 11. 24 cm2 75.4 cm2 12. (2 4) cm2 2.28 cm2 13. 222.8 cm2 14. isosceles right 

15. No. The closest she can come to camp is 10 km. 

16. No. The 15 cm diagonal is the longer diagonal. 

17. 1.4 km; 8 1 

 

2 min 

18. yes 

19. 29 ft 

20. 45 ft 

21. 50 mi 

22. 707 m2 23. 63 and 18 

24. 12 m 

25. 42 

26. No. If you reflect one of the right triangles into 

the center piece, you’ll see that the area of the kite is 

almost half again as large as the area of each of the 

other triangles. 

Extra 

30° 

30° 

30° 

Or students might compare areas by assuming the 

short leg of the 30°-60°-90° triangle is 1. The area 

3 

of each triangle is then 0.87 and the area of 

2 

the kite is 3 3 1.27. 


27. 7 

9 

28. The quarter-circle gives the maximum area. 

Triangle: 

s 

____ 

2 

A 1 

45° 

Square: 

2 

 

4 

2 s 

s 

1_ 

2 

s 

A 1 

2 s 1 2 

s 

2 

s 

4 

Quarter-circle: 

___ 2s 

 

s 1 

4 2r 

r 2s 

 

 

A 1 

4 2 s 

 

s 2 

 

 

s2 

 

4 

29. 1.6 m 

30. 

s 

s 

s 

____ 

2 

2 

2 

45° 

1_ 

2 

s 

s 

 

2 

s 2 

 

 

31. 4; 0; 10. The rule is n 

 

2 if n is even, but 0 if n is 

odd.

32. 4 in./s 12.6 in./s 

33. true 

34. true 

35. False. The hypotenuse is of length x2. 

36. true 

37. false; AB x (x2 1 ) 2 (y 

2 y1 ) 2 

38. False. A glide reflection is a combination of a 

translation and a reflection. 

39. False. Equilateral triangles, squares, and regular 

hexagons can be used to create monohedral 

tessellations. 

40. true 

41. D 

42. B 

43. A 

44. C 

45. C 

46. D 



1 

2 

ABCD is 

a rectangle 

Given 

ABCD is 

a parallelogram 

Definition of 

rectangle 

4 D B 

3 

Definition of 

rectangle 

 

DA CB 

Definition of 

parallelogram 

6 

AC AC 

Same segment 

48. 

W B 

8-ball 

49. 34 cm2 ; 22 42 27.7 cm 

50. 40 

 

3 

cm2 41.9 cm2 51. about 55.9 m 

52. about 61.5 cm2 53. 48 cm 

54. 322 ft2 5 

DAC BCA 


N 

S 

7 

ABC CDA 

SAA Congruence 

Conjecture 

A 

Cue ball 


E 




10 


LESSON 10.1 

1. polyhedron; polygonal; triangles 

2. PQR, TUS 

3. PQUT, QRSU, RPTS 

4. QU ,PT,RS 

5. 6 cm 

6. GYPTAN 

7. point E 

8. GE,YE,PE,TE,AE,NE 

9. 13 cm 

10. D 

11. L 

12. C 

13. G 

14. B 

15. H 

16. E 

17. A 

18. J 

19. J 

20. M 

21. H 

22. I 

23. 


24. 

25. 

26. 

27. true 

28. False. This statement is true only for a right 

prism. 

29. true 

30. true 

31. False. It is a sector of a circle. 

32. true 

33. false; counterexample: 

34. true 

35. true 

x 

x 

2x 

2x


Possible answers include that the number of lateral 

faces of an antiprism is always twice the number 

for the related prism; that the number of vertices is 

the same for each related prism and antiprism; and 

that the number of edges for a prism is three times 

the number of faces, while for an antiprism, the 

number of edges is twice the number of faces. 

37. Answer should include the idea that the 

painting “disappears” into the view out the window. 

Students might also note the effect created by the 

cone-shaped tower appearing similar to the road 

disappearing into the distance. 

38. 8 

36. (Lesson 10.1) 

39. 60 

40. 30 

41a. yes 

41b. yes 

41c. no 

41d. yes 

Triangular Rectangular Pentagonal Hexagonal n-gonal 

prism prism prism prism prism 

Lateral 3 4 5 6 ... n 

faces 

Total 5 6 7 8 ... n 2 

faces 

Edges 9 12 15 18 ... 3n 

Vertices 6 8 10 12 ... 2n 

Triangular Rectangular Pentagonal Hexagonal n-gonal 

antiprism antiprism antiprism antiprism antiprism 

Lateral 6 8 10 12 ... 2n 

faces 

Total 8 10 12 14 ... 2n 2 

faces 

Edges 12 16 20 24 ... 4n 

Vertices 6 8 10 12 ... 2n 

 

 

 

 




LESSON 10.2 

1. 72 cm 3 2. 24 cm 3 3. 108 cm 3 

4. 160 cm 3 502.65 cm 3 

5. 36 cm 3 113.10 cm 3 

6. 324 cm 3 1017.88 cm 3 


8. 960 in 3 9. QT cubic units 


2 

12 

12 in. 

4 in. 

8 in. 

12 

24 in. 

3 

8 

T 

11. 2x 3 12. 3r 3 13. 13x 3 

14. Margaretta has room for 0.5625 cord. She 

should order a half cord. 

15. 170 yd 3 16. 5100 lb 17. 11,140 

18. The volume of the quilt in 1996 was 4000 ft 3 . 

The quilt panels were stacked 2 ft 8 in. high. 

19. 

7. (Lesson 10.2) 

12 


Q 

20. 

21. true 

22. false 

23. 

r 

r 

24. possible solution: prism 

Salt crystal 

r 

25. approximately 1.89 m 

26. 12 62 

Information about Height Right triangular Right rectangular Right trapezoidal Right 

base of solid of solid prism prism prism cylinder 

H 

h 

b 

b 6, b 2 7, H 20 a. V d. V g. V j. V 

h 8, r 3 480 cm 3 960 cm 3 1040 cm 3 180 cm 3 

b 9, b 2 12, H 20 b. V e. V h. V k. V 

h 12, r 6 1080 cm 3 2160 cm 3 2520 cm 3 720 cm 3 

b 8, b 2 19, H 23 c. V f. V i. V l. V 

h 18, r 8 1656 cm 3 3312 cm 3 5589 cm 3 1472 cm 3 

H 

h 

b 

H 

h 

b 2 

b 

H 

r

LESSON 10.3 

1. 192 cm3 2. 84 cm3 263.9 cm3 3. 150 cm3 4. 60 cm3 5. 84 cm3 263.9 cm3 6. 384 cm3 1206 cm3 7. m3 

 

3 

cm3 8. 2 

3 b3 cm3 9. 324x3 cm3 ;29.6% 


11. V 1 

3 M 2H ft3 12. sample answer: 

16 

M 

27 

48 

3 

M 

H 

13. Mount Etna is larger. The volume for Mount 

Etna is approximately 2193 km 3 ,and the volume 

for Mount Fuji is approximately 169 km 3 . 

10. (Lesson 10.3) 

Information about Height Triangular Rectangular Trapezoidal 

base of solid of solid pyramid pyramid pyramid Cone 

b 

H 

h 

14. 78,375 grams 

15. 48 in3 16. 4 units3 17. 144x3 cm3 18. 40,200 gal; 44 h 40 min 

19. 71 ft3 20. 403 barrels 

21a. 163 cm2 21b. 96 cm2 21c. 803 cm2 21d. 2413 120 cm2 22. A 

b 6, b2 7, H 20 a. V d. V g. V j. V 

h 6, r 3 

120 cm3 240 cm3 260 cm3 60 cm3 b 9, b2 22, H 20 b. V e. V h. V k. V 

h 8, r 6 

240 cm3 480 cm3 240 cm3 2480 

 

3 

cm3 b 13, b2 29, H 24 c. V f. V i. V l. V 

h 17, r 8 

884 cm3 1768 cm3 2856 cm3 512 cm3 D' 

H 

b 

h 

D 

3 1 

C 

2 

4 

B 

D'' 

Possible answer: From the properties of reflection, 

1 3 and 2 4. m1 m2 90°, so 

m3 m4 90°, and m1 m2 m3 

m4 180°. Therefore D, C,and D are collinear. 

23a. Y (a c, d) 

23b. Y (a c, b d) 

23c. Y (a c e, b d f ) 

b 

b 2 

H 

h 

H 


r 



LESSON 10.4 

1. 58.5 in3 2. 323 cm3 55.43 cm3 3. 15 cm 

4. 11 cm 

5. 5.0 cm 

6. 8. 5 in. 

2 

2 11 in. 63.24 in3 ; 

11 in. 2 

2 8.5 in. 81.85 in3 The short, fat cylinder has greater volume. 

7. 257 ft 3 

8. 4 cm 

9. He must refute the statement. 

10. 1502 lb 

11. 192.4 gal 

12. 13 min 

13. Answers will vary, but r 2 H should equal 

about 14.4 in 3 . 

14. approximately 38 in 3 

15. 100,000 m 3 , or about 314,159 m 3 ; 

16,528 loads 

16. AB EC because the opposite sides of a 

parallelogram are congruent. EC BD because 

the diagonals of a rectangle are congruent. So, 

AB BD because both are congruent to EC. 

Therefore, ABD is isosceles. 


17. 8.2 cm 

18. x 96° 

19. 

20. 

21a. A 

21b. S 

21c. N 

21d. S 

21e. A 

D 

x 

45° 

45° 

x x 

45° 45° 

A x x 

3x 

x B 

C

LESSON 10.5 

1. 675 cm3 2. 36 cm3 113.1 cm3 3. 47 in3 4. 1798.4 g 

5. The gold has mass 2728.5 g, and the platinum 

has mass 7529.8 g. The solid cone of platinum has 

more mass. 

6. 1.5 cm 

7. 10.5 g/cm3 ;silver 

8. 8000 cm3 9. The volume of the medallion is 160 cm3 .Yes,it 

is gold, and the Colonel is who he says he is. 

10. 679 cm3 11. approximately 193 lb; 22 fish 

9 

12. 3 2 

13. flowchart proof: 

 

QR SP 

Given 

R S 

AIA 

SM MR 

Given 

RMQ SMP 

ASA 

 

MQ MP 

CPCTC 

QMR PMS 

Vertical Angles 

M is the midpoint of PQ 

Definition of midpoint 

14. 15 sides 

15a. (1, 3) 

15b. (x 1) 2 (y 3) 2 25 

16. 58; 3n 2 




1. 36 cm 3 113.1 cm 3 

LESSON 10.6 

2. 

6 cm3 0.533 cm3 3. 9 

 

32 

cm3 0.884 cm3 4. 720 cm3 2262 cm3 5. 30 cm3 94.3 cm3 6. 3456 cm3 10,857 cm3 7. 18 m3 56.6 cm3 8. No. The volume of the ice cream is 85.3 cm3 , 

and the volume of the cone is 64 cm3 . 

9. only 20 scoops 

10. 148 

 

3 

m3 , or about 155 m3 11. They have the same volume. 

12. 9 in. 

13. 3 cm 

14. 8192 

 

3 

cm3 8579 cm3 15. 18 in3 , or about 57 in3 16. No. The unused volume is 16 cm3 ,and the 

volume of the golf ball is 10.6 cm3 . 


17. approximately 15,704 gallons; 53 days 

18. lithium 

19. 31 ft 

20. ABCD is a parallelogram because the slopes of 

CD and AB are both 0 and the slopes of BC and 

AD are both 5 

B (–3, –5) 

y 

3 . 

C (3, 5) D (9, 5) 

A (3, –5) 

x 

21. They trace two similar shapes, except that the 

one traced by C is smaller by a scale factor of 1:2. 

22. The line traces an infinite hourglass shape. Or, 

it traces the region between the two branches of a 

hyperbola. 

23. w 110°, x 115°, y 80°

LESSON 10.7 

1. V 972 cm3 3054 cm3 S 324 cm2 1018 cm2 2. V 0.972 cm3 3.054 cm3 S 3.24 cm2 10.18 cm2 3. V 1152 cm3 3619 cm3 S 432 cm2 1357 cm2 4. S 160 cm2 502.7 cm2 5. V 256 

cm3 268.1 cm3 3 

6. S 144 cm 2 452.4 cm 2 

7. Area of great circle r 2 . Total surface area 

of hemisphere 3r 2 . Total surface area of 

hemisphere is three times that of area of great 

circle. 

8. 2 gal 

9. V 1 

2 4 

3 (1.8)3 1 

2 (1.8)2 (4.0) 

10.368 m 3 32.57 m 3 ; S 1 

2 4(1.8) 2 

1 

2 2(1.8)(4.0) 13.68 m 2 42.98 m 2 

10a. approximately 3082 ft 2 

10b. 13 gal 

10c. approximately 9568 bushels 

11. 153,200,000 km 2 

12. The total cost is $131.95. He will stay under 

budget. 

13. 1.13% 

14. 150 cm 3 471.2 cm 3 

15. 1 

4 

19a. (Lesson 10.7) 

16. 1 

2 

17. 3 

4 

18. The ratio gets closer to 1. 

19a and 19b. See tables below. 

20. AB CB and AD CD by the definition of 

rhombus, and BD BD because it is the same 

segment; therefore ABD CBD by SSS. By 

CPCTC, 2 3 and 1 4, which shows 

that BD bisects both ABC and ADC.Because 

all four sides of the rhombus are congruent, a 

similar proof can be used to show that ABC 

ADC and thus that A and C are both 

bisected by diagonal AC. 

21a. AB CB by the definition of rhombus and 

BE BE because it is the same segment. 1 2 

by the Rhombus Angles Conjecture. Therefore, 

AEB CEB by SAS. 

21b. AE CE by CPCTC, so BD bisects AC. 

21c. 3 4 by CPCTC. Also, 3 and 4 form 

a linear pair, so they are supplementary. Because 

two angles that are congruent and supplementary 

are right angles, 3 and 4 are right angles. 

21d. Because 3 and 4 are right angles, the 

diagonals are perpendicular.You still need to show 

that AC bisects BD. Use a proof similar to that 

given in 21a to show that AEB AED.Then, 

by CPCTC, BE DE, which shows that AC 

bisects BD. 

n 1 2 3 4 5 6 . . . n ... 200 

f(n) 2 1 4 7 10 13 ... 3n 5 ... 595 

19b. (Lesson 10.7) 

n 1 2 3 4 5 6 . . . n ... 200 

f(n) 0 1 

1 

3 

2 5 

3 2 5 3 7 

... n 

n 

1 199 

 

1 

... 201 





1. h A 

b 

P 2h 

2. b 2 

or b P 

 

2 

h 

3. r 3V 

H 

 

4. b c 2 a2 

2 • SA 

5. a l 

P 

6. y2 mx2 x1 y1 or y2 mx2 mx1 y1 7. v d 

t 

The original formula gives distance in terms of 

velocity and time. 

9 

8. F C 5 

32 

The original formula converts degrees Fahrenheit to 

degrees Celsius. 

2 

9. L g T 

2 

The original formula gives the period of a 

pendulum (time of one complete swing) in terms 

of length and acceleration due to gravity. 

10b. (Using Your Algebra Skills 10) 


10a. V F E 2 

10b. See table below. 

11a. The corresponding radii are approximately 

3.63 cm, 3.30 cm, 3.02 cm, and 2.79 cm. 

11b. The corresponding heights are approximately 

9.32 cm,10.49 cm,11.61 cm,and 12.70 cm. 

11c. The corresponding volumes are approximately 

129 cm 3 , 120 cm 3 , 111 cm 3 , and 104 cm 3 . 

11d. Answers will vary. Sample answer: The cone 

with slant height 10 cm has the widest radius, so a 

scoop of ice cream is least likely to fall off, and that 

cone also has the greatest volume. 

11e. Answers will vary. Sample answer: The cone 

with slant height 13 cm has the greatest height, so 

the cone appears bigger even though it has the same 

surface area as the other cones; that cone also has 

the smallest radius and smallest volume, so it could 

hold less ice cream and still appear to be a bigger 

cone. 

m1 m2 m3 f f 

12a. A 

5 

12b. average of 60: 31; average of 70: 56; average 

of 80: 81; average of 90: 106 (impossible) 

Pentahedron Hexahedron Octahedron Decahedron Dodecahedron 

Number of faces 

5 6 8 10 12 

Number of edges 8 10 12 16 20 

Number of vertices 5 6 6 8 10


1. They have the same formula for volume: V BH. 

2. They have the same formula for volume: V 1 

3 BH. 

3. 6240 cm3 4. 1029 cm3 3233 cm3 5. 1200 cm3 6. 32 cm3 7. 100 cm3 314.2 cm3 8. 2250 cm3 7069 cm3 9. H 12.8 cm 

10. h 7 cm 

11. r 12 cm 

12. r 8 cm 

13. 960 cm3 14. 9 m 

15. 851 cm3 16. four times as great 

17a. Vextra large 54 in3 Vjumbo 201.1 in3 Vcolossal 785.4 in3 17b. 14.5 times as great 

18. Cylinder B weighs 8 

 

3 times as much as cylinder A. 

19. 2129 kg; 9 loads 

20. H2r. Vsphere 

0.524. Thus, 52.4% of 

Vbox 

the box is filled by the ball. 

21. approximately 358 yd3 22. No. The unused volume is 98 in3 ,and the 

volume of the meatballs is 32 in3 

. 

23. platinum 

24. No. The ball weighs 253 lb. 

25. 256 lb 

26. approximately 3 in. 

4 r 3 

3 

 

(2r) 3 

27. 

8 m3 0.23 m3 33 

 

32 

28. 160 cubic units 





11 



1. 3 

; 3 

8 5 

AC 

3 D 5 D 

2. CD 

, C 

5 BD 

, B 

8 BC 

8 

 

13 

3a. 3 

1 

9 

3b. 1 

4. a 6 

5. b 16 

6. c 39 

7. x 5.6 

8. y 8 

9. x 12 

10. z 6 

11. d 1 

12. y 5 

13. 318 mi 

14. 2.01 

15. 12 ft by 15 ft 


16. almost 80 years old 

17a. true 

17b. false; 3 

 

6 

1 ,but 

2 3 1 6 2 

17c. true 

17d. true 

17e. false; 3 

 

6 

1 ,but 3 

2 2 

1 

6 

17f. true 

18a. arithmetic: 10, 25, 40, 65; geometric: 10, 20, 

40, 80 or 10, 20, 40, 80 

18b. arithmetic: 2, 26, 50, 74; geometric: 2, 10, 50, 

250 or 2, 10, 50, 250 

18c. arithmetic: 4, 20, 36, 52; geometric: 4, 12, 36, 

108 or 4, 12, 36, 108 

19a. Add the numbers and divide by 2. 

19b. Possible answer: Multiply the numbers and 

take the positive square root of the result. 

19c. c ab; This formula holds for all positive 

values of a and b. 

19d. c 2 • 50 100 10; 

c 4 • 36 144 12

1. A 

2. B 




LESSON 11.1 

6. Figure A is similar to Figure C. Possible answer: 

If and , then . 

7. AL 6; RA 10; RG 4; KN 6 

8. No; the corresponding angles are congruent, 

but the corresponding sides are not proportional. 

9. NY 21; YC 42; CM 27; MB 30 

10. Yes; the corresponding angles are congruent, 

and the corresponding sides are proportional. 

11. x 6 cm, y 3.5 cm 

12. z 10 2 

 

 

A B 

 

B C 

A C 

 

3 

cm 

13. Yes, the corresponding angles are congruent. 

Yes, the corresponding sides are proportional.Yes, 

AED ABC. 

9 9 

14. m 2 

cm 4.5 cm, n 4 cm 2.25 cm 

15. 3 

1 ; 9 

 

1 

16. Yes, they are similar. 

6 

4 

2 

y 

y 

Y (2, 5) 

R (2, 2) 

(0, 6) 

(1, 1) 

5 

17. Possible answer: Assuming an arm is about 

three times as long as a face, each arm would be 

about 260 ft. 

18. Possible answer: Not all isosceles triangles are 

similar because two isosceles triangles can have 

different angle measures. A counterexample is 

shown at right. Not all right triangles are similar 

because they can have different side ratios, as in a 

triangle with side lengths 3, 4, and 5 and a triangle 

with side lengths 5, 12, and 13. All isosceles right 

triangles are similar because they have angle 

measures 45°, 45°, and 90°, and the side lengths have 

the ratio 1:1:2. 

19. 36 20. bc 21. d 

c 

22. Possible answers: Jade might get 1825 

 

4475 

of the 

profits, or $2,773.18, and Omar might get 2650 

 

4475 

of 

the profits, or $4,026.82. Or they take out their 

investments and they divide the remaining $2,325: 

Jade, $1825 $1,162.50 $2,987.50; and Omar, 

$3,812.50. 

23a. 23b. 

60° 

Y' (6, 15) 

R' (6, 6) 

(5, 6) 

O (6, 2) 

(7, 3) 

x 

O' (18, 6) 

24. approximately 92 gallons 

x 

60° 




LESSON 11.2 

1. 6 cm 

2. 40 cm; 40 cm 

3. 28 cm 

4. 54 cm; 42 cm 

5. No, 37 

 

30 

35 

. 

28 

6. Yes, MOY NOT by SAS. 

7. Yes, PHY YHT because YH 12 and 

20 

 

15 

16 

 

12 

12 

 

9 (SSS).Yes, PTY is a right triangle 

because 202 152 25 2 . 

8. TMR THM MHR by AA. 

x 15.1 cm, y 52.9 cm, h 28.2 cm 

9. Yes, QTA TUR and QAT ARU. 

QTA QUR by AA; 6 2 

 

3 cm 

10. 24 cm; 40 cm 

11. Yes, THU GDU and HTU DGU; 

52 cm; 42 cm 

12. SUN TAN by AA; 13 cm; 20 cm 

13. Yes, RGO FRG and GOF RFO. 

GOS RFS by AA; 28 cm; 120 cm 

14. 20 cm; 21 cm 

15. x 50, y 9 


16. r R R2 1 2 

17. She should order approximately 919 lb every 

three months. Explanations will vary. 

18. 448 

19. The corresponding angles are congruent; the 

ratio of the lengths of corresponding sides is 3 

1 ;the 

dilated image is similar to the original. 

20. Yes, ABCD ABCD.The ratio ofthe 

. The ratio of the areas is 4 

perimeters is 2 

1 

y 

D' C' 

A' B' 

x 

2 1 

 

2 1 


22. The statue was about 40 ft, or 12 m, tall. To 

estimate, you need to approximate the height of a 

person (or some part of a person) in the picture, 

measure a part of the statue in the picture, calculate 

the approximate height of that statue piece, and 

assume that the statue has the same proportions 

as the average person. 

1 .

LESSON 11.3 

1. 16 m 

2. 4 ft 3 in. 

3. 30 ft 

4. 10.92 m 

5. 5.46 m 

6. Thales used similar right triangles. The height of 

the pyramid and 240 m are the lengths of the legs of 

one triangle; 6.2 m and 10 m are the lengths of the 

corresponding legs of the other triangle; 148.8 m. 

7. 90 m; R and O are both right angles and 

P is the same angle in both triangles, so 

PRE POC by AA. 

8. 300 cm 

9. 

3 cm 

t 

2 cm 

20 cm 

y 

h 

d 

x 

45 cm 

30 cm 

Possible answer: Walk to the point where the guy 

wire touches your head. Measure your height, h; 

the distance from you to the end of the guy wire, x; 

and the distance from the point on the ground 

directly below the top of the tower to the end of the 

guy wire, y. Solve a proportion to find the height of 

the tower, t: h 

 

t x .Finally,use y the Pythagorean 

Theorem to find the length of the guy 

wire: t2 y2 

. 

10. The triangles are similar by AA (because the 

ruler is parallel to the wall), so Kristin can use the 

length of string to the ruler, the length of string to 

the wall, and the length of the ruler to calculate the 

height of the wall; 144 in., or 12 ft. 

11. MUN MSA by AA; x 31 2 

3 . 

12. BDC AEC by AA; y 63. 

6 

13. GHF FHK GFK by AA; h 18, 13 

9 4 

x 7, 13 

y 44. 13 


A 1 

(8.2)(1.7) 

2 

6.97 cm2 

A 1 

(3)(4.6) 

2 

6.9 cm2 

3 cm 

15. 5 2 

3 

16. 

D 

Given: Parallelogram ABCD 

Show: AB CD and AD BC 

AD BC 

CPCTC 

1.7 cm 

4 

BD BD 

Same segment 

A 

3 

Given 

8.2 cm 

4.6 cm 

ABCD is a parallelogram 

AD BC 

Definition of 

parallelogram 

2 4 


ABD CDB 

ASA 

17a. 4.6.12 

17b. 3.12.12 or 3.12 2 

1 

C 

2 

B 

 

AB CD 

Definition of 

parallelogram 

1 3 


AB CD 

CPCTC 




18. The golden ratio is 1+ 5 , or approximately 

2 

1.618. Here is one possible construction: 

A 

mB 90° 

Construct BC 1 

2 AB 

AC 

2 5 AB 

AX AD 

2 5 AB 1 

2 AB 

AX 5 1 

 

2 

AB 

AB 

2 

AX 

5 1 

1+ 

2 5 

 

Therefore, X is the golden cut. 


D 

X 

C 

B 

19a. Answers will vary. 

19b. Possible answer: The shape is an irregular 

curve. 

19c. Answers will vary. Possible answers: As the 

circular track becomes smaller, the curve becomes 

more circular; as the track becomes larger, the 

curve becomes more pointed near the fixed point. 

As the rod becomes shorter, the curve becomes 

more pointed near the fixed point; as the rod 

becomes longer, the curve becomes more like an 

oval. As the fixed point moves closer to the traced 

endpoint, the curve becomes more pointed near 

the fixed point; as the fixed point moves closer to 

the circular track, the curve begins to look like a 

crescent moon.

1. 18 cm 

2. 12 cm 

3. 21 cm 

4. 15 cm 

5. 2.0 cm 

6. 126 cm2 ; 504 cm2 7. 16 cm 

8. 60 cm 

9. 4 4 

 

9 

cm 

p b 

10. 

q 

; q 

LESSON 11.4 

11. 63 cm 

12. 6 cm 

13. x 3 1 

3 cm, y 5 13 

3 

cm, z 8 2 

 

3 

cm 

14. B (3, 5), R 13 4 ,7 ; k 7 

 

h 

 

4 

15. 2 

1 

 

16. 

H 

L 

E 

O V M 

A 

Consider similar triangles LVE and MTH with 

corresponding angle bisectors EO and HA . To 

show that the corresponding angle bisectors are 

proportional to corresponding sides, for example 

EO EL 

HA 

, HM 

show by AA that LOE MAH, 

EO EL 

and then you can show that HA 

. HM 

You 

know that L M. Use algebra to show that 

LEO MHA. 

T 

17. a 

b 1 c 

 

d 

1 

a 

 

b 

b 

b c d 

 

d 

 

d 

a b c d 

b 

d 

18. The ratio will be the same as the ratio for the 

original rectangle. The ratio is 1 

 

2 if it can be divided 

like this: 

It might be any ratio if divided like this: 

19. Yes, by the SSS or the SAS Similarity 

Conjecture. 

20a. 2a b 

20b. 2a b 

20c. all values of a and b 

20d. no values of a and b 

21. AB 3 cm, BC 7.5 cm 




LESSON 11.5 

1. 18 cm2 2. 18 3. 5; 10 

4. 27 cm2 7. 

1 

5. 49 

6. 1:3 

m 

n2 2 

or m 2 

n 

8. Possible answer: Assuming the ad is sold by 

area, Annie should charge $6000. 

9. 5000 tiles 

11. 

9 

10. 1 

12a. a(x) 2x 

Area 

x in cm 2 

1 2 

2 4 

3 6 

4 8 

5 10 

6 12 

12b. A(x) 2x 2 

Area 

x in cm 2 

1 2 

2 8 

3 18 

4 32 

5 50 

6 72 

12c. The equation for a(x) is linear, so the graph is 

a line. The equation for A(x) is quadratic, so the 

graph is a parabola. 

13. Possible proof: The area of the first rectangle 

is bh. The area of the dilated rectangle is rh rb,or 

r2bh. The ratio of the area of the dilated rectangle 

to the area of the original rectangle is r2bh 

, bh 

or r2 . 

14. 16 

, 

3 

20 

 

3 

15. 8 

16a. (i) 40°; 50°; 40° (ii) 60°; 30°; 60° 

(iii) 22°; 68°; 22° 

Conjecture: similar; right triangle 

s h 

16b. (i) h 

(ii) 

x (iii) h b 

 

m 

or a 

p h 

Conjecture: 

h q 

 

16c. h pq 

Area in cm 2 

Area in cm 2 

10 

5 

70 

60 

50 

40 

30 

20 

10 


y 

y 

5 

5 

x 

x 

9 

16d. x 9, y 16, 12 

12 

, 

16 

h (9)(16 ) 

144 12 

17. 7.1 m 14.2 m 14.2 m 6.5 m 

18. 

42 m 

L 

10 m 

E 

O V M 

A 

Consider similar triangles LVE and MTH with 

corresponding altitudes EO and HA . Show 

EO EL 

that LOE MAH, then HA 

. HM 

You know 

that L M. Because EO and HA are altitudes, 

LOE and MAH are both right angles, and 

LOE MAH. 

EO EL 

So, LOE MAH by AA. Thus HA 

, HM 

which shows that the corresponding altitudes 

are proportional to corresponding sides. 

19. x 92°; Explanations will vary but should 

reference properties of linear pairs, isosceles 

triangles, and alternate interior angles formed 

by parallel lines. 

20. 105.5 cm2 21. 60 ft2 22. true 

23. Two pairs of angles are congruent, so the 

triangles are similar by the AA Similarity 

Conjecture. However, the two sets of corresponding 

sides are not proportional 6 

0 

 

80 

105 

,so 135 

the 

triangles are not similar. 

24. 

Top 

Front 

42 m 

Right side 

Top: 6 square units; front: 4 square units; right side: 

4 square units. The sum of the areas is half the 

surface area. The volume of the original solid is 

8 cubic units. The volume of the enlarged solid is 

512 cubic units. The ratio of volumes is 64 

. 

H 

T 

1

1. 1715 cm 3 

LESSON 11.6 

2. 16 cm, 4 cm; 768 cm3 2412.7 cm3 ; 

12 cm3 37.7 cm3 ; 64 

 

1 

3. 3 25 

; 1 ; 

5 27 

1500 cm3 4. 1944 ft3 6107.3 ft3 ; H 24 

3 

64 

; 

27 

32 ft 

5. 125 

 

27 

6. 2:5 7. 2 

 

3 

8. $1,953.13 

9. 2432 lb 

10. Possible answer: No, a 4-foot chicken, similar 

to a 14-inch 7-pound chicken, would weigh 

approximately 282 pounds 1 43 

7 

483 

. x It is unlikely 

that the legs of the giant chicken would be able to 

support its weight. 

11. Possible answer: Assuming the body types of 

the African goliath frog and the Brazilian gold frog 

are similar, the gold frog would weigh about 

0.0001 kg, or 0.1 g. 

12. surface area ratio 16 

,volume 1 ratio 64 

.The 1 

dolphin has the greater surface area to volume 

ratio. 

1 

13. The ratio of the volumes is .The 27 

ratio ofthe 

surface areas is 1 

9 . 

14. 

800 

600 

400 

200 

x 1 2 3 4 5 

Surface area in cm 2 22 88 198 352 550 

Volume in cm 3 6 48 162 384 750 

y 

Volume 

5 

Surface area 

x 

S(x) 22x 2 and V(x) 6x3 . 

Possible answer: The surface area equation is 

quadratic, so the graph is a parabola, and the 

volume equation is cubic, so the graph is a cubic 

graph. 

15. Possible proof: The volume of the first 

rectangular prism is lwh. The volume of the second 

rectangular prism is rl rw rh,or r3lwh.The ratio 

of the volumes is r3lwh 

, lwh 

or r 3 . 

16. 9,120 m3 or approximately 28,651 m3 s s s2 s 

17. 22 

4 

or 

4 

 

18. yes, because 182 242 302 (Converse of the 

Pythagorean Theorem) 

19a. Possible answer: Fold a pair of corresponding 

vertices (any vertex in the original figure and the 

corresponding vertex in the image) together and 

crease; repeat for another pair of corresponding 

vertices; the intersection of the two creases is the 

center of rotation. 

19b. Possible answer: Draw a segment (a chord) 

between a pair of corresponding vertices and 

construct the perpendicular bisector; repeat for 

another pair of corresponding vertices; the 

intersection of the two perpendicular bisectors 

is the center of rotation. 

20e. Label the third vertex C. Construct segment 

CD, which bisects C. AD 

 

DB 

2x 

 

3x 

2 

3 ,or 2:3 




LESSON 11.7 

1. 5 cm 2. 33 1 

 

3 

cm 

3. 45 cm 4. 21 cm 

5. 28 cm 6. no 

7. José’s method is correct. Possible explanation: 

Alex’s first ratio compares only part of a side of the 

larger triangle to the entire corresponding side of the 

smaller triangle, while the second ratio compares 

entire corresponding sides of the triangles. 

8. yes 

9. 6 cm; 4.5 cm 

10. 13.3 cm; 21.6 cm 

11. yes; no; no 

12. yes; yes; yes 

13. 32 cm; 62 cm 

14. 

15. 

E 

16. Extended Parallel/Proportionality Conjecture 

17. You should connect the two 75-marks. By the 

Extended Parallel/Proportionality Conjecture, 

drawing a segment between the 75-marks will 

75 

form a similar segment that has length 

100 

of the original. 

18. 2064 cm3 6484 cm3 19. possible proof: 

a b 

 

c d 

I J 

ad cb 

ad ab cb ab 

a(d b) b(c a) 

a(d b) 

 

ab 

b(c a) 

 

ab 

d b c a 

b 

a 


F 

, or 75%, 

So two pairs of corresponding sides of XYZ and 

XAB are proportional. X X, so 

XYZ XAB by the SAS Similarity Conjecture. 

Because XYZ XAB, XAB XYZ. 

Hence, AB YZ by the Converse of the Parallel 

Lines Conjecture. 

20. Set the screw so that the shorter lengths of the 

styluses are three-fourths as long as the longer 

lengths. 

21. x 4.6 cm, y 3.4 cm 

22. x 45 ft, y 40 ft, z 35 ft 

23. 343 

 

729 

24. She is incorrect. She can make only nine 8 cm 

diameter spheres. 

25. 6x2 ;24x2 ;54x2 26. 1 

3 r 

27a. Possible construction method: Use the 

triangle-and-circle construction from Lesson 11.3, 

Exercise 18, to locate the golden cut, X, of AB. 

Then use perpendicular lines and circles to create a 

rectangle with length AB and width AX. 

27b. Possible construction method: Construct 

golden rectangle ABCD following the method from 

27a. For square AEFD, locate EF by constructing 

circle A and circle D each with radius AD. Repeat 

the process of cutting off squares as often as 

desired. For the golden spiral from point D to point 

E, construct circle F with radius EF; select point D, 

point E, and circle F and choose Arc On Circle from 

the Construct menu. 


A 

S 

P 

D 

B 

R 

Q 

C 

Given: Circumscribed quadrilateral ABCD, with 

points of tangency P, Q , R, and S 

Show: AB DC AD BC

Use segment addition to show that each sum of 

lengths of opposite sides is composed of four lengths: 

AB DC (AP BP) (DR CR) and AD 

BC (AS DS) (BQ CQ). Using the Tangent 

Segments Conjecture, AP AS, BP BQ, 

CR CQ, and DR DS, and the four lengths in 

each sum are equivalent. Here are the algebraic 

steps to show that the whole sums are equivalent. 

AB DC (AP BP) (DR CR) Segment 

addition. 

(AS BQ) (DS CQ) Substitute AS 

for AP, BQ 

for BP, DS for 

DR, and CQ 

for CR. 

(AS DS) (BQ CQ) Regroup the 

measurements 

by common 

points of 

tangency. 

AB DC AD BC Use segment 

addition to 

rewrite the 

right side as 

the other sum 

of opposite 

sides. 

29. 





1. x 24 2. x 66 

3. x 6 4. x 17 

5. 6 cm; 4.5 cm; 7.5 cm; 3 cm 

6. 4 1 

 

6 

cm; 71 

2 

cm 7. 13 ft 2 in. 

8. It would still be a 20° angle. 

9. 

K P 

L 

10. Yes. If two triangles are congruent, then 

corresponding angles are congruent and 

corresponding sides are proportional with ratio 

1 , 1 so the triangles are similar. 

11. 15 m 12. 4 gal; 8 times 

13. Possible answer: You would measure the 

height and weight of the real clothespin and the 

height of the sculpture. 

Wsculpture 

W clothespin 

Hsculpture 

Hclothespin 

3 

If you don’t know the height of the sculpture, you 

could estimate it from this photo by setting up a 

ratio, for example, 

Hperso 

n Hsculpture 

Hp 

Hsc 

erson’ 

s photo 

ulpture’s 

photo 


14. 9:49 

15. 5 25 

; 1 

4 64 

16. $266.67 

17. 640 cm3 18. 32; 24; 40; 126 

19. 841 coconuts 

20a. 1 to 

4 to 1 

2 ,or 4 to to 2 

20b. 3 to 2 to 1 

20c. Answers will vary. 

21. Possible answer: If food is proportional to 

1 

body volume, then 8000 

of the usual amount of food 

is required. If clothing is proportional to surface 

1 

area, then 400 

of the usual amount of clothing is 

required. It would take 20 times longer to walk a 

given distance. 

22. The ice cubes would melt faster because they 

have greater surface area.



LESSON 12.1 

1. 0.6018 

2. 0.8746 

3. 0.1405 

4. 11.57 

5. 30.86 

6. 62.08 

7. sin A s 

;cos 

t 

A r 

t ;tan A s 

 

r 

8. sin 4 

;cos 

5 

3 ;tan 

5 

4 

3 

7 

9. sin A ;cos 25 

A 24 

7 

;tan 

25 

A ; 24 

sin B 24 

7 

;cos 

25 

B ;tan 25 

B 24 

 

7 

10. 30° 

11. 53° 

12 

12. 30° 

13. 24° 

14. a 35 cm 

15. b 15 cm 

16. c 105 yd 

17. d 40° 

18. e 50 cm 

19. f 33° 

20. g 18 in. 

21. approximately 237 m 

22. x 121 ft 

23. 6.375 

24. 2.2 

25. 16-inch pizza 

26. box of ice cream 

27. 83 cm 13.9 cm 

28. V 288 ft 3 905 ft 3 , S 144 ft 2 452 ft 3 




LESSON 12.2 

1. 9 cm 

2. 64° 

3. 7 cm 

4. 24 m 

5. 22 in. 

6. 49° 

7. 127° 

8. 30° 

9. 64 cm 







16a. approximately 974 m 

16b. approximately 1007 m 

16c. Yes, the height of the balloon would be 1 m 

less because you don’t have to account for the 

tripod. The distance to a point under the balloon 

would not change. 


17. 45° 

18. 60° 

19. 60° 

20a. 15.04 

20b. 2.05 

20c. 5.2304 

20d. 2.644 

21. x 3.5, y 9 1 

 

7 

9.14 

22. The block has a volume of 90 cm3 but it 

displaces only 62.8 cm3 of water. So not all of the 

block is under water, which means it floats. 

23. CZ AX BY 

24a. decreases, approaching 0 

24b. decreases, approaching 0 

24c. remains 90° 

24d. increases, approaching (AO) 2 

24e. decreases, approaching 1 

24f. decreases, approaching 1 

25. R(8, 7) and E(3, 9), or R(4, 3) and 

E(1, 1)

LESSON 12.3 

1. 329 cm 2 

2. 4 cm 2 

3. 11,839 cm 2 

4. 407 cm 2 

5. 35 cm 

6. 17 cm 

7. 30 cm 

8. 56° 

9. 45° 

10. 66° 

11a. approximately 2200 m 

11b. approximately 1600 m 

11c. approximately 1400 m 

12. The other two walls were 300 ft and 

approximately 413 ft. The area was approximately 

45,000 ft 2 . 


14. 2366 cm; approximately 41° 

15. approximately 5° 

16. 93 

 

8 

cm3 , or approximately 6 cm3 17. Because AB CD , A D by the AIA 

Conjecture. Because D and B intercept the 

same arc, D B. Therefore A B by the 

transitive property. So ABE is isosceles by the 

Converse of the Isosceles Triangle Conjecture. 

18. Draw line through the two points. Then fold 

the paper so that the two points coincide. Draw 

another line along the fold. These two lines contain 

the diagonals of the square. Now fold the paper so 

that the two lines coincide. Mark the vertices on 

the other line. 

19. AC 60 cm, AE 93.75 cm, AF 117 cm 

20. Box 1; about 1.2 in. longer 




LESSON 12.4 

1. 32 cm 

2. 47 cm 

3. 341 cm 

4. 74° 

5. 64° 

6. 85° 

7. approximately 43 cm 

8. approximately 30° 

9. approximately 116° and 64° 

10. approximately 6 min 

11. approximately 87.8 ft 



14. The ladder at approximately a 76° angle is not 

safe. The ladder should be at least 6.5 ft and no 

more than 14.3 ft from the base of the wall. 

15. 

c 

A b 

a 

sin A a 

c 

cos A b 

c 

tan A a 

b 

sinA 

cos 

A 

a 

c 

c a 

b 

b 

tan A 

a 

c 

 

b 

c 

16a. PR x3 

16b. Since x 2 x3 2 (2x) 2 ,TPR is a right 

triangle by the Converse of the Pythagorean 

Theorem. Therefore mTPR 90°. 

17a. increases 

17b. decreases 

17c. increases then decreases 

18a. The base perimeters are equal. 

18b. The cone has the greater volume. 

18c. The cone has the greater surface area. 

19a. a translation right 10 units; 

(x, y) → (x 10, y) 

19b. a rotation 180° about the origin; 

(x, y) → (x, y) 




x 

x 

x 

x 

120° 120° 

60° 60° 

2x 

x 

22. Answers will vary from simple (circles and 

segments) to quite complex. See below for a more 

detailed answer. 

Students might begin with the simplest case where 

all factors are equal. With circles of the same radius 

and with endpoints of the segment traveling in the 

same direction, at the same speed, and starting 

from the same relative position, the midpoint will 

trace a circle equal in size to those constructed. 

Changing the size of both circles will change the 

size of the circle traced, as will changing the 

starting positions of the endpoints. The maximum 

radius of the traced circle is limited by the radius of 

the constructed circles; the smallest observable 

trace is a single point. 

With all other factors equal and the endpoints of 

the segment traveling in opposite directions 

around the circles, the midpoint will trace a 

segment. Changing the relative speeds of the 

endpoints, or the relative radii of the circles, will 

produce more complex polar curves, which 

students might describe as flowers, radii, or 

“spirograph-like.” Changing the distance between 

the centers has no effect on the shape of the trace, 

only its position. 

y 

y 

2y 

y 

120° 120° 

60° 60° 

2y 

y

LESSON 12.5 


2. approximately 142 mi/h 

3. approximately 51 km 


5a. approximately 9.1 km 

5b. approximately 255° 


7. approximately 42 ft 

8. They must dig at an angle of approximately 

71.6° and dig for approximately 25.3 m. 

9. approximately 168 km/h 

10. 

6 cm 

10a. apothem 9.23 cm; area 277 cm2 10b. leg 9.7 cm; area of triangle 27.7 cm2 ; 

area of decagon 277 cm2 10c. They are the same. 

11. approximately 108 cm3 12. Sample answer: Any regular polygon can be 

divided into n congruent isosceles triangles, each 

with a vertex angle of 360° 

. 

n 

The altitude from the vertex angle of an isosceles 

triangle bisects the angle and the opposite side. 

This altitude is also an apothem of the regular 

polygon. 

s_ 

2 

360° 

___ 

n 

360° 

___ 

2n 

a 

360° 

___ 

2n 

s_ 

2 

For simplicity, let 360 

. 2n 

Now use the tangent 

ratio to find the length of the apothem. 

tan 

s 

a 2tan 

Now use the area formula for a regular polygon. 

A 1 

2 aP 

A 1 

2 s 

2tan 

ns 

ns2 

A 4tan 

where 360 

 

2n 

13. approximately 12.8 m at an angle of 

approximately 48° 

14. 15363 cm3 2660 cm3 s 

2 

 

a 

15. The area increases by a factor of 9. 

–6 

(–3, 0) 

y 

(0, 2) 

A x 

(1, 0) 6 

1 1_ 

2 2 1 

A 3 1_ 

2 6 9 

–6 (0, –6) 

16. 5 15 

 

4 

5 

12 

17a. decreases then increases 

17b. does not change 

18a. a reflection across the x-axis; 

(x, y) → (x, y) 

18b. a reflection across the y-axis; (x, y) → (x, y) 





1. 

2. 

–2 

10 

8 

6 

4 

2 


All three graphs are V-shaped, consisting of two 

parts that have slopes of 1 and 1. Graph g is a 

translation of graph f right 5 units. Graph h is a 

translation of graph f down 5 units. 

3. 

y 

–4 

y 

2 

x 

f(x) x2 –2 

–4 –2 

–4 

2 4 

–6 

2 

2 

y 

–2 2 4 

–2 

–4 

–6 

y 

6 

4 

2 

–4 –2 2 4 

x 

–2 

–4 

h(x) x3 y 

–6 –4 –2 

–2 

–4 

–6 

g(x) x 

x 

2 

2 

4 

y 

f(x) x 2 

g(x) x 5 

–5 5 10 

x 

–2 

–4 

A vertical stretch by a factor of 1 reflects the 

graph across the x-axis. 


2 

x 

g(x) x 

f(x) x 

4 

6 

x 

6 

4 

2 

–2 

–2 

–4 

–6 

y 

h(x) x 5 

h(x) x 3 

2 

x 

4. 

The graph of y 2x 1 is the same as the 

graph of p(x). The slope of a line is equal to 

the vertical stretch of the linear function. The 

y-intercept is equal to ah k. 

5. a vertical stretch by a factor of 2 and a vertical 

translation down 5 units 

–3 

6. a horizontal translation right 5 units and a 

vertical translation up 2 units 

6 

4 

2 

–2 

y 

–5 

2 

–2 

–4 

–6 

–8 

y 

2 

7. a vertical stretch by a factor of 1 

 

2 (a vertical 

shrink), a horizontal translation right 2 units, 

and a vertical translation down 5 units 

1 

–6 

y 

–4 –2 

–2 

3 

g(x) (x 5) 3 2 

2 

2 

–2 

–4 

–6 

y 

p(x) 2(x 3) 5 

x 

f(x) 2x 3 5 

6 8 

4 6 8 

h(x) 

1 

2 x 2 5 

x 

x 

x 

8. f(x) |x 3| 1; a horizontal translation 

right3unitsandaverticaltranslationdown1unit 

9. f(x) 3x 2 ; a vertical stretch by a factor of 3 

10. f(x) (x 3) 2 ; a vertical stretch by a factor 

of 1 (a reflection across the x-axis) and a horizontal 

translation right 3 units 

11. f(x) 2|x| 3; A vertical stretch by a factor 

of 2 and a vertical translation up 3 units 

12. f(x)3(x1) 2 3; A vertical stretch by a factor 

of 3, a horizontal translation left 1 unit, and a 

vertical translation down 3 units

13. f(x) 1 

2 (x3)2 2; A vertical stretch by a factor 

of 1 

 

2 (a vertical shrink), a horizontal translation 

right 3 units, and a vertical translation up 2 units 

14. Both graphs have the same “hills and valleys” 

shape, but one is a horizontal translation of the 

other. 

p(x) sin(x) 

–360 

15. a horizontal translation; a 1, h 90°, k 0 

16. The dashed function in each graph is the parent 

function f(x) sin(x). 

–360 

–360 

2 

2 

–2 

2 

–2 

q(x) cos(x) 

y 

y 

y 

p(x) 2sin(x) 

360 

360 

r(x) = –sin(x) 

360 

x 

x 

x 

–360 

Var ying a in the equation of the sine function 

causes the same types of vertical stretches as with 

other functions. 

2 

–2 

y 

360 

q(x) 

1_ 

2 

sin(x) 

x 

17. Possible answer: f(x) 2sin(x 90°) 1; 

a vertical stretch by a factor of 2, a horizontal 

translation right 90°, and a vertical translation 

down 1 unit 





1. 0.8387 

2. 0.9877 

3. 28.6363 

4. a 

b ; c a 

; 

b c 

8 

5. ; 

17 

15 

8 

; 17 

15 

6. s; t; s 

t 

7. 33° 

8. 86° 

9. 71° 

10. 1823 cm2 11. 15,116 cm3 12. Yes, the plan meets the act’s requirements. The 

angle of ascent is approximately 4.3°. 

13. approximately 52 km 

14. approximately 7.3° 

15. approximately 22 ft 


17. approximately 2973 km/h 

18. 393 cm2 19. 30 cm 

20. 78° 

21. 105 cm 

22. 51° 

23. 759 cm2 24. approximately 25 cm 

25. 72 cm2 26. approximately 15.7 cm 

27. approximately 33.5 cm2 28. approximately 10.1 km/h at an approximate 

bearing of 24.5° 

29. False; an octahedron is a polyhedron with 

eight faces. 

30. false 

31. true 

32. true 


33. False; the ratio of their areas is m 

34. true 

35. false; tangent of T 

36. true 

37. true 

38. true 

39. true 

40. False; the slope of line 2 is 

m 

41. false 

42. B 43. C 44. A 45. D 

46. B 47. B 48. A 49. B 

50. C 

54. 

51. D 52. C 53. A 

100 

cm3 3 

55. 28 cm3 56. 30.5 cm3 57. 33 

58. (x, y) → (x 1, y 3) 

59. w 48 cm, x 24 cm, y 28.5 cm 

60. approximately 18 cm 

61. (x 5) 2 (y 1) 2 9 

62. Sample answer: Each interior angle in a regular 

pentagon is 108°. Three angles would have a sum of 

324°, or 36° short of 360, which would leave a gap. 

Four angles would have a sum exceeding 360° and 

hence create an overlap. 


64. 30 ft 

65. 4 cm 

66. 

B 

A 

D 

C 

66a. mABC 2 mABD 

66b. Possible answers: BD is the perpendicular 

bisector of AC. It is the angle bisector of ABC,it 

is a median, and it divides ABC into two 

congruent right triangles. 

1 . 

. 

n2 2 

length of leg opposite T 

 

length of leg adjacent to T



LESSON 13.1 

1. A postulate is a statement accepted as true 

without proof. A theorem is deduced from other 

theorems or postulates. 

2. Subtraction: Equals minus equals are equal. 

Multiplication: Equals times equals are equal. 

Division: Equals divided by nonzero equals are equal. 

3. Reflexive: Any figure is congruent to itself. 

A 

C 

ABC ABC 

Transitive: If Figure A is congruent to Figure B and 

Figure B is congruent to Figure C, then Figure A is 

congruent to Figure C. 

If , then . 

P Q R S X Y P Q X Y 

If PQ RS and RS XY, then PQ XY. 

Symmetric: If Figure A is congruent to Figure B, 

then Figure B is congruent to Figure A. 

If 

then 

X 

Y Z 

L 

B 

M N 

 

 

L 

M N 

X 

Y Z 

, 

. 

13 

If XYZ LMN, then LMN XYZ. 

4. reflexive property of equality; reflexive 

property of congruence 

5. transitive property of congruence 

19. (Lesson 13.1) 

Given 

20. (Lesson 13.1) 

6. subtraction property of equality 

7. division property of equality 

8. Distributive; Subtraction; Addition; Division 

9. Given; Addition property of equality; 

Multiplication property of equality; Commutative 

property of addition. 

10. true, definition of midpoint 

11. true, Midpoint Postulate 

12. true, definition of angle bisector 

13. true, Angle Bisector Postulate 

14. false, Line Intersection Postulate 

15. false, Line Postulate 

16. true, Angle Addition Postulate 

17. true, Segment Addition Postulate 

18. 

• That all men are created equal. 

• That they are endowed by their creator with 

certain inalienable rights, that among these are life, 

liberty, and the pursuit of happiness. 

• That to secure these rights, governments are 

instituted among men, deriving their just powers 

from the consent of the governed. 

• That whenever any form of government becomes 

destructive to these ends, it is the right of the 

people to alter or to abolish it, and to institute new 

government, laying its foundation on such 

principles and organizing its powers in such 

form as to them shall seem most likely to effect 

their safety and happiness. 



? 

 

AO and BO are radii 

AOB is isosceles 

1 ? 

2 

AO BO 

3 

Definition of circle 

Definition of ? 

 

isosceles 

? 

 

 

 

 

1 

1 ? 2 

2 

m n 

3 

3 ? 4 

? Given 

Corresponding 

? Corresponding 

Angles Postulate 

Angles Postulate 

 






23. (2n 1) (2m 1) 2n 2m 2 

2(n m 1) 

24. (2n 1)(2m 1) 4nm 2n 2m 1 

4nm 2n 2m 2 1 

2(2nm n m 1) 1 

25. Let n be any integer. Then the next two 

consecutive integers are n 1 and n 2. The sum 

of these three integers is (n) (n 1) (n 2) 

n n 1 n 2. Combining like terms: 3n 3 

3(n 1), which is divisible by 3. 

26. 299 m 

27. sphere, cylinder, cone; cylinder, cone, sphere; 

cone, cylinder, sphere 

21. (Lesson 13.1) 

1 AC BD 

1 Construct angle 

bisector BD 

Angle Bisector 

Postulate 


2 ? AD BC 

4 

ABC 

3 

? 

 

 

? 

 

Given 

Given 

AB BA 

Reflexive property 

of congruence 

22. (Lesson 13.1) 

2 

28. 30 ft 

29. FG 26 and DG 23 because ABGF 

and BCDG are parallelograms. Triangle FGD is 

right (mFGD 90°) by the Converse of the 

Pythagorean Theorem because 26 2 23 2 

62 .But mFGB 128° and mDGB 140° by 

conjectures regarding angles in a parallelogram. 

So, mFGD 92° because the sum of the angles 

around G is 360°. So, mFGD is both 90° and 92°. 

30. x 54°, y 126°, a 7.3 m 

31a. 

4 

31b. 1 

4 

31c. 

 

2 

1 

? 

? 

? ? 

Given 

CBD 

BCD A C 

3 5 6 

ABD BAD 

Definition 

SAS Congruence CPCTC 

of angle bisector Postulate 

4 BD BD ? 

? 

? 

 

AB CB 

BAD 

Reflexive property 

of congruence 

? 

 


Postulate 

? ? 

5 

D C 

? 

 

CPCTC

LESSON 13.2 

1. Linear Pair Postulate 

2. Parallel Postulate, Angle Addition Postulate, 

Linear Pair Postulate, Corresponding Angles 

Postulate 

3. Parallel Postulate 

4. Perpendicular Postulate 

5. 

2 1 

Use the definition of supplementary angles and the 

substitution property to get m1 m1 180°. 

Then use the division property and the definition 

of a right angle. 

6. 

1 2 4 

3 

Use the definition of supplementary angles and the 

transitive property to get m1 m2 m3 

m4. Then use the substitution property and the 

subtraction property to get m1 m4. 

7. 

1 2 

Use the definition of a right angle and the 

transitive property to get m1 m2. Then use 

the definition of congruence to get 1 2. 

8. 

3 1 

3 

Use the VA Theorem and the transitive property to 

get 1 3. Therefore the lines are parallel by the 

CA Postulate. 

9. 

3 2 

3 

2 

1 

1 

1 

Use the VA Theorem and the CA Postulate to get 

1 3 and 2 3. Then use the transitive 

property to get 1 2. 

2 

2 

10. 

2 

3 

Use the VA Theorem and the transitive property to 

get 3 4. Therefore the lines are parallel by the 

Converse of the AIA Theorem. 

11. 

3 Use the Linear Pair Postulate and the definition of 

supplementary angles to get m1 m3 180°. 

Then use the CA Postulate and the substitution 

property to get m1 m2 180°. 

12. 

3 3 1 

Use the Linear Pair Postulate and the definition of 

supplementary angles to get m1 m3 

m1 m2. Then use the subtraction property 

and the Converse of the AIA Theorem to get 

1 2 . 

13. 

A 

1 

B 2 

3 

4 

2 

2 

1 

3 

1 

3 

Construct a transversal across lines 1 and 2 ;it 

will intersect line 3 by the Parallel Postulate. Use 

the Interior Supplements Theorem and the 

definition of supplementary angles to get m1 

m2 180°. Use the CA Postulate and the 

substitution property to get m1 m3 180°. 

Therefore 1 3 by the Converse of the Interior 

Supplements Theorem. 

1 

1 

1 

1 

2 

3 

2 

2 

2 




14. 

Use the definition of perpendicular lines and the 

transitive property to get m1 m2. Therefore 

lines 1 and 2 are parallel by the Converse of the 

AIA Theorem. 

15. 

2 

1 

By the Triangle Sum Theorem, m1 m2 

m3 180°. By the definition of a right triangle, 

1 is a right angle. By the definition of right 

angle, m1 90°. Using the subtraction property, 

m2 m3 90°, so by the definition of 

complementary angles, 2 and 3 are 

complementary. 

16. Linear Pair 

Post. 

VA Thm. 

1 

Converse of 

AIA Thm. 

Converse of 

AEA Thm. 

CA Post. 


3 17. 1066 cm 3 

2 

3 

1 

2 

18. Area (ft 2 ) 

Bottom 6 

2 sides 9 

Back and front 6 

2 rooftops 8 1 

2 

2 gable ends 2 

Total 31 1 

2 

Plywood (4 by 8) 32 

The area is less than that of one sheet of plywood. 

However, it is impossible to cut the correct size 

pieces from one piece. This answer assumes that 

the bottom of the dog house is included. If the 

bottom is not included, and the gables can be cut 

separately from the rectangular part of the front 

and back, then the pieces can be cut from a single 

sheet of plywood. 

19. A(6, 10), B(2, 6), C(0, 0); mapping 

rule: (x, y) → (2x 8, 2y 2)

LESSON 13.3 

1. Case 1: P is collinear with A and B. Use the Line 

Intersection Postulate to show that P and E are the 

same point and use the definitions of bisector and 

midpoint to get AP BP. 

Case 2: P is not collinear with A 

and B. Use the SAS Congruence Postulate to get 

AEP BEP. Then use CPCTC to get AP BP. 

A 

2. Case 1: P is collinear with A and B. Use the 

definitions of congruence and midpoint to show 

that P is the midpoint of AB.Then use the 

definition of perpendicular bisector. 

A 

C 

P 

E 

D 

P 

E 

Case 2: P is not collinear with A and B.Draw 

midpoint E and PE. Use the SSS Congruence 

Postulate to get AEP BEP. Then use CPCTC 

and the Congruent and Supplementary Theorem 

to prove that AEP and BEP are both right 

angles. Therefore PE is the perpendicular bisector 

of AB by the definitions of midpoint and 

perpendicular. 

3. Use the reflexive property and 

C 

the SSS Congruence Postulate to get 

ABC BAC. Therefore, 

A B by CPCTC. 

4. Use the reflexive property and 

the ASA Congruence Postulate to 

get ABC BAC. Then use 

CPCTC and the definition of 

isosceles triangle. 

5. 

B 

C A 

P 

B 

B 

A B 

A B 

Draw line BA. 

Use the Isosceles Triangle Theorem to get 

PAB PBA. Use the Angle Addition 

Postulate and the subtraction property to get 

BAC ABC. Then use the Converse of the 

C 

Isosceles Triangle Theorem (CB CA), the reflexive 

property, and the SSS Congruence Postulate to get 

ACP BCP. Therefore,ACP BCP by 

CPCTC. 

6. 

C m 

A 

Use the Line Intersection Postulate and the 

Perpendicular Bisector Theorem to get AP BP 

and BP CP. Then use the transitive property and 

the Converse of the Perpendicular Bisector 

Theorem to prove that point P is on line n. 

7. 

C 

Use the Line Intersection Postulate and the Angle 

Bisector Theorem to prove that Q is equally distant 

from AB and AC and from AB and BC. Then use the 

transitive property and the Converse of the Angle 

Bisector Theorem to prove that point Q is on line n. 

8. 

B 

A 

Use the Linear Pair Postulate and the definition of 

supplementary angles to get m3 m4 180°. 

Then use the Triangle Sum Theorem and the 

transitive property to get m1 m2 

m3 m3 m4. Therefore, m1 

m2 m4 by the subtraction property. 

9. D 

A 

n 

m 

A B 

1 

3 4 

2 

1 

2 

B 

Q 

n 

P 

 

3 4 

C 

C 

 

Use the Triangle Sum Theorem and the addition 

property to get mA m1 m3 mC 

m4 m2 360°. Then use the Angle Addition 

Postulate and the substitution property to get 

mA mABC mC mCDA 360°. 

10. Use the definitions of median 

and midpoint to get BM 1 BC 2 and 

AN 1 

C 

N M 

AC. 2 Then use the multiplication 

property and the substitution 

A B 

property to get AN BM . By the 

reflexive property, the Isosceles Triangle Theorem, 

and the SAS Congruence Postulate, ABN 

BAM. Therefore, BN AM by CPCTC. 

B 




11. 

A B 

Use the Angle Addition Postulate and the 

definition of angle bisector to get 

mPAB 1 mCAB 2 and 

mQBA 1 mCBA. 2 Then use the Isosceles 

Triangle Theorem, the multiplication property, and 

the substitution property to get PAB QBA. 

By the reflexive property and the ASA Congruence 

Postulate, ABP BAQ. Therefore, AP BQ 

by CPCTC. 

12. 

C 

A B 

Use the Isosceles Triangle Theorem, the Right 

Angles Are Congruent Theorem, and the SAA 

Theorem to get ABT BAS. Therefore, 

AS BT by CPCTC. 

13. 

C 

A D B 

median → angle bisector 

Use the definitions of median, midpoint, and 

isosceles triangle, the reflexive property, and the 

SSS Congruence Postulate to prove that 

ADC BDC. Then use CPCTC and 

the definition of angle bisector. 

C 

Q P 

T S 

A D B 

C median by CPCTC and the definitions of midpoint 

and median. 

altitude → median 

Use the Right Angles Are Congruent Theorem, the 

Isosceles Triangle Theorem, and the SAA Theorem 

to get ADC BDC. Therefore, CD is the 


angle bisector → altitude 

Use the definitions of angle bisector and isosceles 

triangle, the reflexive property, and the SAS 

Congruence Postulate to get ADC BDC. 

Then use CPCTC, the Linear Pair Postulate, and 

the Congruent and Supplementary Theorem to 

prove that ADC and BDC are both right 

angles. Therefore CD is the altitude by the 

definitions of perpendicular and altitude. 

14. x 6, y 3 

15. 21.9 m 

16. first image: (0, 3); second image: (6, 1) 

17. BC FC makes ABCF a rhombus, so its 

diagonals are perpendicular. mFGC 90°, so 

mCFG mFCG 90°. FD GE, so by 

subtraction and transitivity m2 mFCG. 

1 FCG by AIA, so 1 2. 

18a. 18b. 18c. 

19. One possible sequence: 

1. Fold A onto B and crease. Label as 1 . Label the 

midpoint of the arc M. 

2. Fold line 1 onto itself so that M is on the crease. 

Label as 2 . 

M is the midpoint of AB 3 

4 

, and 2 is the desired 

tangent. 

 

 

3 

 

 

6 

3 

4 

A 

C 

A D B 

1 M 

2 

B 

20. mBAC 13° 

21a. B 

21b. B 

22a. x 1 

(a 

2 

c), y 1 (a 

2 

b), z 1 (b 

2 

c) 

22b. w 1 

(a 

2 

b), x 1 (b 

2 

e), y 1 (e 

2 

d), 

z 1 

(d 

2 

a)

1. 

A 

LESSON 13.4 

Use x to represent the measures of one pair of 

congruent angles and y for the other pair. Use the 

Quadrilateral Sum Theorem and the division 

property to get x y 180°. Therefore, the 

opposite sides are parallel by the Converse of the 

Interior Supplements Theorem. 

2. D C 

A 

Use the AIA Theorem, the reflexive property, and 

the SAS Congruence Postulate to get ADC 

CBA. Then use CPCTC and the Converse of the 

AIA Theorem to get AB DC . 

3. D C 

A 

Use the definition of rhombus, the reflexive 

property,and the SSS Congruence Postulate to get 

ABC ADC. Then use CPCTC and the 

definition of angle bisector to prove that AC bisects 

DAB and BCD. Repeat the steps above using 

diagonal DB. 

4. D C 

A 

Use the definition of parallelogram to get AD BC 

and AB DC . Then use the Interior Supplements 

Theorem. 

5. D C 

A 

D C 

1 

7 

8 

1 

2 

4 

3 

2 

6 5 

4 

3 

B 

4 

1 

B 

B 

3 2 

B 

B 

Use the reflexive property and the SSS Congruence 

Postulate to get ABD CDB. Then use 

CPCTC and the Converse of the AIA Theorem to 

get AB CD and AD CB. Therefore,ABCD is a 

rhombus by the definitions of parallelogram and 

rhombus. 

6. 

D 

A 

Use the Converse of the Opposite Angles Theorem 

to prove that ABCD is a parallelogram. Then use 

the definition of rectangle. 

7. D 

C 

A 

Use the definition of rectangle to prove that ABCD 

is a parallelogram and DAB CBA. Then use 

the Parallelogram Opposite Sides Theorem, the 

reflexive property, and the SAS Congruence Postulate 

to get DAB CBA. Finish with CPCTC. 

8. D 

C 

A 

Use the Parallelogram Opposite Sides Theorem, 

the reflexive property, and the SSS Congruence 

Postulate to get DAB CBA. Repeat the above 

steps to get ADC CBA and DAB BCD. 

Then use CPCTC and the transitive property to get 

DAB ABC BCD ADC. Finish with 

the Four Congruent Angles Rectangle Theorem. 

9. D 

C 

A E 

Use the Parallel Postulate to construct DE CB. 

Then use the Parallelogram Opposite Sides 

Theorem and the transitive property to prove that 

AED is isosceles. Therefore, A B by the 

Isosceles Triangle Theorem, the CA Postulate, and 

substitution. 

10. D 

C 

A 

C 

B 

B 

B 

B 

B 

Use the Isosceles Trapezoid Theorem, the reflexive 

property, and the SAS Congruence Postulate to get 

DAB CBA. Then AC BD by CPCTC. 




11. 

A 

2 1 

D 

4 

3 

Use the Parallelogram Opposite Angles Theorem, 

the multiplication property, and the definition of 

angle bisector to get 1 3. Then use the 

Converse of the Isosceles Triangle Theorem, 

the definition of isosceles triangle, and the 

Parallelogram Opposite Sides Theorem to get 

AB BC DC AD . 

12. 

W Z 

X P Y 

Use the Converse of the Angle Bisector Theorem 

to prove that WY is the angle bisector of Y. In 

like manner, WY is the angle bisector of W. 

Therefore, WXYZ is a rhombus by the Converse 

of the Rhombus Angles Theorem. 

13. Linear Pair Postulate 

CA Postulate 

Interior Supplements Theorem 

Parallelogram Consec. Angle Theorem 


B 

C 

Q 

17. (Lesson 13.4) 

parallelogram 

2 diagonals 

2 bisectors 

of sides 

isosceles 

trapezoid 

14. 

ASA Congruence 

Postulate 

Parallelogram 

Diagonal Lemma 

Opposite Sides 

Theorem 


IT Theorem 

Opposite Angles 

Theorem 

Converse of the Rhombus 

Angles Theorem 

Line Postulate Angle Addition 

Postulate 

Double-Edged 

Straightedge Theorem 


Postulate 

IT Theorem 

Converse of the Angle 

Bisector Theorem 

15a. A 

15b. S 

15c. S 

15d. N 

16. 2386 ft 2 


18. V 1 V 2 has length 12.8 and bearing 72.6°. 

19a. B 

19b. A 

20a. 19° 

20b. 52° 

20c. 52° 

20d. 232° 

20e. 19° 

Name Lines of symmetry Rotational symmetry 

none 2-fold 

trapezoid none none 

kite 1 diagonal none 

square 4-fold 

rectangle 2 bisectors of sides 2-fold 

rhombus 2 diagonals 2-fold 

1 bisector of sides none

LESSON 13.5 

1. D: Paris is in France; Tucson is in the U.S.; 

London is in England. Bamako must be the capital 

of Mali. 

2. C: The “Sir” in part A shows that Halley was 

English; Julius Caesar was an emperor, not a 

scientist; Madonna is a singer. Galileo Galilei must 

be the answer. 

3. No, the proof is claiming only that if two 

particular angles are not congruent, then the two 

particular sides opposite them are not congruent. It 

still might be the case that a different pair of angles 

are congruent and that therefore a different pair of 

sides are congruent. 

4. Yes, this statement is the contrapositive of the 

conjecture proved in Example B, so they are 

logically equivalent. 

5. 1. Assume the opposite of the conclusion; 

2. Triangle Sum Theorem; 3. Substitution property 

of equality; 4. 0°; Subtraction property of equality 

6. Assume ZOID is equiangular. Use the definition 

of equiangular and the Four Congruent Angles 

Rectangle Theorem to prove that ZOID is a 

rectangle. Therefore ZOID is a parallelogram, which 

creates a contradiction. 

7. Assume CD is the altitude to AB. Use the 

definitions of altitude, median, and midpoint, the 

Right Angles Are Congruent Theorem, and the 

SAS Congruence Postulate to get ADC BDC. 

Therefore AC BC, which creates a contradiction. 

8. Assume ZO ID. Use the Opposite Sides 

Parallel and Congruent Theorem to prove that 

ZOID is a parallelogram, which creates a 

contradiction. 

9. Given: Circle O with chord AB and perpendicular 

bisector CD 

Show: CD passes through O 

Assume CD does not pass through O. Use the Line 

Postulate to construct OB and OA and the 

Perpendicular Postulate to construct OE. Then use 

the Isosceles Triangle Theorem, the Right Angles 

Are Congruent Theorem, and the SAA Theorem to 

get OEA OEB. From CPCTC and the 

definition of midpoint, prove that E is the midpoint 

of AB, which creates a contradiction. 

C 

O 

10. a 75°, b 47°, c 58° 

11. 42 ft3 132 ft 3 

12a. 

12b. 1 3 

13a. A 

13b. N 

13c. S 

13d. S 

13e. A 

 

3 

3 

 

4 12 

B 

D 

E 

A 




1. 

LESSON 13.6 

Case 1 The same: Use the Inscribed Angle Theorem 

and the transitive property to get A B. 

B 

A 

B 

Case 2 Congruent: Use the multiplication 

property to get 1 

2 mYZ 1 

2 mWX . Then follow 

the steps in Case 1. 

2. D 

Use the Inscribed Angle Theorem, the addition 

property, and the distributive property to get 

mA mC 1 

2 mBCD 

mDAB 

.Then use 

the definition of degrees in a circle, the substitution 

property, and the definition of supplementary 

angles to get A and C are supplementary. 

Repeat the steps using mB and mD to get B 

and D are supplementary. 

3. 

Use the Line Postulate to construct AD . Then use 

the AIA Theorem, the Inscribed Angle Theorem, 

and substitution to get AC BD . 

4. 

T 

C 

R 

A 

A 

A 

C 

W 


Z 

W 

X 

Y 

B 

C 

E 

X 

B 

D 

Use the Cyclic Quadrilateral Theorem, the 

Opposite Angles Theorem, and the Congruent and 

Supplementary Theorem to get R, C, E, and 

T are right angles. 

5. 

P 

Use the Line Postulate to construct OS ,OT, and 

OP. Then use the Tangent Theorem, the Converse 

of the Angle Bisector Theorem, and the SAA 

Theorem to get OSP OTP. PS PT by 

CPCTC. 

6. 

D 

Use the Line Postulate to construct AD . Then use 

the Inscribed Angle Theorem, the addition 


m2 m3 1 

2 mAC mBD . Therefore, 

m1 1 

2 mAC mBD by the Triangle Exterior 

Angle Theorem and the transitive property. 

7. 

P 

B 

B 

S 

2 

O 

D 

O 

a 

3 

1 

E 

b 

A 

A 

1 

T 

C 

C 

Intersecting Secants Theorem: The measure of an 

angle formed by two secants intersecting outside a 

circle is half the difference of the measure of the 

larger intercepted arc and the measure of the 

smaller intercepted arc. Use the Triangle Exterior 

Angle Theorem and the subtraction property to get 

x b a. Then use the Inscribed Angle Theorem, 

the substitution property, and the distributive 

property to get x 1 

2 mBD mAC . 

2 

x

8. 

Use the Inscribed Angle Theorem to get 

mBDC 1 

2 mBEC .By the definition of 

semicircle, mBEC 

mBDC 

180°, so by the 

division property, mBDC 90°. By the 

definition of right angle, BDC is a right angle. 

Because only definitions, properties, and the 

Inscribed Angle Theorem are needed to prove this 

conjecture, it is a corollary of the Inscribed Angle 

Theorem. 

9. 

10. 

B 

Line 

Postulate 

SSS 

Congruence 

Postulate 

IT 

Theorem 

E 

Linear Pair 

Postulate 

A 


IT Theorem 

D 


Postulate 

Converse of the Angle 

Bisector Theorem 

Exterior Angle 

Sum Theorem 

Inscribed Angle 

Theorem 

Cyclic Quadrilateral 

Theorem 

CA 

Postulate 

C 

SAA 

Theorem 

VA 

Theorem 

Angle Addition 

Postulate 

Tangent Segments 

Theorem 

Parallel 

Postulate 

AIA 

Theorem 

Triangle Sum 

Theorem 


Postulate 

Right Angles Are 

Congruent Theorem 

Tangent 

Theorem 

Parallelogram 

Diagonal Lemma 

Opposite Angles 

Theorem 

IT 

Theorem 

Parallelogram Inscribed 

in a Circle Theorem 

Angle Addition 

Postulate 


Postulate 

SAS Congruence 

Postulate 

Midpoint 

Postulate 

Perpendicular 

Postulate 

Right Angles 

Are Congruent 

Theorem 

Arc Addition 

Postulate 

Congruent and 

Supplementary Theorem 

11. A(4.0, 2.9), P(1.1, 2.8) 

12. 3; 1; 9 

13. BC,AB,AC,CD ,AD 

14. 32.5 

15. As long as point P is inside the triangle, 

a b c h. 

Proof: Let x be the length of a side. The areas of the 

three small triangles are 1 

 

2 xa, 1 

2 xb, and 1 

2 xc. The area 

of the large triangle is 1 

 

2 xh. So, 1 

2 xa 1 

2 xb 1 

2 xc 

xh. Divide both sides by 1 x. So, a b c h. 

1 

2 

h 

a 

P 

b 

c 

16a. 27° 

16b. 47° 

16c. 67° 

16d. 133° 

16e. cannot be determined 

16f. cannot be determined 

17. 

x 

A 

O M P 

B 

Steps: 

1. Construct OP. 

2. Bisect OP. Label midpoint M. 

3. Construct circle with center M and radius PM. 

4. Label the intersection of the two circles A and B. 

5. Construct PA and PB. PA and PB are the 

required tangents. 

OAP and OBP are right angles because they are 

inscribed in semicircles. 

2 




1. 

T 

B H 

A 

LESSON 13.7 

Use the Right Angles Are Congruent Theorem, 

CASTC, and the AA Similarity Postulate to get 

BT 

TH 

BHT GEV. Therefore, GV 

VE 

by CSSTP. See 

the Solutions Manual for the family tree. 

2. A 

G 

S L M 

Use the definitions of median and midpoint, the 

Segment Addition Postulate, the substitution 

property, and the multiplication property to get 

BY 1 BI 2 and SL 1 SM. 2 Then use CASTC, 

CSSTP, and the SAS Similarity Theorem to get 

BYG SLA. Therefore, BG 

GY 

 

SA AL 

by CSSTP. See 

the Solutions Manual for the family tree. 

3. C 

F 

1 3 

A B 

P 

Use the definition of angle bisector, the Angle 

Addition Postulate, and the substitution property to 

get mACB 2m1 and mDFE 2m2. Then 

use CASTC and the AA Similarity Postulate to get 

APC DQF. Therefore, AC 

CP 

 

DF 

FQ 

by CSSTP. 

4. C 

D E 

1 

2 

A B 

Use the CA Postulate and the AA Similarity 

Postulate to get CDE CAB. Then use CSSTP 

and the Segment Addition Postulate to get CD DA 

 

CD 

 

CE EB 

DA 

EB 

. CE 

Therefore, CD 

CE 

by algebra. 

5. C 

D E 

1 

2 

A B 

2 4 

D E 

Q 

DA 

EB 

Use the addition property to get CD 

1 

CE 

1. 

Then use algebra and the Segment Addition 

CA 

CB 

Postulate to get CD 

. CE 

Therefore ABC 

DEC by the SAS Similarity Theorem, 1 2 

by CASTC, and DE AB by the CA Postulate. 


G 

V 

E 

B Y I 

L 

6. 

Use the Right Angles Are Congruent Theorem, the 

reflexive property, and the AA Similarity Postulate 

to get ADC ACB and ACB CDB. 

Therefore, ADC ACB CDB by the 

transitive property of similarity. 

7. 

C 

Use the Three Similar Right Triangles Theorem to 

get ADC CDB. Then use CSSTP to get 

x h 

h 

y . 

8. 

A 

A 

Draw the altitude to the hypotenuse, then use the 

ratios given by the Three Similar Right Triangles 

a c 

yields a2 c2 cd. 

Theorem. In particular, c d 

a 

b c 

Now look at the other small triangle and use d b 

to get b2 cd. 

9. 

C 

Begin by constructing a second triangle, right 

triangle DEF (with E a right angle), with legs of 

lengths a and b and hypotenuse of length x. The 

plan is to show that x c, so that the triangles 

are congruent. Then show that C and E are 

congruent. Once you show that C is a right angle, 

then ABC is a right triangle. 

10. H 

L 

Y 

a 

d 

C 

1 2 

D 

h 

x D y 

b a 

c 

B A 

c 

b 

c – d 

P E 

B 

B 

D F 

x 

Use the Pythagorean Theorem to write expressions 

for the lengths of the unknown legs. Show that the 

expressions are equivalent. The triangles are 

congruent by SSS or SAS. 

a 

E 

G 

b

11. 

CA 

Postulate 

12. 

13. 

Perpendicular 

Postulate 

Segment Duplication 

Postulate 

Parallel 

Postulate 

CA 

Postulate 

AA Similarity 

Postulate 

SAS 

Congruence 

Postulate 

SAS Similarity 

Theorem 

SSS Similarity 

Theorem 

Three Similar 

Right Triangles 

Theorem 

Pythagorean 

Theorem 

Segment Addition 

Postulate 

Parallel/Proportionality 

Theorem 

AA Similarity 

Postulate 

Right Angles 

Are Congruent 

Theorem 

SSS 

Congruence 

Postulate 

AA Similarity 

Postulate 

14. approximately 1.5 cm or 6.5 cm 

15a. C 

15b. A 

15c. A 

15d. A 

15e. C 

16a. The vectors are diagonals of your quadrilateral. 

16b. A 180° rotation about the midpoint of the 

common side; the entire tessellation maps onto 

itself. 

17a. a 180° rotation about the midpoint of any 

side 

17b. possible answer: a vector running from each 

vertex of the quadrilateral to the opposite vertex 

(or any multiple of that vector) 

18a. 33° 

18b. 66° 

18c. 57° 

18d. 62° 

18e. PSQ and PTQ are 90° by the Tangent 

Theorem and so are supplementary. So, SPT 

and SQT must also be supplementary by the 

Quadrilateral Sum Theorem. Therefore, opposite 

angles are supplementary, so it’s cyclic. 

18f. Because PS is a tangent, mPSQ 90° 

(Tangent Theorem). Because mPSQ 90°, SQ 

must be a tangent (Converse of the Tangent 

Theorem). 

 

19a. 12 

1 

19b. 

3 

2 

3 

 

6 

4 1 

20a. CDG CFG by SAA; GEA GEB 

by SAS; DGA FGB by the HL Theorem 

20b. CD CF and DA FB by CPCTC; CD 

DA CF FB (addition property of equality). 

Therefore, CA CB, and ABC is isosceles. 

20c. The figure is inaccurate. 

20d. The angle bisector does not intersect the 

perpendicular bisector inside the triangle as 

shown, except in the special case of an isosceles 

triangle, when they coincide. 

21. 173 cm, 346 cm, 20 stones. Draw the trapezoid 

and extend the legs until they meet to form a 

triangle. Use parallel proportionality to find the 

rise. The span is twice the rise. Use inverse tangent 

to find the central angle measure. Divide into 180° 

to find the number of voussoirs. 





1. B(a, 0) 

2. C(a b, c) 

3. Ca b, a2 b2 

, Db, a2 b 


T (0, b) 


2 

0 0 0 

slope of RE a 

0 

a 

0 

b 0 

slope of CE a 

a 

b 

0 

(undefined) 

slope of TC b 

b 0 

 

a 0 

a 

0 

slope of TR b 

0 

 

0 0 

b 

0 

(undefined) 

Opposite sides have the same slope and are 

therefore parallel by the parallel slope property. 

Two sides are horizontal and two sides are vertical, 

so the angles are all congruent right angles. RECT 

is an equiangular parallelogram and is a rectangle 

by definition. 


T (0, 0) 

y 

y 

R (0, 0) 

Z 

b ( , 

c 

2 2) 

X 

a ( , 0 

2 ) 

I (b, c) 

C (a, b) 

x 

E (a, 0) 

Y 

a + b ( , 

c ) 

2 

R (a, 0) 

2 

x 

Let X be the midpoint of TR. 

0 0 0 

X a , 2 2 a 

2 ,0 

Let Y be the midpoint of RI . 

b 0 c b c 

Y a , 2 2 a , 2 2 

Let Z be the midpoint of TI . 

0 c 0 

Z b , 2 2 b 

2 , c 2 

X, Y, and Z are the midpoints of TR,RI , and TI , 

respectively, by the coordinate midpoint property. 

So XY,YZ, and ZX are midsegments by definition. 


y 

P (c, d) 

T (0, 0) 

A (a – c, d) 

R (a, 0) 

0 0 0 

slope of TR a 

0 

a 

0 

d 0 d 

slope of RA a c a 

c 

d d 0 

slope of AP a 

2c 

a 2c 

0 

slope of PT d 

0 

 

c 0 

d 

c 

TR and AP have the same slope and are parallel by 

the parallel slope property. So TRAP has only one 

pair of parallel sides and is a trapezoid by definition. 

PT (c ) 0 2 (d 0) 2 c2 d2 

RA (a c a) 2 (d 0) 2 (c) 2 d2 

c2 d2 

The nonparallel sides of the trapezoid have the 

same length. So trapezoid TRAP is isosceles by 

definition. 

7. 

y 

U (0, a 3) 

E Q 

(–a, 0) (a, 0) 

Show: EQU is equilateral 

EQ 2a 

EU (a 0) 2 a3) 

(0 2 

a2 a 3 2 

4a2 

2a 

UQ (a ) 0 2 0 a 

a2 a 3 2 

4a2 

2a 

EQ EU UQ 

EQU is equilateral 

x 

x 

 

3 2

8. Task 1: Given: A rectangle with both diagonals 

Show: The diagonals are congruent 

Task 2: y 

T (0, b) 

Task 3: Given: Rectangle RECT with diagonals RC 

and TE 

Show: RC TE 

Task 4: To show that two segments are congruent, 

you use the distance formula to show that they 

have the same length. 

Task 5: RC (a ) 0 2 (b 0) 2 a2 b2 

 

TE (a ) 0 2 (0 b) 2 a2 b2 

So RC TE because both segments have the same 

length. Therefore the diagonals of a rectangle are 

congruent. 

9. Task 1: Given: A triangle with one midsegment 

Show: The midsegment is parallel to and half the 

length of the third side 

Task 2: y 

T (0, 0) 

R (0, 0) 

Z 

c ( , ) 

2 d 2 

I (c, d) 

C (a, b) 

x 

E (a, 0) 

Y 

a + c ( , ) 

R (a, 0) 

Task 3: Given: TRI and midsegment YZ 

Show: YZ TR and YZ 1 

2 TR 

Task 4: To show that two segments are parallel, use 

the parallel slope property. The segments are 

horizontal, so to compare lengths, subtract their 

x-coordinates. 

Task 5: 

0 0 0 

Slope of TR a 

0 

a 

0 

Slope of YZ 0 

The slopes are the same, so the segments are parallel 

by the parallel slope property. 

TR a 0 a 

a c c a 

YZ 2 

2 

2 

1 a 

2 

1 

2 TR 

0 

 

a 

 

2 

d 

 

2 

d 

2 

 

a c c 

2 

2 

2 

d 2 

x 

So the midsegment is half the length of the third 

side. Therefore the midsegment of a triangle is 

parallel to the third side and half the length of the 

third side. 

10. Task 1: Given: A trapezoid 

Show: The midsegment is parallel to the bases 

Task 2: y 

One possible set of the coordinates for TRAP is 

shown in the figure. By the coordinate midpoint 

property, the coordinates of M are b 

2 , c 2 and of N 

a d c 

are , . 2 2 

Task 3: Given: Trapezoid TRAP with midsegment 

MN 

Show: MN TR 

Task 4: To show that the midsegment and bases 

are parallel, you need to find their slopes. 

Task 5: 

slope of MN 

c c 

2 

2 

0 

0 

a d a d b 

2 

 

2 

0 0 0 

slope of TR a 

0 

a 

0 

b 

2 

The slope of PA is d 

c c 0 

 

b 

d b 

0. The slopes are 

equal, therefore the lines are parallel. 

11. Task 1: Given: A quadrilateral in which only 

one diagonal is the perpendicular bisector of the 

other 

Show: The quadrilateral is a kite 

Task 2: 

y 

T (–a, 0) 

P (b, c) A (d, c) 

M N 

T (0, 0) 

I (0, b) 

M(0, 0) 

E (0, c) 

R (a, 0) 

K(a, 0) 

Task 3: Given: Quadrilateral KITE with diagonal 

IE, which is the perpendicular bisector of diagonal 

TK 

x 

x 




Show: KITE is a kite 

Task 4: To show that a quadrilateral is a kite, you 

use the distance formula to show that only two 

pairs of adjacent sides have the same length. 

Task 5: 

KI (0 ) a 2 b ( 0) 2 a2 b2 

IT (0 a)) ( 2 (b 0) 2 a2 b2 

TE (0 a)) ( 2 (c 0) 2 a2 c2 

EK (a ) 0 2 (0 c) 2 a2 c2 

 

Adjacent sides KI and IT have the same length and 

adjacent sides TE and EK have the same length, 

and because ⏐b⏐ ⏐c⏐ the pairs are not equal in 

length to each other. Therefore, KITE is a kite by 

definition. Therefore, if only one diagonal of a 

quadrilateral is the perpendicular bisector of the 

other diagonal, then the quadrilateral is a kite. 

12. Task 1: Given: A quadrilateral with midpoints 

connected to form a second quadrilateral 

Show: The second quadrilateral is a parallelogram 

Task 2: y 

( ) 

d 

, 

e 

2 2 

G 

Q (0, 0) 

D (d, e) 

P 

a ( , 0 

2 ) 

b + d 

, 

c + e 

2 2 

A (b, c) 

L ( ) 

U (a, 0) 

Task 3: Given: Quadrilateral QUAD with 

midpoints P, R, L, and G 

Show: PRLG is a parallelogram 

Task 4: To show that a quadrilateral is a 

parallelogram, we need to show that opposite 

sides have the same slope. 

c 

Task 5: slope of PR b 

e 

slope of RL d a 

c 

slope of LG b 

e e 

2 

0 

2 e 

slope of GP d a 

d a 

2 

d 

 

2 

a 

2 

c 

 

2 

 

 

e c e 

2 

2 

 

b 

 

2 

d 

c 

2 

 

 

c e c e 

2 

2 2 

 

b d a b d a 

2 

2 2 

d 

 

2 

b 

2 

b 

2 

c 

2 

0 

 

a b 

2 

a 

2 


a + b 

, 

c 

2 2 

R ( ) 

x 

Opposite sides PR and LG have the same slope, and 

opposite sides RL and GP have the same slope. So 

they are parallel by the parallel slope property, and 

PRLG is a parallelogram by definition. Therefore 

the figure formed by connecting the midpoints of 

the sides of a quadrilateral is a parallelogram. 

13. Task 1: Given: An isosceles triangle with the 

midpoint of the base connected to the midpoint of 

each leg, to form a quadrilateral 

Show: The quadrilateral is a rhombus 

Task 2: y 

Task 3: Given: Isosceles triangle ABC with 

midpoint of base, E, and midpoints of legs, D and 

F, connected to form quadrilateral ADEF. 

Show: ADEF is a rhombus 

Task 4: You need to show that all the sides of 

ADEF have the same length. 

Task 5: By the distance formula, 

2 

2 

AD a a 2 

 

h h 2 a 2 

2 

h 2 

 

2 

 

h 

2 

2 

h a 2 

2 

h 2 

AF 3 a 

 

2 

a 2 

 

 

DE a a 

2 

 

2 

 

0 h 

2 2 

2 

2 

 

a 2 

h 2 2 

 

2 

2 

EF 3 a 

 

2 

a 2 

h 2 

2 

0 a 2 

h 2 

 

a 

D ( , 

h 

2 2) 

a2 h2 

 

2 

a2 h2 

 

2 

a2 h2 

 

2 

a2 h2 

 

2 

A (a, h) 

B (0, 0) E (a, 0) 

3a 

F ( , 

h 

2 2) 

C (2a, 0) 

AD AF DE EF by the transitive property 

of equality. 

Therefore, ADEF is a rhombus by the definition of 

a rhombus. 

x


1. False. The quadrilateral could be an isosceles 

trapezoid. 

2. true 

3. False. The figure could be an isosceles trapezoid 

or a kite. 

4. true 

5. False. The angles are supplementary but not 

necessarily congruent. 

6. False. See Lesson 13.5, Example B. 

7. true 


9. congruent 

10. the center of the circle 

11. four congruent triangles that are similar to the 

original triangle 

12. an auxiliary theorem proven specifically to 

help prove other theorems 

13. If a segment joins the midpoints of the 

diagonals of a trapezoid, then it is parallel to the 

bases. 

14. Angle Bisector Postulate 

15. Perpendicular Postulate 

16. Assume the opposite of what you want to 

prove, then use valid reasoning to derive a contradiction. 

17a. Smoking is not glamorous. 

17b. If smoking were glamorous, then this smoker 

would look glamorous. This smoker does not look 

glamorous, therefore smoking is not glamorous. 

18. False. The parallelogram is a rhombus. 

19. False. Possible counterexample: 

140 

20 

40 40 

140 20 

20 20 

20. False; x 360° 2b so x 90° only if 

b 135°. 

x 

a a 

b 

x 540 (180 2b) 

x 360 2b 

b 

21. True (except in the special case of an isosceles 

right triangle, in which the segment is not defined 

because the feet coincide). 

Given: IsoscelesABC with 

altitudes AD and BE;AC BC;ED 

Show: ED AB 

EAB DBA by the Isosceles 

Triangle Theorem. AEB 

BDA by the definition of 

altitude and by the Right Angles 

Are Congruent Theorem. 

AB AB by the reflexive property of congruence, 

so AEB BDA by SAA. AE BD by 

CPCTC. By the definition of congruence, AC BC 

and AE BD, so EC DC by the subtraction 

property of equality and the Segment Addition 

Postulate. By the division property of equality, 

EC 

AE 

D 

C 

E 

D 

X 

A 

B 

C ,so BD ED divides the sides of ABC 

proportionally. Therefore ED AB by the 

Converse of the Parallel/Proportionality Theorem. 

22. true 

Given: Rhombus ROME with diagonals RM and 

EO intersecting at B 

E 

R 

2 

1 

4 B 

3 

O 

M 

Show: RM EO 

Because diagonals bisect the angles in a rhombus, 

the diagonals are angle bisectors. 

Statement Reason 

1. 1 2 1. Rhombus Angles 

Theorem 

2. RO RE 2. Definition of rhombus 

3. RO RE 3. Definition of 

congruence 

4. RB RB 4. Reflexive property of 

congruence 

5. ROB REB 5. SAS Congruence 

Postulate 

6. 3 4 6. CPCTC 

7. 3 and 4 are 7. Definition of linear 

a linear pair pair 

8. 3 and 4 are 8. Linear Pair Postulate 

supplementary 

9. 3 and 4 are right 9. Congruent and 

angles Supplementary 

Theorem 

10. RM EO 10. Definition of 

perpendicular 




23. true 

D E 

1 3 

A F B 

BAD DCB by the Opposite Angles Theorem. 

m1 1 

 

2 mBAD 1 

2 mDCB m2 by 

the definition of angle bisector. AB DC by the 

definition of parallelogram. 2 and 3 are 

supplementary by the Interior Supplements 

Theorem. 

1 and 3 are supplementary by the 

substitution property. 

AE FC by the Converse of the Interior 

Supplements Theorem. 

24. Use the Inscribed Angle Theorem, the addition 


mP mE mN mT mA 

1 

2 mTN mAT mPA mEP mNE .Because 

there are 360° in a circle, mP mE mN 

mT mA 180°. 

25. T 

R 

Assume mH 45° and mT 45°. Use the 

Triangle Sum Theorem, the substitution 

property, and the subtraction property to get 

mH mT 90°, which creates a 

contradiction.Therefore mH45° or mT45°. 

26. Y 

M 

T R 

Use the definition of midpoint, the Segment 

Addition Postulate, the substitution property, and 

the division property to get MY 

 

TY 1 

2 and YS 

 

YR 

1 

2 . 

Then use the reflexive property and the SAS 

Similarity Theorem to get MSY TRY. 

Therefore, MS 1 TR 2 by CSSTP and the 

multiplication property and MS TR by 

CASTC and the CA Postulate. 

27. D Y 

T 

1 3 

Z 

4 2 

O P 

Use the Line Postulate to extend ZO and DR. Then 

use the Line Intersection Postulate to label P as the 

intersection of ZO and DR . DYR POR by the 


2 

S 

C 

H 

R 

SAA Theorem; thus DY OP by CPCTC. Use 

the Triangle Midsegment Theorem and the 

substitution property to get TR 1 

 

2 (ZO DY). 

Also, use the Triangle Midsegment Theorem to 

get TR ZO. 

28a. The quadrilateral formed when the 

midpoints of the sides of a rectangle are connected 

is a rhombus. 

28b. 

H 

Use the Right Angles Are Congruent Theorem and 

the SAS Congruence Postulate to get AEH 

BEF DGH CGF. Then use CPCTC to 

prove that EFGH is a rhombus. 

29a. The quadrilateral formed when the 

midpoints of the sides of a rhombus are connected 

is a rectangle. 

29b. 

I 

L 

D 

A 

P 

O 

M 

Y X 

By the Triangle Midsegment Theorem both PM 

and ON are parallel to LJ , and both PO and MN 

are parallel to IK. Because LJ and IK are 

perpendicular, we can use corresponding angles 

on parallel lines to prove that the lines that are 

adjacent sides of the quadrilateral are also 

perpendicular. 

30a. The quadrilateral formed when the midpoints 

of the sides of a kite are connected is a rectangle. 

30b. By the Triangle Midsegment 

I 

Theorem both PM and ON are 

parallel to LJ , and both PO and 

MN are parallel to IK. Because 

L 

P M 

J 

LJ and IK are perpendicular, we can 

use the CA Postulate to prove that 

O N 

the lines that are adjacent sides of 

the quadrilateral are also perpendicular. 

K 

31. Use the Line Postulate 

to construct chords DB and 

C 

AC. Then use the Inscribed 

Angles Intercepting Arcs 

P 

B 

Theorem and the AA 

Similarity Postulate to get 

APC DPB. Therefore, 

A 

O 

by CSSTP and the multiplication 

property, AP PB DP PC. 

K 

G 

E 

N 

C 

B 

J 

F 

D

Geo_Book_Answers

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?