Discovering Geometry An Investigative Approach [PDF]

q. 120°. ANSWERS TO EXERCISES. 105. A n sw ers to Exercises. A n sw ers to Exercises. CHAPTER 9 • CHAPTER. CHAPTER 9

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Answers to Exercises CHAPTER 9 • CHAPTER

9

CHAPTER 9 • CHAPTER

LESSON 9.1

b a b a c

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.

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°. T1 23 . T2

ANSWERS TO EXERCISES

105

Answers to Exercises

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  21ab  (b  a)2 c 2  2ab  b 2  2ab  a 2 c 2  a2  b2

19. 54 cm2 20. 3632.4.3.4 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°.

Answers to Exercises

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 cm2 14.  14.1 ft 15.  17.9 cm2 16. 102 m 17a.  1442 cm2 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.

106

ANSWERS TO EXERCISES

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. 1 0 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  21a. 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

A

T

M C

U

24. 790 square units 25.

26. 19

B

USING YOUR ALGEBRA SKILLS 9

1.  6 2. 5 3. 18 2 4. 147 5. 8 6. 2 3 7. 3 2 8. 2 10 9. 5 3 10.  85 11. 4 6 12. 24 13. 12 5 14. 28 15. 6 23 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 3 2.

18. Possible answer: A right triangle with legs of lengths 1 and 3 units has a hypotenuse of length  units. 10 10

1

3

A right triangle with legs of lengths 6 and 2 units has a hypotenuse of length 40   10  10 10 .   2 40 2 6

19. Possible answer: A right triangle with legs of lengths 2 and 3 units has a hypotenuse of length 13 units. 

1

2 1

18

3 13

3

3

5

2

12 . Because y is twice as 20. x  3, y   long as x, y  2x, so  12  2 3. 2 2 2 21. 2  3  7 7

3

1 2

2 20

2

4

ANSWERS TO EXERCISES

107

Answers to Exercises

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 5.   2

LESSON 9.3

72 2 cm 13 cm 10 cm, 5 3 cm 10 3 cm, 10 cm 34 cm, 17 cm 72 cm 12 3 cm 50 cm, 100 cm 16 cm2 1 1 10.  ,  2  2  1. 2. 3. 4. 5. 6. 7. 8. 9.



30°

x

3 2x



45°

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 12   3 2  22 and 42   48 2  82.

Answers to Exercises

16. c 2  x 2  x 2 Start with the Pythagorean Theorem. c 2  2x 2 Combine like terms. c  x 2 Take the square root of both sides. 17. 169 3 m2  292.7 m2 18. 390 m2 19. 3   1

2

3

x

2

x

15°

45° x

20. Construct an isosceles right triangle with legs of length a, construct a 30°-60°-90° triangle with legs of lengths a and a  3 , and construct a right triangle with legs of lengths a  2 and a  3.

1 2

4

3

3

8

a

1 2

2

3 4

13. possible answer:

4

4

14. Possible answer: Use 22  12   5 2 and 62  32   45 2. 5

a 3

1

45

5

3

4 3

6

15a. CDA, AEC, AEB, BFA, BFC 15b. MDB, MEB, MEC, MFC, MFA

108

a 5

a 2

21. 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. 73 22. 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°.

1 2

a 3

Areas: 4.5 cm2, 8 cm2, 12.5 cm2. 4.5 

1

2

2a

a

a

3

4

a 2

ANSWERS TO EXERCISES

24. 80°

2

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  27 3  180 cm2  2 226.8 cm ; surface area of cylinder  78 cm2  245.0 cm2 36 cm  20.8 cm 7.  3  8. 48.2 ft; 16.6 lb 9a. 160 ft-lb 9b. 40 lb 9c. 20 lb 10. about 4.6 ft 2 2 2 2 11. 2

2

4

2

2 2

2

2

12.

13. 12 units 14. 18 2 cm  and CB . ABC  ADC  15. Draw radii CD 90°. For quadrilateral ABCD, 54°  90°  mC    126° but 90°  360°, so mC  126°. BD 126°  226°  360°. 16. 4, 4 3 17. SAA 18. orthocenter 19. 115°  20. PO

2

4

2

2

2

4

4

Answers to Exercises

2

2

ANSWERS TO EXERCISES

109

LESSON 9.5

1. 5 units 4. 354 m 7. rectangle

2. 45 units 5. 52.4 units

3. 34 units 6. isosceles

y 10 A

8

B

6 4 2

D 2

C

4

6

x

8 10

8. parallelogram



E2

4

x

F

H 8

Answers to Exercises

6

G

10

9. kite y L 4 2

K 8

x

I 2 2

J

10. square y

120°

8 P M



3 1 19. 2, 2 20. k   2 , m   6 6 12 21. x  , y   3 3   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)  180° angle of a regular hexagon is    120°, and 6 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.

y 2 2

11b. Circle A: (x  1)2  ( y  2)2  64; Circle B: x 2  (y  2)2  36 11c. (x  h)2  (y  k)2  (r)2 12. (x  2)2  y 2  25 13. Center is (0, 1), r  9. 14. (x  3)2  (y  1)2  18 15a.  14 units 15b.  176  4 11 units 2  (y  y )2  (z  z )2 15c. (x  x )       2 1 2 1 2 1 16.  86.5 units 17. 14 units 18. n  (n  2)  3  (n  3)  (n  1)

60° 60°

120°

4

120°

O N 4 2

11a.

2

60° 60°

x

4

y

y

(x, y) (1, –2)

x

(x, y) Circle A

110

ANSWERS TO EXERCISES

(0, 2) x

Circle B

120°

LESSON 9.6

18 cm2  56.5 cm2 (8  16) m2  9.1 m2 456 cm2  1433 cm2 120 cm2  377.0 cm2 32  32 3  m2  45.1 m2 (25  48) cm2  30.5 cm2 64 7. 3  16 3 cm2  39.3 cm2 8. (240  18) m2  183.5 m2 9. 102 cm 10. Possible proof:  and AN  Given: Circle C with tangents AM   A  Show: AM N 1. 2. 3. 4. 5. 6.





14. 12  6 3  cm  22.4 cm 15.  77 cm  8.8 cm 16. 76 cm 17. Inscribed circle: 3 cm2. Circumscribed circle: 12 cm2. 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:

M

C A

, NC , and AC  to the diagram as shown. Add MC 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. Solving for AM and AN, AM   (AC)2  (MC)2 and 2 2  and NC  are AN   (AC)  (NC) . Because MC 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,   AN . AM 11. 18 m 12. 324 cm2  1018 cm2 13.  3931 cm2

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. 10 in.

10 in.

d  20 in.

10 in.

21b. Possible answer: Measure the circumference with string and divide by . 5 22. 6

ANSWERS TO EXERCISES

111

Answers to Exercises

N

CHAPTER 9 REVIEW

1. 2. 3. 4.

20 cm 10 cm obtuse 26 cm 3 1  5. 2, 2 6. 1, 1 2 2   7. 200 3 cm2  346.4 cm2 8. d  12 2 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; 821 min 18. yes 19. 29 ft 20.  45 ft 21. 50 mi 22.  707 m2 23. 6 3 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.

Answers to Exercises

 



7 27. 9 28. The quarter-circle gives the maximum area. Triangle:

s ____ 2

45°

s



45°

s ____ 2

s 1 A  2   2 

s

2

s  4  2 

Square: _1 s 2

_1 s 2

1 1 s2 A  2s  2s  4 Quarter-circle: s

2s ___ ␲

1 s  4  2r 2s r   1 2s 2 s 2 A  4    

 

s 2 s2     4

Extra

30°

29. 1.6 m 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.

112

ANSWERS TO EXERCISES

31. 4; 0; 10. The rule is n2 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 x 2. 36. true 2 37. false; AB  x (x 2   (y2   y1) 2 1)   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 47. See flowchart below.

48.

N 8-ball W

B

E Cue ball A S

49. 34 cm2; 22  4 2  27.7 cm 40 50. 3 cm2  41.9 cm2 51. about 55.9 m 52. about 61.5 cm2 53. 48 cm 54. 322 ft2

Answers to Exercises

47. (Chapter 9 Review) 1

ABCD is a rectangle Given

2

ABCD is a parallelogram Definition of rectangle

4

D  B Definition of rectangle

3   DA  CB

Definition of parallelogram

5

DAC  BCA AIA Conjecture

7

ABC  CDA SAA Congruence Conjecture

6   AC  AC

Same segment

ANSWERS TO EXERCISES

113

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