Back to Lesson 12-1

Chapter 12

9. a. First, apply a size transformation of magnitude 0.9 and center O, where O is a point not on Arim. Then apply a glide reﬂection so that the image created is facing Arim and on the line.

Similarity Lesson 12-1

(pp. 718–725) A

1.

b. No, because that will produce a ﬁgure that is the composite of a size transformation of positive magnitude and a rotation, not the reﬂection required to change orientation.

A B B

O

C

C

DE = 10. a. Center E, k = ___ 0.5

CE CE = b. Center E, k = ___ 2 DE SS + 8 OF  = __ 9= ____ 11. 1.5 = ______ 6 8 OF

2. O 3.

A

SS + 8 = 12 SS = 4 cm B

O C

FS  = ___ 1.5 FS FS = ___ 1.5 4

C

B

F S = 6 cm A

12. a. O

4. 48

50 = F X  = __ b. ____ 2.5

OA = 5. k = ___ 0.75 OA

c.

A

A

13. By the theorem on page 720, k(AB) = A'B' and k(BC ) = B'C'.

D B B

C

20 FX X O = ____ 2.5 XO 18 = ___ 2.5 XO 18 = XO = ___ 7.2 2.5

k(AB) k(BC)

A'B' = _____ = ___ AB . Therefore, ____

D

B'C'

C

6. a. __4 · 9 = 12 cm 3

b. m∠I = m∠O = 180 - 55 - 85 = 40 7. angle measure; betweenness; collinearity 8.

BC

14. Construct coordinate axes so that the center of the size transformation is at the origin, and let (a, b) be a point on the ﬁgure. Then applying Sk to the ﬁgure maps (a, b) to (ka, kb), and applying ka __ , kb = (a, b). Thus, the S__K1 maps (ka, kb) to __

(k

k

)

composite is the identity transformation. 15. a. (x – 4)2 + (y + 2)2 = 25

A

A

O

b. The center of the image is (3 · 4, 3 · –2) = (12, –6) and the radius is 3 · 5 = 15. The equation is (x - 12)2 + (y + 6)2 = 225. c. (x - 4)2 + (y + 2)2 = 225

237

Geometry Solution Manual

Chapter 12, Lesson 12-1

Back to Lesson 12-1

16. a. M  = (4 · –1, 4 · 3, 4 · 6) = (–4, 12, 24) b. N  = (4 · 2, 4 · 7, 4 · –6) = (8, 28, –24) c. M N   = √(8 - –4)2 + (28 - 12)2 + (–24 - 24)2

= √ 122 + 162 + (–48)2  = 52 = √2704 MN = √ (2 - –1)2 + (7 - 3)2 + (–6 - 6)2 2 + 2 + = √3 4 (–12)2 = √ 169 = 13. So M N  = 4 · MN. 17. the circle 18. Because ABD and ACD are both right triangles, AB = AC is given and AD = AD by the Reﬂexive Property of Equality, ABD  ACD by the HL Congruence Theorem. Therefore, by the CPCF Theorem, BD = CD, and because there are two pairs of congruent, adjacent sides, ABDC is, by deﬁnition, a kite. 19. 24 · 360° = 8640° x2 __

x2 __ x x2 __ 3 3·x·x x 6 __ __ ______ __ 20. __ x = 6 ÷ 3 = 6 · x = 3·2·x = 2 __ 3

190 - 30 = 21. a. _______ 8 cm 20

b. 30 + 12 · 8 = 126 cm c. Answers vary. Sample: no, because growth typically occurs in short bursts that are not constant throughout life and growth slows greatly by age 20. 22. The image of the circle (x – h)2 + (y – k)2 = r 2 under a size transformation with center (0, 0) and magnitude a is (x – ha)2 + (y – ka)2 = (ar)2. The image of (x – h)2 + (y – k)2 = r 2 under a size transformation with center (h, k) and magnitude a is (x – a)2 + (y – a)2 = (ar)2.

238

Geometry Solution Manual

Chapter 12, Lesson 12-1

Back to Lesson 12-2

Similarity

–1

Lesson 12-2

(pp. 726–730)

–1

y

17. The magnitude of both S and T is k, so the side lengths of ABC are multiplied by k in each size transformation. Thus, by the SSS Congruence Theorem, the two images are congruent.

3. ratios; equal 4. a. 12, 11 b. x, y c. x 5. a. xy = 12 · 11 = 132 b. Answers vary. Sample: x = 2, y = 66, and x = 4, y = 33 60 1 = __ 6. ___ x 1.9 x = 1.9(60) = 114 in. b d d = __ __ b. c ab

8. EF BC AB = ___ 9. ___ FG GH 15 12 = __ __ x 10

20. Let x = cube’s edge length. x 3 = 8h

12x = 10 · 15 x = 12.5

3 3 8h = 2 · √ h x = √

(

11. By the Means-Extremes Property, wz = xy. x _z Divide both sides by wy to get __ w = y. 1944 = 12. ____ 0.75 = __3 4

None of the given sizes have this ratio, so the answer is E. 36 A'B' = ____ ____ 4 A'B'

(A'B' )2 = 144 A'B' = 12 AC = ____ BC = ____ AB Sample: ____ A'C'

B'C'

2 __

2 2 3 3 S.A. = 2 · √ h · 6 = 24 √ h or 24h3

u = __ w __ u w _y __ _y 10. __ v x , v = z, x = z

2592

18. Sample: Suppose A = (0, 0) and B = (1, 1). Then the distance between A and B is √2. Also, R(A) = (0, 0) and R(B) = (12, 2), and the 37 , distance between R(A) and R(B) is 2 √ so R does not preserve distance. 19. Assume that there are only ﬁnitely many numbers divisible by 2. Let N be the greatest number divisible by 2. Then, 2N is also a number that is divisible by 2, and 2N is greater than N. This contradicts the assumption that there is a greatest number that is divisible by 2, so there must be inﬁnitely many numbers that are divisible by 2.

a = __ c 7. a. __

15.

y

w = _z __ = _x __ w _ 16. Answers vary. Sample: __ y x, w z, z = x

2. quotient

14.

√

± 5 x = _______ 2

1. D

13.

√ +

± 1 4 x = __________ 2

A'B'

)

( )

21. a. The area is multiplied by k2. b. The area is multiplied by k2. 22. a. Two ﬁgures are congruent if each of them is the image of the other under an isometry. b. An isometry is a transformation that can be created by composing reﬂections. 23. The harmonic mean of two numbers is the reciprocal of the arithmetic mean of the reciprocals of the two numbers. One application is to calculate the average cost of shares of stock purchased over a period of time.

FG · √2 · √ 2 = BA 2 · FG = BA FG = __ 1 ___ 2 BA +1 x____ 1 = __x 1

x(x + 1) = 1 x2 + x - 1 = 0 239

Geometry Solution Manual

Chapter 12, Lesson 12-2

Back to Lesson 12-3

Similarity Lesson 12-3

18. Let h = horizontal length. 7h = __ 10 12

(pp. 731–737)

1. a. The letters are similar because the d is the image of the b under a reﬂection and size transformation. 36 = __ 3 b. __ 48 4 48 = __ 4 __ c. 36 3

12 h = 10 · __ ≈ 17.14 7

19. Yes, the rectangles must be similar because corresponding angles are congruent (all angles are right angles) and corresponding lengths are 3 cm 2 cm = ____ . proportional _____

( 4 ft

2. Two ﬁgures are similar if and only if there is a composite of size transformations and reﬂections mapping one onto the other.

)

BC AB = ____ ____ AB  B C  15 = ____ 20 __ 24 B C 

20.

15 · B C  = 20 · 24

3. a. false

20 · 24 = 32 B C  = ______ 15

b. true 4. Yes, the ﬁgures are similar because LMNK = S2.3  rm(ABED). 5. GF = 6.8 in.; LK = 2.3 · 6.8 = 15.64 in. 6. HJ = 230 ÷ 2.3 = 100 mm; DE = 100 mm 7. ∠ABE and ∠LKN

21. 10(x - 2) = 14x 10x - 20 = 14x –20 = 4x x = –5 ___

22. Because ABC is a right triangle, BC is the diameter of the circle. 0 = __ 0 + 3 _____ 3 __ , 7+ ,7 center = _____ 2 2 2 2 radius = distance from center to (0, 0) = 3 2 + __ 7 2 = __ 29   __

(

8. CM = 2.3 · 4 = 9.2

) (

)

√( 2 ) ( 2 ) √ 2

9. yes

The equation is x - __32

(

10. yes 11. No; Figure I and Figure III in Question 4 are similar, but AB ≠ LK. 12. No; Figure I and Figure II in Question 4 are similar, but oppositely oriented. 13. PA = 4.8

) + ( y - __72 ) 2

2

29 = __ . 2

23. r = __d = 11.95 cm 2

leather thickness: 1.5 mm = 0.15 cm 11.95 - 0.15 3 ≈ 96.28% ratio of air to volume: __________ 11.95 ratio of leather to volume: 100% - 96.28% = 3.72%

(

)

24. 8; rectangle, rhombus, isosceles trapezoid, square, parallelogram.

14. m∠O = 80; m∠A = 119 15.

6 ft

5.5 3.7 = ___ ____ XZ 2.25

3.7 · XZ = 2.25 · 5.5 XZ ≈ 3.34 16. 0.93 · 100 = 93 in. 17. a. Let x = actual distance from Houston to San Antonio. 5 = ___ x __ 6 240

x = 200 mi 6 in. = _____ 1 in. = __________ 1 1 = _______ b. ______ 240 mi

240

40 mi

40 · 5280 ·12

Geometry Solution Manual

25. true

2,534,400

Chapter 12, Lesson 12-3

Back to Lesson 12-3

26. There are comparable properties for similarity.

Proof of the Reﬂexive Property of Similarity: Let F be any ﬁgure. We need to show that F is similar to itself. To do that, reﬂect F over the line m, apply size transformation S with center C and magnitude k, then apply size transformation S' with center C and magnitude __1 , and reﬂect k

that image back over m. rm  S_k1  Sk  rm (F) = F, so there is a composite of reﬂections and size transformations mapping F onto itself. Thus by the deﬁnition of similarity, F ∼ F. Proof of the Symmetric Property of Similarity: If F ∼ G then there is a composite of reﬂections and size transformations that maps F onto G. To map G back onto F, reﬂect G back over the same lines and apply size transformations with reciprocal magnitudes in reverse order. This means F is the image of G under a composite of reﬂections and size transformations, so G ∼ F. Proof of the Transitive Property of Similarity: If F ∼ G then there is a composite of reﬂections and size transformations that maps F onto G. If G ∼ H then there is a composite of reﬂections and size transformations that maps G onto H. The composite of these composites is another composite of reﬂections and size transformations which maps F onto H, so F ∼ H.

241

Geometry Solution Manual

Chapter 12, Lesson 12-3

Back to Lesson 12-4

Similarity Lesson 12-4

10. a. Let x = area of ABCDE. (pp. 738–744)

1. Answers vary. Sample: Let R be a 3 cm by 5 cm rectangle. R has area 15 cm2 and perimeter 16 cm. Let R  be a 30 cm by 50 cm rectangle. R  has area 1500 cm2 and perimeter 160 cm. Thus, when each dimension is multiplied by 10, the area of the image is multiplied by 100 and the perimeter is multiplied by 10. 2. Answers vary. Sample: Let C be a 2 cm by 2 cm by 2 cm cube. C has volume 8 cm3 and surface area 24 cm2. Let C  be a 0.6 cm by 0.6 cm by 0.6 cm cube. C  has volume 0.216 cm3 and surface area 2.16 cm2. Thus, when each dimension is multiplied by 0.3, the volume is multiplied by 0.027 and the surface area is multiplied by 0.09. 3. yes

( )

2

= __9 4

b. 1.986 49 7 2= __ __ 16 4 3 343 7 __ = ___ 64 4

( ) ( )

OC 16. ___

= __1

(OA )

2

2

OC 2 = __ 1 ____ 2 102

OC 2 = 50  = 5 √ OC = √50 2

15 __ 8 __ 5 __ 8

15 __ = __ ·8=3 8

5

The ratio of their volumes is 33 = 27. b. The ratio of their surface areas is 32 = 9. 7. a. __5 9 __ b. 5 9

17. No; because T only changes one coordinate, it is not a size transformation. 10 = __ h 18. __ 16

50 500 ___ h = 16 = 31.25 ft

19. a plane 20. a.

5 __ 9

( )

2

9. A = πr 2, therefore A = π(r  )2 = π(15r)2 = 225πr 2 = 225A

Geometry Solution Manual

( __23)

3

8 = __ 27

8 The second sphere’s volume is __ times the 27 ﬁrst sphere’s volume.

25 = __ 81

8. V = wh, therefore V  = wh = (k )(kw)(kh) = k3 wh = k3V

242

d · √2 = 8 8 √ 2 8 = ____ = 4 √ 2 d = ___

k ≈ 3.946

e. 1

c.

 14. k 2 = 2, so k = √2 = Let d diagonal of smaller square.

2

d. __7 4

6. a.

13. Increasing the length by 40% is the same as multiplying the area by 1.4. Increasing the width by 20% is the same as multiplying the area by 1.2. New ﬂoor area = 2100 · 1.4 · 1.2 = 3528 ft2

3 __ 2

4

c.

12. 33 = 27 times

2 4013 15. a. k 3 = ____ 512  4013 ≈ 1.986 k = 3 ____ 512

5. a. __7 b.

()

√ 2

4. a. __3 2 __ b. 3 2 c.

192 = __ 64 8 2 = __ ___ x 25 5 25 · 192 ______ x = 64 = 75 cm2 NJ NJ b. ___ = ___ = __8 5 6 AE 48 __ NJ = 5 = 9.6 cm √ r (smaller circle) 25 11. _____________ =____= __5 7 r (larger circle) √ 49

b.

( __23 )

2

= __4 9

The second sphere’s surface area is __49 times the ﬁrst sphere’s surface area.

Chapter 12, Lesson 12-4

Back to Lesson 12-4

21. a. Two ﬁgures F and G are congruent ﬁgures if and only if G is the image of F under an isometry. b. Two ﬁgures are oppositely congruent if they are congruent and have opposite orientation. 22. a. true b. False. The shapes might not intersect. Or, if one rectangle is rotated, the intersection can be a triangle, trapezoid, octagon, or other shape. 23. a. Answers vary. Sample: The ratio of similitude between a model toy car and a car is 40. b. Answers vary. Sample: If the model toy car weighs about 600 g, then we estimate the car will weigh 600 · 403 = 38,400,000 g = 38,400 kg. c. Answers vary. Sample: No, a car actually weighs signiﬁcantly less.

243

Geometry Solution Manual

Chapter 12, Lesson 12-4

Back to Lesson 12-5

1 18. 60 · __

Similarity Lesson 12-5

(pp. 745–749)

1. He was the tallest man on record, standing 8 ft 11.1 in. tall 2. 12; 1728

(16 ) ≈ 0.234 lb 19. 115.5 · ( __2 ) = 18.48 5 2

2

20. a. Two ﬁgures F and G are not similar if there is no composite of size transformations and reﬂections that map one onto the other. b.

1 ____ 3. __ ; 1 12 1728

4. 144; 1443 = 2,985,984 5. The giantess is 53 or 125 times the weight of the woman. 6. The giantess’s index ﬁnger is 5 times the length of the woman’s index ﬁnger. 7. The giantess’s footprint is 52 or 25 times the area of the woman’s footprint.

21. Size transformations of magnitude k = 1 can be composites of reﬂections. 22. ﬁgure not to size A

A

8. The giantess’s waist circumference is 5 times the woman’s waist circumference. 9. The amount of weight her legs can support is proportional to the cross-sectional area of her bones and muscles; three times her own weight is just too much for her bones and muscles to support. 10. An elephant has a disproportionately greater volume and therefore a greater amount of weight to support than a mosquito does relative to the cross-sectional area of its legs.

C

B B

P

C

23. If, in two triangles, two sides and the included angle of one are congruent to two sides and the included angle of the other, then the triangles are congruent. 24. m∠YXZ = __1 (38) = 19, m∠XZY = __1 (160) = 80, 2

1 ___ 11. ____ , 1

2

m∠XYZ = 180 - 19 - 80 = 81

1000 100

12. false 30 = 13. k = __ 1.5; 6 · (1.5)3 = 20.25 kg 20

14. 12.99 - 8 = 4.99 14 4.99 · __

( 12 )

58 15. __

( 42 )

2

2

+ 8.00 ≈ \$14.79

· \$1500 ≈ \$2860.54

7 ft 6 in. = _____ 90 in. = __ 9 16. ______ 8 6 ft 8 in. 80 in. 3 9 250 __ ≈ 356 lb 8

( )

28  ≈ 0.529 √___ 100 100 ___ ≈ 1.89 Earth to Mars: k = √ 28

17. a. Mars to Earth: k =

b. Mars to Earth: 0.5293 ≈ 0.148

Earth to Mars: 1.893 ≈ 6.75

244

Geometry Solution Manual

Chapter 12, Lesson 12-5

Back to Lesson 12-5

25. a.

Record Holder Yang Lian Qiu Hongxia Chen Yanqing Liu Haixia Oxana Slivenko Svetlana Podobedova

Ratio 4.521 4.264 4.328 4.079 4.000 3.813

b. For the most part, the ratio decreases as the weight class of the weightlifter increases. c. Answers vary. Sample: Weightlifters in higher weight classes can lift more than lifters in lower weight classes, but the lifters in lower weight classes are able to lift proportionately more than lifters in higher classes. This is because of The Fundamental Theorem of Similarity: if Yang weighs m pounds and can lift nm pounds (n > 1), and Chen weighs 2 __ km pounds (k > 1), then Chen can lift k3 mn pounds, because the amount one can lift is proportional to the cross-sectional area of one’s bones and muscles. So Yang’s ratio n , thus Yang’s ratio is is n and Chen’s ratio is __ k greater than Chen’s.

245

Geometry Solution Manual

Chapter 12, Lesson 12-5

Back to Lesson 12-6

Similarity Lesson 12-6

(pp. 750–755)

1. a. Yes, it would be similar. The conversion factor to change the measure from feet to inches or centimeters is the ratio of similitude in each case. b. No, it would not be congruent to either triangle because the lengths are not in the same units. 2. If the three sides of one triangle are proportional to three sides of a second triangle, then the triangles are similar. 3. a. true

b. false

YE ___ 4. ___ , ES

QR PR = ___ ___ MO NO QR 20 = ___ __ 15 13 52 __ QR = 3

triangles is __12 , so the ratio of the areas is __14 . 18. 0.5 L = 500 mL 500r 3 = 750, so r ≈ 1.1497 and r 2 ≈ 1.31. About 1.31 times more paper is needed for the label on the larger bottle. 19. False. Suppose the edges change from 1 cm to 6 cm. Then the area increases from 1 to 36.

b. No; orientation is not reversed in size changes.

( __3553 ) - ( __2153 ) = __2853 √

a 2 = __ a2 = 1  + ___ 21. __1 a √2 2a2 + 2a2 + __

2

2

11. 3 ÷ __1 = 18; 4 ÷ __1 = 16; 6 ÷ __1 = 18 4

17. The ratio of similitude for the sides of the two

20. a. a size transformation with factor k 2

2 +  9. √15 362 = 39

6

Midsegment of a Triangle Theorem, so the

16. One of the quadrilaterals could be a rectangle, and the other could be a parallelogram that is not a rectangle.

8. √ 102 - 82 = √ 36 = 6

2

2

and DC = __12 BC. Further, DE = __12 AB by the

Theorem, BCA ∼ ECD.

m∠N = 180 - 14.25 - 53.13 = 112.62 m∠P = 53.13 m∠Q = 112.62 m∠R = 14.25

10.

15. Because D and E are midpoints, EC = __1 AC

ratio of similitude __12 . Then, by the SSS Similarity

Yes; 4.5; BCA ∼ FED MN MO PQ 20 __ = ___ 15 4 16 __ PQ = 3

14. 450% · 3.4 = 4.5 · 3.4 = 15.3 The angles will all be 60°, and the side lengths will be 15.3 cm.

three sides of the triangles are proportional with

MA AD 9 ≠ ____ 11.5 5. __ ; not similar 10 15.3 13.5 = FE = ____ 6. ___ 4.5 3 BC 31.5 = DE = ____ ___ 4.5 7 AC 36 = DF = __ ___ 4.5 8 BC PQ PR = ___ 7. ___

13. By the SSS Similarity Theorem, you can construct a triangle using a ruler with side lengths 12, 9, and 4.8 that will be similar to the piece of land. Then the measure of the angles can be determined by measuring them with a protractor, and the corresponding angle measures will be congruent because the triangles are similar.

3

False, because the ratios are not equal.

(

)

√2  a = ___ 9a2 2 + __ 2a 4 4

2

(

2

)

2

22. SSS, ASA, SAS, HL, AAS, SsA

12. By the distance formula, JK = 3 √ 2 , KB = 6, and JB = 3 √ 10 . Also, LI = √ 2 , IE = 2, and

23. Answers vary. Sample: Yes, the two triangles can be similar. One possible ratio of similitude

10 . Therefore, the ratio of similitude LE = √ is 3, and by SSS Similarity, JKB ∼ LIE.

48 and the remaining side of the would be __ 7

is __87 , so the other side of the ﬁrst triangle 72 . second triangle is __ 7

246

Geometry Solution Manual

Chapter 12, Lesson 12-6

Back to Lesson 12-7

BC AB = ___ 8. We are given ∠B  ∠Y and ___ . Applying

Similarity Lesson 12-7

(pp. 756–763)

1. Apply a size change to a triangle to show its image is congruent to its preimage. 2. The SSS Similarity Theorem, the AA Similarity Theorem, and the SAS Similarity Theorem: The SSS Similarity Theorem states that 2 triangles are similar if the 3 sides of the ﬁrst triangle are proportional to 3 sides of the second triangle. The AA Similarity Theorem states that 2 triangles are similar if 2 angles of the ﬁrst triangle are congruent to 2 angles of the other. The SAS Similarity Theorem states that 2 triangles are similar if the ratios of 2 pairs of corresponding sides are equal and the included angles are congruent. 3. a. yes b. AA Similarity Theorem c. IJM ∼ FLM 4. a. yes b. SAS Similarity Theorem c. NPR ∼ NOQ 5. a. yes b. SSS Similarity Theorem c. JKL ∼ XWV 6. a. no b. BZY cannot be similar to either CZY or BCY because it shares only one congruent angle with each, and CZY cannot be similar to BCY because they only share one congruent angle. 7. 5', 8" = 68", 3' = 36", 13', 4" = 160" 68 = ___ x __ 36 160 68 · 160 x = ______ ≈ 302 in. ≈ 25 ft 36

XY YZ BC AB ___ ___ = a size change of k = to XYZ, we have XY YZ

X Y  = k · XY = AB and Y Z  = k · YZ = BC. Also, because size transformations preserve angle measure, ∠Y  ∠Y  = ∠B by transitivity. Thus, ABC  X Y Z  by the SAS Congruence Theorem. Thus, ABC can be mapped onto XYZ by a composite of size changes and reﬂections, so ABC ∼ XYZ. x 9. __5 = _____ + 8

16 8 5 _____ x = ·824 = 15 ft

10. a. AEC and BDC are right triangles with ∠C as a common acute angle.Therefore, by the AA Similarity Theorem, AEC ∼ BDC. b. BD; DC c. 5' 6" = 5.5' x = __ 15 ___ 5.5 4 15 · 5.5 = x = ______ 20.625 ft or 20 ft 7.5 in. 4 x 68 = __ 11. ______ 21 · 12 19 252 · 68 ______ x = 19 ≈ 902 in. ≈ 75 ft 2 in.

12. a. ∠LTR  ∠DTC, because they are vertical angles, and ∠RLT  ∠CDT because ___ ___  AB CD and these are alternate interior

angles. Therefore, by the AA Similarity Theorem LRT ∼ DCT. b. Let x = distance from thumb to object in inches. 3 in. = _____ 24 in. _____ x 16 in. 16 · 24 ______ x = 3 = 128 in.

Distance from face to object: 128 in. + 16 in. = 144 in. = 12 ft 28.8 = __ 48 = 13. a. ____ 2.4 12

20

b. SAS Congruence Theorem 14. a. ABE ∼ DFE, ABE ∼ CFB, DFE ∼ CFB b. __4 = __x 6

8 32 16 __ x = 6 = __ 3

247

Geometry Solution Manual

Chapter 12, Lesson 12-7

Back to Lesson 12-7 28 __

3 c. __4 = __ y 6

28 y = 6 · __ ÷ 4 = 14 3 16 = __ 28 d. AB = CD = 4 + x = 4 + __ 3

3

15. If two isosceles triangles have congruent vertex angles with measure x, then all of their remaining 180 - x . By the AA angles have measure ______ 2

Similarity Theorem, the two triangles are similar. Conclusions

16.

Justiﬁcations

___

1. A, BD bisects ∠CBE.

1. Given

2. ∠CBD  ∠DBE

2. deﬁnition of bisection 3. The angles intercept the same arc. 4. AA Similarity Theorem

3. ∠BEC  ∠CDB

4. BFE ∼ BCD 17.

Conclusions 1. m∠STU = 90, m∠SVT = 90 2. ∠STU  ∠SVT

3. ∠S  ∠S 4. STU ∼ SVT

Justiﬁcations 1. Given

22. By the convenient location for a rectangle, L = (–a, b), M = (a, b), N = (a, –b), and O = (–a, –b). Thus, E = (0, b), F = (a, 0), G = (0, –b), and H = ___ 0). Therefore, the slope of GE is undeﬁned, (–a,___ ___ is vertical while the slope of___ FH is___ 0 and so GE___ thus FH is horizontal. Therefore, GE ⊥ FH. 23. 3h 24. Yes. It is given that___ LMNK is a parallelogram, ____P LM , O is the midpoint of MN, is the midpoint of ___ ___ ___ and KO and KP intersect LN at Q and R. By the Vertical Angles Theorem and the Parallel Lines Theorem, ∠LQP ∠NQK and ∠NLM  ∠LNK. Thus, by the AA Similarity Theorem, QLP ∼ ___ QNK. Because P is the midpoint of LM, we know the ratio of similitude is 2, so 2LQ = QN. By the same reasoning, ∠ORN  ∠KRL and ∠KLN  ∠LNM. Again by the AA Similarity Theorem, NRO ____ ∼ LRK. Because O is the midpoint of MN, the ratio of similitude is 2, so LR = 2RN. So LN = LQ + QN = 3LQ, and LN = RN + LR = 3RN. Thus, LQ = RN, and by substitution, LQ = QR = RN.

2. Reﬂexive Property of Equality, Angle Congruence Theorem 3. Reﬂexive Property of Congruence 4. AA Similarity Theorem

15 √ 3 15 18. __ and _____ 2

2

3

19. a. 500 = 125,000,000 times as much weight b. No; the area of a cross section of its muscles

would only be 250,000 times larger, so the 1 of its own larger ant would be able to lift __ 10 weight.

20. a. true b. false ___

___

; CD ; AB  CD; therefore, there is no size 21. AB transformation S such that S(A) = B and S(C) = D.

248

Geometry Solution Manual

Chapter 12, Lesson 12-7

## Chapter 12 Solutions.pdf

The harmonic mean of two numbers is the. reciprocal of the arithmetic mean of the. reciprocals of the two numbers. One application. is to calculate the average cost of shares of stock. purchased over a period of time. 239 Geometry Solution Manual Chapter 12, Lesson 12-2. Page 3 of 12. Chapter 12 Solutions.pdf. Chapter ...

#### Recommend Documents

Chapter 12 Timeline.pdf
*Saint Apollinaris, First Bishop of Ravenna,. apse of Saint' Apollinare Nuovo. *The Second Coming (Christ Between Angels. and Saints), apse of San Vitale.

Chapter 12 notes.pdf
... was a problem previewing this document. Retrying... Download. Connect more apps... Try one of the apps below to open or edit this item. Chapter 12 notes.pdf.

chapter 12.pdf
chromosome contains half as much genetic material as at the start of anaphase II. Telophase II: The nucleus reforms and the spindle fibers break down. Each cell undergoes cytokinesis, producing. four haploid cells, each with a unique combination of g

AIFFD Chapter 12 - Bioenergetics - GitHub
The authors fit a power function to the maximum consumption versus weight variables for the 22.4 and ... The linear model for the 6.9 group is then fit with lm() using a formula of the form ..... PhD thesis, University of Maryland, College Park. 10.

Chapter 12- Conservation Easements.pdf
There was a problem previewing this document. Retrying... Download. Connect more apps... Try one of the apps below to open or edit this item. Chapter 12- Conservation Easements.pdf. Chapter 12- Conservation Easements.pdf. Open. Extract. Open with. Si

Zinn chapter 12.pdf
State Department list, "Instances of the Use of United States Armed Forces Abroad ... Nicaraguan canal is built, the island of Cuba ... will become a necessity.

Chapter 12 - National Interagency Fire Center
SUPPRESSION CHEMICALS AND DELIVERY SYSTEMS. CHAPTER 12 ..... constraints, congested area, life and property concerns or lack of ground. 10.

Chapter 12 VARIABILITY OF THE MARINE ITCZ OVER ...
ley cell. It regulates the hydrologic cycle over the tropical continents and in- ...... (McIntyre and Molfino 1996) best explained by an equatorward shift of the .... mechanisms (e.g., the strength of the thermohaline circulation or the solar.

Chapter 12 VARIABILITY OF THE MARINE ITCZ OVER ...
Our strategy relies ... sive southward migration of the ITCZ during the last ~7,000 years, in accord .... Figure 12-2 illustrates this effect using weekly SST data.

Chapter 12 - The Scientific Revolution and the Englightment.pdf ...
As the spheres moved, they made. music. Beyond the planets was a ring of fixed stars. This marked the end of. the universe. Beyond that were God and heaven.

BHS Harmony Foundation Chapter Statement of Policy Rev. 12-31 ...
BHS Harmony Foundation Chapter Statement of Policy Rev. 12-31-2015.pdf. BHS Harmony Foundation Chapter Statement of Policy Rev. 12-31-2015.pdf. Open.

Chapter 12 Photon Monte Carlo Simulation
... for viewing the trajectories is called EGS Windows [BW91]. ..... EGS-Windows - A Graphical Interface to EGS. NRCC Report: ... 1954. [Eva55]. R. D. Evans.