49. Lesson 1.4 (pp ) RS m ADC 5 m ADB 1 m CDB. 53. a. m DEF 5 m ABC b. m ABG 5 1 } 2. c. m CBG 5 1 } 2. d.

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1 This section of the book provides step-b-step solutions to eercises with circled eercise numbers. These solutions provide models that can help guide our work with the homework eercises. The separate Selected nswers section follows this section. It provides numerous answers that ou can use to check our own answers. hapter Lesson. (pp. 5 8) Plane PQS and plane HGS intersect at ] RS legged table ma rock from side to side because four points are not necessaril coplanar. 3-legged table would not rock because three points determine a unique plane. Lesson. (pp. 4) 3. (0, ) (, ) (4, ) (, 6) So, } ù } (6) a mi b. d 5 rt r(38) r ø 49.6 mi/h Lesson.3 (pp. 9 ) 5. SM 5 MU SU 5 SM MU Length: 3 (4) 5 7 units oordinate of midpoint: } } 5 }. 49. House Librar 5.7 km.85 km Lesson.4 (pp. 8 3) School 5. not her n a me f or is. The angle is a right angle because it is labeled with a red square. 3. m 5 m m a. m EF 5 m 5 38 b. m G 5 } p m 5 } (3 8) c. m G 5 } p m 5 } (3 8) d. m EG 5 } p m EF 5 } (3 8) Lesson.5 (pp. 38 4) 9. m m m 5 90 m and are a linear pair. 47.Neither, because the sum of their measures is greater than 808. Lesson.6 (pp ) 3. The polgon is a quadrilateral because it has 4 sides. It is equiangular but not equilateral, so it is not regular. 9. decagon is sometimes regular, because all of its sides and all of its angles can be congruent, but the don t have to be. 3 3.T h e p ol g on h a s 3 s i d e s, s o i t i s a t r i a n g l e. It appears to be regular. WORKE-OUT SOLUTIONS WS

2 WORKE-OUT SOLUTIONS Lesson.7 (pp. 5 56) 7. P 5 a b c d 5 } bh 5 } (7 )( 30 ) d. 3 ft p d } 9 ft ø.44 d 4. 5 lw 5 45(30) P 5 l w 5 (45) (30) 5 50 You need to cover 350 square ards with grass seed, and ou need 50 d p } 3 ft feet of fencing. d hapter Lesson. (pp ) 7. 3,, 48, 9, Each number in the pattern is four times the previous number. The net number is ountereample: ( 5) Þ onjecture: In 004, more than 7 trillion messages will be sent. The number of messages has increased each ear for 7 ears prior to 004. If this pattern continues, the total will eceed 7 trillion in 004. Lesson. (pp. 8 85). False, a polgon could have 5 sides without being a regular pentagon. ountereample: 7. Fa l s e, it i s not m a r k e d t h at ] PQ and ] ST intersect at a right angle, so ou do not know that the are perpendicular You c a n s h o w t h a t t h e s t a t e m e nt i s f a l s e b finding a countereample: swimming is a sport, but the participants do not wear helmets. Lesson.3 (pp ) 7. I f a r e c t a n g l e h a s f ou r e q u a l s i d e l e n g t h s, then it is a regular polgon. 7. I f t h e b a k e r s r e v e nu e i s g r e a t e r t h a n i t s costs, ou will get a raise.. e d u c t i v e r e a s on i n g ; t h e c on c l u s i on i s reached b using laws of logic and the facts about school rules and what ou did that da. Lesson.4 (pp. 99 0) 7. Line p intersects line q at point H. 3. Fa l s e ; t w o i nt e r s e c t i n g l i n e s a r e a l w a s coplanar. For eample, consider a railroad crossing sign. 3. P o s t u l a t e 7: I f t w o l i n e s i nt e r s e c t, t h e n t h e i r intersection is eactl one point. Lesson.5 (pp. 08 ) 9. 3( ) 5 9 Given istributive Propert Subtraction Propert of Equalit ivision Propert of Equalit. If 5 0, then P 5 l w Given P w 5 l Subtraction Propert of Equalit P } w 5 l ivision Propert of Equalit When P 5 55 and w 5 : l 5 55 } meters Lesson.6 (pp. 6 9) 7. If F > J and J > L, then F > L. 5. ottage Snack ike rcade shop rentals Kite shop. Statements Reasons. ] TV bisects UTW. Given. >. efinition of angle bisector 3. > 3 3. Given 4. > 3 4. Transitive Propert of ngle ongruence WS

3 Lesson.7 (pp. 7 3) 5. FGH > WXZ; WXZ is a right angle because , so the are congruent b the Right ngles ongruence Theorem. 3. Using the Vertical ngles ongruence Theorem: Statements Reasons. } JK } JM, } KL } ML,. Given J > M, K > L. J is a right angle;. efinition of L is a right angle. perpendicular segments 3. M is a right angle; 3. Right ngle K is a right angle. ongruence Theorem 4. } JM } ML, } JK } KL 4. e f i n it i on of perpendicular segments hapter 3 Lesson 3. (pp. 50 5). The pairs of corresponding angles are and 5, and 6, 3 and 7, and 4 and If two lines are not coplanar, then the never intersect. o not intersect: b. ecause the bars are parallel, corresponding angles between the bars and the foot are congruent. ecause the bod and the foot are parallel, the bars act as transversals, and so the alternate interior angles are congruent. (See diagram.) This forces the foot to sta parallel with the floor. Lesson 3.3 (pp ). Yes; lternate Eterior ngles onverse 9. e c a u s e t h e a lt e r n a t e i nt e r ior a n g l e s a r e congruent, ou know that the top of the picnic table is parallel to the ground. 37. It i s g i v e n t h a t 3 and 5 are supplementar. 3 is supplementar to 4 because the form a straight angle. 4 > 5 b the Supplementar ngles ongruence Theorem. Therefore, min b the lternate Interior ngles onverse. Lesson 3.4 (pp ) 7. m 5 } } } 3. Line : m 5 } } 7 5 } 3 Line : m 5 } } } WORKE-OUT SOLUTIONS Intersect: 3 5. The arm is skew to a telephone pole. Lesson 3. (pp ) 5. If m , then m 5 08, b the lternate Eterior ngles Theorem. 9. orresponding ngles Postulate 39. a. lwas congruent: and 4, and 5, and 8, 4 and 5, 4 and 8, 5 and 8, 3 and, 3 and 7, 3 and 6, and 6, and 7 and ecause m p m 5 } 3 p 3 }, the lines are perpendicular. a b (0, 4) c 3 (0, ) (0, 6) (3, 0) Line b is the most steep because the absolute value of its slope is the greatest. Line c is the least steep because the absolute value of its slope is the least. WS3

4 WORKE-OUT SOLUTIONS Lesson 3.5 (pp ) 7. P(5, 4), m m b 4 5 4(5) b b 5 6 n equation of the line is T he s lop e of a l i ne pa r a l le l t o 5 3 is m 5. 5 m b 5 (0) b b 5 n equation of the line is das since age 4; 5 weight (in kg) The slope,., represents the rate of weight gain, in kilograms per da. The -intercept, 000, represents the weight, in kilograms, at age 4. Lesson 3.6 (pp ) 9. Line f is parallel to line g because the are both perpendicular to line d. 3. Slope of parallel lines: m } Line through point (5, 3): 5 m b 3 5 4(5) b b Perpendicular line through point (0, 0): 5 m b 0 5 } 4 ( 0 ) b b } 4 Intersection of two lines: } (4) 7 5 istance from (4, ) to (0, 0): d 5 Ï }} (4 0) ( 0) ø Sample answer: The shortest distance is the length of the perpendicular segment. hapter 4 Lesson 4. (pp. 4) 9. The triangle has 3 congruent sides and 3 congruent angles, so it is an equiangular equilateral triangle (30) ecause the triangle has a right angle, it is a right triangle. 6 in Each side of the triangle is inches. The measure of each angle is Lesson 4. (pp. 8 3) 9. nlnm > nzyx 5. the Triangle Sum Theorem, m M the Third ngles Theorem, 8 5 m M The length, height, and depth have to be the same. Lesson 4.3 (pp ) 7. True; } E > } G, } EF > } GF, and } F > } F b t h e Refleive Propert. So, nef > ngf b the SSS ongruence Postulate , 5 8, 5 Ï }}} ( 4) ( 6) 5 0 E 5 6, EF 5 Ï }} (3 5) ( ) 5 8, F 5 Ï }} (5 3) (7 ) 5 0 the SSS Postulate, n > nef. 5. ecause } WY > } VY a nd } YZ > } YX, t he n }WZ > } VX. t he Re f l e i v e P r op er t, }WV > } VW. It i s g i v e n t h a t } WX > } VZ. So, b the SSS ongruence Postulate, nvwx > nwvz. Lesson 4.4 (pp ) 3. There is enough information given. WS4

5 9. F H G F E ecause } FH > } H > } > } E, t he n } F > } E. ecause } > } > } GF > } FE, t he n } > } GE. lso, nf and neg are right triangles. So, nf > neg b the Hpotenuse Leg ongruence Theorem. 3. SS ongruence Postulate Lesson 4.7 (pp ) 5. If > E, then } E > } b the onverse of ase ngles Theorem b the ase ngles Theorem and b the orollar to the Triangle Sum Theorem. So, and (5) a. Statements Reasons Lesson 4.5 (pp. 5 55) 5. Yes; S ongruence Postulate 9. F > L 7. S ongruence Theorem Lesson 4.6 (pp ) 9. Let J represent the midpoint of } KM. Use the SSS ongruence Postulate to prove n JKN > n JMN. Then use the SS ongruence Postulate to prove n JKL > n JML. ecause n JKL > n JML, >. 3. Statements Reasons. X > Z, U > T,. Given }ZY > } XY. ntyz > nuyx. S ongruence Postulate 3. VYX > Z T, 3. Eterior ngle WYZ > X U Theorem 4. VYX > Z T, 4. Substitution WYZ > Z T 5. VYX > WYZ 5. Simplification 3. ; Once ou show that XWZ > YZW b the lternate Interior ngles Theorem and that } WZ > } ZW a nd Y > X b the Refleive Propert and the Right ngle ongruence Theorem, ou can show that nwyz > nzxw b the S ongruence Theorem. This means that } WY > } ZX.. >. Given > E > E. } > } > } > } E. onv e r s e of a s e ngles Theorem 3. > 3. ase ngles Theorem 4. > 4. lternate Interior ngles Theorem 5. > 5. Substitution 6. n > n 6. S ongruence Postulate b. n, ne, nefg, n, nef c. E, EFG,, EF Lesson 4.8 (pp ). (, ) ( 4, ) 3. (3, ) (, ) (, 3) (6, 4) (3, 0) (7, ) (, ) (3, 0) O m O > m O This is not a rotation. 39. From to, use a 908 clockwise rotation. From to, use a 908 counterclockwise rotation. 4 WORKE-OUT SOLUTIONS WS5

6 hapter 5 Lesson 5. (pp ) 9. } YJ > } JX > } LK. (p, q) q 5. onnect the three houses with line segments. Fold our house to Mike s house, Mike s house to Ken s house, and Ken s house to our house. The folded lines are the perpendicular bisectors. You should meet where the intersect. WORKE-OUT SOLUTIONS (0, 0) p (p, 0) 5 Ï }} (p 0) (q 0) 5 Ï } p q 5 Ï }} (p p) (0 q) 5 Ï } p q 5 p m 5 q 0 } p 0 5 q } p m 5 0 q } p p 5 q } p m } p M 0 p }, 0 q } 5 p }, q } M p p }, q 0 } 5 3p }, q } M 0 p }, 0 0 } 5 (p, 0) n is not a right triangle because the angle measurements depend on the values of p and q. n is isoceles because W 0 } 6, 0 } 6 5 W(3, 3) V 6 } 8, 6 } 0 5 V(7, 3) m WV } m OH } }WV i } OH WV , OH WV 5 } OH Lesson 5. (pp ) 5. e c a u s e LK 5 LM, L lies on the perpendicular bisector of } KM b t h e onverse of the Perpendicular isector Theorem. 7. Theorem 5.4, G 5 G 5 G 5. Lesson 5.3 (pp ) 7. Not e nou g h i n f or m a t ion i s g i v e n. 5. No, b e c a u s e ou do not k now t h at t he blue line segments are perpendicular to the sides of the angle. 9. uild the fountain where the angle bisectors meet, which is equidistant from the 3 sides. Lesson 5.4 (pp. 3 35) 5. G 5 } 3 E G 5 } 3 (5) G 5 0. ecause nxyw > nzyw, } YW i s a perpendicular bisector, an angle bisector, a median, and an altitude. 39. rea of large triangle 5 } ( 9 )( 3 ) rea of dashed triangle 5 } ( 3 )( 3 ) rea of red triangle in. The altitudes were used. Lesson 5.5 (pp ) 9. } KL, } JL, } JK, and J, K, L b Theorem No, because 3 6 >/ a. Using the Triangle Inequalit Theorem: 489 > 565 < > 76 < 054 value of does not satisf > 054. b. No, 5 40 does not satisf > 76. c. > 76; < 054 d. ecause m < < m <, < 489. ecause m < m 3, < 565. WS6

7 Lesson 5.6 (pp ) 5. TR < UR b the Hinge Theorem, because TQR < USR. 7. m 5 m because both triangles are congruent using the SSS ongruence Postulate. 3. E,,,, hapter 6 Lesson 6. (pp ) 5. 6 L } 5 6 L 000 ml } 0 ml 0 ml p } } L } RS } 5 64 } ZW 4 5 } 8 ST } 3 5 } 48 WX 3 5 } 3 UT } 5 64 } YX 4 5 } 8 RU }ZY 3 5 } } 3 The ratios are not equal, so the corresponding side lengths are not proportional, and the polgons are not similar. 9 ft long } 78 ft long 5 3 } 6 5 ft wide } 36 ft wide 5 5 } 36 The ratios are not equal, so the corresponding side lengths are not proportional, and the surfaces are not similar. 7. } c } (c 5) c Total amount of trail mi 5 36 } } 5 c ups Peanuts: 5 9 } cups hocolate chips: 9 } } } 5 c ups Raisins: 4 9 } } } 5 c ups Lesson 6. (pp ) 5 8 cups.. } 5 E } EF 6 } 5 4 } 7 6(7) So, inch } 5} inches 5 (50) 5 00 ards 50 ards ards inch.5 inches } 5 } 5 3.(.5) 5 8 miles 3. miles miles Lesson 6.3 (pp ) 3. > L, > M, and > N } 5 } LM MN 5 } NL Lesson 6.4 (pp ) 9. Yes; H > J, and m F 5 48 b the Triangle Sum Theorem, so F > K. So, ngfh, nlkj b the Similarit Postulate. 3. the Triangular Sum Theorem, m YZX and m U So, nxyz, nuyw b the Similarit Postulate ll angles of an two 3 equilateral triangles are congruent, so the triangles are similar b the Similarit Postulate Lesson 6.5 (pp ) 3. } 5 } 5 E 0 5 } 3 } 5 } 8 EF 5 } 3 } 5 } F 8 5 } 3 ecause } E 5 } EF 5 } F, n, nef. The scale factor of n to nef is 3 }. 7. Yes, because E } WY 5 9 } } 5 F } XW 5 5 } }, and > W. nef, nwyx, and the scale factor is 3 }. 3. To use the SS Similarit Theorem, ou need > E. 3 3 WORKE-OUT SOLUTIONS WS7

8 Lesson 6.6 (pp ) 5. Yes, b Theorem 6.5 because LK } KJ 5 8 } 5 5 LM } MN 5 } } } 5 } Let 5 the distance along Universit venue from th street to Washington Street. 00 } 5 } , ards Lesson 7. (pp ) 7. (Ï } 6 ) The triangle is a right triangle Ï } 43 ø > 6 Ï } > 4 6 Ï } 43 4 ø > 30 The segment lengths form a triangle. WORKE-OUT SOLUTIONS Lesson 6.7 (pp. 4 45) 5. (, ) (.5,.5) (, ) L(.5,.5) (6, ) M(9,.5) (6, 3) N(9, 4.5) L. The dilation is an enlargement. : (, 0), (, ), and (3, ) : (6, 0), (3, 6), and (9, 6) For each verte (, ) of Figure, the corresponding verte of Figure is (3, 3). The scale factor is The scale factor is hapter 7 5 ft } 6 ft 5 5 }, or. 5. N M ecause (6Ï } 43 ) > 4 30, the segment lengths form an obtuse triangle. 37. a. Let b. ecause , n is a right triangle. c. 6 3 Lesson 7.3 (pp ) } } h 07.5h h ø 53.7 ft z } } 6 6z 5 79 z ø 45.6 Lesson 7. (pp ) ø is about 9. inches. 7 m 8 m h 8 m 7 m 8 h 5 7 h 5 5 h 5 5 rea 5 } bh 5 } (6 )(5) 5 0 m 3 3. ecause , the hpotenuse has length Let c represent the hpotenuse of the large triangle c c Ï } c } 5 }.5 Ï } ø.06 ft.5 Ï 4.5 Lesson 7.4 (pp ) 5. hpotenuse 5 leg p Ï } 3 Ï } 5 p Ï } Ï } 3 5 Ï } (3) p h h feet WS8

9 Lesson 7.5 (pp ) 5. tan 5 opp. } adj. to 5 48 } 0 5 } tan 5 opp. } adj. to 5 0 } } ø tan 78 5 opp. } adj. 5 } 5 ø h 8 ft 788 tan h } 8 8(4.7046) 5 h Lesson 7.6 (pp ) h ø 555 feet 5. sin 5 opp. } hp. 5 8 } 53 ø sin E 5 opp. E } hp } 53 ø cos X cos Y 5 38 adj. to X } hp. 5 3 } 6 5 } adj. to Y } hp. 5 3Ï} 3 } 6 5 Ï} 3 } ø ft sin 38 5 } } ø 37 ft sin 38 Lesson 7.7 (pp ) 5. tan 5 4 } } m 5 tan 3.5 ø () 5 5 () 5 5 tan 5 9 } m 5 tan 0.75 ø m m ø 53.8 The side lengths are 9 units, units, and 5 units. The angle measures are 908, about 36.98, and about The angle of depression is tan 7 } ø tan ø hapter 8 Lesson 8. (pp ) 9. (n ) p n 5 n 5 3 The polgon is a 3-gon.. Pentagon; (5 ) p Heagon; (6 ) p Lesson 8. (pp. 58 5) 9. m m m m b 5 9 b 5 0 and 5a 5 5 a a. PQRS is a parallelogram, so } PQ > } RS. RS 5 PQ 5 3 inches b. PQRS is a parallelogram, so Q > S. m S 5 m Q c. ecause P and Q are supplementar, 808 m Q 5 m P. s m Q increases, m P must decrease. s m Q increases, m P and m R decrease, which also makes PR increase and QS decrease because the side lengths do not change. Lesson 8.3 (pp ) 5. Theorem 8.7, because both pairs of opposite sides are congruent ; So, } > }. Slope of } } Slope of } 5 } (4, 4) (, 4) (0, ) (8, ) Slopes are equal, so } i }. } > } and } i }, so is a parallelogram. WORKE-OUT SOLUTIONS WS9

10 WORKE-OUT SOLUTIONS 3. a. Quadrilateral EFJK: ecause } EG > } KH, and F and J are midpoints of } EG and } KH, t he n } EF > } FG > } KJ > } JH. ecause } EF > } KJ and } EK > } FJ, EFJK is a parallelogram. Quadrilateral FGHJ: ecause } FG > } JH and } FG > } GH, FGHJ is a parallelogram. Quadrilaterl EGHK: ecause } EG > } KH and } EK > } GH, EGHK is a parallelogram. b. lthough the angles ma change, } EG is alwas congruent to } KH and } EK i s a l w a s congruent to } GH. So, EGHK is alwas a parallelogram. Therefore } EG i s a l w a s parallel to } KH. Lesson 8.4 (pp ) 7. J 5 M L 5 K You can sketch a rhombus that has the diagonals shown. So, the statement is onl sometimes true. 5. T h e q u a d r i l a t e r a l i s a s q u a r e b e c a u s e i t i s a parallelogram with four congruent sides and four right angles. 55. Use the tape measure to measure the diagonals. If the are euqal, the region is a rectangle. Lesson 8.5 (pp ). ecause E and H are right angles, }EF i } HG. } EH i s n o t p a r a l le l t o } FG, so EFGH is a trapezoid. 9. Theorem 8.9, because } EF > } EH, t he n F > H m G m E m F m H m G m G Lesson 8.6 (pp ) 3. Propert ~ Rectangle Rhombus Square Kite Trapezoid ll sides are > 5. Trapezoid; it has eactl one pair of parallel sides ( } PS and } QR ) trapezoid is outlined. hapter 9 Lesson 9. (pp ) 7. (, ) ( 4, 6) P(, 3) P9(, 9) Q(, ) Q9(5, 8) R(3, ) R9(7, 5). rule is (, ) ( 5, ) Ï }} (4 ) ( ) 5 Ï } Ï }} ( 4) (3 ) 5 Ï } 3 5 Ï }} (5 4) ( ) 5 Ï } Ï }} (0 ) ( 3) 5 Ï } 5 the SSS ongruence Postulate, n > n999, so the translation is an isometr From cabin to ski lodge: 7 0, , 8 X P X P R R WS0

11 Lesson 9. (pp ) F G F 4 3G 5 F 8 5 G Translation Polgon Image matri matri matri G Reflection Polgon Image matri matri matri 3. F 0 0 GF G 5 F f. 3gF.5G 5 f.() 3(.5)g 5 f6.9g 3. Equipment matri: ost matri: Mice s Keboards ollars Lab Mice 0 Lab F 5 0 G s 3 KeboardsF 5G Total cost: F GF0 3 5G5F970G 840 Lab : $840; Lab : $970 Lesson 9.3 (pp ) Reflection in the -ais Lesson 9.4 (pp ) 3. (a, b) (a, b) J(, 4) J9(, 4) K(5, 5) K9(5, 5) L(7, ) L9(7, ) M(, ) M9(, ) L K M J J M G F 5 4G Rotation Polgon Image matri matri matri 5. F 0 0GF 5 4 K L WORKE-OUT SOLUTIONS WS

12 9. Rot at ion of 9 0 8, because it would take four rotations for the blade to reach its original starting point. Lesson 9.5 (pp. 6 65) 7. P(, 4) R(7, ) G H J G9 H9 J9 G 9. } F G 5 F 0 3 Scale Polgon factor matri H H Image matri WORKE-OUT SOLUTIONS R (7, 3) (6, 5) P (, ) (6, 0) P (, ) R (7, 3) (6, 5) 7. Sample answer: } 9 i ] 0 and } 9 i ] in. 9 in. 9 in. Translate 9 inches to the right, reflect in the -ais. Lesson 9.6 (pp. 6 64) 7. Yes; a rotation of 78 maps the figure onto itself. 3. ; there are no lines of smmetr, but a rotation of 908 maps the figure onto itself. 3. There are 4 lines of smmetr, so n 5 4. n(m ) (m ) m The angle between the mirrors is 458. Lesson 9.7 (pp ) 7. G G H E E F F G G 3 5. k 5 0; 0 5 } J J The length of the dragonfl seen through the magnifing glass is 940 mm, or 94 cm. hapter 0 Lesson 0. (pp ) ] 7. E i s a t a n g e nt l i ne b e c a u s e i t i nt e r s e c t s t h e circle at onl one point. 9. No; because 9 5 Þ 8, n is not a right triangle and } i s not p e r p end ic u l a r t o } mi E 3959 mi 3959 mi E 5 E,000 mi (3959,000) 5 08,098,000 ø 4,46 The length of } and } i s a b out 4,4 6 miles. Lesson 0. (pp ) 5. i s a m i n or a r c. m me No; LP and MN h a v e t he s a me me a s u r e, but are arcs of circles that are not congruent. 3. The measure of each arc is WS

13 Lesson 0.3 (pp ) 7. ecause } LN is a diameter and } LN } PM, } LN bisects } PM. So, ecause } a nd } are equidistant from the center, } > }. So, } and } mu s t b e c on g r u e nt i n or d e r f or and t o b e c on g r u e nt. Lesson 0.4 (pp ). JMK and JLK intercept the same arc, so JMK > JLK. lso, MKL and MJL intercept the same arc, so MKL > MJL I f a r i g ht t r i a n g l e i s i n s c r i b e d i n a c i r c l e, i t s hpotenuse is a diameter of the circle, so the length of the hpotenuse is twice the radius. Lesson 0.5 (pp ) 3. m 5 (658) } ( 308 ( 30)8) } ( 8) Let and E represent the points shown in the diagram. ecause } E is a diameter, m E } m 8 5 } (m E me ) 8 5 } ( ) E 308 Lesson 0.6 (pp ) (6) (9 7) Given: hords } and } intersect in the interior of the circle. Prove: E p E 5 E p E Statements Reasons. m E > m E. Vertical ngles Theorem. m > m. Theorem ne, ne 3. Similarit Postulate 4. E } E 5 E } E 4. e f i n it ion of similarit 5. E p E 5 E p E 5. ross Products Propert Lesson 0.7 (pp ) 7. enter (50, 50); radius 0 ( h) ( k) 5 r ( 50) ( 50) 5 0 ( 50) ( 50) r 5 Ï }} (0 0) (6 0) 5 6 (h, k) 5 (0, 0) ( h) ( k) 5 r ( 0) ( 0) Outside edge: (h, k) 5 (0, 0); r 5.4 ( 0) ( 0) Edge of hole: (h, k) 5 (0, 0); r ( 0) ( 0) ( 0) E WORKE-OUT SOLUTIONS WS3

14 WORKE-OUT SOLUTIONS hapter Lesson. (pp ) b 8 b b } bh 5 } (4)(8 ) 5 6 square units 3. rea 5 rea of Parallelogram rea of left triangle rea of right triangle 5 (8)(3) } ( 9 )(3 ) } ()(3 ) cm 37. Triangular plot: 5 } bh 5 } (4)(5) d 300 d p } min 5 30 min 0 d It takes ou 30 minutes to mow the triangular plot. Rectangular plot: 5 bh 5 4(36) d 864 d p } min min 0 d It takes ou 86.4 minutes to mow the rectangular plot. Lesson. (pp ) 9. 5 } d d 5 } (8 )( ) 5 89 square units 7. 5 } h(b b ) } (0 )[ 0] m } d d 5 } ( 5)( 8 ) 5 0 square millimeters Sample answer: Lesson.3 (pp ) 7. The ratio of the lengths of corresponding sides is 7 } 9, or 7 : 9. The ratio of perimeters is 7 : 9. The ratio of areas is 7 } } 8, or 49 : 8. Red area } lue area 5 49 } 8 Red area } } 8 Red area 5 0,90 } 8 ø 7 square inches 7. M P 4 ft N 5 ft U R T d S d rea of MNPQ 5 } (4)( 5) 5 75 ft rea of RSTU 5 8 ft Ratio of areas 5 75 ft } 8 ft 5 5 } } Ratio of corresponding lengths 5 5 } 4 } d 5 5 } d ft; 5 } d 5 5 } d 5 0 ft The lengths of the diagonals of RSTU are 5.6 feet and 0 feet. 7. New patio Eisting patio ft Ratio of areas ft.5 ft 360 ft } 50 ft 5 36 } } 5 Ratio of corresponding lengths 5 6 } 5 Let represent the distance between the long parallel sides of the new patio. 5 mm 5 mm } } feet 8 mm 8 mm WS4

15 Lesson.4 (pp ) 3. rc length of LM }} 5 m LM } πr } 5 } 608 πr 3608 r (3608) } π(608) r 5 70 } 60π ø a. circle graph is appropriate because the data values add up to 00%. b. us: central angle (3608) Walk: central angle 5 0.5(3608) Other: central angle 5 0.0(3608) Walk 5% 5. Perimeter 5 p Length of each straight section p 5 (6) p 908 } 3608 p π(3) 5 3π ø.4 units Length of each quarter circle 3 5. T h e l e n g t h f e e t 8 i nc he s r e p r e s e nt s t he circumference of the trunk. The can substitute 8 } 5 65 } 3 for in the equation 5 πd and solve for d. 65 } 3 5 πd d ø 7 feet Lesson.5 (pp ) 7. 5 πr πr 54 } π 5 r 7 ø r The radius is about 7 meters. Shaded area 5 rea of rectangle p rea of semicircle us 65% Other 0% c. Sample answer: In the circle graph, r ø.5 cm. rea of bus sector } 3608 p (πp.5 ) ø 4.6 cm rea of walk sector } 3608 p (πp.5 ) ø.8 cm rea of other sector } 3608 p (πp.5 ) Lesson.6 (pp ) } P 5 7s 5 7(9) 5 63 units m EF } 7 m EG 5 } 360 } } 7 8 tan 80 } } a ø 0.7 cm WORKE-OUT SOLUTIONS 5 6(6) p 808 } 3608 (πp 3 ) π ø 7.7 The shaded area is about 7.7 square meters. 5 } ap a } tan 80 } } 4.5 } tan } ( 6 3 ) ø 94.3 square units. E a G F 9 WS5

16 37. pothem 5 a centimeters u 5 } 3608 } } a 5 tan u }. 5 tan tan.58 s tan cm cm u hapter Lesson. (pp ). F V 5 E n 5 8 n The cross section is a circle. rea of octagon 5 } a p ns WORKE-OUT SOLUTIONS rea of silver border 5 } (. )( 8 )(.4 t a n. 58) ø 4.77 square centimeters 5 rea of octagon rea of circle 5 } a p ns πr ø 4.77 π( ) ø.63 square centimeters Lesson.7 (pp ) 3. P(K is on } ) 5 } E , or 6.5% 9. P(Point lies in shading) P(You miss call) () }} () 5 5 } 8, 5 rea of shaded region }} rea of large triangle } (6)(7) } } ()(4) 5 } 4, 0.5, or 5% Time from 7:00 7:0 }} Time from 7:00 8:00 0 min } 60 min 5 } 6 ø 0.67 The probabilit that ou missed our friend s call is about 6.7%. ross Section 3 5. T h e r e a r e 6 v e r t i c e s a r ou n d each heagonal face, and there are no other vertices. Each heatgonal face has 6 edges, and there are 6 additional edges between the square faces. So, there are (6) 5 vertices and (6) edges. heck: F V 5 E Lesson. (pp ) 7. ase of prism: 8 m Ï } 6.75 meters 3 m 8 m rea of base 5 5 } ( 3 )(Ï } 6.75 ) ø.79 S 5 Ph ø (.79) (8 8 3)(9.) ø square meters 9. S 5 πr πrh 5 π(0.8) π(0.8)() ø 4.07 square inches WS6

17 3. a. S 5 (6 p 6) (6 p ) (6 p ) The minimum amount of wrapping paper needed is 360 square inches. b. The area of the net of the bo is larger because there are flaps that fold over and overlap. c. You would want more than 360 square inches of wrapping paper so that ou could wrap the paper around the bo and fold it down to fit; this causes sections of paper to overlap. Lesson.3 (pp ) 7. rea of base 5 } ap S 5 } Pl 5 } ( 6.9 )( 5 p 0) } ( 50 )( 0 ) square millimeter. l 5 4 l 5 Ï } 7 inches in. Lateral area 5 πrl 5 π()(ï } 7 ) ø.95 square inches 4 in. 9. The net represents a regular square pramid. S 5 } Pl 5 6 } (4 p 6)(6) 5 08 square centimeters Lesson.4 (pp. 8 85) 7. rea of base 5 } (7 )(0 ) 5 35 in. Volume 5 h 5 35(5) 5 75 in. 3. Volume 5 πr h 5 π(5 )(6) ø in a. Visualize the concrete block broken into five parts: 4 in in. 4.5 in. 8 in. 8 in. 8 in in. in. 4 in. 4 in..5 in. 8 in..5 in. Total volume 5 (5.75 p p 8) 8 in in. in. 4 in. 8 in. 3(.5 p 4 p 8) 5 70 in. 3.5 in. 8 in. b. Volume of block 5 Volume of large prism Volume of holes 5 (5.75 p 8)(8) (4 p 4.5)(8) 5 70 in. 3 c. The volumes found in parts (a) and (b) are equal. Lesson.5 (pp ) 3. rea of base cm 7. V 5 } 3 h h 5 } 3 (5)( 6 ) 5 50 cm 3 r cm V 5 } 3 π r h sin h } 5 h 5 5 sin 548 ø.4 cm cos r } 5 5 } 3 π(8.8 )(.4) ø cm in. h 6 in. r 5 5 cos 548 ø 8.8 cm h 5 6 V 5 } 3 πr h h 5 4Ï } in. 5 } 3 π( )(4Ï } ) ø 3.70 in. 3 WORKE-OUT SOLUTIONS WS7

18 Lesson.6 (pp ) 3. S 5 4πr 5 4π(4 ) ø 0.06 ft 3. V 5 } 4 3 πr 3 5 } 4 3 π(40 3 ) ø 68,08.57 mm πr 4,855 5 πr r ø 3956 miles Surface area of Western Hemisphere 5 } (4πr ) 5 } (4π)(3956 ) ø 98,33,448 mi Lesson.7 (pp ) 3. Radii: 7 } 4 Heig ht s : 6 } } 5 The clinders are not similar because the ratios of corresponding linear measures are not equal. WORKE-OUT SOLUTIONS 9. Surface area of }} 5 } 3 Surface area of 500 }} Surface area of 5 9 } Surface area of ø m Volume of } Volume of 5 33 } } Volume of 5 7 } Volume of ø 7. m 3 7. Let represent the amount to be added to the smaller bowl and represent the amount to be added to the larger bowl. } 5 33 } 4 3 } } The smaller bowl needs 7 fluid ounces to be added. WS8