27. chord 28. radius 29. diameter 30. secant 31. tangent 32. secant 33. EG **** is a chord (as is EF **** ); 34. **& MN is a chord (as is **** JL );

Transcription

1 HAPTER hapter Opener hapter Readiness Quiz (p. 588). D; F D; Lesson.. heckpoint (p. 590). ED **** is a chord (as are JK **** and FH **** ); ^&( JK is a secant; ^&( A is a tangent; FH **** is a diameter; FG **** is a radius (as is GH **** ); G is the center; is a point of tangenc.. (0, 4). Guided Practice (p. 59). Sample answer: radius chord diameter. 3. E 4. D 5. A F 8. (3, 3) 9. (3, 0) and (3, 6) 0. (0, 3), (3, 0), or (6, 3). (3, 0) and (0, 3). (3, 3) and one of the following: (3, 0), (6, 3), (3, 6). Practice and Applications (pp ) 3. r d (5) 7.5 cm 4. r d (6.5) 3.5 in. tangent 5. r d (3).5 ft 6. r d (8) 4 m 7. d r (6) 5 in. 8. d r (6) 4 ft 9. d r (8.7) 7.4 m 0. d r (4.4) 8.8 cm. chord. tangent 3. diameter 4. radius 5. point of tangenc 6. center of the circle 7. chord 8. radius 9. diameter 30. secant 3. tangent 3. secant 33. EG **** is a chord (as is EF **** ); EG ^&( is a secant; EF **** is a diameter; E **** is a radius (as are F **** and G **& ); D is a point of tangenc. 34. **& MN is a chord (as is **** JL ); ^&( MN is a secant; **** JL is a diameter; KR **** is a radius (as are KJ **** and KL **** ); U is a point of tangenc. 35. LM **** is a chord (as is PN **** ); LM ^&( is a secant; PN **** is a diameter; QR **** is a radius (as are QP **** and QN **** ); K is a point of tangenc. 36. FA ^&( and ^&( E are secants. 37. An two of GD **&, H ****, FA ****, and E **** are chords. 38. Yes; the diameter is the longest chord and must pass through the center of the circle. Since H **** does not pass through the center, it is shorter than the diameter. 39. Yes, a line can be drawn through points K, G, and J that will be tangent to the circle. 40. The center of A is (, ); the center of is (6, ). 4. The length of the radius of A is units; the length of the radius of is units. 4. (4, ) 43. A: (3, ); : (3, 3); intersection: (3, 0); tangent line: -ais 44. A: (3, 3); : (, 3); intersection: (0, 3); tangent line: -ais 45. r ; d AE E A 47. A 4 4 A 8 A A.8 A. Standardized Test Practice (p. 593) 48. D 49. G A E(4, 3) D opright McDougal Littell Inc. Geometr, oncepts and Skills 85 hapter Worked-Out Solution Ke

2 hapter continued. Mied Review (p. 593) 50. HL ongruence Theorem; **** **** and A **** D **** in right triangles A and D, sot A T D b the HL ongruence Theorem. 5. SSS ongruence Postulate; three sides of T JKL are congruent to three sides of T PQR, sot JKL T PQR b the SSS ongruence Postulate. 5. ASA ongruence Postulate; ap at, PR **** TR ****, and aprq atrs, sot PQR T TSR b the ASA ongruence Postulate. 53. slope of A **** slope of **** (, 3) (5, 3) 5 slope of D **** D(4, 0) 5 4 A(0, 0) 4 slope of DA **** The opposite sides of AD have equal slopes, so the are parallel. So, AD is a parallelogram. 54. slope of QR **** 5 5 (0, 5) 0 0 R(, 5) slope of RS **** 5 4 slope of PS **** 0 4 P(, ) S(4, ) slope of QP **** 5 0 The opposite sides of PQRS have equal slopes, so the are parallel. So, PQRS is a parallelogram.. Algebra Skills (p. 593) Lesson.. Activit (p. 594). You get the same results for each circle: MP NP and mamp manp 90.. The lengths are equal. 3. The angle is a right angle.. heckpoint (p. 597). D 5. D Guided Practice (p. 598). ^&( A is tangent to, and point is the point of tangenc.. 90 ; If a line is tangent to a circle, then it is perpendicular to the radius drawn at the point of tangenc. 3. No; 5 5 7,soTAD is not a right triangle and A **** is not perpendicular to D ****. Therefore, D **** is not tangent to. 4. A D 5. A D 4 6. A D 0 5. Practice and Applications (p ) 7. A A 8. A A 4 r 5 5 r 7 6 r 5 5 r 89 r 9 r 64 r 3 r 8 9. A A 0. A AD 0 r r 65 r 5 r 5 86 Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

3 hapter continued. A AD A AD ( ) ( )( ) ( 4) ( 4)( 4) ( 7) ( 7)( 7) ( ) ( )( ) A A 4 r ( r) 6 r 4 r r r 6 r 4 4r r 6 4 4r 4r 3 r 8. A A r (8 r) 44 r 64 8r 8r r 44 r 64 6r r r 80 6r 5 r 9. A A r (9 r) 44 r 8 9r 9r r 44 r 8 8r r r 360 8r 0 r 0. No; A A So, T A is not a right triangle and A **** is not perpendicular to A ****. Therefore, A **** is not tangent to.. No; A A So, T A is not a right triangle and A **** is not perpendicular to A ****. Therefore, A **** is not tangent to.. A **** AD ****, **** D **** 3. aa aad, aa ada, ada aa 4. T A T AD 5. L J M 6. Yes. Eplanations ma var. Sample answer: **** JM KM **** since tangent segments from the same point are congruent. **** LJ LK **** since all the radii of a circle are congruent. LM **** LM **** b the Refleive Propert of ongruence. Then T JLM T KLM b the SSS ongruence Postulate. 7. Let R represent the center of Earth. A AR R A 3960 (,500 3,960) A 5,68,600 70,93,600 A 55,50,000 A 5,977 miles A 5,977 miles 8. F F K 3960 F ,68,600 F 5,683,84.04 F F 39.8 E E E E 5,68,600 5,68,679. E 79. E 9 FE E F miles opright McDougal Littell Inc. Geometr, oncepts and Skills 87 hapter Worked-Out Solution Ke

4 hapter continued. Standardized Test Practice (p. 600) 9. D; EF EG G; PR RS PS r 36 (r 8) r 96 r 8r 8r 34 r 96 r 36r r r 7 r. Mied Review (p. 600) 3. es; 5 4, 5 4, and no; es; 3 3 5, 3 5 3, and es; 9 8 5, 9 5 8, and no; no; (5) (4) 39. (38) 9. Algebra Skills (p. 600) 40. m m 4. m m m 7 ( 3) ( ) m 4 ( 7) 6 3 Lesson.3.3 heckpoint (pp ). ms 58, mefs 58 ;. ms 58, mds 7 ; es no 3. mds 7, mdes ( ) 88 7 ; es 4. mfe t , mf t ; es 5. Arc length of Ar 0 π() 4 π 4.9 in Arc length of DFE t 80 π(4) 4π.57 ft Arc length of MNs π(6) 3π 9.4 cm.3 Guided Practice (p. 604). major arc: AD t or AD t; minor arc: Ar or Dr; semicircle: AD t. Sample answer: A D 3. The measure of an arc is the degree measure of the related central angle (or minus the measure of the related central angle), while an arc length is a portion of the circumference of a circle. 4. mrsr mrps t PQR t is a semicircle, so mpqr t Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

5 hapter continued 7. mqs r mqsp t mqtr t mqrr Arc length of Ar π() 4 π.40 d 9. Arc length of DEs 00 π(6) 0 π 0.47 cm 3. Arc length of FGH t 40 π(5) 0 π(5) 5 5 π m.3 Practice and Applications (pp ) 3. PQr; mpqr DEs; mdes LNr; mlnr AD t; mad t WXY t; mwxy t GFH t; mgfh t minor arc 0. minor arc. semicircle. minor arc 3. major arc 4. semicircle 5. major arc 6. major arc 7. maa mar maa mar maa mar mklr mmns mlnk t mmkn t NJK t is a semicircle, so mnjk t mamql 80 (60 55 ) mml r majqn makql mjmr mlnr If two cities differ b 80 on the wheel, then it is 3:00 P.M. in one cit when it is 3:00 A.M. in the other cit. 43. No; the circles are not congruent. 44. Yes; aad ae since the are vertical angles; mads mes, so ADs Es. 45. Yes; UWs and XZs are arcs of congruent circles with the same measure. 46. No; F is not the center of the circle, so ou cannot determine the measures of JKs and GHs Length of Ar π(3) 3 π.36 cm 4 opright McDougal Littell Inc Length of Ar π(7) 7 π 7.33 in Length of Ar 0 π(0) π 0.94 ft Length of Ar π(4) π.09 cm Length of Ar π() 5π 5.7 m 5. Length of Ar 50 π(6) 5π 5.7 in. 53. No; the have the same arc length onl if the two circles are congruent circles. 54. Arc length π(5588) 45 cm.3 Standardized Test Practice (p. 607) 55. ; length of As π(8) 9.6 ft.3 Mied Review (p. 607) 56. sin sin sin 56 9 sin ; 5.0 ; cos 56 4 cos cos 56 9 cos sin sin ; cos cos Algebra Skills (p. 607) km km ft in. 7 in. in. 6 0 in in ar 7 ds ft 3 3 ft 9 ft ft 7 ft oz 4 oz 4 oz lbs 8 6 oz 8 oz Quiz (p. 607). tangent. secant 3. diameter 4. chord 5. radius 6. point of tangenc Geometr, oncepts and Skills 89 hapter Worked-Out Solution Ke

6 hapter continued 7. QP RP 5 8. QP RP 9. PQ PR Length of Ar π(3) 7 π 3.67 cm 6. Length of Ar 50 π(7) 3 5 π 8.33 m 6 5. Length of Ar π(0) 5 π 4.36 ft 8 Lesson.4.4 heckpoint (pp ). JM HM. SR SN NR 3. ED A HG FG 5. mazwy mavwu 5 ( 0) Guided Practice (p. 60). E **** is a diameter.. PQs and RS r are congruent arcs. 3. PT RT 4. mamhn majhk 6 ( 5) DE A 8.4 Practice and Applications (pp. 6 6) 6. No; A **** is not perpendicular to D ****,soa **** is not a diameter of the circle. 7. No; A **** does not bisect D ****,soa **** is not a diameter of the circle. 8. Yes; A **** is a perpendicular bisector of D ****,soa **** is a diameter of the circle. 9. A DE 0. maa made 7 35 ( 6) 9. D A **** A **** (given) and As s (If two chords are congruent, then their corresponding minor arcs are congruent.) 3. As Ds (given) and **** A D **** (If two minor arcs are congruent, then their corresponding chords are congruent.) 4. **** A D **** (given) and As Ds (If two chords are congruent, then their corresponding minor arcs are congruent.) aaq aqd b the definition of the measure of a minor arc. 5. DF FE 0 6. mds mds mds mes 0 8. mar meds med t meds mds ( 0) ( 30) AF F 0. maa made ( 45) Answers ma var, but the re-creation should be based on the method shown in Eample on page The searcher is constructing a chord of the beacon s circle and the perpendicular bisector of the chord, which is a diameter of the circle. locating the midpoint of the diameter, the searcher locates the center of the circle, which is the location of the beacon..4 Standardized Test Practice (p. 6) 3. a. In a circle, if two chords are congruent, then their corresponding minor arcs are congruent. b. mads mes (5 40) (0 0) c. mads (5 40) (5 0 40) (50 40) 0 ; mes mads 0 d. mds mads mes maes Mied Review (p. 6) 4. mdes mas ms 80 mas mds Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

7 hapter continued 6. maes 80 mdes 7. md t ms mds ma t mas ms Algebra Skills (p. 6) Lesson.5.5 Activit (p. 63) 9. made t mads mdes Step 3. Angle measures will var, but the measures of all three angles will be equal.,. Answers will var, but in each case, marts marus marvs marps. 3. The measure of an inscribed angle is half the measure of the corresponding central angle..5 Guided Practice (p. 67). aa, a, a, ad. aa and a, a and ad 3. maljk mkls 4. makjl mkml t 0 mkls 40 mkls 5. maljk mlmk t 90 mkml t 80 mkml t 6. (30 ) 05 mlmk t 5 0 mlmk t ; 95; z (50 ) z Practice and Applications (pp ) 9. maa mas 0. maa mas.5 heckpoint (pp ) (0 ) (8 ). maa ms (90 ) 45. madef mdfs (60 ) makmp mknp t 0 mknp t 40 mknp t 4. Theorem.8, the triangle is a right triangle. So, 90 and Theorem.8, the triangle is a right triangle. So, 90 and 90 so Theorem.8, the triangle is a right triangle. So, 90 and ; 90; ; maa mas. malmn mlns (80 ) 90 (68 ) mapqr mprs 4. mauts mus s (34 ) 67 (38 ) 9 5. maa ms 6. maa ms 3 ms 64 ms 78 ms 56 ms 7. maa ms 8. marst mrut t 4 ms 8 ms 0 mrut t 40 mrut t opright McDougal Littell Inc. Geometr, oncepts and Skills 9 hapter Worked-Out Solution Ke

8 hapter continued 9. mapqn mpns 0. maxyz mxwz t 50 mpns 00 mpns 03 mxwz t 06 mxwz t. maea mes. made mes 47 mes 94 mes 3. maaed made made 80 maaed maaed maad mads ( ) mads 5. maad mads (06 ) mads 06 mads 6. mdes mda r mas mes mdes mdes 60 (94 ) Yes; Sample answer: maa 47 made (from E. ) and made maa (vertical angles), so T A T DE b the AA Similarit Postulate. 8. T A is an inscribed triangle and **** A is a diameter, so T A is a right triangle with diameter **** A ; 90; T KLM is an inscribed triangle and KM **& is a diameter, so T KLM is a right triangle with diameter KM **&; 90; T PQR is an inscribed triangle and PR **** is a diameter, so T PQR is a right triangle with diameter PR **** ; 90; Sample Answer: Position the verte of the tool on the circle and mark the two points at which the sides intersect the circle; draw a segment to connect the two points, forming a diameter of the circle. Repeat these steps, placing the verte at a different point on the circle. The center is the point at which the two diameters intersect ; 80; ; Yes; both pairs of opposite angles are right angles, which are supplementar angles. 36. Yes; both pairs of opposite angles of an isosceles trapezoid are supplementar. 37. No; if a rhombus is not a square, then the opposite angles are not supplementar. 38. Yes; both pairs of opposite angles are right angles, which are supplementar angles..5 Standardized Test Practice (p. 69) 39. ; maad maa (80 ) H; Mied Review (p. 69) ma ; 8 sin 44 A A sin A sin 44 A.5; tan 44 A 8 A tan A tan 44 A Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

9 hapter continued 48. maj ; sin 35 J K sin 35 JK 6.3 JK; cos 35 K L cos 35 KL 9.0 KL 49. mar ; tan 50 R 5P 5 tan 50 RP 6.0 RP; sin 40 R 5Q RQ sin RQ sin 40 RQ Algebra Skills (p. 69) () () () ( 4)( 4) ( 4)( 4) (6)( ) () 3() () 3 () Lesson Geo-Activit (p. 60) Step. maae 50 Step 3. The measure of aae is 50 for ever student. Step 4. Answers will var, but the measure of aae will alwas be mar mds. Step 5. The measure of an angle formed b intersecting chords of a circle is equal to half the sum of the measures of the intercepted arcs..6 heckpoint (pp. 6 6). mar mds. mar mds (90 70 ) (60) (66 70 ) (36) mar mds 7 ( 99 ) PN PL PM PK 5. A D E RV VT SV VU 7. FH HJ EH HG G GE DG GF 9. KN NM LN NJ Guided Practice (p. 63). Points and E lie inside the circle.. ma mads ms 3. ma mads ms (55 65 ) (0 ) ma ms mads (0 68 ) (78 ) 39 (88 88 ) (76 ) opright McDougal Littell Inc. Geometr, oncepts and Skills 93 hapter Worked-Out Solution Ke

10 hapter continued.6 Practice and Applications (pp ) D. A. ms mads 3. mds mas (34 6 ) (96) (75 5 ) (00) mas mds (30 96 ) (6) mds mar mas mds 55 (89 ) 59 (70 ) mads ms 9 ( 7 ) Yes; Sample answer: If two chords intersect and the measure of each angle formed is the same as the measure of the arc intercepted b the angle, then the circle is divided into two pairs of congruent arcs since vertical angles are congruent. Suppose one pair of arcs has measure and an angle intercepting the arc is a. Then ma ( ) ( ). The angle formed b the chords has the same measure as its intercepted arc, so it is a central angle. 9. N N N ms mads 80 ( 3 ) 80 (54) mas mds 80 (8 00 ) 80 (8) ms mads 80 (90 50 ) 80 (40) EA E E ED; If two chords intersect inside a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord..6 Standardized Test Practice (p. 65) 7. a b. maa mas mdes (00 80 ) (80 ) 90 c. maes mas mds mdes maes maes 6 maes 44 d. Yes; Sample answer: b the Vertical Angles Theorem, aa aed. Also, A 4 E 9 3 D so the two sides that include the congruent angles are proportional. Thus, T A T ED b the SAS Similarit Theorem. 94 Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

11 hapter continued.6 Mied Review (p. 65) 8. 3 b c 9 b c b 9 56 c b a 3 4 a 9 96 a 87 a Algebra Skills (p. 65) c Quiz (p. 65) (3 ) ( 63) The measure of an inscribed angle is half the measure of its intercepted arc; maa mas ma mads ma 58 ; mad ms maa 4 6. The opposite angles of an inscribed quadrilateral are supplementar; ma mad ; maa ma mas mds 35 ( 07 ) Technolog Activit (p. 66). Product will var, but in each case, JH JG JK JL.. The relationship JH JG JK JL is true. 3. no 4. If ^&( JL and ^&( JG are secants of a circle intersecting at a point J outside the circle and points K, L, G, and H are points on the circle as shown in the diagram on page 66, then JH JG JK JL. 5. Measures will var, but in each case, makjh mlgs mkhs. If secants intersect outside a circle, then the measure of the angle formed is half the positive difference of the measures of the intercepted arcs. Lesson.7.7 heckpoint (pp ). r. r ( h) ( k) r ( ( 4)) ( ( 6)) 5 4. ( 4) ( 6) 5 7. mas mds (80 44 ) (4) 6 opright McDougal Littell Inc. Geometr, oncepts and Skills 95 hapter Worked-Out Solution Ke

12 hapter continued radius, center (, 5); 5. radius 8, center (0, 5);.7 Guided Practice (p. 69).. radius, center (0, 0); 3. radius 4, center (, 0); ( h) ( k) r ( h) ( k) r ( 0) ( 0) ( ) ( 0) 4 4 ( ) 6 4. radius, center (, ); ( h) ( k) r ( ( )) ( ) ( ) ( ) 4.7 Practice and Applications (pp ) 5. ; radius, center (3, 0); 6. A; radius, center (0, 0); ( 3) ; radius, center ( 3, 0); ( 3) 4 8. radius 6, center (0, 0); 9. radius, center (0, 0); 0. radius 7, center (, 6);. radius 4, center (4, 3);. radius 5, center (5, ); 3. radius 6, center (, 3); 6. radius, center ( 3, ); ( h) ( k) r ( ( 3)) ( ) ( 3) ( ) 4 7. radius, center (0, ); 8. radius, center (3, 3); ( h) ( k) r ( h) ( k) r ( 0) ( ) ( 3) ( 3) ( ) 4 9. radius.5, center (0.5,.5); ( h) ( k) r ( 0.5) (.5).5 ( 0.5) (.5) radius 4, center (, );. radius 6, center (0, 0); ( h) ( k) r r ( ) ( ) 4 6 ( ) ( ) ( h) ( k) r ( 0) ( 0) ( h) ( k) r ( 4) ( 0) 4 ( 4) 6 4. ( h) ( k) r ( 3) ( ( )) ( 3) ( ) 4 5. ( h) ( k) r ( ( )) ( ( 3)) 6 ( ) ( 3) ( h) ( k) r ( ( 3)) ( 5) 3 ( 3) ( 5) 9 7. ( h) ( k) r ( ) ( 0) 7 ( ) Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

13 hapter continued 8. ( ) ( 3) 4 (0 ) (0 3) 4 ( ) ( 3) R(0, 0) is outside the circle. 9. ( ) ( 3) 4 ( ) ( 4 3) 4 0 ( ) 4 4 A(, 4) is inside the circle. 30. ( ) ( 3) 4 (0 ) ( 3 3) 4 ( ) X(0, 3) is on the circle. 3. ( ) ( 3) 4 (3 ) ( 3) K(3, ) is outside the circle. 3. ( ) ( 3) 4 ( ) ( 4 3) 4 ( ) ( ) M(, 4) is inside the circle. 33. ( ) ( 3) 4 ( ) ( 5 3) 4 0 ( ) T(, 5) is on the circle. 34. ( ) ( 3) 4 ( ) (0 3) D(, 0) is outside the circle. 35. ( ) ( 3) 4 (.5 ) ( 3 3) 4 (0.5) Z(.5, 3) is inside the circle. 36. circle A: 3 9; circle : ( 5) ( 3).5 ( 5) ( 3) 6.5; circle : ( ) ( 5) ( ) ( 5) Tower A transmits to J; tower transmits to K; towers and transmit to L; no tower transmits to M; tower transmits to N. 38. Sample answer: The student did not subtract the coordinates of the center from and in the equation and did not square the radius. The equation should be ( ) ( ) ( h) ( k) r 40. ( h) ( k) r (4 ) (6 ) r (5 3) ( ) r 3 4 r 0 r 9 6 r 4 r 5 r r 5 r ( 3) ( ) 4 ( ) ( ) 5.7 Standardized Test Practice (p. 63) 4. D; ( ( 3)) ( ) ( 3) ( ) 4 4. H; ( ( 3)) ( 0) 5 ( 3) 5 ( 3 3) Mied Review (p. 63) 43. P P (, 3) (, 3) (3, 3); Q Q (6, ) (6, ) (8, ); R R (4, ) (4, ) (6, ); S S (, 0) (, 0) (, 0) opright McDougal Littell Inc. Geometr, oncepts and Skills 97 hapter Worked-Out Solution Ke

14 hapter continued 44. P P (, 3) ( 4, 3 ) ( 3, 4); Q Q (6, ) (6 4, ) (, ); R R (4, ) (4 4, ) (0, ); S S (, 0) ( 4, 0 ) ( 5, ) 45. P P (, 3) (, 3 ) (0, ); Q Q (6, ) (6, ) (5, 0); R R (4, ) (4, ) (3, 3); S S (, 0) (, 0 ) (, ) 46. P P (, 3) ( 3, 3 6) (4, 9); Q Q (6, ) (6 3, 6) (9, 7); R R (4, ) (4 3, 6) (7, 4); S S (, 0) ( 3, 0 6) (, 6) 47. reduction; k P P enlargement; k P P Algebra Skills (p. 63) Lesson.8.8 Geo-Activit (p. 634) Step 4. 0 ; es, ou can turn the paper clockwise or counterclockwise. Step 5. rectangle: 80 clockwise or counterclockwise; square: 90 or 80 clockwise or counterclockwise.8 heckpoint (pp ). no. es; it can be mapped onto itself b a clockwise or counterclockwise rotation of 80 about its center. 3. es; it can be mapped onto itself b a clockwise or counterclockwise rotation of 45, 90, 35, or 80 about its center. 4. A (0, 0), (0, 3), ( 4, 3).8 Guided Practice (p. 636). A center of rotation is the fied point about which a figure is turned when it is rotated.. A figure in a plane has rotational smmetr if the figure looks the same after it is rotated 80 or less. 3. Yes; it can be mapped onto itself b a clockwise or counterclockwise rotation of 80 about its center. 4. Yes; it can be mapped onto itself b a clockwise or counterclockwise rotation of 80 about its center. 5. no 6. A clockwise rotation of 60 about P maps R onto S. 7. A counterclockwise rotation of 60 about P maps R onto Q. 8. A clockwise rotation of 0 about Q maps R onto W. 9. A counterclockwise rotation of 80 about P maps V onto R..8 Practice and Applications (pp ) 0. Yes; it can be mapped onto itself b a clockwise or counterclockwise rotation of 90 or 80 about its center.. Yes; it can be mapped onto itself b a clockwise or counterclockwise rotation of 7 or 44 about its center.. no A A 3. The wheel hub can be mapped onto itself b a clockwise or counterclockwise rotation of 7 or 44 about its center. 4. The wheel hub can be mapped onto itself b a clockwise or counterclockwise rotation of 36, 7, 08, 44, or 80 about its center. 5. The wheel hub can be mapped onto itself b a clockwise or counterclockwise rotation of 45, 90, 35, or 80 about its center. 98 Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

15 hapter continued R S 9. A 8. Y X 9. X W W P P Z Z A Y P T S A A D D P T R 30. A(, ) A (, ) (, 4) ( 4, ) (5, ) (, 5); The coordinates of the image of the point (, ) after a 90 clockwise rotation about the origin are (, ). Z 6 A A 6 X 0. S. R P. D **** 3. LH **** 4. T FGL 5. T MA 6. GE **** T 6 J K J R J(, ) J (, ) K(, 4) K (4, ) L(3, 4) L (4, 3) M(3, ) M (, 3); The coordinates of the image of the point (, ) after a 90 clockwise rotation about the origin are (, ). 6 E E M L D F S L M K 6 D F T D(, 4) D (4, ) E(, 0) E (0, ) F(5, ) F (, 5); The coordinates of the image of the point (, ) after a 90 clockwise rotation about the origin are (, ). W Z Y X Z P W Y X X(, 3) X (, 3) O(0, 0) O (0, 0) Z(3, 4) Z ( 3, 4); The coordinates of the image of the point (, ) after a 80 rotation about the origin are (, ). 3. The design has rotational smmetr about its center; it can be mapped onto itself b a clockwise or counterclockwise rotation of The design has rotational smmetr about its center; it can be mapped onto itself b a clockwise or counterclockwise rotation of 90 or Yes. The image can be mapped onto itself b a clockwise or counterclockwise rotation of 80 about its center. 34. Yes; the answer would change to a clockwise or counterclockwise rotation of 90 or 80 about its center. If ou disregard the colors of the figures, then, for eample, the green fish in the middle map onto the orange fish and the orange fish map onto the green fish. 35. The center of rotation is the center of the circle. 36. Yes; this piece could be hung upside down because the image can be mapped onto itself b a clockwise or counterclockwise rotation of 80 about its center..8 Standardized Test Practice (p. 638) 37. D; in a 90 clockwise rotation, (, ) (, ). 38. H O O X.8 Mied Review (p. 639) 39. A bh ft 40. A bh cm Z 4 opright McDougal Littell Inc. Geometr, oncepts and Skills 99 hapter Worked-Out Solution Ke

16 hapter continued 4. A h(b b ). (0)(6 0) 8 A (0)(6) 80 m.8 Algebra Skills (p. 639) Quiz 3 (p. 639). center (, 6), radius 5. ( h) ( k) r ( 0) ( ( 4)) 3 ( 4) A A(, 4) A ( 4, ) (4, 4) ( 4, 4) (4, ) (, 4) D(, ) D (, ); The coordinates of the image of the point (, ) after a 90 counterclockwise rotation about the origin are (, ). 6 D D A A A(, 0) A (0, ) (4, ) (, 4) (3, 3) ( 3, 3); The coordinates of the image of the point (, ) after a 90 clockwise rotation about the origin are (, )..8 Technolog Activit (p. 640) 7. Yes; it can be mapped onto itself b a clockwise or counterclockwise rotation of 90 or 80 about its center. 8. Yes; it can be mapped onto itself b a clockwise or counterclockwise rotation of 80 about its center. 9. no 0. A 6 A 4 A(, 4) A (, 4) (, ) (, ) (, 3) (, 3); The coordinates of the image of the point (, ) after a 80 rotation about the origin are (, ).. T A is a rotation of T A about point P.. AP A P 3. The measure of aapa is twice the measure of the acute angle formed at the intersection of lines m and k. 4. The measure of aapa changes so it is still twice the measure of the acute angle formed at the intersection of lines m and k. 5. es hapter Summar and Review (pp ). A secant is a line that intersects a circle in two points.. A polgon is inscribed in a circle if all of its vertices lie on the circle. 3. A line in the plane of a circle that intersects the circle in eactl one point is called a tangent. 4. If the endpoints of an arc are the endpoints of a diameter, then the arc is a semicircle. 5. An inscribed angle is an angle whose verte is on a circle and whose sides contain chords of the circle. 6. A chord is a segment whose endpoints are points on a circle. 00 Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

17 hapter continued 7. A rotation is a transformation in which a figure is turned about a fied point. 8. diameter 9. point of tangenc 0. chord. center. tangent 3. secant 4. A A 5. A A 0 8 r 6 4 r r r 36 r 00 r 6 r 0 r 6. A A 7. A AD 7 5 r r 64 r 8 r 8. A AD A AD mdes mes mds. maes 80 mde s mae t maes mes ms 80 mds mar md t ms opright McDougal Littell Inc. 5. mda t mar Arc length π(3) 5 π.6 in Arc length 0 π(8) 6 π 6.76 cm 3 8. Arc length 00 π(5) 5 π 8.73 m 9 9. A ED A D 3. maa made DF FE mes mdr mes mads maa mas 36. maa mas (06 ) maa mad t 00 mad t 00 mad t 39 mas 78 mas ; ; ; mas mds 4. mas mds (80 0 ) (90) (60 30 ) (90) mas mds 80 (00 0 ) 80 (0) Geometr, oncepts and Skills 0 hapter Worked-Out Solution Ke

18 hapter continued 47. ( ) ( 5) 3 ( ) ( 5) ( ( 4)) ( ( )) 4 ( 4) ( ) ( 5) ( ( )) 7 ( 5) ( ) radius 3, center ( 4, ); 5. radius 4, center (, 3); 5. radius 5, center (0, 0); 53. A A 54. P F hapter Test (p. 646). is the center.. A A D is a point of tangenc. 5 A 0 **** A or F **** is a chord. 65 A 400 ^&( A is a secant 5 A F **** or **** is a radius. 5 A F **** is a diameter 3. SR RV 4. ms mas mar ms 6. mad t mds 80 mas mae t mds mdes Arc length 5 π() 3 π 4.09 in Arc length π(5) 5 π 6.54 m 0. A D (88 ) H 44 G H P ; 55. F G N K ( 6) ( ( 3)) 5 ( 6) ( 3) 5 L M P M L 5. K N 56. Yes; it can be mapped onto itself b a clockwise or counterclockwise rotation of 90 or 80 about its center. 0 Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

19 hapter continued The coordinates of the point (, ) after a 90 counterclockwise rotation about the origin are (, ). A(0, 3) A ( 3, 0); (3, ) (, 3); (, ) (, ) 9. ( 30) ( 30) 30 ( 30) ( 30) 900 hapter Standardized Test (p. 647). D 70. G; Arc length π(8) 9.77 cm 3. A 4. H; D; G; ma muvs mstr (95 67 ) (6 ) 8 7. ; ( 7) ( ( )) 8 ( 7) ( ) F; , hapters umulative Practice (pp ). A A. RT RS ST ST 5 ST 3. mastr mastu mautr opright McDougal Littell Inc. 4. mafgh majgf majgh 73 8 majgh 45 majgh 5. M, 3 ( ), 9 5, 4 (, 7) 6. ST ( ) ( ) ( 3 ) ( 5 9 ) ( 4 ) ( 4 ) ; ; ; ; maap madp (6 4) 4 4 7; maap (6(7) 4) (4 4) 8 ; map 80 maap A D ^&( A and D ^&( do not intersect, and the are not parallel. The two lines are skew lines. 3. Since alternate eterior angles are congruent, ou can use the Alternate Eterior Angles onverse to show l m. 4. Since alternate interior angles are congruent, ou can use the Alternate Interior Angles onverse to show l m. 5. Since same-side interior angles are supplementar, ou can use the Same-side Interior Angles onverse to show l m. Geometr, oncepts and Skills 03 hapter Worked-Out Solution Ke

20 hapter continued 6. In a plane, if two lines are perpendicular to the same line, then the are parallel to each other. 7. In an isosceles triangle, the base angles are congruent, and b the Triangle Sum Theorem, the sum of the angles of a triangle is So, the measure of each base angle is A A obtuse 9. **** is the shortest side, so aa is the smallest angle; A **** is the longest side, so a is the largest angle. 0. Yes; aa aj, a al and the included sides, **** A and **** JL are congruent. So, T A T JKL b the ASA ongruence Postulate.. No; ou onl know that GF **** DE **** and GE **** GE **** b the Refleive Propert of ongruence.. Yes; Sample answer: PQ **** RQ **** and PS **** RS **** and their included angles, ap and ar, are congruent. So, T PQS T RQS b the SAS ongruence Postulate. 3. Yes; hpotenuses **& XU and VU **** are congruent and legs XY **** and VW **** are congruent in right triangles T XYU and T VWU. So, T XYU T VWU b the HL ongruence Theorem. 4. A AD ; 8; mak mal mam man ; ; z z 0 8. () (3) length 3 3(6) 8 cm width (6) cm 9. Sample answer: ad aae and ad aae (Refleive Propert of ongruence). Therefore, T D T AE b the AA Similarit Postulate. 30. D AE A bh bh (6)(8) (0 6)(6) (6)(8) (4)(6) m 3. A h(b b ) (8)( 9) (8)(3) 4 ft 33. A d d (6)(0) 80 cm 34. A πr 35. A bh π(8) 84 4h 34π 6 m h 08 d 04 Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.

21 hapter continued 36. S Ph (6 8) ( 6 8)(4) (48) ( 6)(4) 96 (8)(4) in. ; V h (6 8)(4) (48)(4) 9 in S πr πrl 38. S πr πrh π(8) π(8)(7) π(6) π(6)(0) 64π 36π 7π 0π 00π 9π 68 cm ; 603 m ; 8 h 7 V πr h 64 h 89 π(6) (0) h 5 360π h 5; 3 m 3 V 3 πr h 3 π(8) (5) 3 π(64)(5) 30π 005 cm S 4πr 4π() 576π 80 mm ; V 4 3 πr3 4 3 π()3 4 3 π(78) 304π 738 mm sin R leg opposite ar hpotenuse 39 cos R leg adjacent ar hpotenuse 39 tan R l eg opposite ar leg adjacent ar Since sin R , mar sin ; mat 90 mar ; tan 30 7 a a tan a tan 30 a.; sin 30 7 c c sin c sin 30 c ; sin 65 8 r 8 sin 65 r 7.3 r; cos 65 p 8 8 cos 65 p 3.4 p 44. m 9 6 m 8 56 m 377 m 8.4; tan 9 tan ; d 0 9 d d 44 d ; sin 0 sin ; A m Ar πr 75 π(6) 7.5π 3.6 ft 47. mar 75 ; ma t opright McDougal Littell Inc. Geometr, oncepts and Skills 05 hapter Worked-Out Solution Ke

22 hapter continued Arc length π(6).5π 7.9 ft 49. (3 30) ( 7) (9 ) (84 06 ) (90) ( 0) ( ( 3)) 7 ( 3) A A A A 56. A A A (, 5), ( 4, ), (, ) 06 Geometr, oncepts and Skills hapter Worked-Out Solution Ke opright McDougal Littell Inc.