EXTRA PRACTICE. 2. Add 1 to the first number, then 2 to the second number, and n to the nth number; > AC AB BC 12 8 BC 4 BC

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1 EXTRA PRACTICE Chapter (p. 803). Each number is the previous number;. Add to the first number then to the second number and n to the nth number; 6 3. Each number is five times the previous number; 65. Add to the first number to get the fourth number then add to the fourth number to get the seventh number then add to the seventh number to get the tenth number. The numbers in between are s; 0 5. Each number is.5 times the previous number; 6 6. Each number is 3 times the previous number; 6 7. An negative number used is a negative number. 8. For n 3 n n > n n For n n n > n n For n 5 n n > n n For n n n > n n For n n n > n n 9. Sample answer: A B C D 0. Sample answer: P M N. Sample answer: D 3 3 > 3 3 > 0 > > 7776 > > Copright McDougal Littell Inc. A 3 > 3 8 > 6 5 > > > 3 B 5 6 > 6 5 > > 8 > 9 C. Sample answer: 3. Sample answer:. Because C is the midpoint of AE 5. AD AB BC CD Because C is the midpoint of AE So AD BD BC CD 7. Because C is the midpoint of AE R S T AC AE AC AB BC 8 BC BC 8 CD CD CE AE CE CD DE CD 5 7 CD 7 AC AE CE CD DE CD CE DE 5 7 BE CE BC 6 E F G Geometr 77 Etra Practice Worked-out Solutions Ke

2 Etra Practice continued 0. HM ML Because HM ML HM ML. HM ML Because HM ML HM ML. HM 5 (6 6 7 ML Because HM ML HM ML. 3. Verte is Q. The sides are QP and QR. PQR or RQP.. The verte is F. The sides are FG and FE. GFE or EFG. 5. The verte is B. The sides are BA and BC. ABC or CBA. 6. mstr mhjk mhjl mljk mdef mdec mcef mdef mdef obtuse; right; acute; 5 78 Geometr Etra Practice Worked-out Solutions Ke 3. midpoint of 33. midpoint of 3. midpoint of r 7 r muxy m 5 m m 5 m The angle is The complement is 0. P s 7 8 units A s 7 9 square units P units A bh 6 r square units PQ 8 PQ PQ muxy muxy mbxy 37. mbxy muxy 5z 9 7z 3 z z muxy P 3 5 units A bh 3 6 square units Copright McDougal Littell Inc.

3 Etra Practice continued. 6. P l w A lw P s 3 A s 3 8. C d A r Chapter units 88 square units units 9 square units units square units 5. C r units A r square units P l w units A lw square units. If ou read it in a newspaper then it must be true.. If ou have an apple a da then it will keep the doctor awa. 3. If a number is odd then its square is odd.. Inverse: If then. Converse: If then. Contrapositive: If then. 5. Inverse: If ou are not indoors then ou are caught in a rainstorm. Converse: If ou are not caught in a rainstorm then ou are indoors. Contrapositive: If ou are caught in a rainstorm then ou are not indoors. 6. Inverse: If four points are not collinear then the are not coplanar. Converse: If four points are coplanar then the are collinear. Contrapositive: If four points are not coplanar then the are not collinear. 7. Inverse: If two angles are not vertical angles then the are not congruent. Converse: If two angles are congruent then the are vertical angles. Contrapositive: If two angles are not congruent then the are not vertical angles. 8. If two angles are supplementar then the form a linear pair; false. For eample two consecutive angles of a parallelogram are supplementar but the do not form a linear pair. 9. If 6 then 5 7; true 0. If two segments have the same length then the are congruent. If two segments are congruent then the have the same length.. If two angles are right angles then the are supplementar. If two angles are supplementar then the are right angles.. If 0 then 00; if 00 then Conditional statement: If ABC is a right angle then AB BC. Converse: If AB BC then ABC is a right angle. Both of these statements are true so the can form a true biconditional statement.. Conditional statement: If and are adjacent supplementar angles then and form a linear pair. Converse: If and form a linear pair then and are adjacent supplementar angles. Both of these statements are true so the can form a true biconditional statement. 5. Conditional statement: If two angles are vertical angles then the are congruent. Converse: If two angles are congruent then the are vertical angles. The first of these statements is true but the second is false. So the cannot form a true biconditional statement. 6. If we go shopping then we need a shopping list. 7. If we do not stop at the bank then we do not see our friends. 8. If we stop at the bank then we see our friends. 9. We go shopping if and onl if we need a shopping list. 0. If we do not go shopping then we will not see our friends.. We go shopping if and onl if we stop at the bank.. p: It is hot. q: Ma will go to the beach. ~p ~q: If it is not hot Ma will not go to the beach. ~q ~p: If Ma does not go to the beach then it is not hot. Copright McDougal Littell Inc. Geometr 79 Etra Practice Worked-out Solutions Ke

4 Etra Practice continued 3. p: The hocke team wins the game tonight. q: The will pla in the championship. ~p ~q: If the hocke team does not win the game tonight the will not pla in the championship. ~q ~p: If the hocke team does not pla in the championship then the did not win the game tonight.. p: John misses the bus. q: He will be late for school. ~p ~q: If John does not miss the bus then he will not be late for school. ~q ~p: If John is not late for school then he did not miss the bus. 5. AB 6. DF ED 7. AB DF Statements Reasons. P is the midpoint. Given of AC and BD.. AP PC. Definition of midpoint 3. BP PD 3. Definition of midpoint. PD PC.. Given 5. PD PC 5. Definition of segments 6. AP BP 6. Trans. prop. of equalit 7. AP BP 7. Def. of congruent segments Sample answers: RWU QWV 33. B definition of complementar angles m3 m 90 and m m5 90 so m3 m m m5. From the subtraction propert of equalit m3 m5. Therefore b the definition of z z b 36 7b 0 b 6 b 8 5c 9 6c 8 7 c z 80 z 56 z z 36. 3r 5r 6 50 r 5 r 5s 8s s 0 s 37. Statements Reasons. DBE is a rt. angle. Given. mdbe 90. Definition of right 3. DBE 3 3. Angle Addition Postulate. m m3 90. Substitution Prop. of Equalit Vertical arc. 6. m m 6. Definition of 7. m m Substitution Prop. of Equalit 8. and 3 are 8. Def. of complementar complementar Chapter 3 (p. 807). parallel. perpendicular 3. skew. Sample answers: BC HF GE 5. Sample answers: AB BC HG GE 6. Sample answers: GH HF AB AD 7. Sample answers: FHA HAD ADF DFH (These are all the same plane.) 8. corresponding 9. alternate interior 0. alternate eterior. consecutive interior. corresponding 3. Statements Reasons. AB BC. Given. ABC is a right.. Def. of perpendicular lines 3. mabc Def. of right. BD bisects ABC. Given 5. mabd mdbc 5. Def. of bisector 6. mabd mdbc If sides of adj. are then the are complementar. 7. mabd mabd Substitution propert of equalit 8. mabd Distributive propert 9. mabd 5 9. Division propert of equalit. 0; vertical are. 0; consecutive interior are supplementar ; alternate eterior are. 50; and form a linear pair ; linear pair postulate 85; corresponding are. 80 Geometr Etra Practice Worked-out Solutions Ke Copright McDougal Littell Inc.

5 Etra Practice continued 7. 5; vertical are corresponding are. 3. p : ; alternate eterior are. 70; and form a linear pair ; linear pair postulate 8; alternate interior angles are. 0. AE DB; alternate interior angles converse thm. DE CG; Corresponding angles converse thm. No parallel lines 3. Corresponding are. Use corresponding converse thm.. Consecutive interior are supplementar. Use consecutive interior converse thm. 5. Alternate interior are. Use alternate interior converse thm Slope of AB 3 6 Slope of CD Slope of EF 8 0 AB CD EF because their slopes are all equal. 7. Slope of Slope of 8 3 Slope of EF 0 There are no parallel lines because their slopes are not equal Slope of AB undefined Slope of AB CD 3 CD undefined 0 6 Slope of EF 5 undefined 0 AB CD EF because the are all vertical lines. 9. m b 30. m b 5 6 b 3 b 5 b 3 b 9 b b p : m m p p p : 3 5 m m p p 8 6 m m 86 3 p is not perpendicular to p. m b b b 3 b m b 7 b 7 b b Chapter (p. 809) me 0 mf mg 50 00; obtuse mk 60 ml 60; equiangular p : 9 3 p : p : 6 m b 0 5 b 0 0 b 0 b Copright McDougal Littell Inc. Geometr 8 Etra Practice Worked-out Solutions Ke

6 Etra Practice continued Statements Reasons XV ZV. Given 60 VY VW. XV ZV. Def. of congruent segments 0 VY VW mp 0 mq XV VW XW 3. Segment Addition Postulate mr ; right triangle ZV VY ZY. mu 3 60; acute 5. The sum of the measure of an eterior angle and its corresponding interior is 80. So the interior angle measure is The sum of the measures of the interior of a triangle is 80. So the measures of the interior angles are 5 5 and ABC FED 7. ABGH BEFG CDEB AEFH CGFD 8. ABCDEF GHJKLM 9. A F B E C D AB FE BC ED AC FD ms 3 6 mt SSS. Not congruent 5. SAS. ZV VY XW. Substitution propert of equalit 5. XW ZY 5. Substitution propert of equalit 6. XW ZY 6. Def. of congruent segments 7. XY ZW 7. Given 8. YW YW 8. Refleive propert of equalit 9. XYZ ZWY 9. SSS Congruence Postulate 7. Yes; AAS 8. Yes; ASA 9. No 0. Statements Reasons. AD BC. Given. DAE BCE. Alternate interior thm 3. AED CEB 3. Vertical are. AC bisects BD. Given 5. DE EB 5. Definition of segment bisector 6. AED CEB 6. AAS Congruence Postulate. B the ASA Congruence Postulate DAC BCA. Because corresponding parts of triangles are it follows that AB CD.. B the SAS Congruence Postulate EFG HJG. Because corresponding parts of triangles are it follows that GEF GHJ. 3. B the SSS Congruence Postulate RST RQT. Because corresponding parts of triangles are it follows that RQT RST.. Given that BAF FBD AFB BDF because corresponding parts of angles are. 5. Given that CBD BAF BC AB because corresponding parts of angles are. 6. Given that FBD DFE FD DE because corresponding parts of angles are. 7. B the Base Angles Theorem 60. B the Triangle Sum Theorem B the Base Angles Theorem Also Geometr Etra Practice Worked-out Solutions Ke Copright McDougal Littell Inc.

7 Etra Practice continued 9. B the Base Angles Theorem 30. Sample answer: 3. Sample answer: A(5 ) C(5 ) F G GE EF FG GE GE GE Sample answer: 35. Sample answer: B( ) units D( ) 90 5; C(3 ) B(3 ) C D(3 ) A(3 ) 7. HG HE HF 5 8. QY QW QX 7 Using HJE Using SQW HE HJ JE WS WQ SQ 5 HJ WS HJ WS HJ WS GF GH GJ 0 BT 3 BY 3 BY. TX 3 CX TX 3 BY TX 8. AT TZ AT Sample Answer:. Sample Answer: N Q E Q M P D F orthocenter is at Q orthocenter is point Q 5. Sample answer: 6. ZX 7. AC S D C 7 3 A 6 B A 7 B 8. T orthocenter YZ AB R 9. XY AC Chapter 5 (p. 8). BC BA. DC DA 0 3. Point E is on the bisector of AC.. B the Angle Bisector Theorem HG HF 5 5. Point K is on the angle bisector of JEM. 6. DA DC DB 0 Using DGC BC XZ 6 YZ AB GC 0 GC 6 YZ 5 9 GC 8 AC GC 8 6 Copright McDougal Littell Inc. Geometr 83 Etra Practice Worked-out Solutions Ke

8 Etra Practice continued. 3. Shortest: Longest: 6. Smallest: BC AC L. Shortest: Longest: 7. Smallest: DF DE Q 5. Shortest: Longest: 8. Smallest: GJ GH T Largest: M Largest: P Largest: R 9. < 30. < > > < 37. < Chapter 6 (p. 83). Not a polgon. Concave octagon 3. Conve heagon. Concave dodecagon 5. Not a polgon 6. Concave decagon AC AZ 3 5 AZ XY XY; opposite sides of a are.. VYX; opposite of a are.. VY; opposite sides of a are. 3. XT; the diagonals of a bisect each other.. VWY; alternate interior thm VY; opposite sides of a are parallel. 6. YXW and YVW; consecutive angles of a are supplementar. 7. VW and WY; the diagonals of a bisect each other. 8. Yes; both pairs of opposite sides are parallel 9. Yes; both pairs of opposite are No; ou don t know that one angle is supplementar to both consecutive angles.. Sample answer: Midpoint of Midpoint of The diagonals bisect each other so ABCD is.. Sample answer: Midpoint of Midpoint of The diagonals bisect each other so EFGH is. 3. Sample answer: Midpoint of Midpoint of The diagonals bisect each other so RSTU is.. Sample answer: Midpoint of Midpoint of AC 9 6 BD 7 3 EG FG RT SU MP NQ The diagonals bisect each other so MNPQ is. 5. parallelogram rhombus rectangle square 6. rectangle square 7. rectangle square 8. rectangle square 9. rhombus square 30. parallelogram rhombus rectangle square 8 Geometr Etra Practice Worked-out Solutions Ke Copright McDougal Littell Inc.

9 Etra Practice continued 3. Rhombus; P Q S R RP PQRS is not a rectangle because the diagonals QS and RP are not congruent. 33. Square; PQ QR RS SP slope PQ slope QR None of the above; Q 3. P Q S R PQ 6 QR 6 RS 6 SP 6 QS PR PQRS is both a rhombus and a rectangle so it is a square. P R P Q S S R PQ SR PS QR QS PQ 5 QR 5 RS 5 SP 5 PR QS PQRS is both a rhombus and a rectangle 8 so 0it is a square mb 0 ma 70 md8 70 ms 65 mt 5 mu mg me Copright McDougal Littell Inc. Geometr 85 Etra Practice Worked-out Solutions Ke

10 Etra Practice continued Kite because it has two pairs of consecutive congruent sides. 5. Parallelogram or rectangle because it has two pairs of opposite congruent sides. 6. Rectangle or square because it has four right BC AB 5 0 BC AB 65 BC AB 5 CD AD 5 36 CD AD 5 CD AD 39 EF EH 6 EF EH 00 EF EH 0 FG GH 9 FG GH 5 FG GH 5 KM KP 8 KM KP 60 KM KP 80 MN NP 8 6 MN NP 00 MN NP 0 A d d units A b b h units Chapter 7 (p. 85). Z. 90 rotation about the origin 3. Sample answers: QR ZY RP YX PQ XZ. Sample answers: R Y P X Q Z Sample answers: RP A bh 05 5 units 86 Geometr Etra Practice Worked-out Solutions Ke 7. Translation of 8 units to the right; E3 F3 3 G6 3 H6 8. Reflection in the -ais; N P6 Q 0 R3 9. CBA GHJ 0. DEF MNK. FED KNM. HKJ 7. N 3. GFE. MPN 5. NPM 6. M5 N 8. YX PQ XZ RQ YZ P 8 A A A. C 6 M M B 9. Q P P. Graph Then draw AB to find that it crosses the -ais at C Graph Then draw AB to find that it crosses the -ais at C 0. A B A 3 7. A Copright McDougal Littell Inc. 3 Q Q C 3 N

11 Etra Practice continued. Graph Then draw AB to find that it crosses the -ais at C6 0. B A C A 3. Follow the instructions for rotating a figure on page 3. H 0.5 E.5 3 F3.5 6 G H K G. Follow the instructions for rotating a figure on page 3. A.5 B C line a and line b 8. KK JJ JKM R R S A S A R S A V S R V T A T V T V T 5. Follow the instructions for rotating a figure on page 3. J K M 0 N7 6 A A R R R S S R S V S V T V T V A T T TR 38. THG Chapter 8 (p. 87) 5 m 500 cm 5 ft 5 ft.. d ft 5 50 cm 50 cm 5 in 5 in 0 km 0000 m ft in m 900 m measures of angles are cm 7 35 cm 5 cm lengths of sides are 5cm 5 cm 5 cm a 3 3a a 7 K 7 8 d d d 0 8 8d 3 d H G b b b. f c 9 8 6c 3 c 8 9 f 3 8 9f 7 9 9f f Copright McDougal Littell Inc. Geometr 87 Etra Practice Worked-out Solutions Ke

12 Etra Practice continued 0. g g g g u u z 90 u 9 z 0 8. Perimeter of PQRS Perimeter of TVWX Yes; ABC~DEF b the AA Similarit Postulate. 30. Yes; GHJ~KMJ b the AA Similarit Postulate. 3. No; the triangle are not similar because there are not two congruent angles. 3. Z Z Z Z0 36. Yes; ABC ~ DEF b the SSS Similarit Theorem. 37. No 38. Yes; PQR ~ STR b the SAS Similarit Theorem. 39. Similar b SAS Similarit Theorem ACE~BCD Similar b SAS Similarit Theorem MPQ~MNR g 6 g 0g 50 g PS PT SR TQ z CE DF EG FH Geometr Etra Practice Worked-out Solutions Ke. Similar b SAS Similarit Theorem EFG~HJK 8 0. so QS is not parallel to PT so QS PT Chapter 9 (p. 89). ACD~ABC~CBD; AC. FGE~EGH~EFH; HF 3. JLK~JKM~KLM; JL AC DA CB DB A 3 A 6 B 5 B 0 C3 3 C6 6 D D 8 A3 6 A B6 6 B C6 9 C 3 D3 9 D MP 6 MP 6 MP Not a Pthagorean triple 8 UV 500 UV 50 UV Pthagorean triple A 0 A0 0 B5 3 B5 5 C 5 C0 5 D 3 D A A B6 B.5 C 8 C0.5 D8 D QR QR 676 Pthagorean triple r s t QR 576 QR 7 s 5 9 s 65 s 576 s Copright McDougal Littell Inc.

13 Etra Practice continued r s t. r s t 7. 0? 9 5 t r ? 9 69 t r > 35 3 t r 3600 triangle is obtuse r s t. r 60 r s t 8. This is a 5590 triangle so the leg 7 and the hpotenuse leg t r This is a triangle so the hpotenuse shorter leg 6 and the longer leg 3065 t r shorter leg t r This is a triangle so the hpotenuse r 0 0 shorter leg so 0. The longer leg r s t 6. A 3 shorter leg 03. bh s sin S sin U cm 65 s 5 DF 8 DF 6 DF 80 8 FG 6 6 FG 56 FG 9 A bh FG cm right triangle right triangle 39? ? right triangle? 7 9? 9 8 > 30 s 78 s 8 DF 5 triangle is obtuse JL 7 9 JL 9 8 JL 3 A bh JL 39.6 cm not a right triangle. 9? ? < 63 triangle is acute. can t be a triangle 6. can t be a triangle cos S tan S sin T cos T tan T sin cos tan.05 0 sin z cos z tan z cos 65 8 cos tan 65 8 tan cos U 0.67 tan U 5.5 sin V 0.67 cos V tan V u sin 5 u 5 sin u 0.0 v cos 5 v 5 cos v. w tan 38 7 w 7 tan 38 w z 7 cos 38 z.6 z cos 38 Copright McDougal Littell Inc. Geometr 89 Etra Practice Worked-out Solutions Ke

14 Etra Practice continued 37. AC ED a \ b \ AC 8 ED AC 9 ED 53 7 sin ma sin 5 5 md. a \ c \ ma 30 md mb 60 mf 39. MP 0 5. c \ d \ 6 9 MP MP cos mn 6. b \ c \ mn mm PQ \ PQ\ P P P Q Q Q PQ \ PQ \ PQ \ PQ\ Chapter 0 (p. 8). D. E 3. G. C 5. H 6. F 7. A 8. B 9. internal 0. eternal. internal. C 3 3. C 7 r r. The intersection of the two circles is a point of tangenc at The common eternal tangents are the lines 3 and. 6. m AB m DC m AC m ED mcqe maqe m BC m BDC Geometr Etra Practice Worked-out Solutions Ke Copright McDougal Littell Inc.

15 Etra Practice continued The locus of all points that are equidistant from A and B is the bisector of AB which is. 8. The locus of all points 5 units from A is a circle with A as the center and radius 5 which is The locus of all points units from AB is the lines and All points 6 units from B is a circle with center B and radius 6 which is Chapter (p. 83). Sum of measures of interior angles Sum of measures of interior angles Sum of measures of interior angles Sum of measures of interior angles Sum of measures of interior angles Sum of measures of interior angles C 3 r 7 C r C 5 0 r 5 C 7 r 8. Measure of eterior angle Measure of eterior angle 0. Measure of eterior angle. Measure of eterior angle n 3. n Measure of central angle 7. Measure of central angle 8. Measure of central angle 9. Measure of central angle 0. n P 36 8 units A 3 s square units. Measure of central angle P 3 sin units a 3 cos 5 3 A a P square units. Measure of central angle P 50 sin units A a P 0 cos square units n Copright McDougal Littell Inc. Geometr 9 Etra Practice Worked-out Solutions Ke

16 Etra Practice continued 3. Measure of central angle P 67 tan units A ap square units. Measure of central angle P 88 sin units A ap 8 cos square units 5. Measure of central angle P 35 sin units A ap cos square units Perimeter red Area red 3 6. or 3:7 Area blue 7 9 Perimeter blue or 9:9 Perimeter red Area red 3 7. or 3: or 9: Area blue 9 Perimeter blue 3 Perimeter red Area red 8. or :3 or Area blue 3 Perimeter blue :9 9. Ratio of perimeters 5 ratio of side lengths. 8 Ratio of area A in C r in 3. C r 57 3.r 9.07 ft r Length of MN m MN 360 A units units A m BD 360 r units C 7 units C Length of VU m VU r. units r Length of AB m AB units Length of EF m EF 360 C 77. units C C Length of MN m MN r 95.5 units r A m DF 360 r r 5 65 r 360 r r 360 r units Length of SR m SR 360 C r 39.. A r A 5.39 units units 0 9 Geometr Etra Practice Worked-out Solutions Ke Copright McDougal Littell Inc.

17 Etra Practice continued 3.. PPoint is on AB AB or 5% MN 3 5. PPoint is on AD AD or about 67% MN PPoint is on MA MA or about 6.7% MN 6 7. PPoint is on MD MD 0 or about 83.3% MN A mac 360 P r r r r 9.6 or 9.6% P units area of quarter circle area of square 6 area of circle area of triangle area of circle or 68% Chapter (p. 85) 6.7. F 6; V 8; E square 8. triangle r F V E. This is a conve polhedron because each face is a polgon and an two points on the surface can be connected b a segment that lies entirel inside or on the polhedron. It is not regular.. This is not a polhedron because the faces are not polgons. 3. This is a conve polhedron because each face is a polgon and an two points on the surface can be connected b a segment that lies entirel inside or on the polhedron. It is a regular polgon.. F 7; V 0; E 5 5. F 7; V 7; E F V E F V E S B Ph cm S B Ph in. S r rh in. S r rh cm S r rh in. S B Pl in. S r rl cm V r h ft 3 V r h in 3 V r h Bh ft V 3 Bh cm 3 V 3 Bh in S B Ph cm S B Pl cm V Bh in. 3 V Bh cm 3 V Bh r h cm 3 5. V 3 Bh cm 3 V 3 r h in. 3 Copright McDougal Littell Inc. Geometr 93 Etra Practice Worked-out Solutions Ke

18 Etra Practice continued V 3 r h9. V 3 r h mm 87.7 ft 3 S r 3. S r cm 3.7 m V 3 r cm 3 S r in. V in. 3 S 0 ft ft V ft ft 3 S 00 cm cm V 66 3 cm cm 3 V 3 r m 3 S m V m 3 9 Geometr Etra Practice Worked-out Solutions Ke Copright McDougal Littell Inc.

9. AD = 7; By the Parallelogram Opposite Sides Theorem (Thm. 7.3), AD = BC. 10. AE = 7; By the Parallelogram Diagonals Theorem (Thm. 7.6), AE = EC.

9. AD = 7; By the Parallelogram Opposite Sides Theorem (Thm. 7.3), AD = BC. 10. AE = 7; By the Parallelogram Diagonals Theorem (Thm. 7.6), AE = EC. 3. Sample answer: Solve 5x = 3x + 1; opposite sides of a parallelogram are congruent; es; You could start b setting the two parts of either diagonal equal to each other b the Parallelogram Diagonals Theorem