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Geometry: A Complete Course (with Trigonometry) Module D – Instructor's Guide with Detailed Solutions for Progress Tests Written by: Larry E. Collins A T RA ER /2010 12 Quiz Form A Name Class Date Score Unit IV - Triangles Part A - Basic Definitions Lesson 1 - Triangle Parts Lesson 2 - Triangle Types X N P Use the diagram to the right to answer each of problems 1-6. A Y L R Q T M 1. Names the vertices of the triangle. Point L, Point M, Point N _______________________________________ 2. Name the sides of the triangle. LM, MN, LN _______________________________________ 3. Name the angles of the triangle. /NLM, /NML, /MNL _______________________________________ /XMQ (/NMQ), /PNT (/PNM), /RLQ (/RLM) 4. Name the exterior angles of the triangle. /TML (/TMY), /XNR (/XNL), /PLY (/NLY) _______________________________________ 5. For /NLM: a) Name the two sides of the angle LN, LM _______________________________________ b) Name the two side of the triangle that include the angle. LN, LM _______________________________________ NM c) Name the side of the triangle opposite the angle. _______________________________________ 6. For MN: /MNL, /NML a) Name the two angles of the triangle that include this side. ________________________________ b) Name the angle of the triangle which is opposite this side. ________________________________ /NLM © 2006 VideoTextInteractive Geometry: A Complete Course 1 Quiz Form A Name Class Date Score Unit IV - Triangles Part B - Basic Theorems Lesson 1 - Theorem 25: “If you have any given triangle, then the sum of the measures of its angles is 180” Lesson 2 - Theorem 26: “If you have a given exterior angle of a triangle, then its measure is equal to the sum of the measure of the two remote interior angles.” 1. The vertex angle of an isosceles triangle measures 50O. Base angle: _______________ 65O The base angles are congruent. Sketch a figure illustrating this triangle. Then find the measures of the base angles. 50 (x) x + x + 50 = 180 2x = 130 x = 65 (x) 2. The measure of one angle of a triangle is three times the measure of another and the third angle’s measure is equal to their sum. Sketch a figure illustrating this triangle. Then find the measures of all three angles. x + 3x + (x + 3x) = 180 8x = 180 x = 22.5 (3x) (x + 3x) 67.5O First angle: _________ 22.5O Second angle: _________ 90O Third angle: _________ 3x = 67.5 x + 3x = 90 (x) 3. In nABC, the exterior angle at C measures 130O, and m/A = 50O. 80O Measure /B:_________ Sketch a figure illustrating this triangle. Find m/B. A m/ACD = m/A + m/B 130 = 50 + m/B 130 B C 80 = m/B D © 2006 VideoTextInteractive Geometry: A Complete Course 7 Unit IV, Part B, Lessons 1&2, Quiz Form A —Continued— Name C 4. The side AB of nABC is extended to K. The bisector of /A intersects BC and D, and meets the bisector of /CBK outside nABC at P. Also, m/A = 60O and m/ABC = 70O. Sketch a figure 25O illustrating this situation. Find the measure of /P:_______________. m/P is 1/2 m/C 30 How do the measures of /C and /P compare?________________ 30 _________________________________________________________ A _________________________________________________________ First, find m/C: m/A + m/ABC + m/C = 180 60 + 70 + m/C = 180 m/C = 50 Second, find m/BDP m/CAD + m/ACD + m/CDA = 180 30 + 50 + m/CDA = 180 m/CDA = 100 m/BDP = 100 Third, find m /CBP m/ABC + m/CBK = 180 70 + m/CBK = 180 m/CBK = 110 /2 m/CBK = m/CBP = 55 1 • Finally, find m/P m/BDP + m/CBP + m/P = 180 100 + 55 + m/P = 180 m/P = 25 8 © 2006 VideoTextInteractive Geometry: A Complete Course P 50 100 D 100 55 55 70 B K Unit IV, Part B, Lessons 1&2, Quiz Form A —Continued— Name In problems 5-10, find the value of x in each of the accompanying figures, using the given information. A A B 40 D (x) 5. x = ________ 6. 100 x = ________ 34 (x) C D 34 C AB || CD B m/C = 40O m/A + m/B + m/ACB = 180 (Lines ||, Alternate Interior /’s >) m/A + 34 + 90 = 180 m/D = 40O m/A = 56 m/C + m/D + x = 180 40 + 40 + x = 180 x = 100 m/A + x + m/ADC = 180 56 + x + 90 = 180 x = 34 P 7. A A (x) 44 E x = ________ 8. 120 T B 78 x = ________ 42 28 150 R Q (x) 32 D C m/TQR + m/TRQ + m/T = 180 m/BCA + m/ACD = m/BCD m/TQR + m/TRQ + 150 = 180 28 + 32 = m/BCD m/TQR + m/TRQ = 30 60 = m/BCD m/TQR = m/TRQ (so each angle is 15) m/DBC + m/BCD + x = 180 78 + 60 + x = 180 138 + x = 180 x = 42 m/TQR = m/TRQ = m/PQT = m/PRT m/TQR + m/TQP = m/PQR 15 + 15 = 30 m/TRQ + m/TRP = m/PRQ 15 + 15 = 30 m/PQR + m/PRQ + x = 180 30 + 30 + x = 180 x = 120 9. (x) A 90 x = ________ 10. 67 23 x = 67 + 23 x = 90 53 x = ________ (x) C 164 D 164 = m/BAC + m/ACB 164 = m/BAC + 90 B x + m/CAD + m/ACD = 180 x + 37 + 90 = 180 x = 53 74 = m/BAC m/BAC = m/CAD + m/DAB 74 = m/CAD + m/DAB m/CAD = m/DAB m/CAD = 37 © 2006 VideoTextInteractive Geometry: A Complete Course 9 Unit IV, Part C, Lesson 3, Quiz Form B —Continued— Name ⋅ 4. Two corresponding sides of two similar pentagons measure 4 and 6. If the perimeter of the larger pentagon is 15, find the perimeter of the smaller pentagon. _______________ 10 4 x = 6 15 6 ⋅ x = 4 ⋅ 15 4 ⋅ 15 2 ⋅ 2 ⋅ 3 ⋅ 5 = 6 2⋅3 x = 10 x= 5. The sides of a pentagon measure 7, 8, 10, 11, and 12 inches respectively. Find the perimeter of 56 inches a similar pentagon if its longest side measures 14 inches. _______________ Perimeter : 7 +8 +10 +11+12 Compare longest side to longest side. 12 48 = 14 x 14 ⋅ 48 = 12 ⋅ x 672= 12x 672 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 7 8 ⋅ 7 = = = 56 12 2⋅2⋅3 1 P 6. Two similar hexagons have perimeters of 80 and 120 meters respectively. If one side of the1smaller 22 meters hexagon is 15, find the measure of the corresponding side of the larger hexagon. ___________ 2 80 15 = 120 x 120 ⋅ 15 = 80 ⋅ x 120 ⋅ 15 =x 80 2⋅2⋅2⋅3⋅5 ⋅3⋅5 =x 2⋅2⋅2⋅2⋅5 45 1 or 22 = x 2 2 © 2006 VideoTextInteractive Geometry: A Complete Course 29 Unit IV, Part D, Lesson 1, Quiz Form B —Continued— Name J 17. Given: Right nHJK with right angle HJK. JL HK; /K = /HJL H Prove: nJKL ~ nHJL K L STATEMENT REASON 1. Right nHJK with Right Angle HJK. 1. Given 2. JL 2. Given HK 3. /HLJ is a right angle 3. Definition of Perpendicular Lines. 4. /JLK is a right angle 4. Definition of Perpendicular Lines. 5. /HLJ > /JLK 5. Theorem 11 – If you have right angles, then those right angles are congruent. 6. /K > /HJL 6. Given 8. nJKL ~ nHJL 8. Postulate Corollary 12a – If two angles of one triangle are congruent to the two corresponding angles of another triangle, then the triangles are similar. (AA Postulate Corollary) G T J H 18. Given: GH HJ; JK Prove: HT GT = JT KT HJ K STATEMENT 1. GH HJ; JK 2. GH JK HJ REASON 1. Given 2. Theorem 22 – If two lines are perpendicular to a third line, then the two lines are parallel. 3. /GHT > /KJT 3. Theorem 16 – If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 4. /HTG > /JTK 4. Theorem 15 – If two lines intersect, then the vertical angles formed are congruent. 5. nGTH ~ nKTJ 5. Postulate Corollary 12a – If two angles of one triangle are congruent to the two corresponding angles of another triangle, then the triangles are similar. (AA Postulate Corollary) 6. HT GT = JT KT 6. If two polygons are similar, then the measures of corresponding sides are proportional. © 2006 VideoTextInteractive Geometry: A Complete Course 37 Unit IV, Part D, Lesson 2, Quiz Form A —Continued— Name In the figure to the right, AY || EO || RB. Use this figure to complete each of the proportions in problems 11 - 14. 11. 12. YO AE = ? OB YB ? = OB ER ER ? = ____________ D Y AR ? = ____________ O A E R B 13. ? YB = AE YO AR ? = ____________ 14. DY DA = YO ? AE ? = ____________ Using the figure to the right, and the given information in problems 15 and 16 to determine if QT || PS. Answer yes or no. 15. no PR = 30, PQ = 9, RT = 12, RS = 18 Answer: ________ PR SR = PQ ST 30 18 = 9 ST RS – RT =ST 3 ⋅10 3 ⋅6 = 3 ⋅3 1⋅6 10 3 ≠ 3 1 R Q P T S NO! 18 –12=6 30 18 = 9 6 16. RP = 13.5, RQ = 6.3, TR = 4.2, SR = 9.0 RP RS = RQ RT 13.5 9.0 = 6.3 4.2 135 90 = 63 42 yes Answer: ________ 3 ⋅45 2 ⋅45 = 3 ⋅21 2 ⋅21 45 45 = 21 21 YES! © 2006 VideoTextInteractive Geometry: A Complete Course 41 Unit IV, Part D, Lesson 3, Quiz Form B —Continued— Name Use the diagram to the right, and the given information in problems 7 and 8, to find x. 7. nABE ~ nAEC x= 16 = 3 __________ 1 D F 3 4 m/1 = m/2 E BF AB = EB AE 3 4 = 4 AE 4 4 =3 ⋅AE ( )( ) 16=3 ⋅AE 16 = AE 3 8. nCEA ~ nCBE 9 __________ 5 x= B is the midpoint of AC D is the midpoint of CE BD = 4, CE = 6, CB = 5, BE = x + 3 (Note: Use only these segments to solve) DB CB = BE CE 4 5 = x+3 6 x+3 ⋅5=4 ⋅6 ( ) 5x+15=24 5x=9 9 x= 5 © 2006 VideoTextInteractive Geometry: A Complete Course C 2 m/3 = m/4 AB = 4, FB = 3, EB = 4, AE = x 52 B A Unit IV, Part D, Lesson 4, Quiz Form A —Continued— Name M P For problems 3 – 11, use the right triangle shown to find the missing value. Q N 6 3. PQ = 9, QN = 4, MQ = ____________ 4. 9 MQ = MQ 4 ( 27 =x QN = 3, MQ = 9, PQ = ____________ x 9 = 9 3 9 9 =x ⋅3 ( )( ) (MQ )(MQ )=9 ⋅4 (MQ )2 =36 81=x ⋅3 27 =x MQ =6 ( 32 16= PN 5. PM = 12, PQ = 9, PN = ____________ 6. 6 8 = 8 PN 8 ⋅8=6 ⋅ PN 64=6 ⋅PN 64 = PN 6 2 ⋅32 = PN 2 ⋅3 32 = PN 3 9 12 = 12 PN 12 12 =9 ⋅PN ( )( ) 144=9 ⋅PN 16= PN 2 7. PN = 75, PQ = 72, MN = QN = PN - PQ QN =3 ( 3 MN = MN 75 MN MN =3 ⋅75 )( ) (MN )2 =225 15 ____________ 3 MN = 8, QN = 6, PN = ____________ 8. MQ = 4, PQ = 10, QN = MN = 3 ⋅3 ⋅5 ⋅5 10 4 = 4 QN MN = 3 2 ⋅5 2 4 ⋅4=10 ⋅QN 16=10 ⋅QN MN = 3 2 ⋅ 5 2 MN =3 ⋅5 MN =15 16 =QN 10 2 ⋅2 ⋅2 ⋅2 =QN 2 ⋅5 8 =QN 5 MN = 225 M 9. PN = 13, PM = 12, PQ = 144 = 13 ____________ 10. 13 12 = 12 PQ 12 ⋅12=13 ⋅PQ 144=13 ⋅ PQ 144 = PQ 13 PQ QM = QM QN PN -QN = PQ 16 -4=12 12 QM = QM 4 (QM )(QM )=12 ⋅4 Q © 2006 VideoTextInteractive Geometry: A Complete Course 4 ⋅ 4 3 PN = 16, QN = 4, QM =2 ____________ QM 2 =48 56 8 = 5 ____________ ⋅ QM = 2 ⋅2 ⋅2 ⋅2 ⋅3 QM = 2 2 ⋅2 2 ⋅3 QM = 2 2 ⋅ 2 2 ⋅ 3 QM = 2 ⋅2 ⋅ 3 QM =4 3 Unit IV, Part D, Lesson 4, Quiz Form A —Continued— Name A D 13. Given: nABC is a right triangle BD AC DE BC Prove: nABC ~ nDEC B STATEMENT 58 E C REASON 1. nABC is a right triangle 2. BD AC 3. BD is an altitude 4. nADB ~ nBDC 1. 2. 3. 4. 5. /BDC is a right angle 6. nBDC is a right triangle 7. DE BC 8. nDEC ~ nBDC 9. nADB ~ nDEC 5. 6. 7. 8. 9. © 2006 VideoTextInteractive Geometry: A Complete Course Given Given Definition of Altitude Theorem 30 - If you have the altitude to the hypotenuse of a right triangle, then it forms two triangles that are similar to each other, and to the original triangle. Definition of Perpendicular Lines Definition of Right Triangles Given Theorem 30 Postulate Corollary 12c - If two triangles are similar to a third triangle, then the two triangles are similar to each other. Unit IV, Part D, Lesson 4, Quiz Form B —Continued— Name In problems 3 - 5, use the diagram shown, and find the length of the altitude to the hypotenuse. R 3. R P 15 R 4. S 3 Q 5. 12 4 Q 30 S RS = 4 3 _______ QS RS From = RS SP 3 5 SP = _______ P 6 RS SP From = SP SQ 4 RS = RS 12 RS RS =4 ⋅12 P Q 15 SP = SP 3 SP SP =15 ⋅3 ( )( ) ( RS )2 =48 QS = _______ 12 S From ( )( ) (SP )2 =45 SP QS 6 QS = = QS RS QS 24 (QS )(QS )=6 ⋅24 (QS )2 =144 RS = 2 ⋅2 ⋅2 ⋅2 ⋅3 SP = 45 RS = 2 2 ⋅2 2 ⋅3 SP = 9 ⋅5 QS = 144 SP = 9 ⋅ 5 QS =12 RS = 2 2 ⋅ 2 2 ⋅ 3 SP =3 5 RS =2 ⋅2 ⋅ 3 RS =4 3 In problems 6 - 8, use the diagram shown, and find the length of each leg of the right triangle. T R 6. 7. 8. 15 M From 3 N 10 NT = _______ NM = _______ 10 3 RT NT = NT TM 5 NT = NT 20 NT NT =5 ⋅20 From T ( )( ) ( NM )2 =300 NM = 300 NM = 3 ⋅ 100 NM =10 3 R M 32 6R NT = _______ NM = _______ 6 3 ( NT )( NT )=3 ⋅12 ( NT )2 =36 From ( NM = 2 2 ⋅ 3 2 ⋅ 3 NM =2 ⋅3 ⋅ 3 NM =6 3 T 16 NT = _______ NM = _______ 16 3 NT = 2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 NT = 2 8 NT =2 4 NT =16 )( ) ( NM )2 =108 NM = 2 ⋅2 ⋅3 ⋅3 ⋅3 3 8 ( NT )( NT )=8 ⋅32 ( NT )2 =8 ⋅32 9 NM = NM 12 NM NM =12 ⋅9 NM = 108 R RT NT 8 NT = = NT MT NT 32 NT = 36 NT =6 NT = 100 NT =10 15 NM = NM 20 NM NM =15 ⋅20 9 RT NT 3 NT = = NT MT NT 12 ( )( ) ( NT )2 =100 N M N 5 ( 24 NM = NM 32 NM NM =24 ⋅32 )( ) ( NM )2 =24 ⋅32 NM = 2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅3 NM = 2 8 ⋅3 NM = 2 4 3 60 © 2006 VideoTextInteractive Geometry: A Complete Course NM =16 3 Unit IV, Part D, Lesson 4, Quiz Form B —Continued— Name E Use the figure at the right for problems 9 – 12. H Q G 9. EG = 3, EH = 4, HG = 5. EQ = 12 __________ 5 10. 2 or 8 EQ = 4, GH = 10. GQ = __________ GQ EQ GQ 4 x 4 4 ⋅4 VEQG ∼VHEG EQ EG = EH HG EQ 3 = 4 5 3 ⋅ 4 = EQ ⋅ 5 12 = 5 ⋅ EQ EQ HQ 4 = 10 − GQ 4 = 10 - x = x 10 - x = x 2 – 10 x + 16 = 0 ( x - 2)( x - 8 ) = 0 x-2=0 x=2 HG = 15, EG = 9, EQ = VGQE ∼VGEH EQ EG = EH GH 36 5 = 9 EH 15 36 15 ⋅ = 9 ⋅ EH 5 1 36 ⋅ 5 ⋅ 3 = 9 ⋅ EH 5 108 = 9 ⋅ EH 108 = EH 9 9 ⋅ 12 = EH 9 12 = EH ) 16 = 10 x - x 2 12 = EQ 5 11. ( 36 12 . EH = __________ 5 12. EQ = 8, or or x-8=0 x=8 HQ 2 = . HQ = __________ 8 2 QG 1 HQ EQ = EQ QG 2x 8 = 8 x 8 ⋅ 8 = 2x ⋅ x 64 = 2 x 2 32 = x 2 32 = x 16 ⋅ 2 = x 16 ⋅ 2 = x 8 2=x © 2006 VideoTextInteractive Geometry: A Complete Course 61 Name Quiz Form A Class Date Score Unit IV - Triangles Part D - Similarity – Part 2 (Triangles and Their Parts) Lesson 5 - Theorem 31: “If you have a given right triangle, then the square of the measure of the hypotenuse, is equal to the sum of the squares of the measures of the two legs.” (The Pythagorean Theorem) Lesson 6 - Applying Pythagoras to 3-Dimensional Figures For problems 1 – 5, find x in the given figures. Express radicals in simplest form. 2 1. 2 2 ( 3) +5 2. 2 a +b =c 2 x 15 2 =x 2 3 + 25 = x 2 28 = x 2 3 2 2 2 2 a +b =c 11 x 9 + x = 11 2 2 12 x = 40 2 7 =x 2 2 ( 7 ) + ( 11 ) = x 2 7 40 x= 4 ⋅ 10 x = 2 10 a +b =c x x= 2 10 x = _______ 4 ⋅7 = x 3. 2 81 + x = 121 3 28 = x 2 7 =x x = _______ 2 2 4. 2 2 5 2 2 18 = x 2 2 2 2 2 5 + x = 13 13 7 + 11 = x 2 a +b =c x 2 2 25 + x = 169 2 x = 144 x = 12 11 18 = x 3 2 x = _______ 9 ⋅2 = x 12 x = _______ 3 2 =x 5. 5 2 2 x 10 a +b =c 2 2 5 + 10 = x 2 2 2 25 + 100 = x 2 125 = x 2 25 ⋅ 5 = x 5 5 =x x = _______ 5 5 =x © 2006 VideoTextInteractive Geometry: A Complete Course 63 Name Quiz Form A Class Date Score Unit IV - Triangles Part E - Congruence - Part 1 (General Geometric Relationship) Lesson 1 - Definition nXYZ > nUVW Use the pair of triangles to the right, and the given information, for problems 1 – 8. Y X W Z U V 1. Name the three pairs of corresponding angles. _________________________________________ /X corresponds to /U; /Y corresponds to /V; /Z corresponds to /W. Note: /YXZ corresponds to /VUW is also a correct response. However “/YXZ corresponds to __________________________________________________________________________________ /WUV” is an incorrect response since corresponding vertices are not in the correct order. __________________________________________________________________________________ XY corresponds to UV; YZ corresponds to VW; 2. Name the three pairs of corresponding sides. __________________________________________ ZX corresponds to WU. Note: XY corresponds to VU is an incorrect response, technically speaking. __________________________________________________________________________________ Determine whether each statement is correct or incorrect. 3. nYZX > nVWU correct (corresponding parts are congruent) _______________________________________________ 4. nXZY > nUVW incorrect (this could be correct, but we don’t know the specific measures.) _______________________________________________ 5. XY > VU incorrect (because the names do not reflect the corresponding parts, this is _______________________________________________ technically incorrect. However, the two segments are congruent, so the answer could also be “correct”) 6. /Y > /V (these are corresponding angles) correct _______________________________________________ 7. Which statement is correct? b ______________ a) nYZX > nUVW b) nYZX > nVWU c) nZXY > nWVU 8. Which statement is not correct? c ______________ a) XZ > UW b) /Y > /V c) XZ > UV © 2006 VideoTextInteractive Geometry: A Complete Course 71 Unit IV, Part E, Lesson 1, Quiz Form A —Continued— Name A E Use the pair of triangles to the right, for problems 9 – 14. B 9. AD > CE _______________ D 10. BE > BD _______________ /CBE 11. /ABD > _______________ 12. BC > BA _______________ nBEC 13. nBDA > _______________ nADB 14. nCEB > _______________ Given that nMOP > nHAT, complete each statement in problems 15 – 20. 15. /H /M > _______________ 16. AT OP > _______________ 17. m/T m/P = _______________ 18. HT MP > _______________ 19. /O /A > _______________ 20. MO HA > _______________ 21. Sketch and label a pair of obtuse triangles to illustrate the definition of congruent triangles. Mark all six relationships on the triangles. (answers will vary) 72 © 2006 VideoTextInteractive Geometry: A Complete Course C Unit IV, Part E, Lesson 1, Quiz Form A —Continued— Name A 22. Given: AB > AD AC BD AC bisects BD B C D AC bisects /A Prove: nABC > nADC using the definition of congruent triangles. STATEMENT 1. AB > AD 2. AC > AC 3. AC bisects BD 4. BC > DC 5. AC BD 6. /ACB is a right angle 7. /ACD is a right angle 8. /ACB > /ACD 9. AC bisects /A 10. /BAC > /DAC 11. /ABC > /ADC 12. nABC > nADC REASON 1. 2. 3. 4. 5. 6. 7. 8. Given Reflexive Property for Congruence Given Definition of Segment Bisector Given Definition of Perpendicular Definition of Perpendicular Theorem 11 - If you have right angles, then those right angles are congruent. 9. Given 10. Definition of Angle Bisector 11. Corollary 25a - If two angles of one triangle are congruent to two angles of another triangle, then the third pair of triangles are congruent. 12. Two triangles are congruent if and only if there is a correspondence between the vertices such that each pair of corresponding sides and each pair of corresponding angles are congruent. © 2006 VideoTextInteractive Geometry: A Complete Course 73 Name Quiz Form A Class Date Score Unit IV - Triangles Part E - Congruence - Part 1 (General Geometric Relationship) Lesson 2 - Postulate 13: Triangle Congruence Lesson 3 - Congruence Postulate Corollaries For problems 1 – 4, state the given congruence relationship in your own words and label the diagram appropriately. If one leg and acute angle of one right triangle are 1. Postulate Corollary 13c – LA Postulate Corollary: _______________________________________ congruent to the corresponding leg and acute angle of another right triangle, then the triangles are congruent. ___________________________________________________________________________________ C D A N C B F Note: Labeling may vary, but points must correspond M P If two angles and a non-included side of one triangle 2. Postulate Corollary 13a – AAS Postulate Corollary: ______________________________________ are congruent to the corresponding angles and non-included side of another triangle, then the triangles are congruent. ___________________________________________________________________________________ N Note: Labeling may vary, but points must correspond C B A 3. Postulate 13 – ASA Congruence P M If two angles and the included side of one triangle are congruent to Assumption:____________________________________________ the corresponding angles and included side of another triangle, then the two triangles are congruent. ___________________________________________________________________________________ M A Note: Labeling may vary, but points must correspond N B P C the three sides of a triangle are congruent to the three corresponding 4. Postulate 13 – SSS Congruence Assumption: If___________________________________________ sides of another triangle, then the two triangles are congruent. ___________________________________________________________________________________ A M Note: Labeling may vary, but points must correspond B C P N © 2006 VideoTextInteractive Geometry: A Complete Course 79 Unit IV, Part E, Lessons 2&3, Quiz Form A —Continued— Name C F Refering to the triangles to the right, for problems 5 – 10, name the postulate or postulate corollary which would show nABC > nDEF. A 5. /C > /F, AC > DF, BC > EF Postulate 13 – SAS Congruence Assumption ________________________________ 6. /C > /F, /D > /A, AB > DE Postulate Corollary 13a – AAS Postulate Corollary ________________________________ 7. /C and /F are right angles, AB > DE, BC > EF Postulate Corollary 13e – HL Postulate Corollary ________________________________ 8. /C and /F are right angles, /E > /B, DF > AC Postulate Corollary 13c – LA Postulate Corollary ________________________________ 9. /A > /D, AB > DE, AC > DF Postulate 13 – SAS Congruence Assumption ________________________________ 10. /B and /E are right angles, CB > FE, AB > DE Postulate Corollary 13d – LL Postulate Corollary ________________________________ B D B A 11. Given: BD AB; BD AD > CB DC Prove: nABD > nCDB D STATEMENT 1. BD AB 2. BD DC 3. /ABD is a right angle 4. /CDB is a right angle 5. nABD is a right triangle 6. nCDB is a right triangle 7. BD > DB 8. AD > CB 9. nABD > nCDB 80 E C REASON 1. 2. 3. 4. 5. 6. 7. 8. 9. © 2006 VideoTextInteractive Geometry: A Complete Course Given Given Definition of Perpendicular Line Segments Definition of Perpendicular Line Segments Definition of Right Triangle Definition of Right Triangle Reflexive Property for Congruent Segments Given Postulate Corollary 13e – If the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and corresponding leg of another right triangle, then the triangles are congruent. Unit IV, Part E, Lessons 2&3, Quiz Form A —Continued— Name U Repeat problem 13 using a different postulate assumption or postulate corollary. 14. Given: XW UW; XW UQ > VQ Q XV X W Prove: nUWQ > nVXQ V STATEMENT 1. XW UW 2. XW XV 3. UW || XV 4. /WUQ > /XVQ 5. UQ > VQ 6. /WQU > /XQV 7. nUWQ > nVXQ REASON 1. Given 2. Given 3. Theorem 22 - If two lines are perpendicular to a third line, then the two lines are parallel 4. Theorem 16 - If two parallel lines are cut by a transversal, then alternate interior angles are congruent. 5. Given 8. Theorem 15 – If two lines intersect, then the vertical angles formed are congruent. 9. Postulate 13 – If two angles and the included side of one triangle are congruent to the corresponding angles and included side of another triangle, then the two triangles are congruent. (ASA Congruence Assumption) D 15. Given: DG is the perpendicular bisector of EF Prove: nDEG > nDFG E 82 F REASON STATEMENT 1. DG EF 2. /DGE is a right angle 3. /DGF is a right angle 4. nDFE is a right triangle 5. nDFE is a right triangle 6. DG bisects EF 7. EG > FG 8. DG > DG 9. nDEG > nDFG G 1. 2. 3. 4. 5. 6. 7. 8. 9. © 2006 VideoTextInteractive Geometry: A Complete Course Given Definition of Perpendicular Definition of Perpendicular Definition of Right Triangle Definition of Right Triangle Given Definition of Bisector of a Line Segment. Reflexive Property of Congruence Postulate Corollary 13d – If two legs of one right triangle are congruent to the corresponding legs of another right triangle, then the two right triangles are congruent. Unit IV, Part E, Lessons 2&3, Quiz Form B —Continued— Name M Refering to the two triangles to the right, for problems 5 – 7, name the postulate or postulate corollary which would show nMNQ > nRST. R N Q S T 5. /M > /R, NQ > ST, /Q > /T Postulate Corollary 13a – AAS Postulate Corollary ________________________________ 6. MN > RS, MQ > RT, NQ > ST Postulate 13 – SSS Congruence Assumption ________________________________ 7. /N > /S, /Q > /T, NQ > ST Postulate 13 – ASA Congruence Assumption ________________________________ H Y Refering to the triangles to the right for, problems 8 – 13, fill in the blanks with the required angle or side so that the indicated congruence assumption or postulate corollary justifies nXYZ > nGHI. X Z G I XY YZ HI GH 8. SAS Congrunece Assumption; /Y > /H; ________ > ________; ________ > ________. YZ ZX IG HI 9. LA Postulate Corollary; /Y > /H; m/Z = m/I = 90; ________ > ________ or ________ > ________. 10. ASA Congruence Assumption; /X > /G; XY > GH; ________ > ________. XZ GI 11. HL Postulate Corollary; m/Y = m/H = 90; YZ > HI; ________ > ________. YZ XY GH HI 12. AAS Postulate Corollary; /X > /G; /Z > /I; ________ > ________ or ________ > ________. XZ GI 13. HA Postulate Corollary; m/Y = m/H = 90; /X > /G; ________ > ________. 84 © 2006 VideoTextInteractive Geometry: A Complete Course Unit IV, Part E, Lessons 2&3, Quiz Form B —Continued— Name B A 14. Given: AB > DC; AB || DC Q Prove: nABQ > nDCQ C STATEMENT 1. AB > DC 2. AB || DC 3. /B > /C 4. /A > /D 5. nABQ > nDCQ D REASON 1. Given 2. Given 3. Theorem 16 – If two parallel lines are cut by a transversal, then alternate interior angles are congruent. 4. Theorem 16 – If two parallel lines are cut by a transversal, then alternate interior angles are congruent. 5. Postulate 13 – If two angles and the included side of one triangle are congruent to the corresponding angles and included side of another triangle, then the two triangles are congruent. (ASA Congruence Assumption) E 15. Given: nADE is an Isosceles Triangle EC AE; EB DE; AB > DC Prove: nAEC > nDEB A STATEMENT 1. nADE is an isosceles triangle 2. AE > DE 3. EC AE 4. EB DE 5. /AEC is a right angle 6. /DEB is a right angle 7. nAEC is a right triangle 8. nDEB is a right triangle 9. AB > DC 10. BC > CB 11. AC > DB 12. nAEC > nDEB C B D REASON 1. Given 2. Definition of Isosceles Triangle 3. Given 4. Given 5. Definition of Perpendicular Line Segments 6. Definition of Perpendicular Line Segments 7. Definition of Right Triangle 8. Definition of Right Triangle 9. Given 10. Reflexive Property of Segment Congruence 11. Postulate 6 – Ruler – Fourth Assumption – Segment Addition 12. Postulate Corollary 13d – If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and corresponding leg of and their right triangle, then the two right triangles are congruent. (HL Postulate Corollary) © 2006 VideoTextInteractive Geometry: A Complete Course 85 Name Quiz Form A Class Date Score Unit IV - Triangles Part F - Congruence – Part 2 (Applications) Lesson 1 - Overlapping Triangles Lesson 2 - Using the Definiton of Congruence Lesson 3 - Theorem 32 - “If two given triangles are both congruent to a third triangle, then the two given triangles are congruent to each other.” O A 1. Name two pairs of overlapping triangles that appear to be congruent in the figure at the right. E D nDBC and nECB; nABE and nACD ___________________________________________ B C U V W X N Use the figure to the right for problems 2 and 3. 2. Name two different triangles that overlap A O R nVOY and contain OU. _________________________ nUOW, nUOX Q S M P 1 E D B A 3. Name two different triangles that overlap nVOY, nWOY nUOX and contain OY. _________________________ B C U V W Y X A N Use the figure to the right for problem A 4. E AC – BC = AB 29 – 10 = AB 19 = AB AC 3x + 5 3 ⋅8 + 5 B 24 + 5 29 BC x+2 8+2 BD N 10 Q U Q S M S 2 X M O X B A Y V W Y X A B U D C V R T X DB – BC = CD 29 – 10 = CD Q 19 = CD Q P Y C U S U V V W Q W U U E X Y U A Z N V 1 V Z 2 87 © 2006 VideoTextInteractive Geometry: A Complete Course C Q R X U V W A Y X U O B A D Z P C4x – 3U 4 ⋅8 – 3 32 – 3 29 S M W P1 D R E Q AB = CD = 19 Find AB and CD. ___________________________ AC = DB 3x + 5 = 4x - 3 8=x C D O 4. AB = CD, AC = 3x+5, BD = 4x – 3, BC = x + 2. B T R D 19, E Q X T Y Y A A B C YD D D Unit IV, Part F, Lessons 1,2&3, Quiz Form A —Continued— Name O A E Q Use the figure to the right for problem 5. 1 E D 2 X M 5. /1 > /2, m/AEC = 14x – 3, m/DEB = 17x – 15, AE = 4x + 7, U > VnDEB? W DE = 3x + 11, BE = 3x + 7, CE = 5x B– 1. Why isCnAEC P Y X A B D C T properties of nAEC > nDEB by S.A.S. Assumption. Since AE = DE, NCE=BE, and /AEC > /DEB, using the __________________________________________________________________________________ angle addition. __________________________________________________________________________________ R m∠AEC = m∠DEB 14x – 3 = 17x – M 15 12 = 3x 4=x AE 4x + 7 4 ⋅4 +7 16 + 7 A Q DE 3x + 11 3 ⋅ 4 + 11 12 + 11 23 S 23 O A CE 5x – 1 5 ⋅4 – 1 20 – 1O 19 BE Q 3x P +7 3 ⋅4 +7 12 B+ 7 19 S Q U E 1 E 2 Q 1 E D 2 Prove: /NQX > /NPX C U V W Y X C X M Given: MQ > MP; /QMX > /PMX 6. R T U N W V W U P A B Q T STATEMENT X 1. MQ > MP 2.Q/QMX 1. Given > /PMX XS 4. nQMN > nPMN O 5. C.P.C.T.C. > /PNM 6. /QNM C N 6. C.P.C.T.C. V W A Y X B R Z 9. /NQX V > /PNX V A 65 D O VB A C B D E B D T Z X 12 D Y © 2006 VideoTextInteractive Geometry: A Complete Course B C C 8 Z M 10 10 55 A D D B 12 65 55 E 8 C B U X 12 A 5 E A V C C Y T N Z U 55 A F B S C D 88 10 10 X R S A U P 12 O 3 4 E 7. Reflexive Property for Congruence D D C V Y Y 7 C 9. C.P.C.T.C. V Q C A Q U Z A VS Y E D C Assumption - If F 8. Postulate T13 - SAS Congruence two sidesX and the included angle of one triangle are congruent toAthe corresponding sides B and M included angle of another triangle, then the two W triangles are congruent. R U X Q R S 4. Postulate 13 U- SAS Congruence Assumption - If D two sides and the included angle of one triangle B E V U V are congruent to the corresponding sides and 6 5 A C included angle of another triangle, then the two 1 2 triangles are congruent. 1 2 U 8. nQNX > nPNX V V 5. QN > PN Y Z B 7. NX > NXU N U M Q B X V Y 3. Reflexive Property for Congruence P W N 2. Given Q S U REASON Z Y U 3. MN > MN S D C D C R N BV A Y X C Y 8 D E 12 Unit IV, Part F, Lessons 1,2&3, Quiz Form A —Continued— Name B C V U W Y X O B REASON 1. /MRS and /NRS are supplementary 1. Theorem 10 - If the exterior sides of two adjacent angles are opposite rays, then the two angles are supplementary. 2. /PQS and /NQS are supplementary C two adjacent D 10 - If the exterior sides of 2. Theorem angles are opposite rays, then the two angles are U supplementary. V 3. Given 4. /NRS > /NQS 4. Theorem 14 - If two angles are supplementary to the sameXangle or congruent angles, Y then they are Z congruent to each other. 6. NM > NP 6. Given 7. nNQM > nNRP X 7. Postulate Corollary 13a - AAS - If two angles Y and a non-included side of one triangle are R congruent to the corresponding two angles and U V triangle, U non-included side of another then the twoV triangles are congruent. T N 8. /NRP > /NQM U Z V U 5. Reflexive Property for Congruence X V D V B S U D C R Q R Q R T E D S S S A U W R R B E 5 A 1 B M P M Q C C N N N 6 V X Q C S U O S 1 2 2 V Y A 8. C.P.C.T.C Q P A Y X U 5. /N > /N M W W 3. /MRS > /PQS 2 B S P A Q 1 Q S Q STATEMENT E Q M Prove: /NRP > /NQM C T R Given:/MRS > /PQS; NM > NP B A N 8. 2 1 E D T Z X 7 A Y B D C O 3 4 E V M Y D F F C Y T C C X C D 8 12 8 12 8 D U D Z A A Y V C A 90 D 12 E B A 15 D C Z G © 2006 VideoTextInteractive Geometry: A Complete Course 10 10 B 15 A D E H B F B A 15 A 1 E D 2 M Name Quiz Form A B C U V W A Date Class N Y X B D C Score T R Q Unit IV - Triangles Q S S Part F - Congruence M– Part 2 (Applications) P Lesson 4 - Theorem 33 - “If two sides of a triangle Q O B A E A are congruent, then theO angles opposite them Q are congruent.” 1 2 X E Lesson 5 D- Theorem 34 - “If two angles of a triangle M are congruent, thenD the sides opposite them C P A B U V W Y X are congruent.” B C Y U V W A B D X C 1 2 C R W N E T U Complete each of the statements in problems 1-3, refering to the figure to the right. Then state, in U your own words, the theorem you are X using to justify your Y answer. R S M 1. Z Q Q U V P V M Q BD CD If /CBD > /BCD, then __________ > __________. X Y O the sides oppo34 “If two angles of a triangle are congruent, then B A Theorem ______________________________________________ site them are congruent.” ______________________________________________________ U 2. Z SV U V 1 12 5 101 1 2 If m/CBD + m/DBA > m/BCD + m/DCA, C B E U A A Y Y R A Use the diagram to the right, and Q Theorem 33 or S Theorem 34, to complete each statement in problems 4 and 5. D S U V O 12 T X B T C C N 12 D T A D 15 A Z 65 Y E D 55 B B E X A C D 8 Q 55 A C C 8 D 12 8 12 D A C D A Complete Course © 2006 VideoTextInteractive Geometry: A 15 B B A 15 B M R A 10 10 M R D B 65O m/CDB = __________. CB = A C V 12 __________. C B W R V E C V U are congruent.” site them V M ______________________________________________________ U D C 3 D D B Q V A X T D R /ABC /ACB If AB > AC, then __________ > __________. X Y Z then the angles oppo33 - “If two sides of a triangle are congruent, Z Theorem ______________________________________________ X 7 55 B U 5. C B A D 34 - “If two angles of a triangle areUcongruent, then AC V W the Xsides oppoY Theorem ______________________________________________ N site them are congruent.” W ______________________________________________________ 4. V A E AC C D AB then __________ > __________. 3. U Y W D 95 B B A C 7 A C U V W A Y B X Unit IV, Part F, Lessons 4&5, Quiz Form A W —Continued— U B F C X C U Y U Z Z In the figure at the right, nXYZ is isosceles, with XZ > YZ. Also, ZY =W 8, m/Y = 40O, V Y ZX = Y 8 __________. W Y X U E B D Z Y X Z U 12 40O = __________. V V m/X U A C A D 7. V V V and ZU bisects /XZY. Use this information for problems 6-9. 6. V F D X O O D D C Name 3 4 E Z V V 10 10 Y X (Answers will vary) Postulate55Corollary 13 -65 ASA 55 nXZU > nYZU. Why? ___________________________________ B C D B Assumption (or SAS, AAS)A E ___________________________________ 8. 9. C B U E D D E V AU A V B D W S Given: RM > RN R NQ > MQ O SProve: /TNQ > /TMQT R X M 1. GivenQ Y C D A 5. /NRT > /MRT C R B A 5. C.P.C.T.C. CA 8 Y CD B 8. C.P.C.T.C. C 9. /TNQ > /TMQ 9. Theorem G33 - If two sides of a triangle are H congruent, then the angles opposite them are congruent. F E A B 8 D 12 8 12 8. NT >MT © 2006 VideoTextInteractive Geometry: A Complete Course M B 15 7. Postulate 13 - SAS Congruence Assumption - If two sides and the included angle of one triangle A B sides Aand 15 corresponding are congruent to the included angle of another triangle, then the two triangles are congruent. F X 4. Postulate 13 - SSS Congruence Assumption - If three 12 sides of a triangle are congruent to the corresponding sides T of another triangle, then the two triangles are congruent. 7. nNRT > nMRT H T F E 6. Reflexive Property forD Congruence D C O S 6. RT > RT E B 15 W D S 2. Given 8 D 12 8 B D 2 B V C 3. Reflexive Property for Congruence C B 1 C R REASON Z M 4. nRNQ > nRMQ 12 E D A 3. RQ > RQ D D C A 2. NQ > MQ C T B B STATEMENT T A C Q A 1. RM > RN C N R 10. B C B M 96 Y D T A 15 A U X C.P.C.T.C. XU > YU. Why? ______________________________. N Q V Y D D 15 B A U V A V Y X Y Unit IV, Part F, Lessons 4&5, Quiz Form A —Continued— U V U C A Z Name 12 D 10 10 V 55 65 55 In the figure at the right, AC is contained inBplane M, and intersects plane M at point A. C BD A D B E X Also, AC is the perpendicular bisectorC of BD. B N R 12. Prove: nBAC > nDAC Q T 13. Prove: /B > /D 14. Prove: nBDC is isosceles A T B A 2. Given 3. AC is the perpendicular bisector of BD 3. Given 4. /BAC is a right angle X M 6. nBAC is a right triangle 5. Definition of Perpendicular 6. Given 7. nDAC is a right triangle 8. BA D > DA D Y C F C 8. Definition of Segment Bisector 9. nBAC12>nDAC 12 8 D B 15 A 15 8 D 12 9. Postulate Corollary 13d - If the two legs of a right triangle are congruent to the two legs of a right B Atriangle are15congruent to the B two legs of another 10. /B > /D 10. C.PC.T.C 11. BC > DC 11. Theorem 34 - If two angles of a triangle are congruent, then the sides opposite them are congruent. (also C.P.C.T.C.) D E H 12. nBDC is isosceles B F A E 12. A triangle is an isosceles triangle, if and only if, it has at least two congruent sides. © 2006 VideoTextInteractive Geometry: A Complete Course 2 D B 7. Given C 1 C A Z 4. Definition of Perpendicular T 98 A B 5. /DAC is a right angle A C 1. Given 2. BD intersectsS plane M at point A O G B REASON R S 1. AC is contained in plane M C C D W STATEMENT A E B V 8 Y A A R C U D E D M Q E B B D C D U Unit IV, Part F, Lessons 4&5, Quiz Form B —Continued— Q R V W A Y X T B W Name C A M X U R Y X U U V Q Z S Z V V S In the diagram at the right, RQ > RS and TQ > TS. Use this information X Y for problems 8 – 11. V Y C A T 12 D 8. 9. If m/RQS = 6x+10, and m/RSQ = 9x – 17, then m/QRS = U V U 52 __________. O V 10 10 C C D (Note: You might also conclude m/RSQ 8= 60O m/SRQ If QS > RQ, then m/RSQ = __________. 12 B since all three sides of nRQS are congruent.) C C A 15 A B 10. Postulate Corollary 13c N ______________________________________. If /QVT > /SVT, then nQVT > nSVT. Why? 11. Q Theorem 33 T D /TQS > /TSQ. Why? ______________________________________. R E G A 8 55 C D A B A E DC H B M A B D W R F R Q 12. S Given: ST bisects /RTM ST bisects /RSM Prove: /TRM > /TMR X Z M T 1. Given C D 2. Given 12 8 3. /RST > /MST D Y REASON C 2. ST bisects /RSM C 8 D 12 8 12 3. Definition of Angle Bisector D A 15 5. nRTS > nMTS C G 6. RT >MT 7. /TRM > /TMR T A 1. ST bisects /RTM 4. TS > TS O S STATEMENT 100 B V 4. Reflexive for Congruence A 15 B Property B 7. Postulate 13 - ASA Congruence Assumption - If two sangles and the included side of one triangle are congruent to the corresponding angles and D triangle, then the two Eincluded side of another triangles are congruent. H 8. C.P.C.T.C. A A F B 33 - If two sides of a triangle are 9. Theorem congruent, then the angles opposite them are congruent. © 2006 VideoTextInteractive Geometry: A Complete Course 15 B A D 8 Unit IV, Part F, Lessons 6&7, Quiz Form B —Continued— 12 8 12 D NameA A B 15 15 C 5. Given: nABC ~ nDEF BG bisects /ABC; EH bisects /DEF Prove: CG FH = GA HD 1. nABC ~ nDEF 2. AB BC = DE EF 3. BG bisects /ABC; EH bisects /DEF; 108 H A STATEMENT D E G B F REASON 1. Given 2. If two triangles are similar, then corresponding sides are proportional 3. Given 4. AB AG DE DH = ; = BC GC EF HF 4. Theorem 35 - If a ray bisects an angles of a triangle, then it divides the opposite side into two segments that have the same ratio as the two other sides. 5. AB DE = BC EF 6. AG DH = GC HF 7. HF GC = DH AG 8. GC HF = AG DH 5. Switch the means in propotion for statement 2. DE (Multiply by EC on both sides of step 2.) 6. Substitution 7. Multiplication Property for Equality 8. Symmetry Property of Equality © 2006 VideoTextInteractive Geometry: A Complete Course B Name Quiz Form A Class Date Score Unit IV - Triangles Part G - Congruence – Part 3 (Triangle Inequalities) Lesson 1 - Theorem 37 - “If you have a given exterior angle of a triangle, then the measure of that angle is greater than the measure of either remote interior angle.” (Exterior Angle Inequality Theorem) Lesson 2 - Theorem 38 - “In a given triangle, if two sides are not congruent, then the angles opposite those sides are not congruent.” Lesson 3 - Theorem 39 - “In a given triangle, if two angles are not congruent, then the sides opposite those angles are not congruent.” Lesson 4 - Theorem 40 - “In a given triangle, the sum of the lengths of any two sides, is greater than the length of the third side.” For each of problems 1 and 2, write an inequality that relates only x and y. 1. x = 47 + y x>y _________________________ 2. y – x = 25 x<y _________________________ For each of problems 3 and 4, write an inequality that relates only m/1 and m/2 3. m/1 – 160 = m/2 m/1 > m/2 _________________________ 4. m/2 = m/1 + 65 m/1 < m/2 _________________________ 5. In the figure to the right, AB > BC > AC and r = t. Fill in each of the following blanks with >, <, or =. C < y ____________ x (x) > t ____________ y O (r) > r ____________ y O (t) O D (y) O A C B C © 2006 VideoTextInteractive Geometry: A Complete Course 111 17 Unit IV, Part G, Lessons 1,2,3&4, Quiz Form A —Continued— Name In problems 6 and 7, state one or more conclusions that can be drawn from the given information. Give a reason for each conclusion. 6. Given: Quadrilateral RABC with diagonal RB C CB > CR + RB_______ _______ Conclusion: (x) O (t) D O O (y) (r) RB > RA + AB_______ _______ Conclusion: U A R O A B C B S N D G C 12 1 17 17 Theorem 40 - If you have the sum of the measures of two sides of a triangle, then that sum is7greater ____________________________________________________________________ Reason: 1 2 A 88 92 H 0 B E 8 0 2 F 12 than the measure of the third side of the triangle. ____________________________________________________________________ ____________________________________________________________________ t nUNS and nAYD; SU > DA; y > DY;z m/S < m/D xSN A B A R C S A B O (x) O (y) A 10 10 42 41 X 31 32 X X O (r) A Y N D T U AYD UN > _______ (t) Conclusion: P D r 7. Given: A U C x O AY _______ Conclusion: < UN_______ A Reason: 14 14 S of a second C U measure Corollary 39a - If Cthe measure of one angle ofW a triangle is greaterR1than the angle of ____________________________________________________________________ 40 C B Y B 48 G S 17 17 Y N D 46 12 Z 130o E 46 V D 7 8 X 92 2 88 1 the of Ethe side opposite the larger angle is greater2than the measure of theS side ____________________________________________________________________ A triangle, then the measure B F R T H 12 0 0 T opposite the smaller angle. ____________________________________________________________________ t r A P y B 8. What is wrong with this picture? S A Y N D T T X 128 o y x A 10 B 42 41 X 31 32 X 10 E 46 46 Z D R 14 R S S C U remote interior Sangle. (Theorem 37) ____________________________________________________________________ T V R T U 48 14 X Y8 2 12 W ____________________________________________________________________ A A P P 112 N D Y X B 41 2 1 42 10 E Q 7 R © 2006 VideoTextInteractive Geometry: A Complete Course A T T A 10 Y V (with a measure of 128O) should be greater than (not less than) the measure of either The exterior 40 angle S ____________________________________________________________________ 7 F S 130o Y X N D 12 Explain. 1 17 2 1 z A B A P D x A U A U C Y M 4 11 N T Unit IV, Part G, Lessons 1,2,3&4, Quiz Form A —Continued— UA A U A diagram given R information R What conclusion and the C can you makeCfrom the given D in problems 9 - 12, by 37-40. D applying Theorem (x) O (t) O (x) O (t) O O O (y) (y) AC > BC ____________________ (r) O 9. m/1 (r) > m/2 O A A B C C C G C z x A UA A RA 11. B C 17 88 92 H 0 F E 12 Y W 48 1 14 17 7 B NS AD T 14 R X A Y 2 1 A 10 10 B 42 46 A E 41 46 CD 10 42 T S V T TU Y R M X Q 7 2 1 D V U A T T S46 R D 4 1 Y E S R T S T 7 N T 11 M © 2006 VideoTextInteractive Geometry: A Complete Course 46 R 46 46 C D S PR T A Y 46 A P X S P X S T TR P X 31 32 X C U R A U C X S T S R S P A N YD RS 12 R YN D A 46 12 A V Z U 8 P 41 2 F 12 T 130o V 12 2 A 10 0 UA B AS U 48 Z 14 14 1 W 7 8 12 F A U B X 2 z 2 0 N DS 8 8 A 130o10 10 10 10 41 E 42 B 41 B 42 o 31 32 X X 31 32 X y X y m/1 > m/2X 128 o x128/VRT > /VTR ____________________X 12. SR =x ST; VX = VT ____________________ YN D B N DS 40 Y 40 S G Y 7 y r x 12 1 1 17 N D t y r E F U D t 7 88 92 H 0 0 NS D S 12 17 0 88 92 B H C C B G 17 1 2 E B 1 A2 A EH > FH ____________________ B 10. B 17 17 B Name 113 A U A Unit IV, Part G, Lessons 1,2,3&4, Quiz Form A —Continued— Name 130o B S 128 o y x Y N D V 12 S 1 Given: TU > US > VS 8 13.7 X S 2 Prove: ST > SV F R 12 T STATEMENT 3. m/STU = m/TSU 4. US > VS 5. /SUV > /SVU A Y D 10 B 6. m/SUV = m/SVU A 7. m/SUV > m/STU 10 41 42 T X E 8. m/SVU > m/STU 9. ST > SV 46 C 114 46 D U V REASON 1. TU > US P 2. /STU > /TSU A T R 1. Given A Q 2. Theorem 33 - If two sides of a triangle are congruent, then the angles opposite them are congruent 7 4 2 P 3. Definition R Angles of Congruent 1 4. Given 5. Theorem 33 - If two sides of M a triangle are congruent, N T then the angles opposite them are congruent 11 Y of Congruent Angles 6. Definition 7. Theorem 36 - If you have a given exterior angle of a triangle, then the measure of that angle is greater than the measure of either remote interior angle. 8. Substitution 9. Theorem 38 - If the measure of one angle of a triangle is greater then the measure of a second angle of the triangle, then the measure of the side opposite the larger angle is greater than the S measure of the sideT opposite the smaller angle. © 2006 VideoTextInteractive Geometry: A Complete Course Unit IV, Part G, Lessons 1,2,3&4, Quiz Form B (x) —Continued— (r) U A R C D Name (t) O O O (y) O A B C B S G C N D 12 1 17 17 7 In problem 6, state one or more conclusions that can be drawn from the given information. Give a 88 92 2 1 2 reason for each conclusion. A E B F H /D /S > _______ Conclusion: A R C O < m/S___ O z A m/D _______ B 130o N D S T O Reason: B C B P A x D (t) (x)Conclusion: (y) (r) D t Uy r A U C A 12 nUNS and nAYD; SU > DA; SN > DY; UN < AY 6. Given: O 8 0 0 S x Y N D A Y y 10 128 o A B Theorem 38 - If two sides of a triangle are not congruent, then the angles opposite those sides are not congruent 42 ____________________________________________________________________ 41 G C 12 1 VX 31 32 X X 10 E S 40 17 Theorem17(Side-to-Angle Version) Hinge 7 ____________________________________________________________________ 8 1 2 A 0 E A U C 7. X 46 46 48 S Y Z 88 92 2 T V 14 W D R 14 S C F U R T U H 12 ____________________________________________________________________ 0 B t A P Given: nATP D P y r x Prove:z AT > TPS – PA B A A Y N D 40 Y 48 14 W Z 46 R 3. AT > TP – PA 116 41 1. nATP 2. AT + PA > TP 14 U S A 10 X 31 32 X X T X B C 42 R T T STATEMENT 2 1 10 Q 7 M Y REASON E 1. Given 2. Theorem 40 - If you have the sum of the measures of two sides 46 of a D triangle, thenR that sum is greater than the measure of S T the third side of the triangle. 3. Subtraction Property for Inequalities (i.e., Addition Property where the Additive Inverse is added to both sides) © 2006 VideoTextInteractive Geometry: A Complete Course 4 11 G C V 12 S 1 17 17 7 8 X Unit IV, G, Lessons 1,2,3&4, 92 88 Part 2 Quiz Form B E B F R H 12 —Continued— 2 A U Given: PA > PT; m/1 > m/2 8. S Name 0 0 B Prove: NRTD > AR S A Y 2 1 X 31 32 X Z 14 R U B 41 STATEMENT 1. PA > PT 2. m/1 > m/2 S C 3. PR > PR 4. RT > AR 10 42 46 M 4 11 N Y E REASON 1. 2. 3. 4. 46 D R Given Given S Reflexive Property Tfor Congruence Hinge Theorem (Side-to-Angle Version) - If two sides of one triangle are congruent to two sides of a second triangle and the measure of the included angle of the first triangle is greater than the measure of the included angle of the second triangle, then the length of the side opposite the included angle of the first triangle is greater than the length of the side opposite the included angle of the second triangle. A U A T X A Q 7 R T T 10 V U A P P z T T 130o S 128 o y x B Y N D V 12 S 1 7 8 X 2 S R 12 A T T A P 2 1 P X A 10 41 42 10 Q 7 R 9.A What is wrong Twith this picture? Y V U T M 4 11 N Y Theorem 40 - If you have the sum of the measures of two sides of a triangle, then that sum is greater ____________________________________________________________________ E Explain. than the measure of the third side of the triangle. ____________________________________________________________________ 46 C 46 D R S T ____________________________________________________________________ © 2006 VideoTextInteractive Geometry: A Complete Course 117 G C 12 A 17 17 7 88 92 H 1 2 0 0 t D V S1 R7 T 0 8 T 2 A C Y B r S y– 12, useNeach D For problems 10 accompanying figure, and fillU in the blanks with <,P=, or >.P1 x G z 12 A B 1 tV S r > 10. XY _______XZ; XW _______ 7 > 8 17 A X 88 92 CH 0 (x) O (t) 0 F 12R X 40 D O O 48 14 C Y A B W Z 14 R B P 48 14 1 17 17 A 88 92 E A H 10 42 41 7 Y 0 B B 31 32 X 46 D S y W P12 14 41 A T 2 1 U Y R R T X R 46 S T D R S T 2 1 P A Y N D W 14 A 10 41 10 42 T M 4 11 Y E Given: XS > YS; RX > TY S is midpoint of RT Z 46 R U S Prove: m/R > m/T C 46 D R STATEMENT 1. S is midpoint of RT 2. RS > TS 3. RS = TS 4. RX > TY 5. RX = TY 6. XS > YS 7. m/R > m/T 118 Q 7 R T X X 31 32 X 13. S V U A P B 40 R N T T X D E A S T 46 C4 z B E 128 o S y 11 S 10 42 41 R 7S M A 130o Q UV T X x D A 10 X 31 32 X 46 A P Y Y 10 N D E V 42 T Y 2 46 C C N D Z 8 X 12 F 10 S T 0 U C Ut 10 S UT 40 U S < 12. BC _______ DE T B B 46 Y N D 1 2 X < 11. RU _______ US A z M T T B G C A S Y y N D x T 14 X (y) U AR 7 R D S X 31 32 X 2 T A P x S R 12 128 o y S T REASON 1. 2. 3. 4. 5. 6. 7. © 2006 VideoTextInteractive Geometry: A Complete Course V U X 2 F o A130 U C 88 92 H E B A U A YS N D 12 X 2 17 17 Name12 A Part G, Lesson 1,2,3&4, E B F B Unit IV, Quiz Form —Continued— 1 2 A R 8 G C 0 V S B C 1 B Given Definition of Midpoint Definition of Congruence Given Definition of Congruence Given Hinge Theorem (Side-to-Angle Version) - If two sides of one triangle are congruent to two sides of a second triangle and the measure of the included angle of the first triangle is greater than the measure of the included angle of the second triangle, then the length of the side opposite the included angle of the first triangle is greater than the length of the side opposite the included angle of the second triangle. N Unit Test Form A Name Date Class Score Unit IV - Triangles 1. Write true or false for the following statements True (A-2) ____________ a. An equilateral triangle is also an isosceles triangle. (A-2) ____________ False b. A right triangle may also be an acute triangle. False (A-2) ____________ c. All isosceles triangles are equiangular triangles. (A-1) ____________ False d. A right triangle has only one altitude. False (B-1) ____________ e. If nDEF is a right triangle and the measure of one acute angles is 65, then the measure of the other acute angle is 35. (A-2, B-1) /B /C 2. nABC is isosceles with AB = AC. The base angles are ______________ and _______________. (A-2, B-1, F-4) 3. nDEF is isosceles with base angles measuring 25 degrees. The measure of the 130 vertex angle is ______. (F-4) 60 4. nMNQ is equiliateral. m/Q = _______________. (A-1) vertex perpendicular 5. An altitude of a triangle is a segment drawn from a _______________ and _______________ to the opposite side of the triangle. (E-3) hypotenuse 6. The HL Congruence Postulate states that two right triangles are congruent if the ____________ leg and _______________ of one are congruent to the corresponding parts of the other. (B-1) Find the measure of the indicated angles in problems 7 - 9. x x x 7. 35 35 55 x = _______________ A C y t H y E 93 8. x 3535 A D W AH H V V E C W A B E A A U 9. x A D x E EB || DC x 3 AB P D 104 D Ax = _______________ 7 C EB || DC E 38 A 35 A E B D D 60 C A P C 7 B D PM 60 7 A 60 B 45 Q S Q S 90 – 52 = x 45 C 60 B C 6060 B C B 60 B24 A 38 = x 24 B 24 C C C R R D R © 2006 VideoTextInteractive Geometry: A Complete Course C C D D D F N F N R N A 1 3 D 180 – 35 – D 93 = 52 10 7 180 – 38 – 38 = x S 10 Q 104 = x 10 119 CV F U 38 93 x x A38 x = _______________ 90 – 35 = x W 55 = x y 93 35 C C t D 38 3 Unit IV, Unit Test Form A —Continued— Name D Use the figure to the right for problems 16 - 20. 38 (D-2) E EB || DC 93 2 AB = 6, BC = 4. AE = 3, ED = _______________. x x 16. AB AE = BC ED 6 3 = 4 x C (4 )(3 )= (6 )(x ) D 35 35 A A A G 7 M 10 60 A B S Q C 45 45 60 60 C R AB AE C= BC ED 9 x = 6 4 (6 )(x )= (9 )(4 ) C P 7 6 B 17. ABA= 9, BC = 6, ED= 4, AEB = _______________. 24 C B D A 12=6x 2=x W (D-2) E B D D F N R 3 Q E 6x=36D x=6 1 C B (D-2) F H 1 T 1 18. BC =3, AC = 10, AE = E2 , ED = _______________. M K B 3 BC ED = AB AE 3 ED = 1 7 2 R 3 2 D A (7 )( ED ) = 31 73 A 1 B 21 3 ED =1 7 ⋅ ED = 10 (D-2) 1 19. AE = 4, ED = 2, BE = 7, DC = _______________. 2 B (D-2) G 20. (4 )( DC ) = (6 )(7 ) (4 )( DCA) = 42 AD DC = AE EB AE+ ED DC = AE EB 4+2 DC = EB 4 6 DC = 4 7 GAC =F 12, A AB = 8, AE = 1 2 3 _______________, B 6 4 T 2 ⋅2 ⋅2 ⋅114 0 20 x AE= 2 ⋅2 ⋅3 = 3 R AE=6 2 8 12 T3 B B M a 3 C AD = 10 63 D x 6 3 12 8 A 8 10 B N Q 6 5 C A 61 2 42 2 ⋅ 21 21 DC = = = 4 2⋅2 2 1 DC C =10 2 AB AE = AC AD M 10 8 AE = K 12 10 (12 )( AE )= (8 )(10 ) Q M 4 6 E R 8 © 2006 VideoTextInteractive Geometry: A Complete Course C M C K D 121 E Unit IV, Unit Test Form A —Continued— Name D 38 B E EB || DC 93 E x ABC is a right triangle, as shown to the right, For x problems 21-23, triangle with /C as the right angle. DE AB, AE A = 4, ED = 3, AD = 5, and CD = 7. C 4 3 C B 7 A 5 D P P D N Postulate Corollary 12b - If an acute angle of one right triangle is M congruent to an acute State why nADE ~ nABC.________________________________________________________________ 7 60 A angle of another right triangle, then the ________________________________________________________________________________________ 10 right triangles are similar. B 45 S Q (D-1) A 21. ________________________________________________________________________________________ B B C 24 60 R (D-3) 45 60 C D AD AE ED = = AB AC CB 22. Fill in each box to complete the following extended D proportion: F N R Q E 12 23. Find AC: F ____________ H AE A ED = AC BC 4 3 = 12 BC (12 )(3 )= (4 )(BC ) 36= (4 )( BC ) 4 ⋅4 = BC 4 G 2 C 4 D AC = 12 ( AD+ 2 DC or 5+7 1 ) D 9 Find BC: ____________ 15 Find AB: ____________ E 1 3 (D-3) Q A C AE AD = B AC AB 4 5 = 12 AB (12 )(5 )= (4 )( AB ) x 60= (4 )( AB ) 4 ⋅1 15 = AB 4 15= AB 35 35 9 = BC C A For problems 24 and 25, HW bisects /H, as shown in the figure to the right. A 61 (F-6) y A B 1 56 = 10 WA 53.6 = WA C ( ) A M a 63 V C AC U G B (F-6) 10 K 4; 6 25. If HE = 8, HA = 12, and EA = 10, Find EWDand WA. ____________ x B 14 12 8 N B N Q A E (14 )(4 )=(10 )(WA) M W H 5.6 24. If HE = 10, EW = 4, and HA = 14, Find WA. ____________ HE EW = HA WA 10 4 = 14 WA B R HE EW 6 5 HA = WA A x 8 = 8 C 12 10 - x (12 )(x )= (8 )(10 - x ) EW =4 122 3 R 6 C 4 6 WA=10 - 4 WA=6 M 8 E M C K © 2006 VideoTextInteractive Geometry: A Complete Course D T S 12 8 10 WA=10 -E x 12x=80 -8x 20x=80 x=4 D M t 2 C A R K B D D F B 3 x 93 x Unit IV, Unit Test Form A —Continued— Name 35 35 C A A C D t For problems 26–29, nABC is a right triangle with altitude AD W and hypotenuse BC, as shown in the figure to the right. y H (D-4) B A E V 27 26. CD = 3, AD = 9. Find BD. ____________ CD AD = AD BD 3 9 = 9 BD (9 )(9 )= (3 )(BD ) B C F A U Q G B D 81= (3 )( BD ) F H 3 ⋅27 = BD 3 27 = BD R T S E C M (D-4) 2 3 27. CD = 1, CB = 12. Find AC.____________ CD AC = AC CB 1 AC = AC 12 ( AC )( AC )= (1 )(12 ) R K A ( AC )2 =12 B D B A 2 AC = 12 = 4 ⋅ 3 AC =2 3 (D-4) 32 28. AB = 8, BD = 6. Find BC. ____________ 3 BC AB = AB BD BC 8 = 8 6 (8 )(8 )= (BC )(6 ) G E B 10 3 2 ⋅32 = BC 2 ⋅3 32 = BC 3 K 4 S 2 B 5 T 14 x P C R 8 12 4 5 8 5 29. CD = 4, DB = 16, Find AC____________. Find AB____________. Find AD ____________. T 8B CD AC = AC CB CD AC = AC CD+ DB 4 AC = AC 4+16 ( AC )( AC )= (4 )(4 +16 ) ( AC )2 =80 1 M Q 64= ( BC )(6 ) (D-4) A F G DB AB = AB BC DB AB = AB BD+ DC 16 AB D = AB 16 + 4 ( AB )( AB )= (16B)(16+4 ) A ( AB )2 = (16 )( 20 )= 320 AC = 80 = 16 ⋅ 5 AB= 320 = 64 ⋅ 5 AC =4 5 AB=8 5 CD AD = AD DB 4 AD = AD 16 R C ( AD )( AD )= (4E)(16 ) ( AD )2 =64 AD = 64 C D 25 R F 30 AD=8 © 2006 VideoTextInteractive Geometry: A Complete Course 123 C Unit IV, Unit Test Form A —Continued— Name (D-5) No 30. If the measures of the three sides of a triangle are 10, 12, and 15, is the triangle a right triangle? ____ The given triangle is not a right triangle. a + b > c . So, the triangle is an acute triangle. If not, explain. ______________________________________________________________________ 2 2 a +b 2 =c 2 2 38 x 2 E 93 x x 10 2 +12 2 =15 2 35 35 A 100+144=225 244 > 225 C(D-5) A D A C 7 D 26 31.t In rectangle ABCD, as shown to the right, find AC. _______________ 10 Q W ( DCy )2 + ( AD )2 = ( AC )H2 (10 )2 + (24 )2 = ( AC )2 676 = AC E 4 ⋅ 169 = AC V 100+576= ( AC )2 2 ⋅13= AC 676= ( AC )2 26 = AC B A B C 24 C F U (D-5) N A Q G D B E D 2 32. The length of the hypotenuse of a 45-45-90 triangle is _______________ times the length of a leg. (D-5) x R T 93 S 33. The length of the longer leg of a 30-60-90 triangle is x C 35 of the hypotenuse. 35 F 38 H 3 E 2 _______________ xtimes the A (D-5) 38 A 7 1 W PS = ⋅7 or 2 2 x H 7E 3 3 ⋅7 or SQ = V2 2 D A PR. D If PQ = 7, Find PS x D EB A B C A 10 U Q G A E C B Find Find Q 2 DC_______________. K 2 M E R F 60 F D AB = BD = 1 T 1 2 C a A B 1 D A Q 32 C 1 A x B 1 5 14 12 8 B 4 B C R G 2 K A 2 C D x 6 C 8 8 A 10 R M E Q B A © 2006 VideoTextInteractive Geometry: A Complete Course D E D T 124 MR 103 C A B 45 Q 45 4 S D D N start with Q 3 C R A 5 D A P 2 7 60 4 A B D 3N CB_______________. P D C M N 60 G H C G R E T B B S Q C 3 24 S D B 24 1 Find BD_______________. E F B F 60 7 C F 60 7 7 3 10 2 _______________. P 2 Find AD_______________. H Find SQ D S P E C R EB || DC B 35. AB = 1 in the figure to the right.B B B 7 D _______________. B 2 R K A A (D-5) A length M 34. In nPRQ shown to the right, A C QS DEB || DC E M K 61 E C A C D t Unit IV, Unit Test Form A —Continued— y W Name H A E V C A (E-3) 40. U RT; VR > UT; /V > /T Given: VS G D B Prove: RS > US H R STATEMENT T S C REASON 1. VS RT 2. /VSR and /VST are right angles 3. nVSR and nVST are right triangles 4. VR > UT 5. /V > /T 6. nRSV > nUST E M 1. 2. 3. 4. 5. 6. Given Definition of Perpendicular R K Definition of Right Triangles A B Given Given Postulate Corollary 13b - HA Congruence Postulate - If the 38 93 hypotenuse andDacute angleof one right triangle are B congruent x x angle of another to the hypotenuse and corresponding acute right triangle, then the 35 triangles are congruent 35 7. C.P.C.T.C. x 7. RS > US C A D A C D 10 t W y H G E V C (D-1) 41. Given: /A > /D; /B > /E; U of DE G is the midpoint of AB; H is the midpoint Prove: CG = AC FH DF C 24 M Q 3 A G T S C STATEMENT B 10 F NK 4 Q S 2 R F G B A E A T D B F H P R E REASON M 14 x 12 8 T B A C 1. /A > /D 2. /B > /E 3. nABC ~ nDEF 4. G is the midpoint of AB 5. H is the midpoint of DE 6. CG is a median 7. FH is a median CG 1. Given 2. Given D E R K 3. Postulate Corollary 12a - If two angles of one triangle are B congruent to the Btwo corresponding angles of another C triangle, D 25 R then the two triangles are similar. 4. Given B 5. Given A 6. Definition of a Median 2 7. Definition of a Median A D AC 8. FH = DF 8. Corollary 29b - If two triangles are similar, then the measures of corresponding medians are in the same ratio 1 as the measures A G B F of correspondingG sides. E 30 3 C M Q 10 126 © 2006 VideoTextInteractive Geometry: A Complete Course S 2 3 K 4 P C T x B 6 S 5 14 C A 8 1 A C A Unit IV, Unit Test Form A —Continued— y t H 7 D Name 10 B A E B C 24 60 C C D F N A Given: nFDN Uin which /F > /NDQ Q G Prove: nNQD ~ nNDF F H T S E D B R S R (D-1) 42. 60 Q W V 2 D 1 A E C STATEMENT B REASON M 1. nFDN in which /F > /NDQ 2. /N > /N K 3. nNQD ~ nNDF A 1. Given 2. Reflexive Property for Congruence R 3. Postulate Corollary 12a - If two angles of one triangle are congruent to the two corresponding angles of another triangle, D B then the two triangles are similar 38 E EB || DC B x 93 D A C B S T A Prove: ED > CD K N Q x R F B S 2 1 E A 1 2 D 6 B C K 1. Given 34 - If twoFangles R the C of aDtriangle are congruent, then D 25 R 2. Theorem 30 sides opposite them are congruent 3. Given H 4. Given 5. Given M 6. Definition of Perpendicular P A 61 7. Definition of Perpendicular 8. Definition of Right Triangles G K J 9. Definition of Right Triangles M a 10. Postulate Corollary 13e - HI Congruence Postulate - If a 3 hypotenuse and one leg of a right triangle 63 are congruent to the B N Q C hypotenuse and corresponding leg of another right triangle, R then the triangles are congruent. A A 11.C.P.C.T.C. C R A 11. EDC> CD A 2 B M P 10 K 4 1 S B 6 5 T 14 R 8 12 25 M M C R R 83 C 10 M REASON A E F x 2 D R M D x 8 C P 45 Q 45 A C 3. AE > BC 4. DE AE 5. DC BC B 6. /E is a right angle 7. /C is a right angle 8. nAED is a right triangle 9. nBCD is a right triangle 10.nAED > nBCD 3 5 D D A B Q E 12 8 T 7 60 D 6 B 14 E D 1. /1 > /2 2. AD > B BD K G B C 5 T M STATEMENT G 60 R F 4 H C 3 C S M Given: G/1 > 2 /D2; AE > BC; B DE AE; DC BC P S 10 1 B Q C 24 a 60 10 3 A B Q C (E-3) F G C P D A C B 7 G V M E 4 3 61 A A2 D A E 43. x A E W U B x 35 35 D P D A C B P 8 D 6 x 3 Course © 2006 VideoTextInteractive Geometry: A Complete 12 x 8 A B C B 4 6 K 10 E 127 A R Q S N 3 A Unit IV, Unit Test Form A —Continued— D B B B 7 10 60 Name P N A B S Q A 5 D M 7 C 24 C P D A C 45 45 60 60 C R D C Q (F-4) D F N R Given: /ARG > /CGR; DG > DR 44. Q E D Prove: m/3 = m/4 F H 1 3 C 2 D 4 D A 1 A E G 2 C B STATEMENT REASON x 1. /ARG > /CGR 2. m/ARG > m/CGR 3. m/1 + m/3 = m/ARG 4. m/2 + m/4 = m/CGR 5. m/1 + m/3 = m/2 + m/4 6. DG > DR 7. /1 > /2 R C t A 8.2 m/1 = m/2 9. m/3 = m/4 1. 2. 3. 4. 5. 6. 7. M Given Definition of Congruent Angles 35 35 Postulate 7 - Protractor - Fourth Assumption - Angle Addition Postulate 7 - Protractor - Fourth Assumption - Angle Addition A C Substitution D Given Theorem 33 - If two sides of a triangle areWcongruent, then the y H are A congruent.E A 61 angles opposite them 8. Definition of Congruence V C 9. Subtraction Property for Equality M a A B 3 C 1 93 63 B F N A U B N Q x C G D B H R M 10 K B 6 (G-3) D x C 12 8 T B 4 E 8 A B E C M STATEMENT K D REASON D 25 R B F 1. m/B > m/A 2. AC > F BC C D 30 R K R Prove: AC > BC C M 6 10 E C 12 8 A Given: m/B8 > m/A 45. 6 3 5 14 T S 1. Given 2. Corollary 38a - If the measure of one angle of a triangle is A B greater than the measure of a second angle of the triangle, then C of the third side opposite the larger angle is greater the measure than the measure of the side opposite the smaller angle. R R H G E P K S J B D 3 E M Q 10 K 4 S T 2 R F x A 14 x P R A M 12 8 T B A 128 K B C B 5 5 © 2006 VideoTextInteractive Geometry: A Complete Course x 25 B Q A G A F G C y 39 D 7 z N 49 E C 7 Q Unit Test Form B x Name Date Class 35 Score C A t Unit IV - Triangles x H E In all of problems 1–50, choose the letter for the correct response. c 1. A triangle with no congruent sides is _______________. C b. Obtuse d. Isosceles A H E G A R V b. three d. one B T S CC F A U G (A-2) D B K A c An equiangular triangle is also _______________. a. right c. equilateral U D W y b 2. An acute triangle is one with _______________ acute angle(s). 3. A C t (A-2) a. two c. at least one A x 35 V 35 (A-2) a. Equiangular c. Scalene W 93 y R T S b. obtuse d. scalene B H E C M D B (A-2) c 4. nDEG, shown to the right, is a right triangle. DG is called the _______________. a. altitude c. hypotenuse A b. leg d. angle R K B G E (A-1) D d 5. In nABF, shown to the right, BG is a(n) _______________ of the triangle. a. side c. hypotenuse A Q b. leg d. altitude 3 2 G E c 6. An equilaeral triangle must also be _______________. b. obtuse d. right A F G P A Q B T 1 x M 10 3 (A-2) S d 7. In an isoceles triangle the base angles are _______________. a. right c. congruent 4 S (A-2) a. scalene c. acute G B b. acute d. both b and c K 4 T 2 D E 14 x B P R 12 8 T B C R A (D-2) 8. C a An auxilary line is _______________. a. a line added to a figure to help solve a problem c. a hypotenuse D E b. an altitude C B that may be omitted from d. a line aD figure 25 R © 2006 VideoTextInteractive Geometry: A Complete Course C P 30 129 S A D R K y M 7 WA Unit IV, Unit Test Form B H E —Continued— P D A C B 10 B B C 24 A B S Q Name A 60 45 60 C D D B M F (A-1) 60 C R V 45 N A D A 2 61 R b Q 9. In nABC, shown to the right, G/3 is called a(n) _______________. D E C D B a. remote interior angle R ET S b. exterior angle C c. acute a F H G A F EG B 2 3 C D 1 A B D Q M K 3 4 || DC c B 6 10. The HA theorem states that two right triangles are Econgruent ifEB_______________. 93 R K A D (B-1) 7 A 2 10 B a.A 61 c. 45 E C B G E D b. 63 B d. 34 (C-2) A A B (B-1) T 60 C 303 C 12 B 2 B 5 (D-2) 35 25 1 ⋅ BC A 2 F G 1 c. AM = ⋅ AB 2 M 10 G Q N H 4 2 E D B 1 R the d. AN AN T = MN NC B (D-2) a. 6R 12 8 T B R C E M R C 130 D T H A E A B B C KC B M D N P Q B N G K 3E R 8 S J 6 R B C A E G A F G 12 4 x M 10 A E Q 5 2 M M8 or 3 3 x K 3 8 B C C K E BS T 2 D K 4 C y 39 7 14 x P R B A F C R D A B © 2006 VideoTextInteractive Geometry: A Complete Course S T U V C C H R D E 12 8 T F X 30 D 25 R E C R W R S 4 J G H D x b. 2 d. 2 Q V U 25 c. 1.5 A D M A A A A6 d S right, 15. Ifx STC|| PR14in theP figure to the then x10= _______________. 8 C C 6 12 C R B K 61 Q 5 T 3P F 63 6 R C measure of the S A a C S A D G6 K A G E D M b. MN || XBC 3 K 4 W DR B 63 Q MB NC a. MN = C D U 10 M A 8 M C x A K 35 8 B A 8 b 14. If AM = AN in the figure to the right, then _______________. R 93 B 30 M R M N 61 R B D 25 R A 2 E A W Ax 16 A measures B P 1 3 V of the angles of a triangle are in the ratio 2:2:5, then M C R R c K largest angle is _______________. P C S E b. 40O a. 80O D c. 100OB d.C 90O x F C K C Q R H C C 5 a D F E A D 6 4 10 45 y D E M 45 8 A 3 M A R D 25 R t D N y bQ 12. GIf P = x , then =Q_______________. M D 9 9B y 10 P x K 3 4 a. F b. H x T P S E T y 14 9 2 c. d. x M x P 8 yR 13. If the C 1 B 7 x 60 C S 60 C F G F CE 24 C R Q d 11. In nMNQ, shown to the right, find a. _______________ W 8 C C D x 5 S T x a. twoA angles andx the them are congruent B side included between 14 2 x 35 b. each hypotenuse and aP corresponding leg areA congruent R 12B 8 TP B c. eachD hypotenuse and Aa corresponding angle are congruent D B C Q B 10 38 A A 1 M (E-3) 35 1 3 U y H B A E C E 93 U S A b 16. KB || MT. KR = _______________. C A C D a. 20 c. 5 H (D-2) b. 15 d. 14 W AC R G 7 R 8 60 C BF HJ BF GJ c. CF = HJ (C-3) A K B D D 25 R E A 2 D 2 B P G A F G C T 2 1 R B 2 D G D D H T a M Q x R B E C R A D C 8M 10 4 B B K a M 63 C S 6 2 E 6 H 4 3 3 12 T 4 Px P R B R Q 8 10 F G D x 8 BG N F V M K E 30 U A B A x 25 M 6 E R B X D 25 R 8 A 61 b. two rectangles 3 C octagonsQ W Rd. two regular 1 D x 6 5 12 8 T R 8G A a. Postulate 12 - SAS Similarity Assumption E b. Postulate Corollary 12a - AA Similarity Postulate P c. Postulate 12 - SSSC Similarity F S C DC 25 R Assumption 30 d. Definition of Similar Triangles K K B K A C b M because of the ______. C 22. In the figure to the right, DE || BC. So, nABC ~ nADE K D K A 1 C SB A B 5 8 d. = 14 y S10 P a. Postulate Corollary 12a - AA Similarity Postulate S 5 T S T 2b. Postulate 12 - SSS xSimilarity14Assumption C c. Postulate 12 - SAS Similarity R 12 8 Assumption (D-1) A B 10 M T 10 D A b 10 are similar by the _______________. 21. The triangles shown to the right D K P 3 F M Q B 3 P C C U 1 B d 20. Which of these polygons must be similar? _______________ R CM Q 5 G 2 (D-1) M D 3 C A E x 352 D 45 W 45 C (C-3) a. two isosceles triangles A G F G c. two parallelograms with 65DO angles A E 60 C 30 A c. 6 = 8 C 10 A x b. (6)(10) = 8xS R 7 A MA A V c K 19. If in the figure to the right, nABC 3~ nAED,4 then _______________. a. 6 = x 8 10 8 C B R C B Q F BF HJ d. CF = JH JG R BF E 5 N b. CF = GH M H R 6 C D d E 18. IfS nBCF ~ nGHJ, then _______________. a. CFC = GJ BS T y C 12 E B H T Q B C35D P 14 60 R D F b. 25 D d.D 28 B 3 A 10 A 24 E 3 B E D T x t B D E K F A B c 17. KB || MT. MT = _______________. a. 28 U c. 70 (C-3) 3 4H C K V C 3 C B 10 M B 1 B Q M S 2 P A E 60 N x EB || DC QD B x 3 T A E G 35 R D FF G 38 A Use the figure to the right for problems 16 and 17. 35 G Name x (D-2) C 24 R V Unit IV, Unit Test Form B x —Continued— B C D E B A A R T E D R D J F C B A D 25 R C H (C-3) 25 23. nPQR ~ nMNO. PQ = 18, a. 92 3 (C-2) R QR = 12, PR =M16, b. 36 and c. 46 M MN = 12. x B K The perimeter d. 40 R C B a of nMNO is _________. P Q A W R G K S J B D E 3 and 4? _______________ X c 24. What is the geometric mean between b. 3 2 2S a. 3 2 C P 8 U R T c. 1 6 V 2 d. 3 6 A S e. A C 35 16 F x P 131S © 2006 VideoTextInteractive Geometry: A Complete Course 5 25 x B C B C y 39 z25 61 Q 2 M 10 3 Unit IV, Unit Test Form B E —Continued— S A 2 F G T 1 B 3 14 C x P R T (E-3) K 4 Q 12 8 MA 10 Q 3 C B a. 15 5 c. 15 3 S T b. 52 30 Pd. 5 55 5 D x D 25 R C 26. The length of DK, in the figure to the right, is the geometric mean between M (E-3) 12 8 T R B 6 10 C R B M 8 8 C 8 C D x A E 14 6 5 R B b 6 In right triangle CFD, shown to the right, find the length of the altitude CR. ______ 25. B a Name G K 4 E F 30 C A c the lengths of _______. D a. DR and MR b. MK and MD B c. MK and KR d. CD and DR K E C D 25 R F 30 C R D (D-5) A H R c 27. The hypotenuse and one leg of a right triangle are 29 and 21. The other leg measures ___________. a. 1282 c. 20 R b. 36 d. 5 2 P C P A S G K 25 a 28. The legs of a right triangle are 6 and 9. The hypotenuse measures _______________. M M R a. 117 c. 54 b. 3 6 Cd. 117 S A W R A X c 29. A triangle whose sides are 5, 12, and 13 is a(n) _______________. 25 a. obtuse triangle c. right triangle M K R b. acute triangle M T S d. given measure cannot form a triangle R (F-7) x B U V W a 30. A triangle whose sides are 6, 11, andX 15 is a(n) _______________. a. obtuse triangle c. right triangle b. acute triangle V U cannot T measure form a triangle Sd. given (D-5) a 31. The longer leg of a 30-60-90 triangle is 7. The hypotenuse measures _______________. a. 14 3 3 b. 7 3 c. 14 d. 3 7 (D-5) a 32. One side of a square is x. The measure of a diagonal is _______________. b. x 3 c. x 3 2 d. x 2 3 (D-5) c 33. Each side of an equilateral triangle is 18. The measure of an altitude is _______________. 132 B R P (F-7) a. 9 x K b. 9 3 2 c. 9 3 © 2006 VideoTextInteractive Geometry: A Complete Course d. 18 3 C B Q S J (D-5) a. x 2 D V K 4 C Unit IV, SUnit Test Form B T 2 —Continued— x P B t 14 y R x T 93 A x E B C B M x U A C A E a.DnABC > nDEF Bc. nABC > nEDF y C b. nABC > nFED R H E VnEDF d. n BCA > D 25 R 30 T D A FC C B CD B EB 24 C R PA M (E-1) a. /S > /W c. TU > VX H D E BB A b. /T > /W R S d. US > WX C x R G F (E-2) A t a. ASA c. SAS (E-1) C K J B A E A U G S 25 b. AC M M R T x C 2 M A R F G (E-2) 38. T B A 12 W R 3 b. SSS S d. none of2 these K 1 30 A P a B B 14 C A x T U A V 12 8 (E-2) T D 25 R 2 R E c nGHK > nJHK by __________. a. ASA b. SSS D F G 3 B F S R D 25 R T B R 12 8 T M 6 M Q R B R M D8 x F 30 5 C A 14 x 8 C 1061 C M prove nPRQ > a. ASA E b. SSS c.R SAS D 25 R M 25 C F 30 M A 6 K R T A N8 R A C B N R V UM H 4 C D P Q x 25 D K 3 12 G 6 KB R SC J A B 4 C E H E C A A DP Q A F V G x R D B K (F-2) R 8 K U T S C B 6 D E A C H 3 G X 6 10 W B N M K 40. In the figure to the right, given that PR > CRS, and PQ bisectsS RS,X you couldR P b nPSQ _________. D 7 12 63 S 10 C c. SAS P 3 C 6M 63 W 8 S a P E M R C 1 C B K 4 B 10 39 C D K D M 30 8 M E Q 2 (E-2) A C8S 25 x R 39. In the figure to the right, given that GK > JK and HK GJ, you could prove G 5 Q A C 4 D F y C 6 5 T P S A K G 2 D F Q M x 61 1 x B M B E D X Q C 6 M 4 D 45 M C a B The two triangles shown to the right are congruent by _______________. 10 a. ASA c. SAS 10 a 60 R R D AN 8 D A QK A P D 8 3 C R C 61 x A 5 D 45 3 B 7 CA B R 3 C B D 25 52RC 6 M R E B 8 T C 5 C M B 4 3 B C 12 D 25 R E E B 60 CD 1C A 8S 1 14 x 3 A 1 CB A F E c.BAD G P K E B AM N 10 2 EK Q H a. DC A C E 2 B 14 60 R C c B 37. In nADC, the side included by /A and /D is _______________. S T R 7 x Q 3C C EB || DC A B R D 4 3 D B R P 10 24 A F Q R 1 B SK C A S BF A 10 DT P V C AR 2 D Q B M A 2 A D WB H E 4 SA b. SSS G E G R d. Not enough information to tell. D B y P C A D D 2 c 36. The two triangles shown to the right are congruent by ___________. R K C F P 3 G 2 F A x C E R x 35 35 A Q B x M B T 60 B C 10 K 45 1 45 60 14 B D 8 T B E 4 D S DC C G 38 E 93 D E M S 2 Q H T D 3 Q N A b G If nSTU is congruent to nVWX, then _______________. K 7 P Q 60 D F U Q 10 F H S F G B N C R G A C D B S EB || DC 60 8 R D K W 4 Q C E A M 60 E 24 KG DA bDB and _____. d 34. Two statements of t congruence for the two triangles shown are _____ B 35. F D10 F C 7 H E C 35 35 12 38 P QB 3D 6 T C 6 D D A B 8 10 V A (E-1) D S E R B C G A WR 8 C H 12 8 A Name 5 N A DA x U 6 B x F 35 35 10 3 C A x M Q K S J A C B E B C H R 41. In the figure to the right, nABCC> nDEF. Therefore, /CBA > P W R c Q /FED because _____. R P A SX a. they are right triangles b. they are acute triangles 25 V c. CPCTC d. not enough information to U tell T S M M R P C (Appendix A-1) G K A A S J 5 x R B K C B x A x 25 B C B C b 42. The statement “If nABC > nDEF, then /DEF > /ABC” illustrates _______________. a. the reflexive property of congruence T S b. the symmetric property of congruence c. the transitive property of congruence R M 5 5 W X M M K y 39 z 7 W V U X S T U 39 y C F A S R R B D E V © 2006 VideoTextInteractive Geometry: A Complete Course 133 N 49 7