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Mathematics in Action (3rd Edition) 3A Full Solutions 5 Quadrilaterals In △BCD, BD 2 BC 2 CD 2 Quick Review y 15 8 2 (Pyth. theorem) 2 17 Let’s Try (p. 5.3) 1. (a) x 130 180 (int. s, CB // EF) 2. x 50 37 ACD 75 ACD 38 (b) BCD 50 15 65 ∵ ABC BCD 65 ∴ AB // CD (alt. s equal) 2. (a) In △ABC and △CDA, ABC CDA BAC DCA AC CA ∴ △ABC △CDA given alt. s, BA // CD common side AAS (b) ∵ △ABC △CDA ∴ BCA DAC ∴ AD // BC proved corr. s, △s alt. s equal (b) ∵ ∴ ca 70 (corr. s, BA // CD) b 70 180 b 110 (int. s, BA // CD) ∵ △BCE is an equilateral triangle. ∴ EBC ECB 60 prop. of equil. △ ABC BCD (20 60) (40 60) 80 100 180 ∴ AB // DC int. s supp. 4. (a) In △ABC and △ADC, AB AD BAC DAC AC AC ∴ △ABC △ADC b 20 a 20 20 40 (b) ∵ ∴ ∴ 5. (base ∠s, isos. △) (ext. ∠ of △) ∵ CD BD c b 40 (base ∠s, isos. △) ∴ BAC ACD 38 BA // CD alt. s equal 3. Review Exercise 5 (p. 5.5) 1. (a) a 70 (alt. s, AD // BE) (b) In △ABD, ∵ BD AB a 20 ∴ BCD ADE (corr. s, DA // CB) (a) given given common side SAS △ABC △ADC (proved in (a)) (corr. sides, △s) BC DC △BCD is an isosceles triangle. (a) In △ADB, ∵ AB AD ∴ ABD ADB (base s, isos. △) DAB ADB ABD 180 ( sum of △) 56 2ADB 180 2ADB 124 ADB 62 In △ACD, ACD CAD ADB (ext. of △) 28 CAD 62 (c) In △ABC, ∵ BC AC ∴ ABC x 60 x x 180 CAD 34 (base ∠s, isos. △) (∠ sum of △) (b) x 60 BCE ABC (alt. ∠s, DE // AB) yx 60 EAB EAD DAB 34 56 90 AED 90 ∵ EAB AED 90 90 180 ∴ AB // ED (int. s supp.) (d) In △ABC, BC 2 AB 2 AC 2 (Pyth. theorem) x 122 9 2 15 76 5 Quadrilaterals 6. (a) In △BCD, ∴ BC 2 BD 2 (10 2 24 2 ) cm 2 (100 576) cm 2 676 cm 2 CD 26 cm 2 676 cm 2 2 2 ∵ BC 2 BD 2 CD 2 ∴ △BCD is a right-angled triangle, where CBD 90 . (converse of Pyth. theorem) ADB CBD (alt. s, AD // BC) Classwork 90 Classwork (p. 5.9) (a) ∵ ABCD is a parallelogram. ∴ AB DC and BC AD ∴ x 13 and y 22 (b) ∵ BAD ABD ∴ AD BD 24 cm (sides opp. equal s) In △ABD, AB 2 AD 2 BD 2 CG CF (intercept theorem) GD FB y cm 7.2 cm 7.2 cm 12 cm 7.2 y 7.2 12 4.32 (Pyth. theorem) AB 24 2 24 2 cm 1152 cm (or 24 2 cm) Activity Activity 5.1 (p. 5.8) 1. (a) Yes (Reason: alt. s, BC // AD) (opp. sides of // gram) (b) ∵ ∴ ∴ ABCD is a parallelogram. DO BO and CO AO (diags. of // gram) a 9 and b 7 (c) ∵ ∴ ∴ ABCD is a parallelogram. B D and C A (opp. s of // gram) c 68 and d 112 Classwork (p. 5.18) 1. Yes (Reason: opp. sides equal) (b) Yes (Reason: alt. s, BA // CD) 2. Yes (Reason: opp. s equal) (c) Yes (Reason: ASA) 2. 3. No (d) (i) They are equal. (ii) They are equal. 4. Yes (Reason: opp. sides equal and //) (a) Yes (Reason: AAS (or ASA)) 5. Yes (Reason: diags. bisect each other) 6. No (b) OA = OC, OB = OD (Reason: corr. sides, △s) Classwork (p. 5.25) (a) ∵ ABCD is a rhombus. ∴ BC CD AD (property of rhombus) ∴ x y7 Warm-Up Activity (p. 5.24) 1. A: Rhombus B: Square C: Rectangle D: Trapezium 2. rhombus, square, rectangle (b) ∵ ∴ Activity 5.2 (p. 5.47) 1. The length of MN is half of that of BC. 2. ABCD is a rhombus. x 90 (property of rhombus) y 40 (property of rhombus) Yes. It is because MEFN is a rectangle. (or any other reasonable answers) (c) ∵ ∴ To Learn More ∴ To Learn More (p. 5.60) In △ABC, ∵ EF // AC FC EA ∴ (intercept theorem) BF BE x cm 6 cm 12 cm 10 cm 6 x 12 10 7 .2 ABCD is a rhombus. OA OC and OB OD 6 x 3 and y 4 2 (property of rhombus) Classwork (p. 5.28) (a) ∵ ABCD is a rectangle. ∴ ABC 90 (property of rectangle) x 20 90 x 70 BC AD y6 In △BDC, ∵ FG // BD 77 (property of rectangle) Mathematics in Action (3rd Edition) 3A Full Solutions (b) ∵ ∴ ∴ ∴ (b) In △ABC, ∵ AN NB and AM MC 1 ∴ MN BC and NM // BC 2 1 4 y 2 y 8 ABCD is a rectangle. OB OC OA 5 cm (property of rectangle) x5 y 25 10 (c) ∵ ABCD is a rectangle. ∴ OA OD (property of rectangle) ∴ (base s, isos. △) x y In △AOD, (ext. of △) x y 46 2 x 46 x 23 ∴ MNB CBN 180 (b) ∵ ∴ x 68 Quick Practice y 23 Quick Practice 5.1 (p. 5.10) (a) ∵ PQRS is a parallelogram. ∴ PQR PSR (opp. s of // gram) 20 3 x 4 x 1 20 1 4 x 3 x x 19 ADC 90 (property of square) y 90 (b) TQR 3x ABCD is a square. AOD 90 (property of square) ∵ ∴ x 90 ∵ ∴ ABCD is a square. ACD 45 (property of square) ABCD is a square. OA OD OB 7 cm (property of square) x y7 2x 1 3 2x 4 Classwork (p. 5.34) (a) ∵ ABCD is an isosceles trapezium. ∴ x 65 x y 180 65 y 180 y 115 (b) ∵ ∴ x2 ∴ (int. s, AD // BC) OD OB (diags. of // gram) y34 y 1 Quick Practice 5.3 (p. 5.12) ∵ ABCD is a parallelogram. ∴ ABC ADC (opp. s of // gram) 70 ∵ BEFG is a parallelogram. ∴ EBG EFG (opp. s of // gram) 55 ABG ABC EBG ABCD is an isosceles trapezium. BD AC x 42 6 Classwork (p. 5.48) (a) In △ABC, ∵ AM MB and AN NC 1 ∴ MN BC and MN // BC 2 1 x 10 2 5 3 19 57 TQR STQ 180 (int. s, QR // PS) STQ 180 57 123 Quick Practice 5.2 (p. 5.11) ∵ OCED is a parallelogram. ∴ OC DE ( 2 x 1) cm (opp. sides of // gram) and OD CE ( y 3) cm (opp. sides of // gram) ∵ ABCD is a parallelogram. ∴ (diags. of // gram) OC OA y 45 (c) ∵ ∴ ∴ (int. s, NM // BC) x 112 180 Classwork (p. 5.31) (a) ∵ ABCD is a square. ∴ BC AB (property of square) x6 ∴ (mid-pt. theorem) 70 55 15 In △ABG, BAG ABG AGB 180 ( sum of △) BAG 15 140 180 BAG 25 ∵ ABG BAG ∴ △ABG is not an isosceles triangle. (mid-pt. theorem) y 55 (corr. s, MN // BC) 78 5 Quadrilaterals Quick Practice 5.4 (p. 5.19) (a) In EFGH, (4 x 5) 115 (3x 20) 115 360 7 x 255 360 7 x 105 x 15 (b) FEH 4 15 5 65 FGH 3 15 20 65 ∵ EFG EHG and FEH FGH ∴ EFGH is a parallelogram. Quick Practice 5.5 (p. 5.20) ∵ ABCD is a parallelogram. ∴ OA OC and OB OD ∵ OM 2OA 2OC ON BCD BDC DBC 180 ( sum of △) b 36 36 180 b 72 180 b 108 ( sum of polygon) Quick Practice 5.8 (p. 5.28) ∵ PQRS is a rhombus. ∴ PQ QR RS SP and PR QS (property of rhombus) i.e. 4 PQ 68 cm PQ 17 cm In △OPQ, 1 OP PR (property of rhombus) 2 30 cm 2 15 cm OP 2 OQ 2 PQ 2 (Pyth. theorem) opp. s equal diags. of // gram diags. of // gram given OQ 17 2 152 cm 8 cm 1 ∴ Area of △OPQ OQ OP 2 1 8 15 cm 2 2 60 cm 2 ∴ Area of PQRS = 4 area of △OPQ 4 60 cm 2 given ∵ OM ON and OB OD ∴ BNDM is a parallelogram. diags. bisect each other Quick Practice 5.6 (p. 5.21) (a) ∵ ABCD is a parallelogram. ∴ AD BC and ADC ABC BG BC CG AD CG (7 2) cm 9 cm ∴ BG EF ∵ ABC ADC and ADC AEF ∴ ABC AEF ∴ BG // EF ∴ BEFG is a parallelogram. opp. sides of // gram opp. s of // gram 240 cm 2 Quick Practice 5.9 (p. 5.29) ∵ ABCD is a rectangle. ∴ KB KA (property of rectangle) ∴ KBA x (base ∠s, isos. △) (vert. opp. s) AKB 76 In △ABK, BAK ABK AKB 180 (∠ sum of △) x x 76 180 proved given corr. s equal opp. sides equal and // (b) ∵ BEFG is a parallelogram. ∴ EBG EFG (opp. s of // gram) EBG BCD (alt. s, AB // DC) 118 ∴ EFG 118 2 x 104 x 52 ∵ KA KD ∴ KAD KDA y BAD 90 Quick Practice 5.7 (p. 5.27) (a) ∵ ABCD is a rhombus. ∴ AOD 90 (property of rhombus) 5a 90 (property of rectangle) (base s, isos. △) (property of rectangle) x y 90 52 y 90 y 38 a 18 Quick Practice 5.10 (p. 5.30) ∵ ABFE is a rhombus. ∴ BF AE 13 cm ∵ BCDE is a rectangle. ∴ DF BF 13 cm i.e. BD 2 13 cm 26 cm In △BDE, BED 90 (b) ADB 2a 2 18 36 In △BCD, BDC DBC ADB 36 (property of rhombus) 79 (property of rhombus) (property of rectangle) (property of rectangle) Mathematics in Action (3rd Edition) 3A Full Solutions BE 2 DE 2 BD 2 Construct a line TR such that PQ // TR. ∵ PQ // TR and PT // QR ∴ PQRT is a parallelogram. ∴ TR PQ (opp. sides of // gram) (int. s, PS // QR) PQR QPS 180 (Pyth. theorem) DE 26 2 10 2 cm 24 cm ∴ Perimeter of BCDE 2 ( BE DE ) 2 (10 24) cm 120 QPS 180 QPS 60 ∵ PQRS is an isosceles trapezium. ∴ PQ SR 9 cm and RSP QPS ∴ TR SR 9 cm and RSP 60 ∵ △RST is an isosceles triangle. i.e. RTS RST 60 (base s, isos. △) RST RTS TRS 180 ( sum of △) 68 cm Quick Practice 5.11 (p. 5.32) (a) ∵ ABCD is a square. ∴ OA OB OC OD 2 5 cm (property of square) ∴ BD 2 2 5 cm 4 5 cm (b) In △OAB, AOB 90 (property of square) AB OA OB 2 60 60 TRS 180 TRS 60 ∴ △RST is an equilateral triangle. i.e. TS SR 9 cm ∴ QR PT PS TS (15 9) cm 6 cm 2 2 (Pyth. theorem) [(2 5 ) (2 5 ) 2 ] cm 2 2 ∴ 40 cm 2 Area of square ABCD AB 2 40 cm 2 Quick Practice 5.14 (p. 5.41) In △ABE and △CDF, AB CD opp. sides of // gram given AEB CFD 90 alt. ∠s, AB // DC ABE CDF ∴ △ABE △CDF AAS ∴ BE DF corr. sides, △s Alternative Solution ∵ ABCD consists of four congruent right-angled triangles △OAB, △ODA, △OBC and △OCD. 1 ∴ Area of square ABCD 4 2 5 2 5 cm 2 2 40 cm 2 Quick Practice 5.12 (p. 5.33) ∵ ABCD is a square. ∴ AB AD (property of square) ∵ △ADE is an equilateral triangle. ∴ AD AE ∴ AE AB ∴ AEB ABE (base s, isos. △) x BAD 90 (property of square) DAE 60 (prop. of equil. △) In △ABE, ABE BAE AEB 180 ( sum of △) x (90 60) x 180 2 x 150 180 2 x 30 x 15 ABD 45 x y 45 15 y 45 y 30 Quick Practice 5.15 (p. 5.41) In △ABE and △DCE, AB DC ABE DCE 90 ∵ E is the mid-point of BC. ∴ BE CE ∴ △ABE △DCE ∴ AE DE ∵ FD AE and FA DE ∴ AE DE FD FA ∴ AEDF is a rhombus. Quick Practice 5.16 (p. 5.42) ∵ ABCD is a rhombus. ∴ BA// CD and AD // BC In △AEG and △DHG, AEG DHG EG HG AGE DGH ∴ △AEG △DHG AE DH ∴ AG DG In △CHF and △DHG, CFH DGH HG HF DHG CHF ∴ △CHF △DHG CF DG (property of square) Quick Practice 5.13 (p. 5.35) T 80 property of rectangle property of rectangle SAS corr. sides, △s opp. sides of // gram opp. sides of // gram alt. ∠s, BA // CD given vert. opp. ∠s ASA corr. sides, △s corr. sides, △s alt. ∠s, AD // BC given vert. opp. ∠s ASA corr. sides, △s 5 Quadrilaterals ∴ ∵ ∴ ∵ ∴ ∴ CH DH AD DC AE DH 1 DC 2 1 AD 2 DG CF AB BC BE AB AE BC CF BF AEG CFH corr. sides, △s property of rhombus From Quick Practice 5.19 (p. 5.51) In △ABC, ∵ M is the mid-point of BC and AF FC . 1 ∴ BA// MF and FM AB 2 In △DEF, ∵ N is the mid-point of EF and FC CD . 1 ∴ DE // CN and CN DE 2 ∵ BA// DE , BA// MF and DE // CN ∴ MF // CN 1 ∵ FM AB and 2 1 CN DE 2 ∴ AB 2FM and DE 2CN ∵ AB DE ∴ 2 FM 2CN FM CN ∴ FMCN is a parallelogram. From property of rhombus base s, isos. △ Quick Practice 5.17 (p. 5.49) (a) In △XYZ, ∵ P and Q are the mid-points of XY and XZ respectively. 1 PQ YZ ∴ (mid-pt. theorem) 2 1 4 cm YZ 2 YZ 8 cm mid-pt. theorem mid-pt. theorem opp. sides equal and // Quick Practice 5.20 (p. 5.56) ∵ AB // CD // EF // GH and BD DF FH ∴ AC CE EG (intercept theorem) 1 ∴ AC AG 3 1 12 cm 3 4 cm (b) In △XYZ, ∵ P and Q are the mid-points of XY and XZ respectively. ∴ PQ // YZ (mid-pt. theorem) ∴ XPQ XYZ (corr. s, PQ // YZ) 55 In △XPQ, XPQ XQP PXQ 180 ( sum of △) Quick Practice 5.21 (p. 5.58) (a) ∵ AD // EF // BC and AE EB ∴ DF FC (intercept theorem) In △ACD, ∵ AD // GF and DF FC ∴ AG GC (intercept theorem) ∵ DF FC and AG GC 1 ∴ GF AD (mid-pt. theorem) 2 1 10 cm 2 5 cm 55 XQP 38 180 XQP 87 Quick Practice 5.18 (p. 5.50) (a) In △RST, ∵ U and V are the mid-points of ST and RT respectively. 1 UV SR ∴ (mid-pt. theorem) 2 1 5 cm SR 2 SR 10 cm (b) In △RST, ∵ U and V are the mid-points of ST and RT respectively. ∴ SR // UV (mid-pt. theorem) ∴ TSR TUV 68 (corr. s, SR // UV) In △PQR, ∵ S and U are the mid-points of PR and QT respectively. ∴ PQ // SU (mid-pt. theorem) ∴ QPR USR 68 (corr. s, PQ // SU) PQR PRQ QPR 180 ( sum of △) PQR 25 68 180 PQR 87 (b) EG EF GF (14 5) cm 9 cm In △ABC, ∵ AE EB and AG GC 1 ∴ (mid-pt. theorem) EG BC 2 1 9 cm BC 2 BC 18 cm 81 Mathematics in Action (3rd Edition) 3A Full Solutions Quick Practice 5.22 (p. 5.59) (a) ∵ AMCN is a parallelogram. ∴ AN // MC Consider △ABK. ∵ BM MA and MH // AK ∴ BH HK Consider △CDH. ∵ DN NC and KN // HC ∴ KD HK ∴ BH HK KD (b) In △CDH, ∵ DN NC and DK KH 1 ∴ KN HC 2 HC 2 KN ∵ ABCD is a parallelogram. ∴ AB CD In △ABK and △CDH, AB CD ABK CDH BK BH HK KD HK DH ∴ △ABK △CDH ∴ AK CH ∴ AK 2KN 4. ∵ ∴ x6 OC OA intercept theorem y 1 intercept theorem 5. (a) ∵ ∴ ABCD is a parallelogram. (opp.∠s of // gram) ADC ABC 134 ADE 180 ADC (adj.∠s on st. line) 180 134 46 mid-pt. theorem (b) ∵ ∴ opp. sides of // gram proved alt. s, AB // DC Consolidation Corner (p. 5.21) 1. ∵ AB DC 6 cm and AD BC 5 cm ∴ ABCD is a parallelogram. SAS corr. sides, △s 2. Consolidation Corner (p. 5.12) 1. m 55 180 (int. s, AD // BC) ∵ 4. (a) ∵ ∴ n6 3. ∵ ∴ Perimeter of PQRS 28 cm PQ QR RS SP 28 cm 8 x 4 28 (opp. sides of // gram) 8 x 32 b 28 b 2b 28 b 14 C A c 108 given opp. sides equal and // (2 x 2) 2 x ( x 2) (3 x 4) 28 a 24 AB DC given opp. sides equal BAG BEG 30 and (given) AGE ABE 150 ∴ AGEB is a parallelogram. (opp. s equal) ∴ GE AB (opp. sides of // gram) 6 cm ∵ ABCD is a parallelogram. (opp. sides of // gram) DC AB ABCD is a parallelogram. AD BC (opp. sides of // gram) BC EF 5 cm and BC // EF BEFC is a parallelogram. 3. m 125 ∵ ∴ ADE AED 46° (sides opp. equal∠s) AD AE 5 cm (opp. sides of // gram) BC AD 5 cm ∴ 2. (diags. of // gram) y 1 2 Consolidation Corner ∵ ∴ ABCD is a parallelogram. OB OD (diags. of // gram) x24 x4 (b) (opp.∠s of // gram) ABCD is a parallelogram. (opp.∠s of // gram) B D 2h 30 h h 30 C D 180 (int.∠s, BC // AD) k h 180 k 180 30 5. 150 82 PQ [2(4) 2] cm 6 cm QR 2(4) cm 8 cm SR (4 2) cm 6 cm PS [3(4) 4] cm 8 cm ∵ PQ SR and QR PS ∴ PQRS is a parallelogram. opp. sides equal (a) ∵ ABCD is a parallelogram. ∴ BAD BCD 80 and (opp. s of // gram) ABC ADC 5 Quadrilaterals In △CDE, BED CDE DCE 30 80 110 In △ABF, BFD ABF BAF 30 80 110 5. (ext. of △) FDE BFD BED and FBE FDE ∴ BEDF is a parallelogram. (9 6) cm 15 cm i.e. AD 15 cm 6. proved opp. s equal Consolidation Corner (p. 5.36) 2. ∵ ∴ ∴ ABCD is a rhombus. ABD ADB CBD 35 (property of rhombus) In △ABD, BAD ABD ADB 180 ( sum of △) BAD 110 BDE ADE ADB 15 35 50 ∵ EB ED ∴ DBE BDE 50 (base s, isos. △) In △BDE, BED DBE BDE 180 ( sum of △) BED 50 50 180 BED 100 180 BED 80 (property of rhombus) In △PQR, y 180 x z 180 54 54 72 ∵ ∴ BAD 35 35 180 BAD 70 180 PQRS is a rhombus. (property of rhombus) PQ PS w 1 4 w3 x z 54 (Pyth. theorem) PC 10 2 8 2 cm 6 cm BC BP PC (ext. of △) ∵ ∵ ∴ ABCD is a rectangle. DC AB 8 cm , AD BC and BCD 90 (property of rectangle) In △CDP, PC 2 CD 2 DP 2 (b) FBE ABE ABF FDC CDE 1. ∵ ∴ ( sum of △) PQRS is a rectangle. KQ KR (property of rectangle) (base s, isos.△) a KQR 28 In △KQR, b a KQR 28 28 56 (ext. of △) In △PQR, PQR 90 c 180 PQR a 180 90 28 62 3. 4. ∵ ∴ ∵ ∴ Consolidation Corner (p. 5.43) 1. ∵ AB BC ∴ CAB ACB ∵ ADEF is a parallelogram. ∴ DAF DEF ∴ ACB DEF (property of rectangle) ( sum of △) 2. (property of square) c 90 (property of square) PQ PS d 8 (property of square) opp. s of // gram property of rectangle given DF OD FO PQRS is a square. a 45 (property of square) b 90 In △EBO and △FCO, ∵ OA OD and EA DF ∴ EO EA OA given base s, isos. △ OB OC EOB FOC ∴ △EBO △FCO 3. ABCD is an isosceles trapezium. a 65 b3 83 In △BEF and △CED, BFE CDE EBF ECD BE CE ∴ △BEF △CED ∴ FE DE ∵ BE CE and FE DE ∴ BFCD is a parallelogram. ∴ BD FC property of rectangle vert. opp. s SAS alt. s, FA // CD alt. s, FA // CD given AAS corr. sides, △s diags. bisect each other opp. sides of // gram Mathematics in Action (3rd Edition) 3A Full Solutions In △PST, ∵ PV VS and VU // ST ∴ PU UT (intercept theorem) ∵ PV VS and PU UT 1 ∴ VU ST (mid-pt. theorem) 2 1 q 10 2 5 Consolidation Corner (p. 5.52) 1. (a) In △ACD, ∵ AB BC and AE ED 1 ∴ BE CD (mid-pt. theorem) 2 1 x 8.4 2 4.2 ∵ ∴ (mid-pt. theorem) (corr. s, BE // CD) BE // CD y 35 (b) In △ACD, ∵ ABE ACD 55 ∴ BE // CD (corr. s equal) ∵ AB BC and BE // CD (intercept theorem) ∴ AE ED pq ∵ AE ED AD p q 12 ∴ p p 12 p6 (b) In △ABD, ∵ AF FD and BC CD 1 ∴ FC AB (mid-pt. theorem) 2 1 4 x 2 x 8 In △BDE, ∵ BF FE and BC CD 1 ∴ FC ED (mid-pt. theorem) 2 1 4 y 2 y 8 2. 3. In △ABC, ∵ D, E and F are the mid-points of AB, BC and CA respectively. 1 1 ∴ FE AB , DF BC 2 2 1 and DE AC mid-pt. theorem 2 ∴ AD EF and AF ED common side DF FD ∴ △ADF △EFD SSS Consolidation Corner (p. 5.60) 1. ∵ L1 // L2 // L3 and BD DF ∴ CE AC (intercept theorem) 3 cm 2. ∵ AC CE and L1 // L2 ∴ GE BG (intercept theorem) 7 cm 2 3.5 cm q6 ∴ In △ACF, ∵ AB BC and BG // CF ∴ AG GF ∵ AB BC and AG GF 1 ∴ BG CF 2 In △BDG, ∵ BC CD and BG // CE ∴ GE ED ∵ BC CD and GE ED 1 ∴ CE BG 2 1 1 CF 2 2 1 CF 4 ∴ CF 4CE EF CF CE intercept theorem mid-pt. theorem intercept theorem mid-pt. theorem 4CE CE 3CE Exercise Exercise 5A (p. 5.12) Level 1 1. ∵ ABCD is a parallelogram. ∴ AB DC (opp. sides of // gram) x4 (a) In △PQS, ∵ QR RS and RV // QP ∴ PV VS (intercept theorem) ∵ QR RS and PV VS 1 ∴ RV QP (mid-pt. theorem) 2 1 p 8 2 4 AD BC (opp. sides of // gram) y3 2. 84 ∵ ∴ ABCD is a parallelogram. (diags. of // gram) OC OA x2 ∴ OD OB y cm (diags. of // gram) 5 Quadrilaterals BD 3 cm ∴ AD BC 5 z 3z 4 2z 4 z2 ∵ ∴ ABCD is a parallelogram. (diags. of // gram) OB OD y y 3 y 1 .5 3. ∵ ∴ ABCD is a parallelogram. (opp. s of // gram) B D 9. x 30 ∵ ∴ 4. 5. ABCD is a parallelogram. (opp. sides of // gram) AB DC x4 ∴ (opp. sides of // gram) AD BC y 2x 1 2(4) 1 9 ∵ ∴ ∴ 2( x 1) 3( x 1) C D 180 (int. s, BC // AD) (3 y 30) 30 180 3 y 120 y 40 ∵ ∴ 2 x 2 3x 3 x5 ∴ (opp. s of // gram) BCD BAD y 2 y 108 3 y 108 y 36 10. (a) ∵ ABCD is a parallelogram. BAD BCD ∴ (opp. s of // gram) 58 4 x 7 x 10 48 3x x 16 ABCD is a parallelogram. (opp. sides of // gram) OB OD 4 x 2 x 14 3 x 12 x4 OA OC (opp. sides of // gram) (b) DAE 4x 4 16 ∵ ∴ (opp. sides of // gram) 2 y 3y 2 64 DAE AEC 180 (int. s, AD // EC) AEC 180 64 116 y2 6. ∵ ∴ ∵ ∴ 7. ∵ ∴ 11. ∵ ∴ ABCD is a parallelogram. x ADC (opp. s of // gram) 40 ∵ ∴ ABC BAD 180 (int. s, AD // BC) x [(2 x 30) y ] 180 y 180 30 3x 150 3(40) 30 12. (a) ∵ AB AC ∴ ABC ACB (base s, isos. △) In △ABC, BAC ABC ACB 180 ( sum of △) ABCD is a parallelogram. AB DC (opp. sides of // gram) 6 x 2x 3 70 2ABC 180 3x 3 x 1 ∴ 2ABC 110 ABC 55 OB OD (opp. s of // gram) 13 y 10 (b) ∵ ABCD is a parallelogram. ∴ ADC ABC (opp. s of // gram) y3 8. ∵ ∴ WXYZ is a parallelogram. ZY WX 8 cm (opp. sides of // gram) and WZ XY (opp. sides of // gram) Perimeter of parallelogram WXYZ 54 cm WX XY YZ ZW 54 cm 8 cm XY 8 cm XY 54 cm 2 XY 38 cm XY 19 cm 55 ABCD is a parallelogram. (opp. s of // gram) A C 13. ∵ ∴ 60 x 5 x 6 66 6 x x 11 ∵ ∴ 85 ABCD is a parallelogram. EBC AEB (alt. s, BC // AD) 41 EB bisects ABC. ABE EBC 41 Mathematics in Action (3rd Edition) 3A Full Solutions ABC ABE EBC 41 41 82 ADC ABC 82 18. ∵ ∴ 5x 6 2 x 3 3x 9 (opp. s of // gram) x3 ∴ Level 2 14. ∵ ABCD and BEFC are parallelograms. ∴ BCD BAD (opp. s of // gram) 124 and BCF BEF (opp. s of // gram) ∵ ∴ 32 CDE 72 CDE 40 ∵ CED CDE ∴ △CDE is not an isosceles triangle. 7 cm 16. ∵ ∴ (opp. sides of // gram) BCEF is a parallelogram. CBF CEF 56 ∵ BC BG ∴ BCG BGC In △BCG, CBG BCG BGC 180 56 2BCG 180 2BCG 124 BCG 62 ∵ ABCD is a parallelogram. ∴ BAD BCD 62 (diags. of // gram) 19. CED ADE (alt. s, AD // BC) 32 ∵ ABCD is a parallelogram. ∴ ADC ABC (opp. s of // gram) 72 ADE CDE ADC BEFG is a parallelogram. (diags. of // gram) AB AF 7 cm and AE AG 6 cm (diags. of // gram) ABCD is a parallelogram. (opp. sides of // gram) CD BA AD BC 9 cm DE AD AE (9 6) cm 3 cm ∴ OA OC x 3 y 3 x 18 3 y 2 x 18 2 y x6 3 2 (3) 6 3 8 146 BCD BCF DCF 360 (s at a pt.) 124 146 DCF 360 270 DCF 360 DCF 90 ∴ DC is perpendicular to CF. 15. ∵ ∴ ABCD is a parallelogram. AB DC (opp. sides of // gram) 20. ∵ ∴ PXYS is a parallelogram. (opp. s of // gram) SPX SYX 40 In △PZS, PSZ SPZ PZR (ext. of △) PSZ 40 142 (opp. s of // gram) ∵ ∴ (base s, isos. △) PSZ 102 PQRS is a parallelogram. (opp. s of // gram) PQR PSZ 102 ( sum of △) 21. ∵ ∴ ABCD is a parallelogram. BCD BAD 106 DCE 180 BCD 180 106 74 (opp. s of // gram) ∵ ∴ 17. ABF FED (alt. s, AB // EC) 48 ∵ BE bisects ABC. ∴ ABC 2ABF 2 48 CEFG is a parallelogram. EFG DCE 74 DGF DCE 74 (opp. s of // gram) (adj. s on st. line) (opp. s of // gram) (corr. s, GF // CE) FD FG (base. s, isos. △) GDF DGF 74 In △DGF, DFG GDF DGF 180 ( sum of △) ∵ ∴ 96 ∵ ABCD is a parallelogram. ∴ ADC ABC (opp. s of // gram) 96 ADE ADC 180 (adj. s on st. line) ADE 96 180 ADE 84 DFG 74 74 180 DFG 32 86 5 Quadrilaterals DFE DFG EFG ∴ Perimeter of ACED 2( AC AD) 32 74 2(12 10) cm 106 44 cm 22. (a) ∵ ABCD is a parallelogram. ∴ BCD BAD (opp. s of // gram) 60 (alt. s, BC // AD) DEC ADE 24. In △ABD, AB 2 BD 2 AD 2 BD AD AB 2 60 In △CDE, CDE DEC ECD 180 ( sum of △) ∵ ∴ ∵ ∴ CDE 60 60 180 CDE 60 ∵ CDE DEC ECD 60 ∴ CD DE CE ∴ △CDE is an equilateral triangle. AE 2 AB 2 BE 2 AE AB BE 2 (Pyth. theorem) (Pyth. theorem) 2 193 cm AC AE CE ( 193 193 ) cm 2 193 cm Exercise 5B (p. 5.21) Level 1 1. ∵ OA OC 3 cm and OB OD 4 cm ∴ ABCD is a parallelogram. 23. (a) ∵ ABCD is a parallelogram. ∴ GD BG (diags. of // gram) 8 cm and GC AG (diags. of // gram) 6 cm In △AGD, (Pyth. theorem) AD 2 AG 2 GD 2 AB 2 BG 2 AG 2 (diags. of // gram) 7 2 12 2 cm (5 8) cm 13 cm Perimeter of the parallelogram 2( AB BC ) 2(8 13) cm 42 cm (opp. sides of // gram) 252 7 2 cm 24 cm ABCD is a parallelogram. BE DE and AE CE BD 24 cm BE DE 24 cm 2 BE 24 cm BE 12 cm In △AEB, (b) ∵ ABCD is a parallelogram. ∴ AD = BC and AB = DC (opp. sides of // gram) ∵ △CDE is an equilateral triangle. ∴ EC DC DE 8 cm BC BE EC AD 6 2 8 2 cm 10 cm BC AD 10 cm In △AGB, (Pyth. theorem) 2 2. 3. 4. AB 8 2 6 2 cm 10 cm DC AB (opp. sides of // gram) 10 cm ∴ Perimeter of ABCD 2( AB BC ) 2(10 10) cm 40 cm (b) ∵ ACED is a parallelogram. ∴ AC DE (opp. sides of // gram) and AD CE (opp. sides of // gram) AC AG GC (6 6) cm 12 cm 87 given diags. bisect each other AB DC 2.3 cm and BA// CD ∴ ABCD is a parallelogram. given opp. sides equal and // AB DC and AD BC 3 cm ∴ ABCD is a parallelogram. given opp. sides equal ∵ ∵ A B C D 360 A 60 120 60 360 A 240 360 A 120 ∵ A C 120 and B D 60 ∴ ABCD is a parallelogram. ADB CBD AD// BC AD BC and AD// BC ABCD is a parallelogram. 5. ∵ ∴ ∵ ∴ 6. (a) ∵ ∴ AD BC 5a 2 3a 4 2a 6 a3 sum of polygon opp. s equal given alt. s equal given opp. sides equal and // Mathematics in Action (3rd Edition) 3A Full Solutions (b) 7. AB (3 1) cm 4 cm DC (7 3) cm 4 cm ∵ AD BC and AB DC ∴ ABCD is a parallelogram. Method 2: ∵ △AOD △COB ∴ OA OC and OD OB ∴ ABCD is a parallelogram. opp. sides equal Level 2 12. (a) In ABCD, A B C D 360 (a) In ABCD, 143 (3x 11) 143 (2 x 5) 360 ( sum of 280 5 x 360 polygon) (5a 5) (2a 10) 120 5 x 80 x 16 (b) ABC 3(16) 11 37 ADC 2(16) 5 37 ∵ BAD BCD and ABC ADC ∴ ABCD is a parallelogram. 8. (a) In △AGD, GAD ADG BGD CF EF ECF CEF 50 (3a 15) 360 10a 250 a 25 (b) A 5( 25) 5 120 B 2( 25) 10 60 D 3( 25) 15 60 ∴ A C and B D ∴ ABCD is a parallelogram. opp. s equal (ext. of △) (base s, isos. △) 13. (a) ∵ ABCD is a parallelogram. ∴ BC AD opp. sides of // gram ∵ PADQ is a parallelogram. ∴ AD PQ opp. sides of // gram ∴ BC PQ BAD ABC 50 120 170 180 ∴ AD is not parallel to BC. ∴ ABCD is not a parallelogram. P S 180 S 180 P Q R 180 (b) ∵ ABCD is a parallelogram. ∴ BC // AD ∵ PADQ is a parallelogram. ∴ AD // PQ ∴ BC // PQ Also, BC PQ proved in (a) ∴ PBCQ is a parallelogram. opp. sides equal and // int. s, QP // RS int. s, QP // RS Q 180 R ∵ ∴ ∴ 10. ∵ ∴ ∵ ∴ ∵ ∴ ∴ P R given S Q PQRS is a parallelogram. opp. s equal 14. ∵ ABCD is a parallelogram. (opp. sides of // gram) BC AD (given) FC AE BF BC FC AD AE ED BC // AD BF // ED BFDE is a parallelogram. 11. Method 1: ∵ △AOD △COB ∴ AD CB and OAD OCB ∴ AD // BC ∴ ABCD is a parallelogram. opp. s equal (c) ∵ ABCD is a parallelogram. ∴ (opp. sides of // gram) AD BC 18 3b b 2 4b 20 b5 (b) ∵ 9. ( sum of polygon) 10a 110 360 GAD 15 65 GAD 50 i.e. BAD 50 ∵ ∴ given corr. sides, △s diags. bisect each other ∴ ∵ ∴ ∴ 15. ∵ ∴ (opp. sides equal and //) ∵ given corr. sides, △s corr. s, △s alt. s equal opp. sides equal and // ∴ ∵ ∴ 88 AB FC EF FC AB EF AE FD BF FD AE BF ABFE is a parallelogram. PQRS is a parallelogram. PO OR and QO OS X is the mid-point of PO. 1 XO PO 2 Y is the mid-point of OR. 1 OY OR 2 1 PO 2 XO given diags. of // gram given diags. of // gram opp. sides equal diags. of // gram diags. of // gram 5 Quadrilaterals ∵ ∴ XO OY and QO OS XQYS is a parallelogram. PQR MNO 66 diags. bisect each other 16. (a) ∵ △ABC △CDE ∴ BC DE and ACB CED ∵ BC is the angle bisector of ACE. ∴ ACB BCE ∴ BCE CED ∴ CB // DE ∴ BCDE is a parallelogram. given corr. sides, △s corr. s, △s 19. (a) ∵ ABCD is a parallelogram. ∴ AD BC ∵ △ABE and △CDF are equilateral triangles. ∴ AB AE BE and DC DF FC ∵ AB DC ∴ DF BE ∵ ADC CBA FDC ABE 60 ∴ ADF ADC 60 alt. s equal opp. sides equal and // (b) ∵ △ABC △CDE ∴ ABC CDE (corr. s, △s) 54 ∵ BCDE is a parallelogram. ∴ CBE CDE (opp. s of // gram) 54 ABE ABC CBE 54 54 108 17. (a) In △ABO and △CDO, AOB COD OAB OCD 90 BO DO ∴ △ABO △CDO CBA 60 CBE ∴ △AFD △CEB (b) ∵ △AFD △CEB ∴ AF CE AE FC ∴ AECF is a parallelogram. Exercise 5C (p. 5.37) Level 1 1. ∵ ABCD is a rhombus. ∴ AB BC CD DA ∴ x4 vert. opp. s given given AAS (b) ∵ △ABO △CDO ∴ AO CO BO DO ∴ ABCD is a parallelogram. (proved in (a)) opp. sides of // gram opp. sides of // gram opp. s of // gram prop. of equil.△ SAS proved in (a) corr. sides, △s from (a) opp. sides equal (property of rhombus) 2y 2 4 2y 2 proved in (a) corr. sides, △s given diags. bisect each other y 1 3z 5 4 3z 9 z3 18. (a) 2. ABCD is a rhombus. DAB 2 BAC (property of rhombus) 2 30 60 S With the notation in the figure, ∵ △MNO △PQR ∴ MN PQ and MNO PQR PSO PQR ∴ MNO PSO ∴ MN // PQ ∴ MNQP is a parallelogram. ∵ ∴ p DAB (alt. s, AD // EC) 60 given corr. sides, △s corr. s, △s corr. s, NO // QR (int. s, AB // DC) DAB q 180 60 q 180 q 120 3. corr. s equal opp. sides equal and // ∵ ∴ ABCD is a rhombus. a 55 (property of rhombus) b 90 (property of rhombus) In △DOC, OCD CDO AOD (b) ∵ MNQP is a parallelogram. ∴ MNQ MPQ (opp. s of // gram) 112 MNO MNQ ONQ ∵ ∴ 112 46 66 89 (ext. of △) CDO 90 55 35 ADO CDO (property of rhombus) ADC ADO CDO c 35 35 70 Mathematics in Action (3rd Edition) 3A Full Solutions 4. ∵ ∴ ABCD is a rectangle. (property of rectangle) OD OB 2x 2 x 1 ∵ ∴ AC BD (property of rectangle) y 2 2x 2 2(1) 4 UQR PQR PQT 90 35 55 In △UQR, x UQR PRQ 55 45 100 y TQR 180 (ext. of △) (int. s, PS // QR) y 55 180 5. ∵ ∴ y 125 ABCD is a rectangle. AD BC (property of rectangle) 2x 1 7 2x 6 10. BAD ABC 180 (int. s, AD // BC) 2 x 62 180 x3 2 x 118 x 59 ∴ BAD 90 (property of rectangle) In △ABD, ABD ADB BAD 180 ( sum of △) ∵ ∴ 64 y 90 180 3 y 10 62 3 y 72 y 24 154 y 180 y 26 6. ∵ ∴ ∵ ∴ ABCD is a rectangle. (property of rectangle) ABC 90 x 70 90 x 20 OA OB OAB OBA 11. ∵ ∴ ∵ ∴ ∴ (property of rectangle) (base s, isos.△) ADB DBC y 30 12. ∵ ∴ EFGH is a rhombus of side 10 cm. EH 10 cm and EOH 90 (property of rhombus) ∵ EO OG (property of rhombus) 1 ∴ EO EG 2 1 12 cm 2 6 cm In △EOH, (property of square) 2b 7 13 2b 6 b3 8. ∵ ∴ EO 2 OH 2 EH 2 PQRS is a square. (property of square) POS 90 2 y 30 90 2 y 120 ∵ ∴ y 60 OR OQ (property of square) z 1 11 (Pyth. theorem) OH 10 2 6 2 cm 8 cm FO OH (property of rhombus) FH 2 OH 2 8 cm 16 cm 13. (a) ∵ PQRS is a rhombus. ∴ PQS RQS (property of rhombus) 3 x 4 4 x 9 x 13 z 12 9. (alt. s, AD // BC) (ext. of △) PQRS is a square. a 45 (property of square) SR PS ABCD is an isosceles trapezium. BD AC 2 (2 x 1) 7 2x 3 7 2x 4 x2 y 70 In △OAB, z y 70 70 70 140 7. ABCD is an isosceles trapezium. DCB ABC ∵ PQRS is a square. ∴ QRP 45 (property of square) and PQR 90 (property of square) (b) ∵ PQRS is a rhombus. ∴ PRQ PRS and PR QS (property of rhombus) 90 5 Quadrilaterals RQT 4 13 9 43 In △QRT, RQT QRT QTR 180 ( sum of △) 43 QRT 90 180 In △ABC, AC 2 AB 2 BC 2 URS 2 QRT 2 47 94 90 ∴ △SUR is an obtuse-angled triangle. ∴ Peter’s claim is incorrect. ∵ ABCD is a rectangle. OB OA and ABC 90 (property of rectangle) (base s, isos. △) OBA OAB 34 In △OAB, BOC OAB OBA m 34 34 68 In △ABC, ACE BAC ABC n 34 90 124 (ext. of △) 1 72 cm 2 2 72 cm 4 3 2 18 cm or cm 2 2 17. ∵ PQRS is a square. ∴ TPK 45 (property of square) ∵ TP KP ∴ PTK PKT (base s, isos. △) In △PTK, PTK PKT TPK 180 ( sum of △) 2PKT 45 180 2PKT 135 PKT 67.5 ∵ PKQ 90 (property of square) ∴ TKQ PKQ PKT 90 67.5 22.5 (ext. of △) 15. (a) ∵ ABCD is a rectangle. ∴ OA OB OC OD and BCD 90 (property of rectangle) ∴ OBC OCB (base s, isos. △) In △OBC, BOC OBC OCB 180 ( sum of △) 60 2OBC 180 2OBC 120 OBC 60 ∵ OBC OCB BOC 60 ∴ △OBC is an equilateral triangle. i.e. OB BC 5 cm BD OB OD (5 5) cm 10 cm 18. Construct ST such that ST QR , where T is a point on QR. QT PS 3 cm TR QR QT (8 3) cm 5 cm ST PQ 12 cm In △RST, (b) In △BCD, BC 2 CD 2 BD 2 72 cm 2 OE EC 1 OE OC 2 ∴ ∴ 2 72 cm 1 OC AC 2 1 72 cm 2 133 QRT 180 QRT 47 14. ∵ ∴ (Pyth. theorem) AC 6 6 cm 2 (Pyth. theorem) CD 10 5 2 cm 2 75 cm (or 5 3 cm) 16. ∵ ∴ ABCD is a square. OA OB OC OD and ABC 90 (property of square) AB BC 24 cm 4 6 cm RS 2 ST 2 RT 2 (Pyth. theorem) RS 12 2 52 cm 13 cm Perimeter of trapezium PQRS PQ QR RS PS (12 8 13 3) cm 36 cm 91 Mathematics in Action (3rd Edition) 3A Full Solutions Level 2 19. ∵ PQRS is a rhombus. ∴ PRS PRQ (property of rhombus) SRQ PQR 180 (int. s, PQ // SR) 22. ∵ ∴ ∵ ∴ ∴ SRQ 76 180 SRQ 104 1 PRS SRQ 2 1 104 2 52 ∵ TR is the angle bisector of PRS. 1 ∴ PRT PRS 2 1 52 2 26 SPR PRS 52 (property of rhombus) In △TPR, PTR TPR PRT 180 ( sum of △) ABCD is a square. BCD ADC 90 BCE ADE (property of square) 90 60 30 Similarly, ADE 30 ∵ △CDE is an equilateral triangle. ∴ CE DE CD AB BC CD DA (property of square) ∴ CE BC and DE DA ∴ CEB CBE (base s, isos.△) and DEA DAE (base s, isos.△) In △BCE, CEB CBE BCE 180 ( sum of △) 2CEB 30 180 2CEB 150 CEB 75 PTR 52 26 180 Similarly, DEA 75 AEB 360 CEB DEA CED (s at a pt.) 360 75 75 60 150 PTR 78 180 PTR 102 20. ABC ADC (opp. s of // gram) 132 ABK ABC KBC 23. (a) ∵ DBCE is a parallelogram. ∴ BD CE (opp. sides of // gram) 8 cm ∵ ABCD is a rectangle. ∴ AC BD (property of rectangle) 8 cm 132 54 78 ∵ ABCD is a rhombus. ∴ DAC BAC (property of rhombus) (int. s, AB // DC) ADC (DAC BAC ) 180 132 2BAC 180 2BAC 48 BAC 24 AC CE CAE CED (base s, isos. △) 60 In △ECA, ECA CAE CED 180 ( sum of △) ECA 60 60 180 (b) ∵ ∴ In △ABK, BAK ABK AKB 180 ( sum of △) 24 78 AKB 180 102 AKB 180 AKB 78 ∵ ABK AKB ∴ AB AK (sides opp. equal s) i.e. △ABK is an isosceles triangle. 21. ∵ ∴ ∵ ∴ ∴ △CDE is an equilateral triangle. CED ECD EDC 60 (prop. of equil.△) ECA 60 ∴ △ECA is an equilateral triangle. i.e. AE CE 8 cm ∵ ABCD is a rectangle. ∴ ADC 90 (property of rectangle) ∵ AC CE and CD AE ∴ AD DE (prop. of isos. △) 1 AE 2 1 8 cm 2 4 cm PQRS is a square. PRQ 45 (property of square) UVRP is a rhombus. UVP PVR (property of rhombus) (UVP PVR) PRQ 180 (int.s, UV // PR) 2PVR 45 180 2PVR 135 PVR 67.5 24. Construct a line FG such that DC // FG. 92 5 Quadrilaterals ∵ BA // CD // GF and AF // BG ∴ ABGF is a parallelogram. (int. s, AF // BE) ABC 120 180 ABC 60 FGE ABC 60 (corr. s, AB // FG) ∵ CEFD is an isosceles trapezium. ∴ FEG 60 and FE DC Exercise 5D (p. 5.43) Level 1 1. ∵ ABCD is a parallelogram. ∴ ABC ADC ∵ DEFG is a parallelogram. ∴ EFG EDG ∴ ABC EFG 2. (property of rhombus) AB 15 cm In △EFG, EFG FGE FEG 180 ( sum of △) EFG 60 60 180 EFG 120 180 EFG 60 ∴ △EFG is an equilateral triangle. i.e. GE FE 15 cm ∵ DF // CG and DC // FG ∴ CGFD is a parallelogram. ∴ DF CG (opp. sides of // gram) CE GE (23 15) cm 8 cm ABCD is a parallelogram. BC AD AFCE is a square. AF FC CE AE BC FC opp. s of // gram opp. sides of // gram property of square AD AE ED 3. BC BE EF CF AD BC and AB DC ∵ CDF CFD ∴ CD CF ∴ AB BC CD DA ∴ ABCD is a rhombus. 4. In △EAF and △EDG, ∵ ABCD is a rectangle. ∴ AB DC 1 ∵ AF AB and 2 1 DG DC 2 ∴ AF DG EAF EDG 90 EA ED ∴ △EAF △EDG ∴ EF EG i.e. △EFG is an isosceles triangle. 25. (a) ∵ ABCD is a square. ∴ AB BC , ABC 90 and BAC 45 (property of square) ∵ BCE is an equilateral triangle. ∴ BC CE BE and EBC 60 (prop. equil. △) ∵ BA BE ∴ BAE BEA (base s, isos. △) ABE ABC EBC 90 60 30 In △ABE, ABE BEA BAE 180 30 2BAE 180 2BAE 150 BAE 75 CAE BAE BAC ∵ ∴ ∵ ∴ BF opp. s of // gram 5. ( sum of △) 75 45 30 ∵ ACFE is a parallelogram. ∴ CFE CAE (opp. s of // gram) 30 6. (b) CAE AEF 180 (int. s, AC // EF) 30 AEF 180 AEF 150 BEF AEF BEA 7. 150 75 75 ∵ BEA BEF 75 ∴ BE bisects AEF. 93 In △AED and △CFD, DAE DCF DEC DFA AED 180 DEC CFD 180 DFA ∴ AED CFD AD CD ∴ △AED △CFD ∴ AE FC ∵ AB BC ∴ BAC BCA DFA BCA ∴ DAF DFA ∴ AD DF (a) property of square opp. sides of // gram given sides opp. equal s property of rectangle given given property of rectangle given SAS corr. sides, △s property of rhombus given adj. s on st. line adj. s on st. line property of rhombus AAS corr. sides, △s given base s, isos. △ corr. s, DF // BC sides opp. equal s ADC 90 DCE ADC ACD CDE ACE ACD DCE CDE ADC ADE property of rectangle alt. s, AD // BE alt. s, AC // DE Mathematics in Action (3rd Edition) 3A Full Solutions (b) 8. property of rectangle BC AD ∵ AC // DE and AD // BE ∴ ACED is a parallelogram. ∴ AD CE opp. sides of // gram ∴ BC CE i.e. C is the mid-point of BE. ∵ ABCD is a parallelogram. ∴ BCD BAD ∵ BQCP is a rhombus. ∴ CBP BCP In △BCP, BPD CBP BCP In △CFE, CFE CEF ECF 180 45 CEF 45 180 90 CEF 180 CEF 90 ∴ FE EC Level 2 11. (a) ∵ ∴ opp. s of // gram property of rhombus ∵ ∴ ext. of △ 2BCP 2BAD 9. ∵ ABCD is a square. ∴ AB BC CD AD Also, BP BQ In △APD and △CQD, AP AB BP ∴ ∴ property of square given CQ DAP DCQ 90 AD CD ∴ △APD △CQD ∴ DP DQ ∴ △PDQ is an isosceles triangle. property of square proved SAS corr. sides, △s 10. ∵ ∴ ∵ ∴ DP DQ △PDQ is an isosceles triangle. given base s, isos.△ property of rectangle EDA ADC EDC In △ABE and △DCE, AB DC EAB EDC EA ED ∴ △ABE △DCE ∴ BE CE ∴ △EBC is an isosceles triangle. 13. (a) ∴ ∴ opp. sides of // gram opp. sides of // gram opp. sides of // gram SSS 12. ∵ EA ED ∴ EAD EDA BAD ADC 90 ∴ EAB EAD BAD Alternative Solution Join BD. In △BPD and △BQD, BP BQ ∵ ABCD is a square. ∴ PBD QBD 45 BD BD ∴ △ BPD △ BQD ABCD is a parallelogram. (opp. sides of // gram) AD BC and AD // BC AEFD is a parallelogram. (opp. sides of // gram) AD EF and AD // EF BC EF and BC // EF BCFE is a parallelogram. (opp. sides equal and //) (b) In △ABE and △DCF, AB DC AE DF BE CF ∴ △ABE △DCF BC BQ sum of △ given property of square common side SAS corr. sides, △s ABCD is a rectangle. BCD 90 property of rectangle (b) CE and CF are the angle bisectors of ACB and ACD respectively. ACE BCE and ACF DCF BCD 90 BAC ACD ∵ AX and CY bisect BAE and DCE respectively. ∴ BAX EAX and DCY ECY ∴ EAX ECY In △AXE and △CYE, EAX ECY AE CE AEX CEY ∴ △AXE △CYE alt. s, BA // CD BE DE ∵ △AXE △CYE ∴ XE YE BX BE XE diags. of // gram proved in (a) corr. sides, △s DE YE DY ACE BCE ACF DCF 90 2(ACE ACF ) 90 ACE ACF 45 ECF 45 94 property of rectangle proved given SAS corr. sides, △s proved diags. of // gram vert. opp. s ASA 5 Quadrilaterals 14. (a) ∵ ABCD is a square. ∴ AB CB and ABC 90 ABF ABC 180 ABF 90 180 ABF 90 In △ABF and △CBE, AB CB ABF CBE 90 BF BE ∴ △ABF △CBE (b) ∵ △ABF △CBE ∴ BAF BCE In △AGE, AGE GAE AEC In △BCE, CBE BCE AEC △BAX △DCY (proved in (a)(i)) (corr. sides, △s) BX DY and BX // YD BX DY BYDX is a parallelogram. (opp. sides equal and //) In △BAX and △DAX, (common side) AX AX DAX DAC 180 (adj. s on st. line) (b) ∵ ∴ ∵ ∴ property of square adj. s on st. line proved proved given SAS DAX 45 180 DAX 135 (proved) BAX DAX (property of square) AB AD ∴ △BAX △DAX (SAS) ∴ BX DX (corr. sides, △s) Also, BX DY (opp. sides of // gram) and BY DX (opp. sides of // gram) ∴ BX DY BY DX ∴ BYDX is a parallelogram with four equal sides. ∴ BYDX is a rhombus. (proved in (a)) (corr. s, △s) (ext. of △) (ext. of △) 90 BCE AGE GAE AGE 90 ∴ CG AF 15. (a) ∵ ABCD is a square. ∴ BAD 90 ∵ AEFG is a square. ∴ EAG 90 ∵ EAB EAG GAB 90 GAB GAD GAB BAD GAB 90 ∴ EAB GAD (b) In △AEB and △AGD, AB AD AE AG EAB GAD ∴ △AEB △AGD ∴ BE DG 16. (a) (i) 17. (a) In △BPC and △DQA, BPC DQA PBC QDA BC DA ∴ △BPC △DQA property of square property of square (b) ∵ △BPC △DQA ∴ BP DQ and PC QA BA DC AP BP BA DQ DC CQ ∴ AQCP is a parallelogram. property of square property of square proved in (a) SAS corr. sides, △s (ii) ∵ △BAX △DCY ∴ AXB CYD ∴ BX // YD proved in (a) corr. sides, △s corr. sides, △s opp. sides of // gram opp. sides equal 18. (a) ∵ ABCD is a square. ∴ BAD 90 (property of square) BAO BAE OAE 90 OAE ∵ ABCD is a square. ∴ AB BC CD DA BAC BCA and property of square DAC DCA 45 In △BAX, adj. s on st. line BAX BAC 180 BAX 45 180 BAX 135 In △DCY, DCY DCA 180 DCY 45 180 DCY 135 In △BAX and △DCY, AX CY BAX DCY AB CD ∴ △BAX △DCY given opp. s of // gram opp. sides of // gram AAS In △ABO, ABO BAO AOE (ext. of △) ABO (90 OAE) 90 OAE ABO (b) In △ADF and △BAE, ∵ OAE OBA i.e. DAF ABE DA AB ADF BAE 90 ∴ △ADF △BAE ∴ AF BE adj. s on st. line given proved property of square SAS 19. ∵ AB AC ∴ ABC ACB DGB ACB ∴ DBG DGB ∴ DB DG ∵ DG DB CE and DG // CE proved in (a)(i) corr. s, △s alt. s equal 95 property of square property of square ASA corr. sides, △s given base s, isos. △ corr. s, DG // AE sides opp. equal s given Mathematics in Action (3rd Edition) 3A Full Solutions ∴ ∴ ∴ CDGE is a parallelogram. GF FC F is the mid-point of CG. (b) In △FBG, ∵ BD DF and BE EG 1 ∴ DE FG (mid-pt. theorem) 2 1 6 cm 2 3 cm opp. sides equal and // diags. bisect each other Exercise 5E (p. 5.52) Level 1 1. In △ABC, ∵ AD DB and AE EC 1 ∴ DE BC (mid-pt. theorem) 2 1 3 x 2 x6 2. 6. In △ABC, ∵ AD DB and AE EC ∴ DE // BC (mid-pt. theorem) ∴ ACB AED (corr. s, DE // BC) (opp. s of // gram) ABC 108 In △ACE, ∵ AB BE and F is the mid-point of CE. ∴ BF // AC (mid-pt. theorem) ∴ ACB CBF (alt. s, BF // AC) 32 In △ACB, BAC ACB ABC 180 ( sum of △) BAC 32 108 180 BAC 40 x 77 7. 3. In △ABC, ∵ AD DB and AE EC ∴ DE // BC (mid-pt. theorem) ∴ AED ACB (corr. s, DE // BC) x 50 ∴ 4. ∴ ∴ (mid-pt. theorem) ∴ ∴ In △ABC, ∵ AD DB and AE EC ∴ DE // BC (mid-pt. theorem) ∴ AED ACB (corr. s, DE // BC) 45 In △ADE, ADE AED DAE 180 ( sum of △) a 45 60 180 a 75 ∴ 5. 1 DE BC 2 1 14 y 2 y 28 1 DE DB 2 1 b 11 2 5.5 ∵ ∴ ∵ (mid-pt. theorem) (a) In △ABC, ∵ BF (3 3) cm 6 cm FA and BG ( 4 4) cm 8 cm GC 1 ∴ FG AC (mid-pt. theorem) 2 1 12 cm 2 6 cm 96 △ABC is an equilateral triangle. AB BC AC P, Q and R are the mid-points of AB, BC and CA respectively. AP PB BQ QC AR RC 1 1 PQ AC , QR AB 2 2 1 and PR BC mid-pt. theorem 2 PQ QR PR △PQR is an equilateral triangle. 8. In △ABC, ∵ E and F are the mid-points of AC and BC respectively. ∴ AB // EF mid-pt. theorem ∵ E and F are the mid-points of BD and BC respectively. ∴ EF // DC mid-pt. theorem ∵ AB // EF and EF // DC ∴ AB // DC 9. In △ABC, ∵ M and N are the mid-points of AC and BC respectively. 1 ∴ AB // MN and MN AB (mid-pt. theorem) 2 1 MN 12 cm 2 6 cm ∵ AB // CD and AB // MN ∴ MN // CD ∵ MN CD 6 cm and MN // CD ∴ MNDC is a parallelogram. (opp. sides equal and //) ∴ Cathy’s claim is correct. 5 Quadrilaterals (b) In △AFG, ∵ E and C are the mid-points of AF and AG respectively. ∴ EC // FG (mid-pt. theorem) (alt. s, EC // FG) EFG BEF Level 2 10. In △DEF, ∵ FG GD and FC CE 1 ∴ GC DE (mid-pt. theorem) 2 1 a 8 2 4 62 FAG AFG AGF 180 ( sum of △) 52 62 AGF 180 AGF 66 In △ABC, ∵ CF FA and CG 4 cm GB 1 ∴ FG AB (mid-pt. theorem) 2 1 5 b 2 b 10 14. In △ABD, ∵ E and F are the mid-points of AB and BD respectively. ∴ EF // AD ∴ BFE BDA In △BCD, ∵ F and G are the mid-points of BD and BC respectively. ∴ FG // DC ∴ BFG BDC EFG BFE BFG BDA BDC ADC 11. In △BCD, ∵ DG GB and DF FC 1 ∴ GF BC (mid-pt. theorem) 2 1 8 cm 2 4 cm EG EF GF (6 4) cm 2 cm In △ABD, ∵ BE EA and BG GD 1 ∴ EG AD (mid-pt. theorem) 2 1 2 x 2 x4 mid-pt. theorem corr. s, FG // DC 15. (a) In △ABC, ∵ D and F are the mid-points of AB and CA respectively. 1 ∴ DF // BC and DF BC 2 (mid-pt. theorem) ∵ E and F are the mid-points of BC and CA respectively. 1 ∴ FE // AB and FE AB 2 (mid-pt. theorem) ∴ BEFD is a parallelogram. ∴ DFE DBE (opp. s of // gram) 90 ∴ △DEF is a right-angled triangle. 12. In △ACD, ∵ AF FC and AG GD ∴ FG // CD (mid-pt. theorem) ∴ ADC AGF (corr. s, CD // FG) 35 Reflex BCD 360 x (s at a pt.) ∵ ABCD is a quadrilateral. ∴ BAD ABC reflex BCD ADC 360 ( sum of polygon) 50 60 (360 x) 35 360 (b) x 145 13. (a) In △ABC, ∵ D and E are the mid-points of AB and BC respectively. 1 ∴ DE AC 2 1 4 cm AC 2 AC 8 cm ∵ AC CG 8 cm ∴ C is the mid-point of AG. mid-pt. theorem corr. s, EF // AD 1 BC 2 1 8 cm 2 4 cm 1 FE AB 2 1 6 cm 2 3 cm DF 1 FE DF 2 1 3 4 cm 2 2 6 cm 2 Area of △DEF mid-pt. theorem 97 Mathematics in Action (3rd Edition) 3A Full Solutions 18. In △ADC, ∵ AB BD and N is the mid-point of CD. 1 ∴ BN // AC and BN AC 2 ∵ AB AC 1 ∴ BN AB 2 ∵ M is the mid-point of AB. 1 ∴ BM AB BN 2 ∵ AB AC ∴ ABC ACB ∵ NBC ACB ∴ ABC NBC In △CMB and △CNB, BM BN MBC NBC BC BC ∴ △CMB △CNB CM CN ∵ N is the mid-point of CD. 1 ∴ CM CD 2 i.e. CD 2CM 16. (a) In △BFC, ∵ BD DF and CH HF 1 ∴ DH BC (mid-pt. theorem) 2 In △ADE, ∵ AB BD and AC CE 1 ∴ BC DE (mid-pt. theorem) 2 DE 2 BC HE DE DH 2 BC ∴ ∴ In ∵ ∴ 3 BC 2 1 BC DH 2 HE 3 BC 2 1 3 DH : HE 1 : 3 ∴ (b) ∵ 1 BC 2 DH : HE 1: 3 5 cm 1 HE 3 HE 15 cm △CFG, CH HF and CE EG 1 HE FG (mid-pt. theorem) 2 1 15 cm FG 2 FG 30 cm In △BCD, ∵ F and G are the mid-points of BC and DC respectively. 1 ∴ FG // BD and FG BD 2 ∴ EH // FG and EH FG ∴ EFGH is a parallelogram. given given base s, isos. △ alt. s, BN // AC proved proved common side SAS corr. sides, △s Exercise 5F (p. 5.61) Level 1 17. Join BD. In △ABD, ∵ E and H are the mid-points of AB and AD respectively. 1 ∴ EH // BD and EH BD 2 mid-pt. theorem AB // CD // EF and BD DF CE AC (intercept theorem) x3 1. ∵ ∴ 2. In △ACD, ∵ AB BC and BE // CD ∴ AE ED (intercept theorem) y4 3. ∵ ∴ ∵ ∴ ∴ mid-pt. theorem ∵ ∴ mid-pt. theorem QXY QPR 50 (corr. s equal) XY // PR QX XP and XY // PR QY YR (intercept theorem) 1 QY QR 2 1 a 18 2 9 YR QY ba 9 opp. sides equal and // 98 4. ∵ AB // CD // EF // GH and AC CE EG ∴ BD DF FH 5 cm (intercept theorem) DH DF FH x 55 10 5 Quadrilaterals 5. In △BFG, ∵ BE EG and DE // FG ∴ DF BD (intercept theorem) p2 ∵ ∴ 6. (b) In △ABE, ∵ AG GE and DG // BE ∴ AD BD (intercept theorem) 18 cm BD 2 9 cm DE // FG // CA and EG GA (intercept theorem) FC DF q p 2 ∵ ∴ In △ACF, ∵ AB BC and BG // CF ∴ AG GF (intercept theorem) AG GE and AD DB 1 DG BE (mid-pt. theorem) 2 1 8 cm 2 4 cm y3 10. (a) In △PQR, ∵ QC CP and RN NP ∴ CN // QR In △ACB, ∵ AQ QC and MQ // BC ∴ AM BM EF AE AG GF z 12 y 3 12 3 3 6 In △ADE, ∵ AF (3 3) cm 6 cm FE and CF // DE ∴ CD AC (intercept theorem) x 22 4 7. 8. intercept theorem (b) In △ACB, ∵ AQ QC and AM MB 1 ∴ MQ BC mid-pt. theorem 2 In △ABC, ∵ AM MB and MN // BC ∴ AN NC (intercept theorem) ∵ AC 6 cm 6 ∴ AN cm 2 3 cm In △AND, ∵ DP PN and AD // QP ∴ AQ QN (intercept theorem) ∵ AN 3 cm 3 ∴ QN cm 2 1.5 cm 11. In △ABF, ∵ BD DF and BA// DC ∴ AC CF ∵ BD DF and AC CF 1 ∴ CD AB 2 AB 2CD In △BFE, ∵ BD DF and DC // FE ∴ BC CE ∵ BD DF and BC CE 1 ∴ CD EF 2 EF 2CD ∴ AB EF In △ACE, ∵ AF FE and BF // CE ∴ AB BC (intercept theorem) In △ADE, ∵ AF FE and CF // DE ∴ AC CD (intercept theorem) ∵ AB BC ∴ CD 2 AB ∵ BD 12 cm ∴ BC CD 12 cm intercept theorem mid-pt. theorem intercept theorem mid-pt. theorem from and Level 2 12. In △ACE, ∵ AB BC and BD // CE ∴ DE AD (intercept theorem) a3 In △AGJ, ∵ AH HJ and FH // GJ ∴ AF FG (intercept theorem) b4 AB 2 AB 12 cm In △AEG, ∵ AD DE and AF FG 1 ∴ DF EG (mid-pt. theorem) 2 1 2 c 2 c4 3 AB 12 cm AB 4 cm 9. mid-pt. theorem (a) In △AEC, ∵ F and G are the mid-points of AC and AE respectively. ∴ GF // EC (mid-pt. theorem) i.e. DF // BC 99 Mathematics in Action (3rd Edition) 3A Full Solutions 13. ∵ ∴ BA // CF // DE and BC CD AF FE (intercept theorem) x5 ∴ In △ABE, ∵ EF FA and GF // BA ∴ EG GB (intercept theorem) ∵ EF FA and EG GB 1 ∴ FG AB (mid-pt. theorem) 2 1 y 14 2 7 16. (a) In △AMN and △BMN, AMN ABC 90 BMN 180 AMN 90 ∴ AMN BMN AM BM MN MN ∴ △AMN △BMN In △BDE, ∵ BC CD and BG GE 1 ∴ CG DE (mid-pt. theorem) 2 1 z 8 2 4 ∴ AG GE and AB BC 1 BG CE (mid-pt. theorem) 2 1 2 c 2 c4 ∴ corr. s, MN //BC adj. s on st. line given common side SAS 17. (a) ∵ ABD BDE ∴ AB // DF In △ABC, ∵ BE EC and AB // DF ∴ AD CD In △BDF, ∵ DE EF and EC // FB ∴ DC CB (intercept theorem) ba 3 ∵ (mid-pt. theorem) (b) In △ABC, ∵ AM MB and MN // BC ∴ AN NC (intercept theorem) ∵ △AMN △BMN (proved in (a)) ∴ NA NB (corr. sides, △s) ∴ NB NC i.e. △BNC is an isosceles triangle. 14. In △ACE, ∵ AG GE and GB // EC ∴ BC AB (intercept theorem) a3 ∵ 1 PA 2 1 QD 4 cm 12 cm 2 QD 4 cm 6 cm QD 2 cm QE given alt. s equal intercept theorem (b) In △ABC, ∵ BE EC and AD DC 1 ∴ DE AB mid-pt. theorem 2 AB 2 DE ∵ DE EF given ∴ AB DF ∴ ABFD is a parallelogram. opp. sides equal and // DE EF and DC CB 1 CE BF (mid-pt. theorem) 2 1 4 (2 d ) 2 82d d 6 18. (a) (i) 15. In △ABC, ∵ D and E are the mid-points of AB and AC respectively. 1 ∴ DE // BC and DE BC (mid-pt. theorem) 2 1 DE 8 cm 2 4 cm In △APC, ∵ AE EC and PA // QE ∴ PQ QC (intercept theorem) ∵ AE EC and PQ QC ∵ ABCD is a parallelogram. ∴ AD // BC and (diags. of // gram) AE EC ∵ BCEF is a parallelogram. ∴ FE // BC In △ABC, ∵ AE EC and GE // BC ∴ AG GB (intercept theorem) ∴ AG : GB 1 : 1 (ii) In △ABC, ∵ AE EC and AG GB 1 ∴ GE BC (mid-pt. theorem) 2 ∵ FE BC (opp. sides of // gram) ∴ FG GE 100 1 BC 2 5 Quadrilaterals ∴ ∴ 3. FG 1 BC 2 FG : BC 1 : 2 AG GB and FG GE ∴ AFBE is a parallelogram. x 12 50 (property of rhombus) x 38 y 50 (b) ∵ 4. proved in (a) diags. bisect each other 19. Join BC. Produce EF such that it meets BC at a point G. (property of rectangle) OA OB OAB OBA (base s, isos. △) 65 In △ABO, OAB OBA AOB 180 ( sum of △) 65 65 x 180 x 50 ∵ ∴ AD BC y 11 5. (property of rhombus) (property of rectangle) y 90 (property of square) OC OB (property of square) z 5 In △ACB, ∵ CF FA and FG // AB ∴ CG GB ∵ CF FA and CG GB 1 ∴ FG AB 2 In △BDC, ∵ BG GC and EG // DC ∴ BE ED ∵ BG GC and BE ED 1 ∴ EG DC 2 EF EG FG 1 1 DC AB 2 2 1 ( DC AB) 2 6. ABC DCB 79 BAD ABC 180 (int. s, AD // BC) x 79 180 x 101 7. (a) ∵ ABCD is a parallelogram. ∴ OB OD and OA OC (opp. sides of // gram) x 4 and 9 y y 3 2 y 12 y6 intercept theorem mid-pt. theorem intercept theorem mid-pt. theorem (b) ∵ ABCD is a rhombus. ∴ AOB 90 and AB BC CD DA (property of rhombus) In △ABO, AB 2 OA2 OB 2 (Pyth. theorem) Enrichment Topic (p. 5.64) 1. By the property of kite, x 130 AB (9 6) 2 4 2 cm AB AD y 3 .5 2. 32 4 2 cm 5 cm Perimeter of ABCD 4 AB 4 5 cm 20 cm By the property of kite, x 20 y 90 8. Check Yourself (p. 5.69) 1. (a) 2. (b) (c) (d) (e) (f) (g) (h) ∵ AD BE and BE BC given ∴ AD BC ∵ BC BE given ∴ BCE BEC 65 In △BCE, BCE BEC EBC 180 sum of △ 65 65 EBC 180 EBC 50 ∵ ADB DBC 50 ∴ AD // BC alt. s equal ∴ ABCD is a parallelogram. opp. sides equal and // 4 x x 90 (opp. s of // gram) 3 x 90 x 30 101 Mathematics in Action (3rd Edition) 3A Full Solutions 9. ∵ ABCD is a square. ∴ CAD 45 (property of square) ∵ AFDE is a rhombus. ∴ ADF 67 (property of rhombus) In △AGD, AGD ADG GAD 180 ( sum of △) AGD 67 45 180 AGD 68 10. ∵ ∴ ∵ ∴ 4. EPN PDC 90 (corr. s equal) NP // CD EP PD and NP // CD AN NC (intercept theorem) 7 cm 5. In △ABC, ∵ AN NC and AM MB 1 ∴ MN BC (mid-pt. theorem) 2 1 12 cm 2 6 cm ∵ ABCD is a rhombus. ∴ AED 90 (property of rhombus) In △AED, ( sum of △) 2 x 3 x 90 180 5 x 90 x 18 y 2x 2 18 36 (property of rhombus) z 3x 3 18 54 (property of rhombus) ∵ ∴ ∴ ABCD is a rectangle. (property of rectangle) ED EC (base s, isos.△) EDC ECD b In △CDE, ECD EDC CED 180 ( sum of △) b b 80 180 2b 100 b 50 Revision Exercise 5 (p. 5.71) Level 1 1. ∵ ABCD is a parallelogram. ∴ BCD BAD (opp. s of // gram) x 140 In △ACD, (property of rectangle) ADC 90 CAD ACD ADC 180 ( sum of △) a b ADC 180 a 50 90 180 a 40 BAD ADC 180 (int. s, BA // CD) 140 y 180 y 40 ∴ AD BC 6. (opp. sides of // gram) a 3 5 a 8 2. ∵ ∴ ABCD is a parallelogram. OB OD (opp. sides of // gram) BCE ECD 90 (property of rectangle) 60 x 90 y 1 11 y 2 y 10 y5 ∴ OC OA BCE y (alt. s, BC // AD) In △BCF, BCF BFC CBF 180 ( sum of △) y 80 40 180 y 60 x 30 7. (opp. sides of // gram) ∵ ∴ ABCD is a square. (property of square) CED 90 6 x 90 x 15 ∴ AD CD 5y 6 3y 2 2y 4 y2 3x 1 y 3 3x 1 5 3 3x 1 8 3x 9 x3 3. ∵ ∴ ABCD is a rhombus. AD CD (property of rhombus) a37 a4 ∴ AED 90 (property of rhombus) 8. b 30 90 b 60 102 (property of square) ∵ ABCD is a square. ∴ ACD 45 (property of square) ACD DCE 180 (adj. s on st. line) 45 x 180 x 135 5 Quadrilaterals ∴ OA OB 14. ∵ ABCD is a square. ∴ ABD 45 (property of square) ∵ △AEB is an equilateral triangle. ∴ ABE 60 (prop. of equil. △) DBE ABD ABE 45 60 105 (property of square) y4 2 y6 9. In △ACF, ∵ AB BC and AE EF 1 ∴ BE CF (mid-pt. theorem) 2 1 x 9 2 4.5 (prop. of equil.△) EAB 60 (property of square) BAD 90 EAD EAB BAD 60 90 150 ∵ △AEB is an equilateral triangle. ∴ EA AB Also, AB AD (property of square) ∴ EA AD ∴ ADE AED (base s, isos.△) In △AED, EAD ADE AED 180 ( sum of △) In △ADG, ∵ AC ( 4 4) cm 8 cm CD and AF (3 3) cm 6 cm FG 1 ∴ CF DG (mid-pt. theorem) 2 1 9 y 2 y 18 10. ∵ ∴ 150 2ADE 180 ADE 15 (property of square) ADB 45 BDE ADB ADE 45 15 30 AE // BF // CG and AB BC (intercept theorem) EF FG 2x 3 4x 5 2x 8 x4 FG (4 x 5) cm (4 4 5) cm 11 cm ∵ BF // CG // DH and FG 11 cm GH ∴ CD BC (intercept theorem) 15. ∵ AEDF is a rhombus. ∴ AF DF (property of rhombus) ∴ FAD FDA (base s, isos. △) In △ADF, AFD FAD FDA 180 ( sum of △) 122 2FAD 180 2FAD 58 FAD 29 EAD FAD (property of rhombus) 29 (property of rectangle) ADC 90 In △ACD, CAD ACD ADC 180 ( sum of △) 29 ACD 90 180 ACD 61 y5 11. In △BCD, ∵ DF FC and EF // BC ∴ DE EB (intercept theorem) x3 In △ACD, ∵ DF FC and AD// EF ∴ AE EC (intercept theorem) y2 12. ∵ AB AE ∴ ABE AEB (base s, isos. △) In △ABE, BAE ABE AEB 180 ( sum of △) 30 2ABE 180 2ABE 150 ABE 75 ∵ ABCD is a parallelogram. ∴ ADC ABC (opp. s of // gram) 75 16. 13. By the property of kite, CED 90 With the notations in the figure, OA OC and OB OD (property of rhombus) x 90 BAE DAE y 35 103 Mathematics in Action (3rd Edition) 3A Full Solutions 19. (a) ∵ ABCD is a parallelogram. ∴ BAD DCB , ABC ADC , AB CD and AD BC ∵ BE and FD are the angle bisectors of ABC and ADC respectively. ∴ ABE CBE ADF CDF In △AEB and △CFD, ABE CDF AB CD BAE DCF ∴ △AEB △CFD 24 cm 2 12 cm In △AOD, OA OA2 OD 2 AD 2 (Pyth. theorem) OD 20 122 cm 16 cm BD 2 OD 2 2 16 cm 32 cm ∴ The length of another diagonal of the rhombus is 32 cm. 17. (a) ∵ AQDR is a rectangle. ∴ ARD 90 (property of rectangle) In △ADR, (Pyth. theorem) AD 2 AR 2 DR 2 AD AR DR 2 (b) ∵ ABCD is a parallelogram. ∴ AD BC and AD BC ∵ △AEB △CFD ∴ AE CF ED AD AE 2 3 2 4 2 cm 5 cm ∵ ABCD is a rhombus. ∴ AB BC = CD AD 5 cm ∴ Perimeter of ABCD AB BC CD AD (5 5 5 5) cm 20 cm BC CF BF ∴ BFDE is a parallelogram. 20. (a) In △ABP and △CDQ, ABP CDQ ∵ APQ CQP ∴ APB 180 APQ (b) ∵ AQDR is a rectangle. ∴ DQ AR 3 cm and AQ DR 4 cm (property of rectangle) ∵ ABCD is a rhombus. ∴ CQ AQ 4 cm and BQ DQ 3 cm (property of rhombus) ∵ BPCQ is a rectangle. ∴ CP BQ 3 cm and BP CQ 4 cm (property of rectangle) Area of ABPCDR = 6 area of △ADR 1 6 AR DR 2 1 6 3 4 cm 2 2 36 cm 2 proved proved proved ASA proved in (a) corr. sides, △s opp. sides equal and // alt. s, AB // DC given adj. s on st. line 180 CQP adj. s on st. line CQD AB CD ∴ △ABP △CDQ opp. sides of // gram AAS (b) ∵ APQ CQP ∴ AP // QC ∵ △ABP △CDQ ∴ AP CQ ∴ APCQ is a parallelogram. 21. (a) In △AED and CFB, AD CB BAD BCD 90 ∵ △ABE △CDF ∴ BAE DCF EAD BAE BAD 18. (a) ∵ △ABE is an equilateral triangle. ∴ BAE ABE AEB 60 (prop. of equil. △) In quadrilateral ABCD, BAD ABC BCD ADC 360 ( sum of polygon) (60 40) (60 20) 100 ADC 360 100 80 100 ADC 360 ADC 80 (b) BAD 60 40 100 ABC 60 20 80 ∵ BAD BCD and ABC ADC ∴ ABCD is a parallelogram. opp. s of // gram opp. sides of // gram DCF BCD FCB AE CF ∴ △AED △CFB ∴ BF ED property of square property of square given corr. s, △s corr. sides, △s SAS corr. sides, △s (b) ∵ △ABE △CDF ∴ BE FD ∴ BFDE is a parallelogram. ∴ Daniel’s claim is incorrect. opp. s equal 104 given alt. s equal proved in (a) corr. sides, △s opp. sides equal and // (given) (corr. sides, △s) (opp. sides equal) 5 Quadrilaterals BDF FDE 1 ∴ BDF BDE 2 1 144 2 72 DOA 90 In △DOF, AFD DOA BDF 90 72 18 22. (a) ∵ BCHG is a parallelogram. ∴ GB HC 7 cm (opp. sides of // gram) In △AFB, ∵ AE EF and EG // FB ∴ AG GB (intercept theorem) AB AG GB (7 7) cm 14 cm ∵ (b) In △AFB, ∵ AE EF and AG GB 1 EG FB ∴ (mid-pt. theorem) 2 1 2 cm FB 2 FB 4 cm ∵ ABCD is a parallelogram. ∴ BC AD 6 cm (opp. sides of // gram) FC FB BC ( 4 6) cm 10 cm (property of rhombus) (ext. of △) 25. ∵ ∴ ∵ ∴ ABCD is a square. (property of square) DBC 45 BF // CE (property of rhombus) BCE 180 DBC (int.∠s, BF // CE) 180 45 135 ∵ AB BC CD DA (property of square) DC CE EF FD (property of rhombus) ∴ BC CE ∴ EBC BEC (base s, isos. △) In △BCE, BCE EBC BEC 180 ( sum of △) 23. (a) ∵ ABCD is a parallelogram. ∴ AB DC 10 cm and AD BC 12 cm (opp. sides of // gram) MB AB AM 135 2EBC 180 EBC 22.5 DBG DBC EBC 45 22.5 22.5 (10 5) cm 5 cm AN AD ND (12 6) cm 6 cm ∵ AM MB and AN ND 1 ∴ MN // BD and MN BD 2 (mid-pt. theorem) ∴ BDNM is a trapezium. (b) (given) 26. ∵ △BCG is an equilateral triangle. ∴ CG GB BC Also, BC CD (property of square) ∴ CD CG ∴ CDG CGD (base s, isos.△) ∵ BCG 60 (prop. of equil.△) DCB 90 (property of square) ∴ DCG DCB BCG 90 60 30 In △CDG, DCG CDG CGD 180 ( sum of △) 1 BD 2 BD 2MN ∵ Perimeter of BDNM = 38 cm ∴ MN MB BD ND 38 cm MN 5 cm 2MN 6 cm 38 cm 3MN 11 cm 38 cm 3MN 27 cm MN 9 cm MN 30 2CDG 180 CDG 75 (property of square) CDB 45 GDB CDG CDB 75 45 30 Level 2 24. ∵ ABCD is a rhombus. ∴ ADB CBD ABD (property of rhombus) ∴ ADB ABD 1 ABC 2 1 72 2 36 (adj. s on st. line) BDE 180 ADB 180 36 144 (property of rectangle) KB KC KBC KCB (base s, isos.△) 24 In △BCK, BKC 180 KBC KCB ( sum of△) 180 24 24 132 AKD BKC 132 (vert. opp. s) (prop. of equil.△) AKE 60 27. (a) ∵ ∴ 105 Mathematics in Action (3rd Edition) 3A Full Solutions EKD AKD AKE 132 60 72 Perimeter of parallelogram ABCD AB BC CD AD (7.5 15 7.5 15) cm 45 cm (b) ∵ △AKE is an equilateral triangle. ∴ EAK 60 (prop. of equil. △) (alt. s, AD // BC) DAK ACB 24 EAD EAK DAK 29. In △ADQ and △CBP, ADB CBD ADQ 180 ADB 180 CBD CBP AD CB DQ BP ∴ △ADQ △CBP ∴ AQ CP AQD CPB ∴ AQ // PC ∴ APCQ is a parallelogram. 60 24 36 ∵ △AKE is an equilateral triangle. ∴ KA KE Also, KA KD (property of rectangle) ∴ KE KD ∴ KED KDE (base s, isos.△) In △DEK, EKD KDE KED 180 ( sum of △) 72 2KDE 180 KDE 54 KDA KBC (alt. s, AD // BC) 24 EDA KDE KDA 54 24 30 ∵ EAD EDA ∴ △AED is not an isosceles triangle. ∴ Philip’s claim is incorrect. 28. (a) Let CBE a and BCE b. ABE CBE a DCE BCE b ABC BCD 180 2a 2b 180 a b 90 In △BCE, BEC BCE CBE 180 BEC a b 180 BEC 90 180 BEC 90 ∴ △BEC is a right-angled triangle. BC BE CE 2 opp. sides of // gram given SAS corr. sides, △s corr. s, △s alt. s equal opp. sides equal and // ABCDE is a regular pentagon. ABC BCD (5 2) 180 5 108 ∵ AB CB ∴ BAC BCA base s, isos. △ In △ABC, sum of △ ABC BCA BAC 180 108 2BCA 180 2BCA 72 BCA 36 ∵ BCDF is a rhombus. ∴ BCF DCF property of rhombus 108 2 54 FCG BCF BCA given given int. s, AB // DC sum of △ 54 36 18 GCD GCF DCF 18 54 72 BGC GCD 72 FCG BGC 18 72 90 (Pyth. theorem) 2 12 2 9 2 cm 15 cm AD BC 15 cm ∵ DEC BCE and DCE BCE ∴ DCE DEC ∴ DC DE 7.5 cm AB DC 7.5 cm adj. s on st. line 30. ∵ ∴ (b) In △BCE, BC 2 BE 2 CE 2 alt. s, AD // BC adj. s on st. line 31. (a) In △BCE and △BGA, GBE CBA 90 CBE CBG GBE (opp. sides of // gram) (alt. s, DE // CB) CBG CBA GBA ∵ ABCD and BEFG are two identical rectangles. ∴ BA BE and BG BC ∴ △BCE △BGA (sides opp. equal s) (opp. sides of // gram) 106 alt. s, BF // CD property of rectangle SAS 5 Quadrilaterals (b) ∵ △BCE △BGA (proved in (a)) ∴ BCE BGA (corr. s, △s) BPG 180 PBQ BGA ( sum of △ ) in △PED and △QEB, ∵ PD // BQ ∴ DPE BQE PED QEB ED EB ∴ △PED △QEB ∴ PE QE and ED EB ∴ BQDP is a parallelogram. In △PED and △PEB, PED PEB 90 ED EB PE PE ∴ △PED △PEB ∴ PD PB Also, PD BQ and PB QD ∴ PD QD BQ PB ∴ BQDP is a parallelogram with four equal sides. ∴ BQDP is a rhombus. 180 PBQ BCE ∴ 32. (a) (i) BQC BQC 107 In △ABC and △AFE, BAC FAE ABC 90 AFE AC AE ∴ △ABC △AFE (ii) In △ABG and △AFG, ABG AFG 90 ∵ △ABC △AFE ∴ AB AF AG AG ∴ △ABG △AFG ( sum of △ ) common angle property of square given given AAS proved in (i) proved in (i) corr. sides, △s common side RHS 35. (a) In △OCD and △OQP, COD QOP ∵ CO DO and CQ DP ∴ QO CO CQ (b) BAC 45 (property of square) ∵ △ABG △AFG (proved in (a)(ii)) ∴ BAG FAG (corr. s, △s) 1 BAC 2 1 45 2 22.5 In △ABG, AGB BAG ABG 180 ( sum of △) AGB 22.5 90 180 AGB 67.5 33. (a) In △ACE and △DBF, AE DF ∵ OA OD ∴ CAE BDF AC DB ∴ △ACE △DBF ∴ ACE DBF (b) In △ACE, NEF CAE ACE In △DBF, NFE BDF DBF ∴ NEF NFE ∴ NE NF i.e. △NEF is an isosceles triangle. ∴ given given common side SAS corr. sides, △s opp. sides of // gram common angle property of rectangle given DO DP PO CO DO QO PO ∴ △OCD ~ △OQP ∴ OCD OQP ∴ DC // PQ Also, AB // DC ∴ AB // PQ given property of rectangle base s, isos. △ property of rectangle SAS corr. s, △s property of rectangle alt. s, PD // BQ vert. opp. s given AAS corr. sides, △s given diags. bisect each other ratio of 2 sides, inc. corr. s, ~△s corr. s equal property of rectangle (b) In △AOP and △BOQ, PO QO (proved in (a)) AOP BOQ (vert. opp. s) AO BO (property of rectangle) ∴ △AOP △BOQ (SAS) ∴ AP BQ (corr. sides, △s) ∵ ABQP is a quadrilateral with AB // PQ and AP BQ. ∴ ABQP is an isosceles trapezium. ext. of △ ext. of △ 36. In △ABC, ∵ M and N are the mid-points of AB and AC respectively. ∴ MN // BC and 1 MN BC 2 In △BCD, ∵ P and Q are the mid-points of CD and BD respectively. ∴ QP // BC and 1 QP BC 2 ∵ QP MN and QP // MN ∴ MNPQ is a parallelogram. ∴ MQ NP sides opp. equal s 34. With the notation in the figure, 107 mid-pt. theorem mid-pt. theorem opp. sides equal and // opp. sides of // gram Mathematics in Action (3rd Edition) 3A Full Solutions 37. (a) (i) In △APR and△CQR, AR CR ARP CRQ PAR QCR ∴ △APR △CQR (ii) ∵ △APR △CQR ∴ PA QC and BQ QC ∴ PA BQ Also, AP // BQ ∴ ABQP is a parallelogram. 1 AQ h Area of △ APQ 2 Area of △ QCD 1 RQ h 2 AQ RQ AR RQ RQ 2 RQ RQ RQ 3RQ RQ 3 ∴ Area of △APQ : area of △QCD 3 : 1 given vert. opp. s alt. s, AP // QC ASA proved in (i) corr. sides, △s given given opp. sides equal and // (b) In △ADE, ∵ B and C are the mid-points of AD and AE respectively. 1 ∴ BC DE (mid-pt. theorem) 2 DE 2 BC ∵ BQ QC ∴ BC 2 BQ 2 AP ∴ DE 2 BC 4 AP AP 1 DE 4 ∴ AP : DE 1 : 4 39. (a) In △ABD, ∵ AE EB and ∴ DG GB ∵ AE EB and 1 ∴ EG AD 2 In △BCD, ∵ DG GB and ∴ DF FC ∵ DG GB and 1 ∴ GF BC 2 ∴ EF EG GF 38. (a) In △PQR and △DQC, QPR QDC alt. s, PR // CD PQ DQ given PQR DQC vert. opp. s ∴ △PQR △DQC ASA (b) (i) ∵ AP PB and PR // BC ∴ AR RC (intercept theorem) ∵ △PQR △DQC (proved in (a)) ∴ RQ CQ (corr. sides, △s) 1 RC 2 AR RC RQ 1 RC 2 2 ∴ AR : RQ 2 : 1 AD// EG intercept theorem DG GB mid-pt. theorem GF // BC intercept theorem DF FC mid-pt. theorem 1 1 AD BC 2 2 1 ( AD BC ) 2 (b) In △ABC, ∵ AE EB and EH // BC ∴ AH HC ∵ AE EB and AH HC 1 ∴ EH BC 2 1 EG GH BC 2 1 GH BC EG 2 In △ABD, ∵ AE EB and EG // AD ∴ DG GB ∵ AE EB and DG GB 1 ∴ EG AD 2 1 1 ∴ GH BC AD 2 2 1 ( BC AD) 2 (ii) Let h be the height of △APQ and △PQR. Area of △QCD = area of △PQR 108 intercept theorem mid-pt. theorem intercept theorem mid-pt. theorem 5 Quadrilaterals ∴ Multiple Choice Questions (p. 5.77) 1. Answer: C (opp. sides of // gram) BC AD ∵ AD BE ∴ BC BE ∴ BCE BEC (base s, isos. △) 35 In △BCF, FBC BCF AFC 180 100 80 BCD BCE ECD 80 60 140 ∵ △CDE is an equilateral triangle. ∴ EC CD DE Also, AB BC CE EA (property of rhombus) ∴ DE EA and CD CB ∴ EDA EAD and (base s, isos.△) CDB CBD In △AED, AED EDA EAD 180 ( sum of △) (ext. of △) FBC 35 100 FBC 65 i.e. ABC 65 2. Answer: D BCD x (opp. s of // gram) (opp. s of // gram) CBF z BGC y (vert. opp. s) In △BCG, BCG BGC CBG 180 160 2EDA 180 EDA 10 In △BCD, BCD CBD CDB 180 ( sum of △) 140 2CDB 180 CDB 20 ADB EDC EDA CDB 60 10 20 30 ( sum of △) x y z 180 y 180 x z 3. Answer: B ADC 90 In △ACD, (property of rectangle) AC AD CD 2 BCE 180 ABC (int. s, EC // AB) 2 2 Alternative Solution Construct BF such that BF // AD. (Pyth. theorem) AC 2 7 cm 2 2 53 cm ACF 90 and AC CF 53 cm In △ACF, AF 2 AC 2 CF 2 (property of square) (Pyth. theorem) Let EAD a and CBD b. ∵ AE // BC (property of rhombus) ∴ a DAB 100 180 (int. s, AE // BC) DAB 80 a DAB ABF 180 (int. s, AD // BF) AF ( 53 ) ( 53 ) 2 cm 2 10.3 cm (cor. to 3 sig. fig.) 4. Answer: C AEC ECD 180 (int. s, EA // CD) (80 a) (100 CBF ) 180 CBF a (alt. s, AD // BF) ADB CBD CBF ba ∵ △CDE is an equilateral triangle. ∴ EC CD DE Also, AB BC CE EA (property of rhombus) ∴ DE EA and CD CB ∴ EDA EAD a and (base s, isos. △) CDB CBD b ∵ CDE 60 (prop. of equil.△) ∴ EDA ADB CDB 60 a (b a ) b 60 2b 2a 60 b a 30 ∴ ADB b a 30 125 ECD 180 ECD 55 (property of square) BDC 45 In △CDF, CFD FCD FDC 180 ( sum of △) CFD 55 45 180 CFD 80 5. Answer: C ∵ △CDE is an equilateral triangle. ∴ CED ECD EDC 60 (prop. of equil.△) ∵ ABCE is a rhombus. ∴ AEC ABC (property of rhombus) 100 AED AEC CED 100 60 160 ∵ EC // AB (property of rhombus) 109 Mathematics in Action (3rd Edition) 3A Full Solutions 6. (common side) DE DE ∴ △ADE △CDE (SSS) ∴ III must be true. ∴ The answer is D. Answer: A OA OC , OB OD and AOB 90 (property of rhombus) 1 OA AC 2 1 16 cm 2 8 cm In △AOB, OA2 OB 2 AB 2 10. Answer: A For I: AEB BED 180 (adj. s on st. line) AEB 120 180 AEB 60 AB BE BAE AEB (base s, isos. △) 60 In △ABE, ABE AEB BAE 180 ( sum of △) ABE 60 60 180 ABE 60 ∵ ABE AEB BAE 60 ∴ AB BE AE ∴ △ABE is an equilateral triangle. ∴ I must be true. For II: It is not necessary that BCDE is a parallelogram because BE may not be parallel to CD. ∴ II may be false. For III: It is not necessary that ABCE is a rhombus because BC and EC may not equal to AB. ∴ III may be false. ∴ The answer is A. (Pyth. theorem) ∵ ∴ OB 17 2 82 cm 15 cm Area of ABCD = 4 area of △AOB 1 4 OA OB 2 1 4 8 15 cm2 2 240 cm2 7. Answer: B ∵ The interior angles of a rhombus may not be all equal. ∴ It is incorrect that any rhombus must be a regular polygon. 8. Answer: D 11. Answer: B In △ADE, ∵ AB BD and AC CE 1 BC DE (mid-pt. theorem) ∴ BC // DE and 2 1 2.5 cm DE 2 DE 5 cm In △AHI, ∵ AD DH and AE EI 1 ∴ DE // HI and DE HI (mid-pt. theorem) 2 1 5 cm HI 2 HI 10 cm With the notation in the figure, ∵ AD AB ∴ ADB ABD (base s, isos. △) 23 By the property of kite, AED 90 In △AED, a AED ADE 180 ( sum of △) a 90 23 180 a 67 9. 12. Answer: A ∵ P and R are the mid-points of AB and CA respectively. ∴ PR// BC (mid-pt. theorem) APR x and ARP y (corr. s, PR // BC) ∵ P and Q are the mid-points of AB and BC respectively. ∴ PQ // AC (mid-pt. theorem) RPQ ARP y (alt. s, AR // PQ) APQ APR RPQ x y Answer: D For I, by the property of kite, AC BD ∴ I must be true. For II, by the property of kite, BAE BCE ∴ II must be true. For III, in △ADE and △CDE, (given) AE CE (property of rhombus) AD CD 110 5 Quadrilaterals BAD ABC 180 (int. s, AD // BC) (85 CAD) (25 CAD) 180 2CAD 70 13. Answer: A In △ABC, ∵ AQ QC and PQ // BC ∴ AP PB (intercept theorem) ∴ I must be true. ∵ AQ QC and AP PB 1 ∴ PQ BC (intercept theorem) 2 BC 2 PQ ∴ II must be true. For III, in △ACD, ∵ AR RD and AQ QC ∴ QR // CD (mid-pt. theorem) AQR ACD (corr. s, QR // CD) ∵ It is not necessary that AC AD . ∴ ACD ADC may not be true. ∴ AQR ADC may not be true. ∴ The answer is A. CAD 35 2. 64 2DAE 180 DAE 58 (alt. s, AD // BC) ACB DAC 58 In △BCE, (ext. of △) CBE ECB AEB CBE 58 89 CBE 31 3. 14. Answer: D ∵ BG : GC 3 :1 BG 3 ∴ GC BG 3GC In △BGE, ∵ EA AB and AF // BG ∴ EF FG (intercept theorem) ∵ EA AB and EF FG 1 ∴ AF BG (mid-pt. theorem) 2 1 3GC 2 3GC 2 (opp. sides of // gram) BC AD BG GC AF FD 3GC 3GC GC FD 2 5GC FD 2 FD 5 GC 2 ∴ FD : GC 5 : 2 4. Exam Corner Exam-type Questions (p. 5.80) 1. ∵ AD DE ∴ DAE DEA (base s, isos. △) In △AED, ADE DAE DEA 180 ( sum of △) In △BCG and △DCH, ∵ CE and CF divide BCD into three equal parts. ∴ BCG DCH ∵ ABCD is a rhombus. ∴ BC DC ∴ CBG CDH ∴ △BCG △DCH ∴ CG CH CHG CGH x CHG FHG 180 x y 180 111 adj. s on st. line ∵ EB : BC 3 : 2 ∴ Let EB 3x and BC 2 x . In △BEF and △CED, (corr. s, AB // DC) EBF ECD (corr. s, AB // DC) EFB EDC (common angle) BEF CED ∴ △BEF ~ △CED (AAA) FB BE ∴ (corr. sides, ~△s) DC CE FB 3x DC 3 x 2 x FB 3 DC 5 3 FB DC 5 (opp. sides of // gram) AB DC AF FB DC 3 AF DC DC 5 2 AF DC 5 AF 2 DC 5 ∴ AF : DC 2 : 5 ACB CAD (alt. s, AD // BC) DCB 25 ACB 25 CAD ∵ ABCD is a trapezium with AD // BC . ∴ ABC DCB 25 CAD property of rhombus property of rhombus ASA corr. sides, △s base s, isos. △ Mathematics in Action (3rd Edition) 3A Full Solutions 5. Do and Investigate (p. 5.82) (a) In △BCF and △DCF, (property of square) BC DC BCF DCF 45 (property of square) (common side) CF CF ∴ △BCF △DCF (SAS) 1. (b) CDF CBF (corr. s, △s) 77 (property of square) DCF 45 In △CDF, AFD CDF DCF (ext. of △) 77 45 122 6. 7. ∵ AD is a median of △ABC. ∴ BD DC ∵ ADG CED ∴ GD// FC alt. s equal In △AGD, ∵ AE ED and FE // GD ∴ AF FG intercept theorem In △BCF, ∵ CD DB and FC // GD ∴ FG GB intercept theorem ∴ AF FG GB 2. Join BD. ∵ AE EB and AH HD 1 ∴ EH BD and EH // BD 2 ∵ CF FB and CG GD 1 ∴ FG BD and FG // BD 2 ∴ EH FG and EH // FG ∴ EFGH is a parallelogram. (a) In △ACD, ∵ DN NC and AD // ON ∴ AO OC intercept theorem In △ABC, ∵ AO OC and BM MC ∴ BA// MO mid-pt. theorem ∵ AD // BO and BA// OD ∴ ABOD is a parallelogram. 3. (a) yes, rectangle (b) yes, rhombus (b) In △ACD, ∵ DN NC and AO OC 1 ∴ ON AD (mid-pt. theorem) 2 ∵ BO AD (opp. sides of // gram) 1 ∴ ON BO 2 BO 2 ON ∴ BO : ON 2 : 1 (c) yes, square (d) yes, rhombus 112 mid-pt. theorem mid-pt. theorem from and opp. sides equal and //