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1)Since the radius of the circle is 12cm ∴AO=BO=CO=12cm Draw a point D which DO⊥BC ∠OBC=60/2=30 cos30=BD/12 BD=6√3cm 2a)∠A0C=∠ADC(property of rhombus) Reflex ∠AOC=2∠ADC(∠at centre, twice∠s at circumference) ∠AOC+Reflex ∠AOC=360(Reflex ∠s at a point) ∠AOC+2∠ADC=360 ∠AOC+2∠AOC=360 ∠AOC=120 2b)2∠ABC=∠AOC(∠at centre, twice∠s at circumference) 2∠ABC=120 ∠ABC=60 3)∠BAP/∠PAC=arcBP/arcPC ∠BAP=∠PAC ∠BAC=45(property of square) ∠BAP+∠PAC=45 ∠BAP=22.5 ∠BAC=90(property of square) ∠AQB=180-90-22.5=67.5 4)∠ACB/∠ABD=arcAB/arcAD 2∠ABD=∠ACB ∠ABC=90(∠in semi-circle) ∠CAB=180-90-∠ACB(∠ sum of △) ∠CAB=90-2∠ABD (90-2∠ABD)+∠ABD=58(ext.∠s sum of △) 90-∠ABD=58 ∠ABD=32 5)∠BPD=2(65)((∠at centre, twice∠s at circumference) ∠BDP=130 ∠BCD+∠BPD=180(opp.∠ cyclic quad.) ∠BCD=50 9a)∠CED=∠BEA(vert.opp ∠s) ∠ECD=∠EBA(∠s in the same segment) ∠EAB=∠EDC(∠s in the same segment) △ECD～△EBA (AAA) 9b)△ECD～△EBA (AAA) 4/3=6/y(...) 4y=18 y=4.5 10)x=25(alt ∠s,AD//BC) ∠DBC=25(∠s in the same segment) 56+∠DBC+x+y=180(∠ sum of △) y=74 11)∠ABC=40(∠s in the same segment) ∠ACB=(180-∠ABC)/2(∠ sum of △) ∠ACB=70 ∠DCB=90(∠in semicircle) ∠DCA=∠DCB-∠ACB ∠DCA=20 ∠ABD=∠DCA(∠s in the same segment) ∠ABD=20 12a) Let O be the centre of the circle, OR be y, the radius be r PQ⊥AB AR=RB(line from centre prep. to chord bisects chord) AR=10√3/2=5√3 AR^2+OR^2=r^2(Pyth. theorem) 75+y^2=r^2-----(1) y+5=r----(2) From (2): (y+5)^2=r^2 y^2+10y+25=r^2------(3) Put (3) into (1): 75+y^2=y^2+10y+25 y=5-----(4) Put (4)into (2): 5+5=r r=10 The radius of the circle is 10cm y+5=r--------(1) 12b)BQ^2=RB^2+RQ^2(Pyth. theorem) BQ^2=75+25 BQ=10 or -10(rejected) The length of BQ is 10 cm I am too tired. So I don't do la. If you have any difficulties, you can add my msn chanhonpui1991@hotmail.com 2007-01-03 13:53:27 補充： so i will help you 2007-01-06 14:33:38 補充： 2 3a. BA=BCBD=DB∠ABD=∠CBD△BAD≡△ BCDb.△BAD≡△BCD∠BAD=∠BCD∠BAD+∠BCD=180∠BAD=180/2=90c.AD=√(13^2-5^2)=12△BAD≡△ BCDCD=AD=12area of ABCD=2X5X12/2=6025.∠CDB be x∠CDB/∠ODB=1x=∠ODBx+∠ODB=∠ODC∠ODC=2x ∠ODB=∠DBOx=∠DBO∠AOB=∠ODB+∠DBO=2x=∠ODCBO//CD
其他解答:
1) Draw a perpendicular to AB from O and let the foot be D. This perpendicular will divide AB into equal halves, i.e. AD = BD (radius perp. to chord bisects chord) Performing the action in similar way, i.e. draw from O to AC and BC. Then △OAB, △OBC and △OCA are congruent (SSS). ∠OAD = 30 deg and AD = OA cos 30 deg = 6x開方(3) cm So Perimeter of ∠ABC = 6x6x開方(3) = 36開方(3) cm 2)a) Let ∠ADC be x deg, then reflex ∠AOC = 2x deg (angle at centre = twice the angle at circumference) Moreover, since AOCD is a rhombus, obtuse ∠AOC = ∠ADC = x deg. Hence, 2x + x = 360 (obtuse + reflex ∠AOC) x = 120 ∠AOC = 120 deg b) ∠ABC = 60 deg (angle at centre = twice the angle at circumference) 3) Join AC, then ∠BAC = 45 deg (half of angle in square) And also, ∠BAP = ∠PAC (same arc, same angle) So, ∠BAP = 22.5 deg Thus, ∠AQB = 180 deg - 22.5 deg - 90 deg = 67.5 deg 5) Join DP and BP, ∠DPB = 130 deg (angle at centre = twice the angle at circumference) Since DPBC is a cyclic quadrilateral, ∠DPB + ∠BCD = 180 deg So ∠BCD = 50 deg 9)a) ∠AEB = ∠DEC = (angles in the same segment) ∠EDC = ∠EAB = (angles in the same segment) ∠ECD = ∠EBA = (angles in the same segment) So △EBA is similar to △ECD. b) By the properties of same corresponding side ratio for similar triangles: EC/EB = DE/AE y/6 = 3/4 y = 4.5 10) x deg = 25 deg (alt. angles AD//BC) ∠ADB = x deg = 25 deg (angle in the same segment) 25 + 25 + y + 56 = 180 (angles sum in triangle) y = 74. 11) ∠BAC = ∠BDC = 40 deg (angle in the same segment) ∠ABC = ∠ACB = (180 deg- 40 deg)/2 = 70 deg (base angles of isos triangles) ∠BCD = 90 deg (angle in semi circle) ∠DCB = 90 deg - 70 deg= 20 deg ∠ABD = ∠DCB = 20 deg 12) a) Let O be the centre of the circle and join OB. Also let the radius be r. BR = AB/2 = 5開方(3) cm (radius perp. to chord bisects chord) Also OR = r-5 Applying Pyth. Theorem in △ORB, (r-5)^2 + (5開方(3))^2 = r^2 r^2 – 10r + 25 + 75 = r^2 r = 10cm b) QR = 5cm Applying Pyth. Theorem in △QRB, 5^2 + (5開方(3))^2 = BQ^2 BQ = 10cm 14) a) ∠EAC = ∠DAC (common angle) ∠ADC = ∠ABC (angle in the same segment) ∠ABC = ∠ACB (base angles of isos triangle) Thus ∠ADC = ∠ACE Hence, by the sum of triangle’s angles, ∠AEC = ∠ACD Therefore, △ACE ~ △ADC b) By the properties of same corresponding side ratio for similar triangles: AC/AD = AE/AC AD = AC x AC/AE = 324/16 = 20.25cm 15) a) ∠DAC = ∠BAE (common angle) ∠DCA = ∠BEA (angle in the same segment) Hence, by the sum of triangle’s angles, ∠ADC = ∠ABE Therefore, △DAC ~ △BAE b) By the properties of same corresponding side ratio for similar triangles: DA/BA = AC/AE AE = AC x DA/BA = 12 x 7/6 = 14cm Hence DE = 7cm 16a) ∠PBC = ∠PDA (ext. ∠ of cyclic quad.) ∠PCB = ∠PDA (corr. ∠ AD//BC) Thus ∠PBC = ∠PCB and hence △PBC is an isos. triangle. b) ∠PBC = ∠AQC = 68 deg (∠ in the same segment) ∠PCB = ∠PBC = 68 deg So, ∠BPC = 180 - 68 - 68 = 44 deg c) ∠PAD = ∠PBC (corr. ∠ AD//BC) So ∠PAD = ∠PDA and hence △PAD is an isos. triangle and PA = PD DC = PC - PD = PB - PA = 6cm 2007-01-03 23:01:51 補充： 17a) ∠CDP = ∠ABP (ext. ∠ of cyclic quad.)∠DCP = ∠BAP (ext. ∠ of cyclic quad.)∠CPD = ∠APB (common ∠)Thus, △ABP ~ △CDP (3 corresponding angles equal)b) By the properties of same corresponding side ratio for similar triangles:AB/CD = BP/DPBP = AB x DP/CD = 14 x 15/10 = 21cm
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1)Since the radius of the circle is 12cm ∴AO=BO=CO=12cm Draw a point D which DO⊥BC ∠OBC=60/2=30 cos30=BD/12 BD=6√3cm 2a)∠A0C=∠ADC(property of rhombus) Reflex ∠AOC=2∠ADC(∠at centre, twice∠s at circumference) ∠AOC+Reflex ∠AOC=360(Reflex ∠s at a point) ∠AOC+2∠ADC=360 ∠AOC+2∠AOC=360 ∠AOC=120 2b)2∠ABC=∠AOC(∠at centre, twice∠s at circumference) 2∠ABC=120 ∠ABC=60 3)∠BAP/∠PAC=arcBP/arcPC ∠BAP=∠PAC ∠BAC=45(property of square) ∠BAP+∠PAC=45 ∠BAP=22.5 ∠BAC=90(property of square) ∠AQB=180-90-22.5=67.5 4)∠ACB/∠ABD=arcAB/arcAD 2∠ABD=∠ACB ∠ABC=90(∠in semi-circle) ∠CAB=180-90-∠ACB(∠ sum of △) ∠CAB=90-2∠ABD (90-2∠ABD)+∠ABD=58(ext.∠s sum of △) 90-∠ABD=58 ∠ABD=32 5)∠BPD=2(65)((∠at centre, twice∠s at circumference) ∠BDP=130 ∠BCD+∠BPD=180(opp.∠ cyclic quad.) ∠BCD=50 9a)∠CED=∠BEA(vert.opp ∠s) ∠ECD=∠EBA(∠s in the same segment) ∠EAB=∠EDC(∠s in the same segment) △ECD～△EBA (AAA) 9b)△ECD～△EBA (AAA) 4/3=6/y(...) 4y=18 y=4.5 10)x=25(alt ∠s,AD//BC) ∠DBC=25(∠s in the same segment) 56+∠DBC+x+y=180(∠ sum of △) y=74 11)∠ABC=40(∠s in the same segment) ∠ACB=(180-∠ABC)/2(∠ sum of △) ∠ACB=70 ∠DCB=90(∠in semicircle) ∠DCA=∠DCB-∠ACB ∠DCA=20 ∠ABD=∠DCA(∠s in the same segment) ∠ABD=20 12a) Let O be the centre of the circle, OR be y, the radius be r PQ⊥AB AR=RB(line from centre prep. to chord bisects chord) AR=10√3/2=5√3 AR^2+OR^2=r^2(Pyth. theorem) 75+y^2=r^2-----(1) y+5=r----(2) From (2): (y+5)^2=r^2 y^2+10y+25=r^2------(3) Put (3) into (1): 75+y^2=y^2+10y+25 y=5-----(4) Put (4)into (2): 5+5=r r=10 The radius of the circle is 10cm y+5=r--------(1) 12b)BQ^2=RB^2+RQ^2(Pyth. theorem) BQ^2=75+25 BQ=10 or -10(rejected) The length of BQ is 10 cm I am too tired. So I don't do la. If you have any difficulties, you can add my msn chanhonpui1991@hotmail.com 2007-01-03 13:53:27 補充： so i will help you 2007-01-06 14:33:38 補充： 2 3a. BA=BCBD=DB∠ABD=∠CBD△BAD≡△ BCDb.△BAD≡△BCD∠BAD=∠BCD∠BAD+∠BCD=180∠BAD=180/2=90c.AD=√(13^2-5^2)=12△BAD≡△ BCDCD=AD=12area of ABCD=2X5X12/2=6025.∠CDB be x∠CDB/∠ODB=1x=∠ODBx+∠ODB=∠ODC∠ODC=2x ∠ODB=∠DBOx=∠DBO∠AOB=∠ODB+∠DBO=2x=∠ODCBO//CD
其他解答:
1) Draw a perpendicular to AB from O and let the foot be D. This perpendicular will divide AB into equal halves, i.e. AD = BD (radius perp. to chord bisects chord) Performing the action in similar way, i.e. draw from O to AC and BC. Then △OAB, △OBC and △OCA are congruent (SSS). ∠OAD = 30 deg and AD = OA cos 30 deg = 6x開方(3) cm So Perimeter of ∠ABC = 6x6x開方(3) = 36開方(3) cm 2)a) Let ∠ADC be x deg, then reflex ∠AOC = 2x deg (angle at centre = twice the angle at circumference) Moreover, since AOCD is a rhombus, obtuse ∠AOC = ∠ADC = x deg. Hence, 2x + x = 360 (obtuse + reflex ∠AOC) x = 120 ∠AOC = 120 deg b) ∠ABC = 60 deg (angle at centre = twice the angle at circumference) 3) Join AC, then ∠BAC = 45 deg (half of angle in square) And also, ∠BAP = ∠PAC (same arc, same angle) So, ∠BAP = 22.5 deg Thus, ∠AQB = 180 deg - 22.5 deg - 90 deg = 67.5 deg 5) Join DP and BP, ∠DPB = 130 deg (angle at centre = twice the angle at circumference) Since DPBC is a cyclic quadrilateral, ∠DPB + ∠BCD = 180 deg So ∠BCD = 50 deg 9)a) ∠AEB = ∠DEC = (angles in the same segment) ∠EDC = ∠EAB = (angles in the same segment) ∠ECD = ∠EBA = (angles in the same segment) So △EBA is similar to △ECD. b) By the properties of same corresponding side ratio for similar triangles: EC/EB = DE/AE y/6 = 3/4 y = 4.5 10) x deg = 25 deg (alt. angles AD//BC) ∠ADB = x deg = 25 deg (angle in the same segment) 25 + 25 + y + 56 = 180 (angles sum in triangle) y = 74. 11) ∠BAC = ∠BDC = 40 deg (angle in the same segment) ∠ABC = ∠ACB = (180 deg- 40 deg)/2 = 70 deg (base angles of isos triangles) ∠BCD = 90 deg (angle in semi circle) ∠DCB = 90 deg - 70 deg= 20 deg ∠ABD = ∠DCB = 20 deg 12) a) Let O be the centre of the circle and join OB. Also let the radius be r. BR = AB/2 = 5開方(3) cm (radius perp. to chord bisects chord) Also OR = r-5 Applying Pyth. Theorem in △ORB, (r-5)^2 + (5開方(3))^2 = r^2 r^2 – 10r + 25 + 75 = r^2 r = 10cm b) QR = 5cm Applying Pyth. Theorem in △QRB, 5^2 + (5開方(3))^2 = BQ^2 BQ = 10cm 14) a) ∠EAC = ∠DAC (common angle) ∠ADC = ∠ABC (angle in the same segment) ∠ABC = ∠ACB (base angles of isos triangle) Thus ∠ADC = ∠ACE Hence, by the sum of triangle’s angles, ∠AEC = ∠ACD Therefore, △ACE ~ △ADC b) By the properties of same corresponding side ratio for similar triangles: AC/AD = AE/AC AD = AC x AC/AE = 324/16 = 20.25cm 15) a) ∠DAC = ∠BAE (common angle) ∠DCA = ∠BEA (angle in the same segment) Hence, by the sum of triangle’s angles, ∠ADC = ∠ABE Therefore, △DAC ~ △BAE b) By the properties of same corresponding side ratio for similar triangles: DA/BA = AC/AE AE = AC x DA/BA = 12 x 7/6 = 14cm Hence DE = 7cm 16a) ∠PBC = ∠PDA (ext. ∠ of cyclic quad.) ∠PCB = ∠PDA (corr. ∠ AD//BC) Thus ∠PBC = ∠PCB and hence △PBC is an isos. triangle. b) ∠PBC = ∠AQC = 68 deg (∠ in the same segment) ∠PCB = ∠PBC = 68 deg So, ∠BPC = 180 - 68 - 68 = 44 deg c) ∠PAD = ∠PBC (corr. ∠ AD//BC) So ∠PAD = ∠PDA and hence △PAD is an isos. triangle and PA = PD DC = PC - PD = PB - PA = 6cm 2007-01-03 23:01:51 補充： 17a) ∠CDP = ∠ABP (ext. ∠ of cyclic quad.)∠DCP = ∠BAP (ext. ∠ of cyclic quad.)∠CPD = ∠APB (common ∠)Thus, △ABP ~ △CDP (3 corresponding angles equal)b) By the properties of same corresponding side ratio for similar triangles:AB/CD = BP/DPBP = AB x DP/CD = 14 x 15/10 = 21cm
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