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F5 basic properties of circle
1.In the figure,QR is a diameter of the circle.PQ,PS and SR are tangents to the circle at Q,A and R respectively.PR cuts QS at B.PQ = 8cm and SR = 6cm(a)Show that PQ//SR(done)(b)Find the radius of circle(done)(c)Show that AB//PQ(d)Hence find... 顯示更多 1.In the figure,QR is a diameter of the circle.PQ,PS and SR are tangents to the circle at Q,A and R respectively.PR cuts QS at B.PQ = 8cm and SR = 6cm (a)Show that PQ//SR(done) (b)Find the radius of circle(done) (c)Show that AB//PQ (d)Hence find AB http://upload.lsforum.net/users/public/f43861%20002g144.jpg 2.In the figure,TA and TB are common tangents to the two circles.PQ is a common tangent that touches the circles at C and D respectively.If PQ is perpendicular to TB,TQ = 8cm and PQ = 6cm,find the radii if the two circles. http://upload.lsforum.net/users/public/r18594%20003u144.jpg
最佳解答:
(1)(a) ∵ ∠PQR + ∠SRQ = 90° + 90° (tangent ⊥ radius) = 180° ∴ PQ // SR (int. ∠s. supp.) (1)(b) Let S' be the projection of S on PQ SS' 2 + PS' 2 = PS 2 (Pyth. theorem) SS' 2 + (8 - 6)2 = (8 + 6)2 (tangent properties) SS' 2 = 192 SS' = 8√3 cm thus radius = 8√3 / 2 = 4√3 cm (1)(c) Since ?PBQ ~ ?RBS (AAA) QB : BS = 8 : 6 = 4 : 3 QB = 4BS / 3 QS 2 = QR 2 + SR 2 (Pyth. theorem) QS 2 = (8√3)2 + 62 = 228 QS = 2√57 cm QB + BS = 2√57 4BS / 3 + BS = 2√57 BS = 6√57 / 7 cm In ?SAB and ?SPQ, SA : SP = 6 : 14 = 3 : 7 SB : SQ = (6√57 / 7) : 2√57 = 3 : 7 ∠ASB = ∠PSQ (common ∠) ∴?SAB ~ ?SPQ (2 sides proportional, int. ∠ equal) ∴∠SAB = ∠SPQ (corr. ∠s of ~?s) ∴ AB // PQ (corr. ∠s equal) (1)(d) AB : 8 = 6 : 14 (corr. sides of ~?s) AB = 24 / 7 cm (2) Let TA touches O? and O? at A and E respectively, and TB touches O? and O? at B and F respectively, Let r? cm and r? cm be the radii of O? and O? respectively, O?E = O?F = O?D = DQ = FQ = r? O?A = O?B = O?C = CQ = QB = r? PC = PA = PQ - CQ = 6 - r? (tangent properties) PE = PD = PQ - DQ = 6 - r? (tangent properties) PT 2 = PQ 2 + QT 2 (Pyth. theorem) PT 2 = 6 2 + 8 2 = 100 PT = 10 cm TE = TF = PT - PE = 10 - (6 - r?) = 4 + r? (tangent properties) TQ = TF + FQ => 8 = 4 + r? + r? => r? = 2 O?F : O?B = TF : TB (corr. sides of ~?s) 2 : r? = 6 : 8 + r? 16 + 2r? = 6r? r? = 4 therefore, the radii of O? and O? are 2cm and 4cm respectively.
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F5 basic properties of circle
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發問:1.In the figure,QR is a diameter of the circle.PQ,PS and SR are tangents to the circle at Q,A and R respectively.PR cuts QS at B.PQ = 8cm and SR = 6cm(a)Show that PQ//SR(done)(b)Find the radius of circle(done)(c)Show that AB//PQ(d)Hence find... 顯示更多 1.In the figure,QR is a diameter of the circle.PQ,PS and SR are tangents to the circle at Q,A and R respectively.PR cuts QS at B.PQ = 8cm and SR = 6cm (a)Show that PQ//SR(done) (b)Find the radius of circle(done) (c)Show that AB//PQ (d)Hence find AB http://upload.lsforum.net/users/public/f43861%20002g144.jpg 2.In the figure,TA and TB are common tangents to the two circles.PQ is a common tangent that touches the circles at C and D respectively.If PQ is perpendicular to TB,TQ = 8cm and PQ = 6cm,find the radii if the two circles. http://upload.lsforum.net/users/public/r18594%20003u144.jpg
最佳解答:
(1)(a) ∵ ∠PQR + ∠SRQ = 90° + 90° (tangent ⊥ radius) = 180° ∴ PQ // SR (int. ∠s. supp.) (1)(b) Let S' be the projection of S on PQ SS' 2 + PS' 2 = PS 2 (Pyth. theorem) SS' 2 + (8 - 6)2 = (8 + 6)2 (tangent properties) SS' 2 = 192 SS' = 8√3 cm thus radius = 8√3 / 2 = 4√3 cm (1)(c) Since ?PBQ ~ ?RBS (AAA) QB : BS = 8 : 6 = 4 : 3 QB = 4BS / 3 QS 2 = QR 2 + SR 2 (Pyth. theorem) QS 2 = (8√3)2 + 62 = 228 QS = 2√57 cm QB + BS = 2√57 4BS / 3 + BS = 2√57 BS = 6√57 / 7 cm In ?SAB and ?SPQ, SA : SP = 6 : 14 = 3 : 7 SB : SQ = (6√57 / 7) : 2√57 = 3 : 7 ∠ASB = ∠PSQ (common ∠) ∴?SAB ~ ?SPQ (2 sides proportional, int. ∠ equal) ∴∠SAB = ∠SPQ (corr. ∠s of ~?s) ∴ AB // PQ (corr. ∠s equal) (1)(d) AB : 8 = 6 : 14 (corr. sides of ~?s) AB = 24 / 7 cm (2) Let TA touches O? and O? at A and E respectively, and TB touches O? and O? at B and F respectively, Let r? cm and r? cm be the radii of O? and O? respectively, O?E = O?F = O?D = DQ = FQ = r? O?A = O?B = O?C = CQ = QB = r? PC = PA = PQ - CQ = 6 - r? (tangent properties) PE = PD = PQ - DQ = 6 - r? (tangent properties) PT 2 = PQ 2 + QT 2 (Pyth. theorem) PT 2 = 6 2 + 8 2 = 100 PT = 10 cm TE = TF = PT - PE = 10 - (6 - r?) = 4 + r? (tangent properties) TQ = TF + FQ => 8 = 4 + r? + r? => r? = 2 O?F : O?B = TF : TB (corr. sides of ~?s) 2 : r? = 6 : 8 + r? 16 + 2r? = 6r? r? = 4 therefore, the radii of O? and O? are 2cm and 4cm respectively.
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