標題:
急! Properties of Circles 12q15
發問:
請詳細步驟教我計以下四條 : (不要綱址回答) 圖片參考：http://imgcld.yimg.com/8/n/HA00997022/o/701203030014413873444360.jpg
34)(64 - 25)2 + ?2 = 952 ?2 = 952 - 392 ? = √7504 ? = 87(nearest integer) 1)∠ONB = 90° (line joining centre to mid-pt. of chord⊥chord) ∠CBN = 90° (property of rectangle) ∴ BC // ON (alt.∠s equal)∠BCP = ∠CBN = 90° (property of rectangle) ∴ CP // BN (int.∠s supp.) ∴ ∠CPN = ∠BNO = 90° (corr.∠s, CP // BN) CPD is a tangent to circle O. (converse of tangent⊥radius) ∴CPD touches the circle at P. 2)Join OO' , meeting AB at C.∠OCA =∠O'CB (vert. opp. ∠s) ∠COA = ∠CO'B (alt.∠s, OA // BO') ∴ ΔCOA ~ ΔCO'BSo ∠O'BA = ∠OAB = 90° (tangent⊥radius) ∴ AB is a tangent to circle O' (converse of tangent⊥radius) 6)∠CAB = ∠CAB (common) AC : AB = 1 : 1 (given) AE : AD = 1 : 1 (given) ∴ ΔABC ~ ΔADE (S.A.S.)So ∠CED = ∠ACB = ∠ABC (base ∠s, isos.Δ)∴ BCDE is a cyclic quadrilateral. (ext.∠= int. opp.∠)
其他解答:
急! Properties of Circles 12q15
發問:
請詳細步驟教我計以下四條 : (不要綱址回答) 圖片參考：http://imgcld.yimg.com/8/n/HA00997022/o/701203030014413873444360.jpg
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最佳解答:34)(64 - 25)2 + ?2 = 952 ?2 = 952 - 392 ? = √7504 ? = 87(nearest integer) 1)∠ONB = 90° (line joining centre to mid-pt. of chord⊥chord) ∠CBN = 90° (property of rectangle) ∴ BC // ON (alt.∠s equal)∠BCP = ∠CBN = 90° (property of rectangle) ∴ CP // BN (int.∠s supp.) ∴ ∠CPN = ∠BNO = 90° (corr.∠s, CP // BN) CPD is a tangent to circle O. (converse of tangent⊥radius) ∴CPD touches the circle at P. 2)Join OO' , meeting AB at C.∠OCA =∠O'CB (vert. opp. ∠s) ∠COA = ∠CO'B (alt.∠s, OA // BO') ∴ ΔCOA ~ ΔCO'BSo ∠O'BA = ∠OAB = 90° (tangent⊥radius) ∴ AB is a tangent to circle O' (converse of tangent⊥radius) 6)∠CAB = ∠CAB (common) AC : AB = 1 : 1 (given) AE : AD = 1 : 1 (given) ∴ ΔABC ~ ΔADE (S.A.S.)So ∠CED = ∠ACB = ∠ABC (base ∠s, isos.Δ)∴ BCDE is a cyclic quadrilateral. (ext.∠= int. opp.∠)
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