貨運百科

此文章來自奇摩知識+如有不便請留言告知

標題:

中4maths 7C 4,6,7,8

發問:

中4maths 7C 4,6,7,8 [IMG]http://i164.photobucket.com/albums/u4/ming21ki/P1010196.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/P1010199.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/P1010201.jpg[/IMG]

最佳解答:

(4) By Pyth. theorem, O'A = √(42 + 32) = 5 (6) By tangent properties, we have: OR is perp. to RS OP is perp. to PQ MQ is perp. to PQ MS is perp. to RS Also let the point of intersection between PQ and RS be T, then PORT and SMQT are both cyclic quad. Hence x = y = 60° (Ext. angle of cyclic. quad) (7) Join PD, then PD and QE are perpendicular to DE by tangent properties and hence PD//QE. So, ∠DPF = ∠FQE (Alt. ∠s, PD//QE) ∠PDF = ∠QEF (proved) PD = QE (Given) △PDF and △QEF are congurent (ASA) So DF = FE = 8 cm PF = FQ = (1/2)√(62 + 82) = 5 cm (8a) Let r be the radius of the smaller circle. Also, join AB and BQ, then BQ = r and AB = 9 + r Draw a perpendicular from B to AP and let the foot be C, then BC = PQ = 12 cm and AC = 9 - r By Pyth theorem, AB2 = AC2 + BC2 (9 + r)2 = (9 - r)2 + 122 81 + 18r + r2 = 81 - 18r + r2 + 144 36r = 144 r = 4 cm (b) ABQP is in fact a trapezium and therefore its area is: (1/2) × (9 + 4) × 12 = 78 cm2.

其他解答:

PF = FQ = (1/2)√(62 + 82) = 5 cm This is wrong, the correct solution is PF = FQ = √(6^2 + 8^2) = 10 cm and PQ = 2 PF = 20 cm|||||4. OO' = 4-1 = 3cm O'A = (4^2 - 3^2)^1/2 = 5 cm 6. angle ORS = angle OPQ = 90 degree y = 60 degree angle MQP = angle MSR = 90 degree x = 606A560DB68EC65D60