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a_a;" maths 4條 1

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a_a;" maths 4條 1 [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0004-2.jpg[/IMG]

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(4) ∠DAC = 33° (Angle in alt. segment) Since AB = CD, ∠DAC = ∠ACB (Equal chord means equal angle at circumference) ∠ACB = 33° (5) ∠TBA = ∠ATB = 42° (Base angles of isos. triangle) ∠CAT = ∠TBA = 42° (Angle in alt. segment) ∠ACT = 180° - ∠CAT - ∠CTA = 96° (6a) TA = TC and TB = TD (Tangent properties) CD = TC - TD = TA - TB = AB = 12 cm (6b) PA and QB are perpendicular to AB since AB is the common tangent. So drawing a perpendicular from Q to PA and let the foot be E, we have: QE = 12 cm PE = 5 cm By Pyth. theorem, PQ = 13 cm (7) ∠ACF = ∠CFG (Alt. angles AD//EG) ∠FBC = ∠CFG (Angle in alt. segment) ∠FBC = ∠ACF Hence FB = FC since FBC is an isos. triangle (8a) OA and AB are perpendicular, so do OC and CD. Therefore, by Pyth. theorem, AB = r√3 Also since AB//CD, OAB and OCD are in fact similar triangles with side ratio = 1:2 Therefore CD = 2r√3 (8b) r = 2, AB = 2√3 Area of OAB = 2√3 cm2 Therefore area of OCD = 4√3 cm2 since side ratio of 1:2 implies area ratio of 1:4 Finally area of ABDC = 6√3 cm2

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