如图,四边形ABCD,内接于圆O,AD平行BC,弧

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由条件知四边形ABCD为等腰梯形 ∠AOB = ∠COD 令∠1 = ∠AOB；∠2 = ∠AOD；∠3 = ∠BOC；圆半径为R 四弧的等式同乘R得到 2∠1 = ∠2 + ∠3 又 2∠1 + ∠2 + ∠3 = 2π 得∠2 + ∠3 = π 解法一：AD = 2Rtan(∠2/2) = 4 BC = 2Rtan(∠3/2) = 2Rtan((π - ∠2)/2) = 2Rctan(∠2/2) = 6 AD*BC = 4R^2 = 24 R = √6 AD/BC = tan^2(∠2/2) = 2/3 tan(∠2/2) = √6/3 tan(∠3/2) = ctan(∠2/2) = √6/2 梯形的高 h = R(ctan(∠2/2) + ctan(∠3/2)) = √6(√6/2 + √6/3) = 5 梯形面积 = (4+6)*5/2 = 25 解法二：设E是AD的中点,F是BC的中点,连接EO,FO △AOD与△COB是等腰三角形所以EO,FO分别平分∠2,∠3,且△EOD与△COF是直角三角形 ∠2与∠3互补,则∠EOD与∠COF互余,∠COF与 ∠FCO互余所以∠EOD = ∠FCO OD = OC = R △EOD≌△COF OE = CF = 6/2 = 3 OF = ED = 4/2 = 2 EF = OF + OE = 3 + 2 =5 梯形面积 = (4+6)*5/2 = 25