如图，AB是⊙O的直径，弦CD⊥AB，∠CDB=30°，CD=214，则阴影部分图形的面积为为9π329π32．

Question

如图，AB是⊙O的直径，弦CD⊥AB，∠CDB=30°，CD=2(1)/(4)，则阴影部分图形的面积为为

(9π)/(32)

(9π)/(32)

．

Answer 1

如图，假设线段CD、AB交于点E．
∵AB是⊙O的直径，弦CD⊥AB，
∴CE=ED=(9)/(8)．
又∵∠CDB=30°，
∴∠COE=2∠CDB=60°，∠OCE=30°，
∴OE=CE•cot60°=(9)/(8)×(√(3))/(3)=(3√(3))/(8)，OC=2OE=(3√(3))/(4)，
∴S_阴影=S_扇形OCB-S_△COE+S_△BED
=(60π×OC2)/(360)-(1)/(2)OE•EC+(1)/(2)BE•ED
=(9π)/(32)-(1)/(2)OE•EC+(1)/(2)OE•EC
=(9π)/(32)．
故答案是：(9π)/(32)．