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CB//OA CB=1cm 求OABC面積
Aug 21st 2013, 04:21

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sol: 先OB兩點連線,劃分成兩個正三角形 正三角形面積為: √3a^2 / 4,a為邊長 四邊形面積:[ √3*(1^2) / 4]*2= √3/ 2 故答案為(A)...
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