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求區域的面積

Apr 30th 2013, 03:11

切面為正六邊形，其頂點必在立方體的稜中點上，如下圖A,B
檢視圖片
OAB為三角形，因為正方體邊長=2，一半邊長=1，所以AB=√2，OH=(√6)/2，△OAB面積=(√3)/2，得正六邊形面積=6*(√3)/2=3√3。

正方體內切球半徑R=1/2正方體邊長=1
內切球圓心和正六邊形形心同為O，故所切圓必為大圓，此圓面積=πR^2=π

故所求面積= 3√3 - π

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