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[問題]: A vertex of a Tetrahedron is called trirectangular if all edges of the vertex are perpendicular to each other. Let ABCD be a tetrahedron with A trirectangular. Area [ABC] = 15, [ACD]=16, [ABD]=240. What is the area of [BCD]?
問題來源:http://groups.wenxuecity.com/groups/bbs.php?act=bbsview&gid=1731&basecode=968444
Picture was drawn by ca981.
[學一個公式] 有類似戈股定理的關係:
[BCD]^2 = [ABC] ^2+ [ACD]^2 + [ABD]^2
公式證明 (by student99 http://groups.wenxuecity.com/groups/bbs.php?act=bbsview&gid=1731&basecode=968444)
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我想6700417所說的公式可以這樣證明: 假設三個麵積分別是a, b, c. 那麽三個垂直邊則分別是:sqrt(2abc)/a, sqrt(2abc)/b, sqrt(2abc)/c. 所以ABCD的體積是sqrt(2abc)/3. 假設A 是原點,三條垂直邊分別是x, y, z軸。那麽平麵BCD 的方程就是:ax + by + cz = sqrt(2abc), 根據點到平麵的公式,我們可以求出A 到 BCD 的距離:sqrt(2abc)/sqrt(a^2 + b^2 + c^2) 由此得出:[BCD] * sqrt(2abc)/sqrt(a^2 + b^2 + c^2 ) * 1/3 = sqrt(2abc)/3. 所以 [BCD]^2 = a^2 + b^2 + c^2 . |