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[几何] 矩形的三个顶点处的直角三角形的面积为3,5,8,求矩形面积

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其妙 Posted 2014-8-26 21:47 |Read mode
Last edited by hbghlyj 2025-4-3 03:09如图,E、F分别是矩形ABCD 的BC边和CD边上的点,且$\S{ABE}=3$, $\S{ECF}=8$, $\S{ADF}=5$,则矩形ABCD的面积为__

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地狱的死灵 Posted 2014-8-26 22:09
设AD=a,AB=b
BE=6/b,DF=10/a
(b-10/a)(a-6/b)=16,
即可解出ab

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