如图,平行四边形ABCD的对角线AC,BD交于点O,AC⊥AB,AB=2根号5,且AC:BD=2:3(1)求AC的长;(2)求△AOD的面积
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![如图,平行四边形ABCD的对角线AC,BD交于点O,AC⊥AB,AB=2根号5,且AC:BD=2:3(1)求AC的长;(2)求△AOD的面积](/uploads/image/z/4769541-45-1.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B9%B3%E8%A1%8C%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%E7%9A%84%E5%AF%B9%E8%A7%92%E7%BA%BFAC%2CBD%E4%BA%A4%E4%BA%8E%E7%82%B9O%2CAC%E2%8A%A5AB%2CAB%3D2%E6%A0%B9%E5%8F%B75%2C%E4%B8%94AC%EF%BC%9ABD%3D2%3A3%281%29%E6%B1%82AC%E7%9A%84%E9%95%BF%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E2%96%B3AOD%E7%9A%84%E9%9D%A2%E7%A7%AF)
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如图,平行四边形ABCD的对角线AC,BD交于点O,AC⊥AB,AB=2根号5,且AC:BD=2:3(1)求AC的长;(2)求△AOD的面积
如图,平行四边形ABCD的对角线AC,BD交于点O,AC⊥AB,AB=2根号5,且AC:BD=2:3
(1)求AC的长;(2)求△AOD的面积
如图,平行四边形ABCD的对角线AC,BD交于点O,AC⊥AB,AB=2根号5,且AC:BD=2:3(1)求AC的长;(2)求△AOD的面积
作CE∥BD交AB的延长线于E
显然BECD是平行四边形
∴EC=BD,BE=CD=AB
在Rt△AEC中,AE=2AB=4√5,AC:CE=2:3
于是可设AC=2x,CE=3x用勾股定理列方程得到AC=(8/13)√65
平行四边形ABCD的面积=2S△ABC=S△AEC=(1/2)AC ·AE=(160/13)√13
S△AOD=(1/4)平行四边形ABCD的面积=(40/13)√13