设AB=CD=a,AD=BC=b
∵BD⊥EF
∴S⊿ABD∽S⊿BFE
a-1/b=1/a又a+b=9
解得a=3,b=6
矩形ABCD的面积为18
已知直线ABCD的周长为18,E,F分别是边AB,BC上的点,且AE=BF=1,若EF垂直BD,求这个矩形的面积.
已知直线ABCD的周长为18,E,F分别是边AB,BC上的点,且AE=BF=1,若EF垂直BD,求这个矩形的面积.
数学人气:154 ℃时间:2019-10-19 21:10:34
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