矩形ABCD中,AB=2,BC=3,M是BC的中点
∴BM=BC/2=3/2=1.5,在Rt△ABM中,AM=√(AB^2+BM^2)=√(2^2+1.5^2)=2.5
∵∠BAM=90°-∠DAM=∠ADP ∠B=∠AMP=90°
∴△ABM∽△DPA
∴AB/PD=AM/AD
∴PD=AB*AD/AM=2*3/2.5=1.2
即D点到AM的距离就是1.2
矩形ABCD中,AB=2,BC=3,M是BC的中点,连接AM,作DP垂直与AM交AM于点P.求D点到AM的距离
矩形ABCD中,AB=2,BC=3,M是BC的中点,连接AM,作DP垂直与AM交AM于点P.求D点到AM的距离
数学人气:843 ℃时间:2019-08-20 21:04:46
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