已知:CD是RT△ABC斜边的高,AB=5,BC=4.求S△ABC:S△ACD:S△BCD

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已知:CD是RT△ABC斜边的高,AB=5,BC=4.求S△ABC:S△ACD:S△BCD
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已知:CD是RT△ABC斜边的高,AB=5,BC=4.求S△ABC:S△ACD:S△BCD
已知:CD是RT△ABC斜边的高,AB=5,BC=4.求S△ABC:S△ACD:S△BCD

已知:CD是RT△ABC斜边的高,AB=5,BC=4.求S△ABC:S△ACD:S△BCD
角ACB为直角.
AC²+BC²=AB²得:
AB=3
△ABC与△ACD相识得:
CD:BC=AD:AC=AC:AB=4:5
得:CD=(4/5)*3=12/5
AD=(4/5)*4=16/5
S△ACD=CD*AD/2=96/25
△ABC与△BCD相识得:
CD:AC=BD:BC=BC:AB=3:5
得:CD=(3/5)*4=12/5
BD=(3/5)*3=9/5
S△BCD=CD*AD/2=54/25
S△ABC=AC*BC/2=6
S△ABC:S△ACD:S△BCD
=6:96/25:54/25
=1:16/25:9/25
=25:16:9
=5²:4²:3²