Question

Cho tam giác ABC vuông tại A, AB=3cm , AC = 4cm, ph giác AD. Kẻ DE vuông góc AC, DF vuông góc vs AB. Tính S hcn AEDF

Answer 1

c/m AEDF là hình vuông.

\(S_{ADB}+S_{ADC}=S_{ABC}=\frac{1}{2}AB.AC=\frac{1}{2}.3.4=6\left(cm^2\right)\)

\(\frac{S_{ADB}}{S_{ADC}}=\frac{DB}{DC}=\frac{AB}{AC}=\frac{3}{4}\Rightarrow S_{ADB}=\frac{3}{7}S_{ABC}=\frac{3}{7}.6=\frac{18}{7}\left(cm^2\right)\)

\(S_{ADB}=\frac{1}{2}DF.AB\Rightarrow\frac{18}{7}=\frac{1}{2}.DF.3\Rightarrow DF=\frac{12}{7}\left(cm\right)\)

\(S_{AEDF}=DF^2=\left(\frac{12}{7}\right)^2=\frac{144}{49}\left(cm^2\right)\)