Cho gửi dùm ạ <3
Tam giác ABC vuông tại A.
\(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}\)
\(AC^2=CH.BC\Rightarrow CH=\dfrac{AC^2}{BC}\)
\(BH+CH=\dfrac{AB^2}{BC}+\dfrac{AC^2}{BC}\)
\(AH=\dfrac{BC^2}{BC}=BC=20\left(cm\right)\)
AB.AC = AH.BC
\(AB.AC=20^2\) (1) ; \(\dfrac{AB}{AC}=\dfrac{3}{4}\)(2)
Lấy pt 1 nhân pt2 ta được
\(\dfrac{AB.AC.AB}{AC}=\dfrac{20^2.3}{4}\)
\(AB^2=300\Rightarrow AB=3\sqrt{10}\) (cm)
\(AC=\dfrac{20^2}{3\sqrt{10}}=\dfrac{40\sqrt{10}}{3}\)
\(S=\dfrac{AB.AC}{2}=\dfrac{3\sqrt{10}40\sqrt{10}}{\dfrac{3}{2}}=200\)