Giải thích các bước giải:
Ta có:
$AH\perp BC\to BH^2=AB^2-AH^2=\dfrac{324}{25}$
$\to BH=\dfrac{18}5$
Xét $\Delta ABH,\Delta ABC$ có:
Chung $\hat B$
$\widehat{AHB}=\widehat{BAC}(=90^o)$
$\to\Delta BAH\sim\Delta BCA(g.g)$
$\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$
$\to BC=\dfrac{BA^2}{BH}=10$
$\to AC=\sqrt{BC^2-AB^2}=8$
$\to S_{ABC}=\dfrac12AB\cdot AC=24, P_{ABC}=AB+BC+CA=24$