Giải thích các bước giải:
Xét $\Delta AHB,\Delta AHC$ có:
$\widehat{AHB}=\widehat{AHC}(=90^o)$
$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to \Delta AHB\sim\Delta CHA(g.g)$
$\to \dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to AH^2=HB.HC=900\to AH=30$
$\to AB=\sqrt{AH^2+HB^2}=5\sqrt{61}, AC=\sqrt{AH^2+HC^2}=6\sqrt{61}$
Mà $BC=BH+HC=61$
$\to S=\dfrac12AB\cdot AC=915, P=AB+BC+CA=11\sqrt{61}+61$