Giải thích các bước giải:

a.Ta có:

$S_{ABC}=\dfrac12AH\cdot BC=\dfrac12\cdot 15\cdot 7.2=54$

Xét $\Delta AHB,\Delta AHC$ có:

$\widehat{AHB}=\widehat{AHC}(=90^o)$

$\widehat{HAB}=90^o-\widehat{HAC}=\hat C$

$\to \Delta HAB\sim\Delta HCA(g.g)$

$\to \dfrac{HA}{HC}=\dfrac{HB}{HA}$

$\to AH^2=HB.HC$

$\to 7.2^2=HB(BC-HB)$

$\to 7.2^2=HB(15-HB)$

$\to 7.2^2=15HB-HB^2$

$\to HB^2-15HB+7.2^2=0$

$\to HB\in\{5.4, 9.6\}$

$\to HC=BC-HB\in\{5.4, 9.6\}$

Vì $AB<AC\to HB<HC$

$\to HB=5.4, HC=9.6$

$\to AB=\sqrt{AH^2+HB^2}=\sqrt{7.2^2+5.4^2}=9$

$\to AC=\sqrt{BC^2-AB^2}=\sqrt{15^2-9^2}=12$

$\to P_{ABC}=9+12+15=36$

b.Xét $\Delta AHB,\Delta AHC$ có:

$\widehat{AHB}=\widehat{AHC}(=90^o)$

$\widehat{HAB}=90^o-\widehat{HAC}=\hat C$

$\to \Delta HAB\sim\Delta HCA(g.g)$

$\to \dfrac{HA}{HC}=\dfrac{HB}{HA}$

$\to AH^2=HB.HC$

$\to AB^2-HB^2=HB.HC$

$\to 21^2-HB^2=HB\cdot 22.4$

$\to HB\in\{-35, 12.6\}$

Vì $HB>0$

$\to HB=12.6$

$\to BC=BH+CH=35$

$\to AC=\sqrt{BC^2-AB^2}=\sqrt{35^2-21^2}=28$

$\to P_{ABC}=AB+BC+CA=84$

     $S_{ACB}=\dfrac12AB\cdot AC =294$