Ta có: $\dfrac{AB}{AC} = \dfrac{3}{5}$

$\Rightarrow AB = \dfrac{3}{5}AC$

Áp dụng định lý Pytago, ta được:

$BC^2 = AB^2 + AC^2$

$\Leftrightarrow BC^2 = \left(\dfrac{3}{5}AC\right)^2 + AC^2$

$\Leftrightarrow BC^2 = \dfrac{34}{25}AC^2$

$\Rightarrow BC = \dfrac{AC\sqrt{34}}{5}$

Áp dụng hệ thức lượng, ta được:

$AC^2 = CH.BC$

$\Leftrightarrow AC^2 = 12.\dfrac{AC\sqrt{34}}{5}$

$\Leftrightarrow AC = \dfrac{12\sqrt{34}}{5} \, cm$

$\Rightarrow AB = \dfrac{12\sqrt{34}}{5}.\dfrac{3}{5} = \dfrac{36\sqrt{34}}{25} \, cm$

$\Rightarrow BC = \dfrac{12\sqrt{34}}{5}.\dfrac{34}{5} = \dfrac{408}{25} \, cm$

$\begin{cases}P_{ABC} = AB + AC + BC =\dfrac{36\sqrt{34}}{25} + \dfrac{12\sqrt{34}}{5} + \dfrac{408}{25} = \dfrac{408 + 96\sqrt{34}}{25} \, (cm)\\S_{ABC} = \dfrac{1}{2}AB.AC = \dfrac{1}{2}.\dfrac{36\sqrt{34}}{25}.\dfrac{12\sqrt{34}}{5} = \dfrac{7344}{125} = 58,752 \, (cm^2)\end{cases}$

Do $\dfrac{AB}{AC}=\dfrac{3}{5}$

$⇒\dfrac{AB}{3}=\dfrac{AC}{5}=k$
$⇒AB=3k;AC=5k$

$⇒BC=\sqrt[]{AB^2+AC^2}=\sqrt[]{(3k)^2+(5k)^2}=\sqrt[]{34k^2}=\sqrt[]{34}.k$

Áp dụng hệ thức lượng trong tam giác vuông $ABC$ có:

$AC^2=BC.CH$
$⇔25k^2=\sqrt[]{34}.k.12$

$⇒25k=\sqrt[]{34}.12$

$⇒k=\dfrac{\sqrt[]{34}.12}{25}$

$⇒AB=3.\dfrac{\sqrt[]{34}.12}{25};AC=5.\dfrac{\sqrt[]{34}.12}{25};BC=\sqrt[]{34}.\dfrac{\sqrt[]{34}.12}{25}$

Nên chu vi tam giác $ABC=AB+AC+BC=3.\dfrac{\sqrt[]{34}.12}{25}+5.\dfrac{\sqrt[]{34}.12}{25}+\sqrt[]{34}.\dfrac{\sqrt[]{34}.12}{25}=38,71...(cm2)$
$S_{ABC}=\dfrac{AB.AC}{2}=\dfrac{3.\dfrac{\sqrt[]{34}.12}{25}.5.\dfrac{\sqrt[]{34}.12}{25}}{2}=58,752(cm^2)$