∆ABC is right-angled at A and AD ⊥ BC.

In $\triangle A B C$ and $\triangle A D C$, we have:

$\angle B A C=\angle A D C=90^{\circ}$

$\angle A C B=\angle A C D \quad$ (common)

By AA similarity, we can conclude that $\triangle B A C \sim \triangle A D C$.

Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.

$\therefore \frac{\operatorname{ar}(\Delta B A C)}{\operatorname{ar}(\Delta A D C)}=\frac{B C^{2}}{A C^{2}}$

$\Rightarrow \frac{\operatorname{ar}(\Delta B A C)}{\operatorname{ar}(\Delta A D C)}=\frac{13^{2}}{5^{2}}$

$=\frac{169}{25}$

Hence, the ratio of areas of both the triangles is 169 : 25