Question

∆ABC is right-angled at A and AD ⊥ BC. If BC = 13 cm and AC = 5 cm, find the ratio of the areas of ∆ABC and ∆ADC.

Answer 1

In $△ABC\mathrm{and}△ADC$, we have:
$$\begin{gathered} \angle BAC=\angle ADC=90° \\ \angle ACB=\angle ACD\left(\mathrm{common}\right) \end{gathered}$$
By AA similarity, we can conclude that $△BAC~△ADC$.
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.

$$\begin{gathered} \therefore \frac{\mathrm{ar}(∆BAC)}{\mathrm{ar}(∆ADC\hspace{0.17em})}=\frac{B{C}^2}{A{C}^2} \\ \Rightarrow \frac{\mathrm{ar}(∆BAC)}{\mathrm{ar}(∆ADC\hspace{0.17em})}=\frac{{13}^2}{5^2} \\ =\frac{169}{25} \end{gathered}$$

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