ABC is a triangle in which ∠A = 90°, AN ⊥ BC,
Question:

$\mathrm{ABC}$ is a triangle in which $\angle \mathrm{A}=90^{\circ}, \mathrm{AN} \perp \mathrm{BC}, \mathrm{BC}=12 \mathrm{~cm}$ and $\mathrm{AC}=5 \mathrm{~cm}$. Find the ratio of the area of $\triangle \mathrm{ANC}$ and $\triangle \mathrm{ABC}$.

Solution:

Given: In $\triangle \mathrm{ABC}, \angle A=90^{\circ}, \mathrm{AN} \perp \mathrm{BC}, \mathrm{BC}=12 \mathrm{~cm}$ and $\mathrm{AC}=5 \mathrm{~cm}$.

TO FIND: Ratio of the triangles $\triangle A N C$ and $\triangle A B C$.

In ∆ANC and ∆ABC,

$\angle \mathrm{ACN}=\angle \mathrm{ACB}$

$\angle \mathrm{A}=\angle \mathrm{ANC}$

$\angle \mathrm{A}=\angle \mathrm{ANC} \quad\left(90^{\circ}\right.$ each $)$

$\therefore \triangle A N C \sim \triangle A B C \quad$ (AA Similarity)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\therefore \operatorname{Ar} \triangle A N C A r \Delta A B C=A C B C 2 \Rightarrow A r \triangle A N C A r \triangle A B C=5 \mathrm{~cm} 12 \mathrm{~cm} 2$

$\Rightarrow \frac{A r(\Delta A N C)}{A r(\Delta A B C)}=\frac{25}{144}$