Question

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

Answer 1

In $△ANC$ and $△BAC$

$\angle ANC=\angle BAC={90}^{\circ }$

$\angle ACN=\angle BCA$

So by AA similarity, $△ANC\sim △BAC$

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\frac{area\left(△ANC\right)}{area\left(△BAC\right)}=\frac{A{C}^2}{B{C}^2}$

$\frac{area\left(△ANC\right)}{area\left(△BAC\right)}=\frac{5^2}{{12}^2}$

$\frac{area\left(△ANC\right)}{area\left(△BAC\right)}=\frac{25}{144}$

Hence the ratio of areas of $△ANC$ and $△BAC$ is $25:144$