Question

In fig $△ACB\sim △APQ$. If $BC=10cm,PQ=5cm,BA=6.5cm$ and $AP=2.8cm$, find $CA$ and $AQ$. Also find the area $\left(△ACB\right)$ : area $\left(△APQ\right)$.

Answer 1

If two triangles are similar then the sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle.

As $△ACB\sim △APQ$

So $\frac{AC}{AP}=\frac{CB}{PQ}=\frac{AB}{AQ}$

$\frac{AC}{2.8}=\frac{10}{5}=\frac{6.5}{AQ}$

$\frac{AC}{2.8}=2=\frac{6.5}{AQ}$

Hence $AC=2\times 2.8=5.6cm$

Also $AQ=\frac{6.5}{2}=3.25cm$

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

$\frac{area\left(△ACB\right)}{area\left(△APQ\right)}=\frac{A{C}^2}{A{P}^2}$

$\frac{area\left(△ACB\right)}{area\left(△APQ\right)}=(\frac{5.6}{2.8}{)}^2$

$\frac{area\left(△ACB\right)}{area\left(△APQ\right)}=2^2=4$

Hence $\frac{area\left(△ACB\right)}{area\left(△APQ\right)}=4$