Question

Prove that the ratio of the area of two similar triangle is equal to the square of the ratio of their corresponding sides.

Answer 1

Given:- Let $△ABC\sim △PQR$

$AD$ and $PS$ are corresponding medians.

To prove:- $\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}={\left(\frac{AD}{PS}\right)}^2$

Proof:- In $△ABC$

$\because AD$ is median.

$\therefore BD=CD=\frac{1}{2}BC$

Similarly, in $△PQR$

$PS$ is median.

$\therefore QS=RS=\frac{1}{2}QR$

Now,

$△ABC\sim △PQR\left(\text{Given}\right)$

$\angle B=\angle Q.....\left(1\right)\left(\text{Corresponding angles of similar triangles are equal}\right)$

$\frac{AB}{PQ}=\frac{BC}{QR}\left(\text{Corresponding sides of similar triangles are proportional}\right)$

$\Rightarrow \frac{AB}{PQ}=\frac{2BD}{2QS}(\because \text{AD and PS are medians})$

$\frac{AB}{PQ}=\frac{BD}{QS}.....\left(2\right)$

Now, in $△ABD$ and $△PQS$,

$\angle B=\angle Q\left(\text{From}\left(1\right)\right)$

$\frac{AB}{PQ}=\frac{BD}{QS}\left(\text{From}\left(2\right)\right)$

$\therefore △ABD\sim △PQS\left(\text{By SAS Property}\right)$

Therefore,

$\frac{AB}{PQ}=\frac{AD}{PS}.....\left(3\right)(\because \text{Corresponding sides of similar triangles are proportonal})$

Now,

$\because △ABC\sim △PQR$

As we know that ratio of area of similar triangles is always equal to the square of ratio of their corresponding side.

Therefore,

$\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}={\left(\frac{AB}{PQ}\right)}^2$

$\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}={\left(\frac{AD}{PS}\right)}^2\left(\text{From}\left(3\right)\right)$

Hence proved.