Question

Prove: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

Answer 1

Given:$△ABC\sim △PQR$

To prove:$\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{AC}{PR}\right)}^2$

Construction:Draw $AM\perp BC$ and $PN\perp QR$

Proof:area$\left(△ABC\right)=\frac{1}{2}\times base\times height=\frac{1}{2}\times BC\times AM$ ........$\left(1\right)$

area$\left(△PQR\right)=\frac{1}{2}\times base\times height=\frac{1}{2}\times QR\times PN$ ........$\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$ we get

$\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times QR\times PN}$

$=\frac{BC\times AM}{QR\times PN}$ .......$\left(A\right)$

In $△ABM$ and $△PQN$

$\angle B=\angle Q$ as $△ABC\sim △PQR$ and angles of similar triangles are equal.

$\angle M=\angle N$ (both ${90}^{\circ }$)

$△ABM\sim △PQN$ by AA similarity

$\therefore \frac{AB}{PQ}=\frac{AM}{PN}$ (corresponding sides of similar triangles are proportional) .......$\left(B\right)$

From $\left(A\right)$

$\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}=\frac{BC\times AM}{QR\times PN}=\frac{BC}{QR}\times \frac{AB}{PQ}$ from $\left(B\right)$ .........$\left(C\right)$

Now, Given$:△ABC\sim △PQR$

$\Rightarrow \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

Putting in $\left(C\right)$ we get

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

Now, again using $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

$\Rightarrow \frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{AC}{PR}\right)}^2$