Question

Prove that the areas of two similar acute triangles are proportional to the squares of the corresponding sides.

Answer 1

Statement: Ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: Two triangles $ABC$ and $PQR$ such that $△ABC\sim △PQR$

To prove: $\frac{area\left(ABC\right)}{area\left(PQR\right)}={\left(\frac{{AB}^2}{PQ}\right)}^{}={\left(\frac{{BC}^{}}{PQ}\right)}^2={\left(\frac{{CA}^{}}{RP}\right)}^2$

Proof: For finding the areas of the two triangles,we draw altitudes $AM$ and $PN$ of the triangles.

Now, $area\left(ABC\right)=\frac{1}{2}BC\times AM$

And $area\left(PQR\right)=\frac{1}{2}QR\times PN$

So, $\frac{area\left(ABC\right)}{area\left(PQR\right)}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times QR\times PN}=\frac{BC\times AM}{QR\times PN}.....\left(1\right)$

Now, in $△ABC$ and $△PQN$

$\angle B=\angle Q$ (As $△ABC\sim △PQR$)

and $\angle M=\angle N$ (Each$={90}^o$)

so, $△ABM\sim △PQN$ (AA similarity criterion)

Therefore $\frac{AM}{PN}=\frac{AB}{PQ}.....\left(2\right)$

Also, $△ABC\sim △PQR$

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}.......\left(3\right)$

$\frac{area\left(ABC\right)}{area\left(PQR\right)}=\frac{AB}{PQ}\times \frac{AM}{PN}$ (from (1) and (3))

Therefore $=\frac{AB}{PQ}\times \frac{AB}{PQ}={\left(\frac{AB}{PQ}\right)}^2$ (from (2))

Now using (3) we get

$\frac{area\left(ABC\right)}{area\left(PQR\right)}={\left(\frac{{AB}^2}{PQ}\right)}^{}={\left(\frac{{BC}^{}}{PQ}\right)}^2={\left(\frac{{CA}^{}}{RP}\right)}^2$