Question

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

Answer 1

Hint: If two triangles are similar then their corresponding sides are proportional to each other. We can also say the ratio of corresponding sides in two similar triangles are equal.

Complete step-by-step answer:
Let we have two similar triangle ABC and PQR as below:

As we know to find the area we need the height of the triangle. So we can draw a perpendicular AD from A to BC and PS from P to QR.

In triangle ABD and triangle PQS
$\angle B=\angle Q$ $\left\{ \because \Delta ABC\sim \Delta PQR \right\}$
$\angle ADB=\angle PSQ={{90}^{\circ }}$ (By construction)
As two angles are equal so the third angle of both triangles should also be equal.
$\angle BAD=\angle QPS$
So by AAA similarity
$\Delta ABD\sim \Delta PQS$
So we can say the ratio of corresponding sides should be equal. So we can write
$\dfrac{AB}{PQ}=\dfrac{AD}{PS}$ …………………………………………..(i)
Now we can write area of triangle ABC as
$Area(\Delta ABC)=\dfrac{1}{2}\times AD\times BC$ ………………………………..(ii)
$Area(\Delta PQR)=\dfrac{1}{2}\times PS\times QR$…………………………………….(iii)
On dividing equation (ii) and (iii)
$\dfrac{Area(\Delta ABC)}{Area(\Delta PQR)}=\dfrac{AD\times BC}{PS\times QR}$
By using equation (i) we can write
$\dfrac{Area(\Delta ABC)}{Area(\Delta PQR)}=\dfrac{AB\times BC}{PQ\times QR}$……………………………..(iv)
As given $\Delta ABC\sim \Delta PQR$
So this ratio of corresponding sides should be equal. So we can write
$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AC}{PR}$
Hence we can use this value in equation (iv). So we can write
$\dfrac{Area(\Delta ABC)}{Area(\Delta PQR)}=\dfrac{AB\times AB}{PQ\times PQ}={{\left( \dfrac{AB}{PQ} \right)}^{2}}$
Similarly we can write
$\dfrac{Area(\Delta ABC)}{Area(\Delta PQR)}={{\left( \dfrac{AB}{PQ} \right)}^{2}}={{\left( \dfrac{BC}{QR} \right)}^{2}}={{\left( \dfrac{AC}{PR} \right)}^{2}}$
Hence we can say the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

Note: In general area(A) of any triangle is
$A=\dfrac{1}{2}\times base\times height$
That’s why we need to construct perpendicular triangles for height.