Question

Given $\Delta ABC \sim \Delta PQR$, if $\dfrac{{AB}}{{PQ}} = \dfrac{1}{3}$, then find $\dfrac{{ar\Delta ABC}}{{ar\Delta PQR}}$.

Answer 1

Hint: If two triangles are similar, then their corresponding sides are in the same ratio.
Let $\Delta ABC \sim \Delta PQR$
$ \Rightarrow $$\dfrac{{AB}}{{PQ}} = \dfrac{{BC}}{{QR}} = \dfrac{{CA}}{{RP}}$

Complete step-by-step answer:
We are given two triangles $ABC$ and $PQR$ such that $\Delta ABC \sim \Delta PQR$.

We have to find $\dfrac{{ar\Delta ABC}}{{ar\Delta PQR}}$.
For finding the areas of the two triangles, we draw altitude$AM$ and $PN$ of the triangles.
Area of triangle= $\dfrac{1}{2} \times base \times height$
Now, $ar\left( {\Delta ABC} \right) = \dfrac{1}{2} \times BC \times AM$
and $ar\left( {\Delta PQR} \right) = \dfrac{1}{2} \times QR \times PN$
So, $\dfrac{{ar\Delta ABC}}{{ar\Delta PQR}}$ $ = \dfrac{{\dfrac{1}{2} \times BC \times AM}}{{\dfrac{1}{2} \times QR \times PN}}$$ = \dfrac{{BC \times AM}}{{QR \times PN}}$ …. (1)
Now, in $\Delta ABM$ and $\Delta PQN$,
$\angle B = \angle Q$ (As $\Delta ABC \sim \Delta PQR$)
$\angle M = \angle N$ (Each is of $90^\circ $)
So, by AA similarity criteria, we get-
$\Delta ABM \sim \Delta PQN$
$ \Rightarrow \dfrac{{AM}}{{PN}} = \dfrac{{AB}}{{PQ}}$ …. (2)
Also, $\Delta ABC \sim \Delta PQR$
We know that if two triangles are similar, then their corresponding sides are in the same ratio.
So, $\dfrac{{AB}}{{PQ}} = \dfrac{{BC}}{{QR}} = \dfrac{{CA}}{{RP}}$ …. (3)
Therefore, $\dfrac{{ar\Delta ABC}}{{ar\Delta PQR}} = \dfrac{{AB}}{{PQ}} \times \dfrac{{AM}}{{PN}}$ [From (1) and (3)]
$ \Rightarrow \dfrac{{ar\Delta ABC}}{{ar\Delta PQR}} = \dfrac{{AB}}{{PQ}} \times \dfrac{{AB}}{{PQ}}$ [From (2)]
$ \Rightarrow \dfrac{{ar\Delta ABC}}{{ar\Delta PQR}} = {\left( {\dfrac{{AB}}{{PQ}}} \right)^2}$
But, $\dfrac{{AB}}{{PQ}} = \dfrac{1}{3}$
$ \Rightarrow \dfrac{{ar\Delta ABC}}{{ar\Delta PQR}} = {\left( {\dfrac{1}{3}} \right)^2}$
$ \Rightarrow \dfrac{{ar\Delta ABC}}{{ar\Delta PQR}} = \dfrac{1}{9}$

Therefore $\dfrac{{ar\Delta ABC}}{{ar\Delta PQR}}$ is equal to $\dfrac{1}{9}$

Note: An another method to solve this question is a theorem which states that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore, For given $\Delta ABC \sim \Delta PQR$;
$\dfrac{{ar\Delta ABC}}{{ar\Delta PQR}} = {\left( {\dfrac{{AB}}{{PQ}}} \right)^2}$
But $\dfrac{{AB}}{{PQ}} = \dfrac{1}{3}$
$ \Rightarrow \dfrac{{ar\Delta ABC}}{{ar\Delta PQR}} = {\left( {\dfrac{1}{3}} \right)^2}$
$ \Rightarrow \dfrac{{ar\Delta ABC}}{{ar\Delta PQR}} = \dfrac{1}{9}$