Question

$\text{In the given figure,}△ABC\sim △PQR.\text{Then,}\frac{areaof△ABC}{areaof△PQR}\text{equals}$

• A. $\frac{{AB}^2}{{PQ}^2}$
• B. $\frac{{BC}^2}{{QR}^2}$
• C. $\frac{{AC}^2}{{PR}^2}$
• D. All of the above.

Answer 1

The correct option is D All of the above.

We are given two triangles ABC and PQR such that $△ABC\sim △PQR$.

For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles ABC and PQR respectively, as shown below.

Now,
$\text{area of}△ABC=\frac{1}{2}\times BC\times AM\text{and}\text{area of}△PQR=\frac{1}{2}\times QR\times PN$

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

Now, in $△ABM$ and $△PQN$,
$\angle B=\angle Q$ (As $△ABC\sim △$PQR)
and $\angle AMB=\angle PNQ={90}^{\circ }$.

So, $△ABM\sim △PQN$
( By AA similarity criterion)
$\therefore \frac{AM}{PN}=\frac{AB}{PQ}\cdots \left(2\right)$
Also, $△ABC\sim △PQR$ (Given)
So, $\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\cdots \left(3\right)$

From (1) and (3), we get,
$\frac{areaof\left(ABC\right)}{areaof\left(PQR\right)}=\frac{AB\times AM}{PQ\times PN}=\frac{AB\times AB}{PQ\times PQ}=\frac{A{B}^2}{P{Q}^2}$

Now using (3), we get,
$\frac{areaof△ABC}{areaof△PQR}=\frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{A{C}^2}{P{R}^2}$