Question

In the given figure, $△ABC\sim △$PQR and AM and PN are medians of $△$ABC and $△$PQR respectively. Then $\frac{Areaof△ABC}{Areaof△PQR}$ = _____

• A. $\frac{{AB}^2}{{PQ}^2}$
• B. $\frac{{AM}^2}{{PN}^2}$
• C. (a) and (b) above.
• D. $\frac{{PQ}^2}{{QR}^2}$

Answer 1

The correct option is C (a) and (b) above.

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

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

$\frac{AB}{PQ}=\frac{2BM}{2QN}=\frac{BM}{QN}$

$\Rightarrow \frac{AB}{PQ}=\frac{BM}{QN}$

$\angle$ B = $\angle$ Q

$\therefore △ABM\sim △$ PQN (by SAS similarity criterion)

$\Rightarrow \frac{AB}{PQ}=\frac{AM}{PN}$

So , $\frac{areaof△ABC}{areaof△PQR}=\frac{A{B}^2}{P{Q}^2}=\frac{A{M}^2}{P{N}^2}$