Question : In $\triangle$ABC and $\triangle$PQR, $\angle$B = $\angle$Q, $\angle$C = $\angle$R. M is the midpoint of side QR. If AB : PQ = 7 : 4, then $\frac{\text{area($\triangle$ ABC)}}{\text{area($\triangle$ PMR)}}$ is:
Option 1: $\frac{35}{8}$
Option 2: $\frac{49}{16}$
Option 3: $\frac{49}{8}$
Option 4: $\frac{35}{16}$
Correct Answer: $\frac{49}{8}$
Solution : In $\triangle ABC$ and $\triangle PQR$, $\angle B = \angle Q$ $\angle C = \angle R$ ⇒ $\angle A = \angle P$ ⇒ $\triangle ABC ~ \triangle PQR$ In $\triangle PQR$, Since PM is the median, it divides the triangle into two parts of equal areas. Area ($\triangle PMR) = \frac{1}{2} × \text {area} \;(\triangle PQR)$ -----------(i) Since the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides. $\frac{\text{area $(\triangle ABC$)}}{\text{area ($\triangle PMR)$}}$ = $\frac{\text{2×area $(\triangle ABC)$}}{\text{area $(\triangle PQR)$}}$ = $2×(\frac{7}{4})^2$ = $2×\frac{49}{16}$ = $\frac{49}{8}$ Hence, the correct answer is $\frac{49}{8}$.
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