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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