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$\text{It is given that}$ $\Delta ABC\sim \Delta PQR$ , with $\frac{BC}{QR}=\frac{1}{3}.$ $\text{Then, }$ $\frac{ar\left ( PRQ \right )}{ar\left ( BCA \right )}$ $\text{is equal to }$

(a) $9$

(b) $3$

(c) $\frac{1}{3}$

(d) $\frac{1}{9}$

Answers (1)

Answer : [A]

Given : $\Delta ABC\sim \Delta PQR$

$\frac{BC}{QR}=\frac{1}{3}$

It is given that $\Delta ABC\sim \Delta PQR$

$\therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}\; \; \; \; \; \; ...(1)$ $\therefore \frac{ar(PRQ)}{ar(BCA)}=\frac{PR^{2}}{AC^{2}}$

[Q If triangles are similar their areas are proportional to the squares of the corresponding sides.]

$\Rightarrow \frac{ar(PRQ)}{ar(BCA)}=\frac{QR^{2}}{BC^{2}}\; \; \; \; \; \; \; ...(2)$ $\text{{using equation (1)}}$

$\frac{QR}{BC}=\frac{3}{1} \; \; \; \; \; \; \; \; \; \left \{ \because \frac{BC}{QR}=\frac{1}{3} \right \}$

$\text{Put}$ $\frac{QR}{BC}=\frac{3}{1}$ $\text{in (2) we get}$

$\therefore \frac{ar(PRQ)}{ar(BCA)}=\frac{3^{2}}{1^{2}}=\frac{9}{1}=9$

Hence option A is correct.

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