It is given that triangle ABC sim triangle PQ | vedclass

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(B) Given that $	riangle ABC \sim 	riangle PQR$ and $\frac{BC}{QR} = \frac{1}{3}.$We know that the ratio of the areas of two similar triangles is eq

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Given that $	riangle ABC \sim 	riangle PQR$ and $\frac{BC}{QR} = \frac{1}{3}.$We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.Therefore,$\frac{\operatorname{ar}(	riangle PRQ)}{\operatorname{ar}(	riangle BCA)} = \frac{(QR)^2}{(BC)^2} = \left(\frac{QR}{BC}ight)^2.$Since $\frac{BC}{QR} = \frac{1}{3},$ it follows that $\frac{QR}{BC} = \frac{3}{1} = 3.$Thus,$\frac{\operatorname{ar}(	riangle PRQ)}{\operatorname{ar}(	riangle BCA)} = (3)^2 = 9.$

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

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