If triangle ABC sim triangle QRP,frac{operato | vedclass

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(A) Given,$	riangle ABC \sim 	riangle QRP$.We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of thei

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$10$

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(A) Given,$	riangle ABC \sim 	riangle QRP$.We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.Therefore,$\frac{\operatorname{ar}(	riangle ABC)}{\operatorname{ar}(	riangle QRP)} = \frac{BC^2}{RP^2}$.Given that $\frac{\operatorname{ar}(	riangle ABC)}{\operatorname{ar}(	riangle QRP)} = \frac{9}{4}$ and $BC = 15 \, cm$.Substituting the values,we get $\frac{9}{4} = \frac{(15)^2}{RP^2}$.$\frac{9}{4} = \frac{225}{RP^2}$.$RP^2 = \frac{225 	imes 4}{9} = 25 	imes 4 = 100$.$RP = \sqrt{100} = 10 \, cm$.Since $	riangle ABC \sim 	riangle QRP$,the corresponding side to $PR$ is $AC$. However,the question asks for $PR$ based on the similarity ratio. From the similarity $	riangle ABC \sim 	riangle QRP$,we have $\frac{BC}{RP} = \frac{AC}{QP} = \frac{AB}{QR}$.Thus,$RP = 10 \, cm$.

If $\triangle ABC \sim \triangle QRP$,$\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle QRP)} = \frac{9}{4}$,$AB = 18 \, cm$ and $BC = 15 \, cm$,then $PR$ is equal to (in $cm$):

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