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(C) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow PQR$.According to the theorem of areas of similar triangles,

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow PQR$.According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.Therefore,$\frac{	ext{Area}(\Delta ABC)}{	ext{Area}(\Delta PQR)} = \left(\frac{AB}{PQ}ight)^2$.Substituting the given values: $\frac{36}{64} = \left(\frac{12}{PQ}ight)^2$.Taking the square root on both sides: $\sqrt{\frac{36}{64}} = \frac{12}{PQ}$.$\frac{6}{8} = \frac{12}{PQ}$.Simplifying the fraction: $\frac{3}{4} = \frac{12}{PQ}$.Cross-multiplying: $3 	imes PQ = 12 	imes 4$.$3 	imes PQ = 48$.$PQ = \frac{48}{3} = 16$.

If $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow PQR$. If $AB = 12$,$\text{Area}(\Delta ABC) = 36$,and $\text{Area}(\Delta PQR) = 64$,then $PQ = \ldots$

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