Delta PQR sim Delta XYZ for the correspondenc | vedclass

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(D) Given that $\Delta PQR \sim \Delta XYZ$ under the correspondence $PQR \leftrightarrow XZY$.According to the theorem of the ratio of areas of two s

Vedclass · Accepted Answer

$392$

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(D) Given that $\Delta PQR \sim \Delta XYZ$ under the correspondence $PQR \leftrightarrow XZY$.According to the theorem of the ratio of areas of two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.Therefore,$\frac{	ext{Area}(\Delta PQR)}{	ext{Area}(\Delta XYZ)} = \left( \frac{PQ}{XZ} ight)^2$.Given $\frac{PQ}{XZ} = \frac{7}{3}$ and $	ext{Area}(\Delta XYZ) = 72$.Substituting these values,we get $\frac{	ext{Area}(\Delta PQR)}{72} = \left( \frac{7}{3} ight)^2$.$\frac{	ext{Area}(\Delta PQR)}{72} = \frac{49}{9}$.$	ext{Area}(\Delta PQR) = \frac{49}{9} 	imes 72$.$	ext{Area}(\Delta PQR) = 49 	imes 8 = 392$.

$\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XZY$. If $\frac{PQ}{XZ} = \frac{7}{3}$ and the area of $\Delta XYZ$ is $72$,find the area of $\Delta PQR$.

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