In Delta ABC and Delta PQR,angle B = angle Q | vedclass

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(A) Given that in $\Delta ABC$ and $\Delta PQR$,$\angle B = \angle Q$ and $\angle C = \angle R$. By $AA$ similarity criterion,$\Delta ABC \sim \Delta

Vedclass · Accepted Answer

$\frac{49}{8}$

Vedclass · Answer

(A) Given that in $\Delta ABC$ and $\Delta PQR$,$\angle B = \angle Q$ and $\angle C = \angle R$. By $AA$ similarity criterion,$\Delta ABC \sim \Delta PQR$. The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore,$\frac{	ext{area}(\Delta ABC)}{	ext{area}(\Delta PQR)} = \left(\frac{AB}{PQ}ight)^2 = \left(\frac{7}{4}ight)^2 = \frac{49}{16}$. Since $M$ is the mid-point of $QR$,the median $PM$ divides $\Delta PQR$ into two triangles of equal area. Thus,$	ext{area}(\Delta PMR) = \frac{1}{2} 	imes 	ext{area}(\Delta PQR)$. Substituting this into the ratio: $\frac{	ext{area}(\Delta ABC)}{	ext{area}(\Delta PMR)} = \frac{	ext{area}(\Delta ABC)}{\frac{1}{2} 	imes 	ext{area}(\Delta PQR)} = 2 	imes \frac{	ext{area}(\Delta ABC)}{	ext{area}(\Delta PQR)} = 2 	imes \frac{49}{16} = \frac{49}{8}$.

In $\Delta ABC$ and $\Delta PQR$,$\angle B = \angle Q$ and $\angle C = \angle R$. $M$ is the mid-point of side $QR$. If $AB : PQ = 7 : 4$,then $\frac{\text{area}(\Delta ABC)}{\text{area}(\Delta PMR)}$ is:

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