Question

In Figure , QS and PT are parallel, and the lengths of segmentsPQ and QR are as marked. If the area of $△$ $QRS$ is $x$, find the area of $△$ $PRT$ in terms of $x$.

• A. $\frac{3x}{2}$
• B. $\frac{9x}{4}$
• C. $\frac{5x}{2}$
• D. $3x$
• E. $\frac{25x}{4}$

Answer 1

The correct option is E $\frac{25x}{4}$

Given, $QS$ and $PT$ are parallel.

Since $\frac{RQ}{QP}=\frac{2}{3}$, the ratio of $\frac{RS}{ST}=\frac{2}{3}$
Area of triangle $RQS$ is equal to $\frac{1}{2}\times RQ\times RS\times \sin R=x$
Area of triangle $PRT=\frac{1}{2}\times PR\times RT\times \sin R$

$=\frac{1}{2}\times \frac{5}{2}\times QR\times \frac{5}{2}\times RS\times \sin R$

$=\frac{25}{4}\times x$