Question

In a trapezium $PQRS$ with $PQ\parallel SR$ the diagonals $PR$ and $QS$ intersect at $x$, if $PQ=\frac{2}{3}RS$. Find the ratio of areas of triangles $PRQ$ and $RXS$.

Answer 1

Given:$PQ=\frac{2}{3}RS$

Solution:

$\Rightarrow \frac{PQ}{RS}=\frac{2}{3}$

In $△PQR$ and $△RXS$

$\angle SRX=\angle QPX$ (alternate angles)

Also, $\angle RSX=\angle PQX$ (alternate angles)

$\therefore △PQR\cong △RXS$ by AA similarity theorem.

Hence, $\frac{ar\left(△PQR\right)}{ar\left(△RXS\right)}={\left(\frac{2}{3}\right)}^2$

$=\frac{4}{9}$