Question

In the given figure , QP || XY, QX || PR and PY || QR. prove that ar($△$QXR) = ar($△$PQR)

Answer 1

$PQ\parallel XT$ [ Since, $XT$ is a part of $XY$ ]

Similarly, $QX\parallel PT$ [ $PT$ is a part of $PR$ ]

$\therefore$ $PQXT$ is a parallelogram

Similarly, $PQ\parallel SY$

$QS\parallel PY$

$\therefore$ $PQSY$ is a parallelogram.

Now, parallelogram $PQSY$ and parallelogram $PQXT$ lie on the same base $XY$ and between same parallel lines $PQ$ and $XY$.

$\therefore$ $ar\left(PQSY\right)=ar\left(PQXT\right)$ ----- ( 1 )

Now, $△QXR$ and parallelogram $PQXT$ lie on the same base $QX$ and between same parallel lines $QX$ and $PR$

$\therefore$ $ar\left(QXR\right)=\frac{1}{2}ar\left(PQXT\right)$ ----- ( 2 )

Similarly, $△PYR$ and parallelogram $PQSY$ lie on the same base $PY$ and between same parallel lines $PY$ and $QR$

$\therefore$ $ar\left(PYR\right)=\frac{1}{2}ar\left(PQSY\right)$ ------ ( 3 )

From ( 1 ), ( 2 ) and ( 3 )

$\Rightarrow$ $ar\left(QXR\right)=ar\left(PYR\right)$