Question

The adjoining figure shows a trapezium $ABCD$ in which side $AB$ is parallel to side $DC$. $P$ and $Q$ are mid-points of diagonals $AC$ and $BD$ respectively, $CQ$ joined and produced meets $AB$ at point $R$. Then:

• A. $PQ=\frac{1}{2}(AB-DC)$
• B. $PQ=(AB-DC)$
• C. $PQ=\frac{1}{3}(AB-DC)$
• D. $PQ=\frac{1}{4}(AB-DC)$

Answer 1

The correct option is A $PQ=\frac{1}{2}(AB-DC)$

Given: $ABCD$ is a trapezium. $AB\parallel CD$, $P$ is mid point of $AC$ and $Q$ is mid point of $BD$

Now, In $△CDQ$ and $△RBQ$

$\angle QDC=\angle QBR$ .....(Alternate angles)

$\angle CQD=\angle BQR$ ....(Vertically Opposite angles)

$DQ=BQ$ .....($Q$ is mid point of $BD$)

Thus, $△DQC\cong △BQR$ ....(ASA rule)

Hence, by CPCT, $CQ=QR$ and $DC=RB$

Now, In $\Delta ACR$, $P$ is mid-point of $AC$ and $Q$ is mid-point of $RC$

By Mid Point Theorem, we have

$PQ=\frac{1}{2}AR$

$PQ=\frac{1}{2}(AB-BR)$

$PQ=\frac{1}{2}(AB-DC)$ ....(Since, $DC=RB$)