Question

In the figure given below, $ABCD$ and $PQRC$ are rectangles, where $Q$ is the mid-point of $AC$, then

• A. $DP=PC$
• B. $\frac{1}{3}PR=\frac{3}{2}DB$
• C. $DP=PQ$
• D. $DP=\frac{1}{2}AC$

Answer 1

The correct option is A $DP=PC$

Given

$Q$ is the mid-point of $AC$

Also $ABCD$ and $QRCP$ are rectangles

$⟹AD||BC||QP$

Therefore , By basic prportionality theorem

If $Q$ is the midpoint and $AD||PQ$

$\therefore P$ is the midpoint of $CD$

$⟹DP=PC$