Question

In a rectangle $ABCD$, $P$, $Q$, $R$ and $S$ are the midpoints of the sides $AB$, $BC$, $CD$ and $DA$ respectively. Find the area of $PQRS$ in terms of area of $ABCD$.

Answer 1

Given: $ABCD$ is a rectangle, $P,Q,R$ and $S$ are the midpoints of $AB,BC,CD$ and $DA$ respectively. $PQ,QR,RS$ and $SP$ are joined.

To find: the areas $\left(PQRS\right)$ in terms of area $\left(ABCD\right)$

$S$ and $Q$ are the mid-point of $AD$ and $BC$.

Therefore, $SQ$ divides rectangles $ABCD$ into two rectangles equal in area.

Area$\left(ABQS\right)=$ area$\left(SQCD\right)=$$\frac{1}{2}$area$\left(ABCD\right)$

$\Delta SRQ$ and parallelogram (rectangles) $SQCD$ stand on the same base $SQ$ and are between the same parallel.

Therefore, Area$\Delta SRQ=$$\frac{1}{2}$ area$\left(SQCD\right)$

Similarly Area$\Delta SPQ=$$\frac{1}{2}$ area$\left(SABQ\right)$

Adding, area $\Delta SRQ+$ area $\Delta SPQ=$$\frac{1}{2}$ area $\left(SQCD\right)+$$\frac{1}{2}$ area $\left(SABQ\right)$

i.e., area $\left(PQRS\right)=$$\frac{1}{2}$ area $\left(ABCD\right)$

Hence, area $\left(PQRS\right)=$$\frac{1}{2}$ area $\left(ABCD\right)$.