, 2012

Solution 1

Draw a line from $M$ to $T$ on $SR$ so that $MT$ is parallel to $QR$. Then $MTRQ$ is a rectangle. This means that the area of $\triangle MQR$ is half of the area of rectangle $MTRQ$. Thus, the area of $MTRQ$ is $2\times 100 = 200$. Since $M$ is the midpoint of $PQ$ and $PQRS$ is a square, then $T$ is the midpoint of $SR$. This means that the area of $MTRQ$ is half of the area of $PQRS$. Therefore, the area of $PQRS$ is $2 \times 200 = 400$. [[IMAGE0]] Solution 2 Suppose that the side length of square $PQRS$ is $2x$. Since $M$ is the midpoint of $PQ$, then $MQ = \frac{1}{2}(2x)=x$. Since $PQRS$ is a square, then $\triangle MQR$ is right-angled at $Q$. Therefore, the area of $\triangle MQR$ is $\frac{1}{2}(MQ)(QR) = \frac{1}{2}(x)(2x)=x^2$. Since the area of $\triangle MQR$ is 100, then $x^2 = 100$, and so $x = 10$, since $x > 0$. Thus, the side length of square $PQRS$ is $2x=20$ and so the area of square $PQRS$ is $20^2 = 400$.