CEMC Fermat, 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 $2 \times 200 = 400$.

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 $\text{x gt; 0}$.

Thus, the side length of square $PQRS$ is $2x=20$ and so the area of square $PQRS$ is $20^2 = 400$.