Problem 527

Solution 1

Since $PQRS$ is a square, then its diagonal $SQ$ cuts it into two equal areas. Therefore, the ratio of the area of $\triangle PQS$ to the area of square $PQRS$ is $1:2$. $\triangle PQS$ can be viewed as having base $PS$ and height $PQ$. $\triangle MQS$ can be viewed as having base $MS$ and height $PQ$. (This is because $PQ$ is perpendicular to the line containing $MS$.) Since $MS = \frac{1}{2}PS$, then the area of $\triangle MQS$ is one-half of the area of $\triangle PQS$. Since the ratio of the area of $\triangle PQS$ to the area of square $PQRS$ is $1:2$, then the ratio of the area of $\triangle QMS$ to the area of square $PQRS$ is $1:4$. Solution 2 Suppose that the side length of square $PQRS$ is $2a$. Then the area of square $PQRS$ is $(2a)^2 = 4a^2$. Since $M$ is the midpoint of side $PS$, then $PM=MS=a$. Then $\triangle QMS$ can be seen as having base $MS$ and height $PQ$. (This is because $PQ$ is perpendicular to the line containing $MS$.) Since $MS=a$ and $PQ=2a$, then the area of $\triangle QMS$ is $\frac{1}{2}(MS)(PQ) = \frac{1}{2}a(2a)=a^2$. Therefore, the ratio of the area of $\triangle QMS$ to the area of square $PQRS$ is $a^2:4a^2$ which equals $1:4$.