In the diagram, $P Q R S$ is a square and $M$ is the midpoint of $P S$. The ratio of the area of $\triangle Q M S$ to the area of square $P Q R S$ is
(A) $1: 6$
(B) $1: 4$
(C) $1: 3$
(D) $1: 8$
(E) $1: 2$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-051.jpg?height=339&width=361&top_left_y=1386&top_left_x=1294)
Solution 1

Since $P Q R S$ is a square, then its diagonal $S Q$ cuts it into two equal areas.

Therefore, the ratio of the area of $\triangle P Q S$ to the area of square $P Q R S$ is $1: 2$.

$\triangle P Q S$ can be viewed as having base $P S$ and height $P Q$.

$\triangle M Q S$ can be viewed as having base $M S$ and height $P Q$. (This is because $P Q$ is perpendicular to the line containing $M S$.)

Since $M S=\frac{1}{2} P S$, then the area of $\triangle M Q S$ is one-half of the area of $\triangle P Q S$.

Since the ratio of the area of $\triangle P Q S$ to the area of square $P Q R S$ is $1: 2$, then the ratio of the area of $\triangle Q M S$ to the area of square $P Q R S$ is $1: 4$.

## Solution 2

Suppose that the side length of square $P Q R S$ is $2 a$.

Then the area of square $P Q R S$ is $(2 a)^{2}=4 a^{2}$.

Since $M$ is the midpoint of side $P S$, then $P M=M S=a$.

Then $\triangle Q M S$ can be seen as having base $M S$ and height $P Q$. (This is because $P Q$ is perpendicular to the line containing $M S$.)

Since $M S=a$ and $P Q=2 a$, then the area of $\triangle Q M S$ is $\frac{1}{2}(M S)(P Q)=\frac{1}{2} a(2 a)=a^{2}$.

Therefore, the ratio of the area of $\triangle Q M S$ to the area of square

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-100.jpg?height=350&width=377&top_left_y=1863&top_left_x=1514)
$P Q R S$ is $a^{2}: 4 a^{2}$ which equals $1: 4$.

ANswer: (B)