In the diagram, $P Q R S$ is a trapezoid with an area of 12 . $R S$ is twice the length of $P Q$. The area of $\triangle P Q S$ is
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-066.jpg?height=214&width=466&top_left_y=1086&top_left_x=1247)
Solution 1

Since $P Q$ is parallel to $S R$, then the height of $\triangle P Q S$ (considering $P Q$ as the base) and the height of $\triangle S R Q$ (considering $S R$ as the base) are the same (that is, the vertical distance between $P Q$ and $S R$ ).

Since $S R$ is twice the length of $P Q$ and the heights are the same, then the area of $\triangle S R Q$ is twice the area of $\triangle P Q S$.

In other words, the area of $\triangle P Q S$ is $\frac{1}{3}$ of the total area of the trapezoid, or $\frac{1}{3} \times 12=4$.

## Solution 2

Draw a line from $Q$ to $T$, the midpoint of $S R$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-290.jpg?height=233&width=463&top_left_y=504&top_left_x=885)

Since $S R=2(P Q)$ and $T$ is the midpoint of $S R$, then $P Q=S T=T R$.

We consider $P Q, S T$ and $T R$ as the bases of $\triangle P Q S, \triangle S T Q$ and $\triangle T R Q$, respectively.

Using these three segments as the bases, each of these triangles has the same height, since $P Q$ is parallel to $S R$.

Since $P Q=S T=T R$ and these triangles have the same height, then the three triangles each have the same area.

The trapezoid is thus cut into three triangles of equal area.

Therefore, the area of $\triangle P Q S$ is one-third of the area of entire trapezoid, or $\frac{1}{3} \times 12=4$.

ANSWER: (B)