In the diagram, $R$ is on $Q S$ and $Q R=8$.

Also, $P R=12, \angle P R Q=120^{\circ}$, and $\angle R P S=90^{\circ}$.

What is the area of $\triangle Q P S$ ?
(A) $72 \sqrt{3}$
(B) 72
(C) 36
(D) $60 \sqrt{3}$
(E) $96 \sqrt{3}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-094.jpg?height=266&width=618&top_left_y=1672&top_left_x=1054)

Part C: Each correct answer is worth 8.
Since $\angle Q R P=120^{\circ}$ and $Q R S$ is a straight line, then $\angle P R S=180^{\circ}-120^{\circ}=60^{\circ}$.

Since $\angle R P S=90^{\circ}$, then $\triangle S R P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Therefore, $R S=2 P R=2(12)=24$.

Drop a perpendicular from $P$ to $T$ on $R S$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-180.jpg?height=287&width=631&top_left_y=2011&top_left_x=801)

Since $\angle P R T=60^{\circ}$ and $\angle P T R=90^{\circ}$, then $\triangle P R T$ is also a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle. Therefore, $P T=\frac{\sqrt{3}}{2} P R=6 \sqrt{3}$.

Consider $\triangle Q P S$. We may consider $Q S$ as its base with height $P T$.

Thus, its area is $\frac{1}{2}(6 \sqrt{3})(8+24)=96 \sqrt{3}$.

ANSWER: (E)