In the diagram, $\triangle P Q R$ has $\angle R P Q=90^{\circ}, P Q=10$, and $Q R=26$. The area of $\triangle P Q R$ is
(A) 100
(B) 120
(C) 130
(D) 60
(E) 312

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-057.jpg?height=290&width=539&top_left_y=289&top_left_x=1194)
Since $\triangle P Q R$ is right-angled at $P$, we can use the Pythagorean Theorem.

We obtain $P Q^{2}+P R^{2}=Q R^{2}$ or $10^{2}+P R^{2}=26^{2}$.

This gives $P R^{2}=26^{2}-10^{2}=676-100=576$ and so $P R=\sqrt{576}=24$, since $P R>0$.

Since $\triangle P Q R$ is right-angled at $P$, its area equals $\frac{1}{2}(P R)(P Q)=\frac{1}{2}(24)(10)=120$.

ANSWER: (B)