In the diagram, $\triangle P Q R$ is right-angled at $R, P R=12$, and $Q R=16$. Also, $M$ is the midpoint of $P Q$ and $N$ is the point on $Q R$ so that $M N$ is perpendicular to $P Q$. The area of $\triangle P N R$ is
(A) 21
(B) 17.5
(C) 36
(D) 16
(E) 21.5

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-010.jpg?height=287&width=393&top_left_y=1423&top_left_x=1338)
Since $\triangle P Q R$ is right-angled at $R$, then by the Pythagorean Theorem,

$$
P Q^{2}=P R^{2}+Q R^{2}=12^{2}+16^{2}=144+256=400
$$

Since $P Q>0$, then $P Q=20$.

Since $M$ is the midpoint of $P Q$, then $M Q=\frac{1}{2} P Q=10$.

Now $\triangle N M Q$ is similar to $\triangle P R Q$, since each is right-angled and they share a common angle at $Q$.

Therefore, $\frac{N Q}{P Q}=\frac{M Q}{R Q}$ and so $\frac{N Q}{20}=\frac{10}{16}$ which gives $N Q=\frac{20 \cdot 10}{16}=\frac{25}{2}$.

Thus, $R N=R Q-N Q=16-\frac{25}{2}=\frac{32}{2}-\frac{25}{2}=\frac{7}{2}$.

Since $\triangle P N R$ is right-angled at $R$, its area equals $\frac{1}{2} \cdot P R \cdot R N=\frac{1}{2} \cdot 12 \cdot \frac{7}{2}=21$.

ANSWER: (A)