, 2022

Since $\triangle PQR$ is
right-angled at $R$, then by the
Pythagorean Theorem, $$PQ^2 = PR^2 + QR^2 = 12^2 + 16^2 = 144 + 256 = 400$$ Since $PQ>0$, then $PQ = 20$.

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

Now $\triangle NMQ$ is similar to
$\triangle PRQ$, since each is
right-angled and they share a common angle at $Q$.

Therefore, $\$\dfrac{NQ}{PQ} = \dfrac{MQ}{RQ}$$and so$$\dfrac{NQ}{20} = \dfrac{10}{16}\$$ which
gives $NQ = $\text{20}$ 10}{16} =
$\dfrac{25}{2}\$.$

Thus, $RN = RQ - NQ = 16 - $\dfrac{25}{2} = \dfrac{32}{2} - \dfrac{25}{2} = \dfrac{7}{2}\$.$

Since $\triangle PNR$ is right-angled
at $R$, its area equals $\$\dfrac{1}{2} \cdot PR \cdot$ RN = $\dfrac{1}{2} \cdot 12 \cdot \dfrac{7}{2}$ = 21$.