22. The diagram shows an arc $P Q$ of a circle with centre $O$ and radius 8. Angle $Q O P$ is a right angle, the point $M$ is the midpoint of $O P$ and $N$ lies on the arc $P Q$ so that $M N$ is perpendicular to $O P$.
Which of the following is closest to the length of the perimeter of triangle $P N M$ ?
A 17
B 18
C 19
D 20
E 21
22. Consider triangles $M O N$ and $M P N$. Note that $M O=M P=4$ because $M$ is the midpoint of $O P ; \angle O M N=\angle P M N=90^{\circ}$ because $M N$ is perpendicular to $O P$; side $N M$ is common to both triangles. So triangles $M O N$ and $M P N$ are congruent (SAS). Therefore $P N=O N=8$ because $O N$ is a radius of the circle. By Pythagoras' Theorem in triangle $O M N, O N^{2}=O M^{2}+M N^{2}$, so $8^{2}=4^{2}+M N^{2}$. Therefore $M N^{2}=8^{2}-4^{2}=48$. So $M N=\sqrt{48}$.
The perimeter of triangle $P N M$ is $P N+N M+M P=8+\sqrt{48}+4=12+\sqrt{48}$. Now $6.5<\sqrt{48}<7$, since $\sqrt{42.25}<\sqrt{48}<\sqrt{49}$. So $\sqrt{48}$ is closer in value to 7 than it is to 6 . So $12+\sqrt{48}$ is nearer to 19 than it is to 18 .