18. The diagram shows a semicircle with diameter $P Q$ inscribed in a rhombus $A B C D$. The rhombus is tangent to the arc of the semicircle in two places. Points $P$ and $Q$ lie on sides $B C$ and $C D$ of the rhombus respectively. The line of symmetry of the semicircle is coincident with the diagonal $A C$ of the rhombus. It is given that $\angle C B A=60^{\circ}$. The semicircle has radius 10 . The area of the rhombus can be written in the form $a \sqrt{b}$ where $a$ and $b$ are integers and $b$ is prime. What is the value of $a b+a+b ?$
18. 603 Let $O$ be the centre of the semicircle and let $M$ and $N$ be the feet of the perpendiculars drawn from $O$ to $A B$ and $A D$ respectively. Let $G$ be the intersection of the diagonals of the rhombus.
$$
\begin{array}{l}
P O=10 \text { and } \angle O P C=30^{\circ} . \text { So } O C=10 \tan 30^{\circ}=\frac{10}{\sqrt{3}} . \\
M O=10 \text { and } \angle O A M=60^{\circ} . \text { So } A O=\frac{10}{\sin 60^{\circ}}=\frac{20}{\sqrt{3}} .
\end{array}
$$

Therefore $A C=\frac{10}{\sqrt{3}}+\frac{20}{\sqrt{3}}=\frac{30}{\sqrt{3}}=10 \sqrt{3}$.
Hence $A G=5 \sqrt{3}$ and $\angle G B A=30^{\circ}$. So $B G=\frac{5 \sqrt{3}}{\tan 30^{\circ}}=15$.
Therefore the area of triangle $B G A$ is $\frac{1}{2} \times 15 \times 5 \sqrt{3}=\frac{75}{2} \sqrt{3}$.
So the area of the rhombus is $4 \times \frac{75}{2} \sqrt{3}=150 \sqrt{3}$.
Therefore $a=150$ and $b=3$, so $a b+a+b=450+150+3=603$.