21. Triangle $P Q R$ is equilateral. A semicircle with centre $O$ is drawn with its diameter on $P R$ so that one end is at $P$ and the curved edge touches $Q R$ at $X$. The radius of the semicircle is $\sqrt{3}$.
What is the length of $Q X$ ?
A $\sqrt{3}$
B $2-\sqrt{3}$
C $2 \sqrt{3}-1$
D $1+\sqrt{3}$
E $2 \sqrt{3}$
SOLUTION
D

Because $O P$ and $O X$ are radii of the semicircle, we have $O P=O X=\sqrt{3}$.

Because the semicircle touches $Q R$ at $X$, the line $Q R$ is a tangent to the semicircle at $X$ and therefore the radius $O X$ is perpendicular to $Q R$. Therefore $O X R$ is a right-angled triangle.

Because the triangle $P Q R$ is equilateral, $\angle O R X=60^{\circ}$.

Hence
$$
\frac{O X}{O R}=\sin 60^{\circ}=\frac{\sqrt{3}}{2} \text { and } \frac{X R}{O R}=\cos 60^{\circ}=\frac{1}{2}
$$

It follows that
$$
O R=\frac{2}{\sqrt{3}} \times O X=\frac{2}{\sqrt{3}} \times \sqrt{3}=2,
$$
and, therefore,
$$
X R=\frac{1}{2} \times O R=\frac{1}{2} \times 2=1 .
$$

We can now deduce that
$$
Q R=P R=O R+O P=2+\sqrt{3} \text {. }
$$

Therefore
$$
Q X=Q R-X R=(2+\sqrt{3})-1=1+\sqrt{3}
$$