12. A circle with radius 1 touches the sides of a rhombus, as shown. Each of the smaller angles between the sides of the rhombus is $60^{\circ}$.
What is the area of the rhombus?
A 6
B 4
C $2 \sqrt{3}$
D $3 \sqrt{3}$
E $\frac{8 \sqrt{3}}{3}$
Solution
E
Let $P, Q, R$ and $S$ be the vertices of the rhombus.
We leave it to the reader to prove that the diagonals of a rhombus bisect the angles of the rhombus and meet at right angles (see Exercise 12.1).
Let $O$ be the point where the diagonals $P R$ and $Q S$ of the rectangle meet.

We also leave it to the reader to prove that the four triangles $P O Q, Q O R, R O S$ and $S O P$ are congruent (see Exercise 12.1) and that $O$ is the centre of the circle (see Exercise 12.2).

We let $K$ be the point where $P Q$ touches the circle. Then $O K=1$. Because the radius of a circle is at right angles to the tangent at the point where the radius meets the circle, $\angle P K O=90^{\circ}$.
From the right-angled triangle $P K O$ we have
$$
\frac{O K}{O P}=\sin 30^{\circ}=\frac{1}{2},
$$
and therefore, since $O K=1$, it follows that $O P=2$.
Because $\angle P O Q=90^{\circ}$, it follows from the triangle $P O Q$ that $\angle O Q K=60^{\circ}$, and hence
$$
\frac{O K}{O Q}=\sin 60^{\circ}=\frac{\sqrt{3}}{2},
$$
and therefore $O Q=\frac{2}{\sqrt{3}}$.
It now follows that the area of the triangle $P O Q$ is given by
$$
\frac{1}{2}(O P \times O Q)=\frac{1}{2}\left(2 \times \frac{2}{\sqrt{3}}\right)=\frac{2}{\sqrt{3}} .
$$

Therefore, as the rhombus is made up of four triangles each congruent to the triangle $P O Q$, the area of the rhombus is
$$
4 \times \frac{2}{\sqrt{3}}=\frac{8}{\sqrt{3}}=\frac{8 \sqrt{3}}{3} .
$$

Note: See Exercise 12.4 for a way to remember the values of $\sin \left(30^{\circ}\right)$ and $\sin \left(60^{\circ}\right)$.