23. The diagram shows a hexagon. All the interior angles of the hexagon are $120^{\circ}$. The lengths of some of the sides are indicated. What is the area of the hexagon?
A $20 \sqrt{3}$
B $21 \sqrt{3}$
C $22 \sqrt{3}$
D $23 \sqrt{3}$
E $24 \sqrt{3}$
23. A The diagram shows that the hexagon is an equilateral triangle of side 11 which has an equilateral triangle of side 1 removed from the bottom left corner, an equilateral triangle of side 2 removed from the bottom right corner and an equilateral triangle of side 6 removed from the top.
Using the formula $\Delta=\frac{1}{2} a b \sin C$ for the area of a triangle, the area of an equilateral triangle of side $l$ is
$\frac{1}{2} \times l \times l \times \sin 60^{\circ}=\frac{1}{2} \times l^{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} l^{2}$. So the area of the hexagon is $\frac{\sqrt{3}}{4}\left(11^{2}-6^{2}-1^{2}-2^{2}\right)=\frac{\sqrt{3}}{4}(121-36-1-4)=\frac{\sqrt{3}}{4} \times 80=20 \sqrt{3}$.