IMC 2018 Q23

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 \(\Delta = \tfrac{1}{2}ab \sin C\) for the area of a triangle, the area of an equilateral triangle of side \(l\) is
\(\dfrac{1}{2} \times l \times l \times \sin 60° = \dfrac{1}{2} \times l^2 \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{4}l^2\). So the area of the hexagon is
\(\dfrac{\sqrt{3}}{4}\left(11^2 - 6^2 - 1^2 - 2^2\right) = \dfrac{\sqrt{3}}{4}(121 - 36 - 1 - 4) = \dfrac{\sqrt{3}}{4} \times 80 = 20\sqrt{3}\).