H3. The diagram shows a regular dodecagon and a square, whose vertices are also vertices of the dodecagon.
What is the value of the ratio area of the square : area of the dodecagon?
Solution 1
Let the side length of the dodecagon be $x$. Consider one of the four trapezia created by the edge of the square and the edges of the dodecagon.
The interior angle of a dodecagon is $150^{\circ}$, which quickly leads to the fact that the angles at the base of the trapezium are both equal to $30^{\circ}$.
We can then split the trapezium up into a rectangle and two right-angled triangles:
Then, by recognising that each of the triangles in the trapezium is half an equilateral triangle (or by using the well-known values of $\sin 30^{\circ}$ and $\cos 30^{\circ}$ ), we can establish that the height of this trapezium is $x / 2$ and the base is $x+x \sqrt{3}$.
The area of the square is therefore $(x+x \sqrt{3})^{2}=2 x^{2}(2+\sqrt{3})$ and the area of the dodecagon is equal to (the area of the square) $+(4 \times$ the area of the trapezium) which is
$$
2 x^{2}(2+\sqrt{3})+4 \times \frac{x+(x+x \sqrt{3})}{2} \times \frac{x}{2}=3 x^{2}(2+\sqrt{3}) .
$$

We can divide through by the common factor of $x^{2}(2+\sqrt{3})$ to show that the required ratio is $2: 3$.

Solution 2
Since the dodecagon is regular, all its diagonals intersect at the same point (its centre). Call this point $O$. Since the diagonals of the square are two of the diagonals of the dodecagon, they intersect at $O$. Hence $O$ is also the centre of the square.
Let each diagonal have length $2 r$.
Then the dodecagon comprises 12 congruent isosceles triangles radiating from $O$, each with two sides equal to $r$ and vertex angle $30^{\circ}$ :
The area of the dodecagon is equal to $12 \times$ the area of one of these triangles, which is equal to
$$
12 \times\left[\frac{1}{2} \times r \times r \times \sin 30^{\circ}\right] .
$$

Since $\sin 30^{\circ}=\frac{1}{2}$, this simplifies to
$$
\text { Area of dodecagon }=3 r^{2} \text {. }
$$

The square can be split into four congruent isosceles right-angled triangles, with perpendicular sides of length $r$.
The area of the square is equal to $4 \times$ the area of one of these triangles, which is equal to
$$
4 \times\left[\frac{1}{2} \times r \times r\right]
$$
which simplifies to
$$
\text { Area of square }=2 r^{2} \text {. }
$$

The required ratio is therefore $2 r^{2}: 3 r^{2}$ which is equal to $2: 3$.