Calculate the areas of regular dodecagons (twelve-sided polygons) inscribed and circumscribed about a unit circular disk and thereby deduce the inequalities $3\lt \pi \lt 12(2-\sqrt{3})$.
This is a problem on an application of integration. However, I have no idea on how to calculate the areas in the first place. I would greatly appreciate it if anyone could provide any solutions, hints or suggestions.
On
Assume that a regular $n$-agon is inscribed in the unit circle. Then the length of a side is given by $2\sin\frac{\pi}{n}$ and the distance of a side from the centre of the circle is given by $\cos\frac{\pi}{n}$, so the area of a triangle made by two consecutive vertices of the $n$-agon and the centre of the circle is $\sin\frac{\pi}{n}\cos\frac{\pi}{n}=\frac{1}{2}\sin\frac{2\pi}{n}$ and we have:
$$ \pi \cos^2\frac{\pi}{n} \leq n\sin\frac{\pi}{n}\cos\frac{\pi}{n} \leq \pi $$ or: $$ \frac{n}{2}\sin\frac{2\pi}{n}\leq \pi \leq n\tan\frac{\pi}{n}.$$ Now just take $n=12$ to have: $$ 3\leq \pi\leq 24-12\sqrt{3}.$$
Here are some hints to get you started:
You can break up the inscribed $12$-gon into $12$ isosceles triangles. Each isosceles triangle has a vertex angle of $\dfrac{360^{\circ}}{12} = 30^{\circ}$ and legs of length $1$.
Similarly, you can break up the circumscribed $12$-gon into $12$ isosceles triangles. Each isosceles triangle has a vertex angle of $\dfrac{360^{\circ}}{12} = 30^{\circ}$ and legs of length $\dfrac{1}{\cos 15^{\circ}}$.
The area of a triangle with sides $a$ and $b$ and an angle $C$ (between those sides) is $\dfrac{1}{2}ab\sin C$.
Using the above hints, you can calculate the area of the inscribed $12$-gon and the area of the circumscribed $12$-gon. Then, compare these to the area of the unit circle to get the desired inequalities.