Area of a Polygon in a Polygon I'm dealing with a regular polygon with 7 corners. In this polygon is another polygon defined by connecting one point with the two opposite points of the same polygon. 
I made a small sketch of the polygons

I need to calculate the ratio of the area of the big and of small polygon. I know how to calculate the area of the big polygon but I don't know how to get area of the small polygon.
 A: Let $O$ be the centre of the circumscribed circle of our heptagons, and the outer heptagon $P_1P_2P_3P_4P_5P_6P_7$ has circumradius $R$.
Then the distance from $O$ to side $P_1P_2$ is $R\cos(\pi/7)$ (as seen by drawing the isosceles triangle $OP_1P_2$.
Similarly, can you find the distance from $O$ to side $Q_1Q_2=P_3P_7$ and finish it from here?
A: Easy to see that $P_1Q_2P_3P_2$ and $P_1P_7Q_1P_2$ are parallelograms, which says that
$$P_1P_7=Q_1P_2=P_1P_2=P_1Q_2=P_2P_3.$$ 
Now, let $P_1P_4\cap P_6P_2=\{A\}$.
Thus, from $\Delta Q_1Q_2A$ we obtain:
$$Q_1Q_2=2AQ_2\sin\frac{\measuredangle Q_1AQ_2}{2}=2AQ_2\sin\frac{3\pi}{14}.$$
Also, from $\Delta P_1P_2A$ we obtain:
$$P_1P_2=2AP_1\sin\sin\frac{\measuredangle P_1AP_2}{2}=2AP_1\sin\frac{3\pi}{14}.$$
Since $$AQ_2+AP_1=P_1Q_2=P_2P_3=P_1P_2,$$ we obtain:
$$\frac{Q_1Q_2}{2\sin\frac{3\pi}{14}}+\frac{P_1P_2}{2\sin\frac{3\pi}{14}}=P_1P_2,$$ which gives
$$\frac{P_1P_2}{Q_1Q_2}=\frac{1}{2\sin\frac{3\pi}{14}-1},$$ which gives the answer:
$$\frac{1}{\left(2\sin\frac{3\pi}{14}-1\right)^2}.$$
Another way.
Let $R$ be a radius of the circumcircle of $P_1P_2...P_7.$
Thus, $$Q_1Q_2=P_7P_3-2P_1P_2=2R\sin\frac{3\pi}{7}-4R\sin\frac{\pi}{7}$$
and $$P_1P_2=2R\sin\frac{\pi}{7}.$$
Id est, $$\frac{P_1P_2}{Q_1Q_2}=\frac{\sin\frac{\pi}{7}}{\sin\frac{3\pi}{7}-2\sin\frac{\pi}{7}}=\frac{1}{3-4\sin^2\frac{\pi}{7}-2}=$$
$$=\frac{1}{1-2\left(1-\cos\frac{2\pi}{7}\right)}=\frac{1}{2\cos\frac{2\pi}{7}-1},$$ which gives a ratio of areas again.
