Find the area of quadrilateral formed by $4$ (not consecutive) vertices of a $12$-gon inscribed in a circle. A regular $12$ sided polygon is inscribed in a circle of radius $10$. $A,B,C,D,E$ are its consecutive vertices taken in that order. Find the area of quad. $ABDE$.
The angle of $12$-gon is $150^{\circ}$.
Also its side will be $20\sin15^{\circ}$.
That's all I can think of.
 A: Let's put the center $O$ of the circle at $(0,0)$, $C$ at $(0,1)$ in units of the circle's radius. $\widehat{AOC}=\widehat{COE}=60°$, $\widehat{BOC}=\widehat{COD}=30°$. $A$ is at $(1/2,-\sqrt3/2)$, $B$ is at $(\sqrt3/2,-1/2)$, $D$ is at $\sqrt3/2,1/2)$, $E$ is at $(1/2,\sqrt3/2)$.

$AE//BD$: we want the area of a trapezoid, that is the product of the average length of its parallel sides, times the distance between them. That area come as $(\sqrt3/2+1/2)\cdot(\sqrt3/2-1/2)=1/2$, in units of the radius.
Given that the radius is $r=10$, the desired area is $1/2\cdot r^2=50$.
A: It can be easily shown that $\angle B= \angle D= 135^{\circ}$, $\theta=150^{\circ}$, where $\theta$ is the angle of the intersection of diagonals. Now:
$$R\sin B=\frac{d_1}2 \text{ , }  R\sin D=\frac{d_2}2$$
Since the area of a quadrilateral is given by $2A=d_1d_2\sin\theta$, combining it with our above formula gives:
$$A=2R^2\sin B\sin D\sin\theta=2\cdot100\cdot \frac12\cdot \frac12=50$$
Putting on our known values on this general formula, we got the required answer.
