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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.

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@user86418 Its asking the area of quad. ABDE in that order. –  Sawarnik Apr 5 at 14:30
    
@Sawarnik: Ah, thank you. (Spurious comment deleted, and apologies to Googler for my apparent inability to read strings of capital letters....) –  user86418 Apr 5 at 15:00

2 Answers 2

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 $(1/2,\sqrt3/2)$. figure

$AE//BD$: 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 $1/2\cdot r^2=50$.

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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.

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@fgrieu Right, corrected! –  Sawarnik Apr 7 at 19:03
    
@fgrieu Ah! Now? –  Sawarnik Apr 7 at 19:06
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+1 for the idea of computing the area as ${d_1d_2\over2}\sin\theta$ and getting the details right! –  fgrieu Apr 7 at 19:08

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