# find area of darkened figure

We know that length of diameter $AB=6$; $CA=2\sqrt{3}$; We should find area of darkened figure I have tried following :

Since $AB$ is diameter  we  can say $ACB$ is right triangle  so


we can caluclate area of $ABC$, but my question is how calculate area of this triangle inside circle? Please help me.

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Perhaps you could be able to find $\angle ABC$ using some trinometric functions and then use it to find the angle $\angle AOD$, where D is the intersection of $BC$ with the circle. en.wikipedia.org/wiki/Inscribed_angle – Martin Sleziak Jun 22 '11 at 17:45

From the metric data you can compute the angle at $B$ since $\tan B = \sqrt 3/3$. Let $D$ be the intersection of $BC$ with the circle. Since the angle at $B$ is known and $OB=OD$, you can easily compute the area of the circular sector $AOD$ and of the triangle $BOD$.
@user3196: The angle at $B$ is known, as lhf wrote. Its tangent is $1/\sqrt{3}$, so it is $30^\circ$. A little angle chasing shows that therefore $\angle AOD$ is $60^\circ$, so the circular sector has area one-sixth of the area of the circle. You now only need to find the area of $\triangle BOD$. – André Nicolas Jun 22 '11 at 18:18
The area of $\triangle BOD$ is the same as that of $\triangle AOD$, an equilateral triangle of edge 3. – Henry Jun 22 '11 at 18:27
$\triangle ABC$ is half an equilateral triangle making $\angle BCA = \pi /3$ and $\angle CBA = \pi /6$, so if $D$ is the intersection of $BC$ and the circle then $\triangle OAD$ is equilateral, and with the circle radius of 3 you can calculate all the areas.