Area of a Triangle Just From a Circle?

a=AC=AD+DC=ODcot(60)+ODcot(15) $=\sqrt{3}(\frac{1}{\sqrt{3}})+\sqrt{3}(2+\sqrt{3})=4+2\sqrt{3}$ Now, ar(∆ABC)= $\frac{1}{2}a^2sin(120)$ $=12+7\sqrt{3}$ a=12, b=7, c=3 and a+ b+c=22

Let $O$ be the center of the incircle, $D$ be the point of tangency of the incircle with side $BC$ and $E$ be the point of tangency of the incircle with side $AB$ . Now as at most one interior angle of a triangle can be obtuse we know that $\angle BAC = 120^{\circ}$ , and so $\angle ABC = \angle ACB = 30^{\circ}$ .

Next note that $\Delta OEA$ is similar to $\Delta BDA$ , and so $\angle AOE = \angle ABD = 30^{\circ}$ . Then with $|OE| = |OD| = \sqrt{3}$ we have that $|OA| = |OE|\sec(30^{\circ}) = \sqrt{3}*\dfrac{2}{\sqrt{3}} = 2$ .

Thus $|AD| = |AO| + |OD| = 2 + \sqrt{3}$ , and so $|BD| = |AD|\cot(30^{\circ}) = (2 + \sqrt{3})*\sqrt{3} = 3 + 2\sqrt{3}$ .

But the area of $\Delta ABC$ equals

$\dfrac{1}{2}|AD|*|BC| = |AD|*|BD| = (2 + \sqrt{3})(3 + 2\sqrt{3}) = 6 + 4\sqrt{3} + 3\sqrt{3} + 6 = 12 + 7\sqrt{3}$ .

Thus $a + b + c = 12 + 7 + 3 = \boxed{22}$ .

$Let~ ID \bot BC=\sqrt3. ~~BD=DC,\\ \angle ~B=\angle ~C=30^o.~~~\therefore ~Tan\frac 1 2 B=2-\sqrt3.=\dfrac {\sqrt3}{BD}\\ \therefore ~BD=\dfrac{\sqrt3}{2-\sqrt3}=\sqrt3*(2+\sqrt3).\\ AD=TanB*BD=2+\sqrt3. ~~\\ \therefore~ AREA~ of ~\Delta ~ ABC=BD*AD=\sqrt3*(2+\sqrt3)^2.\\ \Delta ~ ABC=12+7\sqrt3 =a+b\sqrt c. ~~\therefore ~a+b+c=22.$