One angle of an isosceles triangle is `120^0` and the radius of its incircle is `sqrt(3)dot` Then the area of the triangle in sq. units is
(a) `7+12sqrt(3)` (b) `12-7sqrt(3)` (c) `12+7sqrt(3)` (d) `4pi`

Explanation

Let `x` are the two equal sides of the isoceles triangle and `a` is the third side.

One of the angle given is `120^@`, so the remaining two equal angles will be `30^@`.

So, from sine law,

`x/(sin30^@) = x/(sin30^@) = a/(sin120^@)`

`=>a = xsqrt3`

Now, `Delta = 1/2*x*x*sin120^@ = sqrt3/4x^2`

Inradius,`r = Delta/s = (sqrt3/4x^2)/((a+2x)/2)`

`=>sqrt3 = (sqrt3x^2)/(2(xsqrt3+2x))`

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