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

Options

A. `7+12sqrt3`
B. `12-7sqrt3`
C. `12+7sqrt3`
D. `4pi`

Explanation

Let ABC be the isosceles triangle with AB = AC and `angleA=(2pi)/3`. Then, it area `Delta` is given by
`Delta=1/2ABxxACxxsin(2pi)/3`
`rArrDelta=(sqrt3)/4x^(2),` ,whereAB=AC =x(say)
Using Sine rule, we have
`(AB)/(sin30^(@))=(BC)/(sin120^(@))=(AC)/(sin30^(@))rArr=(2a)/(sqrt3)rArrx=a/(sqrt3)`
`therefore2s=2x+arArr2s=(2a)/(sqrt3)+arArrs=((2+sqrt3)/(2sqrt3))a`
and, `Delta=(sqrt3)/4xx(a^(2))/3=(a^(2))/(4sqrt3)`
Now, `r=(Delta)/s`
`rArrsqrt3=(a^(2))/(4sqrt3)xx(2sqrt3)/((2+sqrt3)a)rArrsqrt3=a/(4+2sqrt3)rArra+6+4sqrt3`
Hence, `Delta=(a^(2))/(4sqrt3)rArrDelta=((6+4sqrt3)^(2))/(4sqrt3)=12+7sqrt3`