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One angle of an isosceles triangle is 12...

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`

A

`4sqrt(3)`

B

`12-7sqrt(3)`

C

`12+7sqrt(3)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C

Let `AB=AC=a and angleA=120^(@)`
`:.` Area of triangle `=(1)/(2)a^(2)sin120^(@)`
where `a=AD+BD=sqrt(3)tan30^(@)+sqrt(3)cot 15^(@)`
`=1+(sqrt(3))/(tan(45^(@)-15^(@))`

`rArra=1+sqrt(3)((1+tan45^(@)tan30^(@))/(tan45^(@)-tan30^(@)))`
`=1+sqrt(3)((sqrt(3)+1)/(sqrt(3)-1))=4+2sqrt(3)`
`:.` Area of a triangle
`=(1)/(2)(4+2sqrt(3))^(2)((sqrt(3))/(2))=(12+7sqrt(3))` sq units
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