In an equilateral triangle, 3 coins of radii 1 unit each are kept so that they touch each other and also the sides of the triangle. Area of the triangle is

A) 4+2√3B) 6+4√3C) 12+(7√3/4)D) 3+(7√3/4)

Considering any one side,

For convenience, I choose BC

BC=BD+DE+EC

We can prove that BD=DE by congruency
Also, BE= 2*radius of circle= 2(1)=2 units

We need to find BD;
So, In triangle O2BD

Using tan30= O2D/BD

get the value of BDand consequently value of BC

Once the value of BC is achieved, you can easily find the area of the triangle by

Area of an equilateral triangle= (3^1/2)(a^2)/4 where a is the side of an equilateral triangle.