anindyat wrote:

Yea ....fairly long process

Join the intersecting points of circle and triangle to the center of the circle.

So, the angle at center should be 120

Part of the circle constitutes 120 degree at the center so the circle, so the area is [pi * (sqrt3) ^ 2] * 120 /360 = pi

The remaining portion has 2 isosceles triangle, having two sides equal to radius = (sqrt 3)

And the angles are 30 and 30 and 120.

I am sure someone will come up with a better approach

Same approach... same result

Pi*r^2 - 2* (120/360*Pi*r^2 - sqrt(3)/4*r^2)

With:

> r = radius of circle = sqrt(3)/4*4 = sqrt(3)

> Pi*r^2 = the area of the circle

> -2* = the 2 identical areas above each side of the triangle

> 120/360*Pi*r^2 = the area of the sector defined by angle of 120Â° (iscole triangle with r as equal sides)

> - sqrt(3)/4*r^2 = the aera of the isocele triangle of 120Â°

Finally,

r^2*(Pi/3+sqrt(3)/2)= Pi + 3*sqrt(3)/2