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Getting pi + 3 (sqrt 3)/2 But adopted fairly long process, takes around 4 to 5 min any simple approach.. ??

If I guessed it correctly you divided the area in one sector and two triangles.
The area of sector is is pi and what did u get as the area of other two traingles. I am getting the area as r^2/ for a single triangle?
Am i doing something wrong here?

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°

Phew! Thanks guys! Great job! I understood your approach. Few minutes to figure out and few more minutes to calculate... Can we afford this many minutes in the real exam? Hope we don't get such complicated questions.

Its height/2 is the radius of the circle = sqrt(3)

Area of sector with 60 at vertex = Angle/360 * PI * r^2 = PI/2

I'm afraid we could use this formula only when the angle is at the center. Any one?

Can you elucidate this please?

If the angle of the sector is at the centre, we can determine the ratio of the sector to the circle by dividing its angle by 360. Hence, this fraction multiplied by the area of the circle gives the area of the sector.
But the ratio of the sector to the circle cannot be determined if the angle is not at the centre. Hope I'm answering your question.