According to me , the answer shuld be (B)2/5. Pls tell us the OA?

Here is how i solved it:-

Given radius , r =1

First name the center of circle as O,

Draw a perpendicular from the point R to the base of triangle extending till center of circle.

The perpendicular from the vertex of an isosceles triangle upon its base bisects its base. Also will the perpendicular from the center bisects the sides of rectangle since its inscribed in it.

Area of rectantgle = Area of triangle

w x h = 1/2 x w x (height of perp)
= 1/2 x w x (radius - h/2) [ since the center of circle is center of rectangle]

so, the equation is
wh=(1/2 )w(1-h/2)

h = (1/2)(1-(h/2))
solving for h, we get h=2/5

----- Some correct me if I am wrong -----

Excellent!

How did it occur to you that you should draw the perpendicular from R to O?

I did not knew to use the center of the rectangle concept. I was trying hard to some how use the radius given but could not use it.

Good Solution Punit

Rule to know to solve this is that the center of the circle is also the center of an inscribed rectangle/square. So when you draw a line from the vertex of the equilateral triangle to the center O it will bisect the base of the triangle.