We can form a chord connecting the two points where the circle and the square intersect
Call these points B and C
Now triangle BOC is equilateral becuse OB = OC = BC = 1
This means that angle AOB = 60°
And we can find the shaded area between the chord and the edge of the circle thusly :
Area of sector BOC - area of the equilateral triangle =
pi (1)^2 (60/360) - (1/2)*(1)^2 * √ 3 / 2 =
[ pi/6 - √ 3/4 ]
Now the height of the equlteral triangle can be found as
area = (1/2)base * height
√ 3/4= (1/2)(1) * height
2√ 3/4 = height = √ 3/2
If we connect AO we can call the intersection of this segment and the side of the square Q
And note that QO = 1/2
And the remaining rectagular shaded area has a width of 1
And its height = height of the equlateral triangle - OQ = √ 3/2 - 1/2 = [ √ 3 - 1] / 2
So the remaining rectagular shaded area = (1) [ √ 3 - 1 ] / 2 = [ √ 3 - 1 ] / 2
So.....the total shaded area = [ pi/6 - √ 3/4] + [√ 3 - 1] / 2 =
pi/6 + √ 3/2 - √ 3/4 - 1/2 =
[ pi/6 + √ 3/4 - 1/2] units^2 ≈ .46 units^2