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This problem attracted several correct answers anonymously submitted by students at Madras College . This solution is included because of its brevity:
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AD = 1 unit, which is the radius of the circle centre
A.
The coordinates of P are (AD \cos 60^{\circ} , AD \sin
60^{\circ} ) = ({1\over2}, {\sqrt{3}\over2})
Similarly coordinates of Q are (AD \cos 30^{\circ} , AD
\sin 30^{\circ} ) = ( {\sqrt{3}\over2},{1\over2}) and angle PAQ
is 30^{\circ}.
Area of square PQRS = ({1\over2} - {\sqrt{3}\over2})^2 +
({\sqrt{3}\over2} - {1\over2})^2 = 2 - \sqrt{3}
Area of a segment PQ = [{1\over12} area of circle, centre
A radius 1 unit] - [Area of Triangle APQ ]
Using the formula {1\over12}\pi r^2 for the 30 degree
sector and {1\over2} ab\sin C for the area of the triangle we
get:
Area of a segment PQ = {1\over12} \pi.1.1 -
{1\over2}.1.1.{1\over2} = {\pi\over12} - {1\over4} \par Bounded
area = area of square PQRS + area of 4 equal segments
Bounded area = 2 - \sqrt{3} + 4 ({\pi\over12} - {1\over4}) =
1 + {\pi\over3} - \sqrt{3}
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