Chetan Mandayam Nayakar
Last Activity: 14 Years ago
The circle is divided into 3 regions- a semicircle of area pR2/2, the region whose area is sought, and, a region of a circle of radius R, bounded by ,a chord whose mid point is R/2 from the centre, and the smaller arc of the circle. Consider a chord whose midpoint is between R/2 and R away from the centre. Call this distance ‘r’. It is easy to see that the area of this region is equal to a = ∫2√(R2-r2) dr from r=R/2 to R . Let r = Rcos(theta). a = ∫2R2 sin2(theta)d(theta).from theta = zero to (pi/3). sin2(theta) = (1-cos2(theta))/2
Thus a = R2((p/3) – ( √3/4)). So the area which the problem asks is
pR2/2 - R2((p/3) – ( √3/4)) = R2((π/6) +( √3/4))
p=π everywhere
