BITSAT 2015 Paper

Let AB be the chord of length √2 . Let O be the centre of the circle and let OC be the perpendicular from O on AB.
Then, AC = BC =

√2

2

=

1

√2

In ΔOBC, we have
OB = BC cosec 45°
=

1

√2

×√2 = 1
∴ Area of the circle = π(OB)2 = π sq units