IBPS RRBs Officer Scale I and II 13 Sep 2015 Paper

80. (3)

AB and CD are diameters.
Let, AB = 2x cm
∴ Radius of circle = x cm.
∴ Area of the shaded region
= (2x)2–πx2
=(4x2–πx2)sq.cm.
According to the question,
4x2–πx2=42
⇒ 4x2–

22x2

7

=42
⇒

28x2−22x2

7

=42
⇒ 6x2=42×7
⇒x2=

42×7

6

=7×7
∴ x = 7 cm.
∴ Diameter of circle
= 2 × 7 = 14 cm