Question

Area of the circle in which a chord of length $\sqrt { 2 }$ makes an angle $\frac { \pi } { 2 }$ at the centre is.

• A. $\frac { \pi } { 2 }$
• B. $2 \pi$
• C. $\pi$
• D. $\frac { \pi } { 4 }$

Answer 1

Answer: (C)

$\pi$

Answer 2

Let AB be the chord of length $\sqrt { 2 }$, O be centre of the circle and let OC be the perpendicular from O on AB. Then

$A B = \sqrt { 2 }$

$A C = B C = \frac { \sqrt { 2 } } { 2 } = \frac { 1 } { \sqrt { 2 } }$

In $\triangle O B C , O B = B C \operatorname { cosec } 45 ^ { \circ } = \frac { 1 } { \sqrt { 2 } } \cdot \sqrt { 2 } = 1$

$\therefore$Area of the circle $= \pi ( O B ) ^ { 2 } = \pi$.