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

Option: 1

\pi / 2


Option: 2

2 \pi


Option: 3

\pi


Option: 4

\pi / 4


Answers (1)

best_answer

Let \mathrm{AB} be the chord of length \mathrm{\sqrt{2}, O} be the centre of the circle and let \mathrm{O C}  be the perpendicular from \mathrm{O } on \mathrm{AB }. Then \mathrm{A C=B C=\sqrt{2} / 2=1 / \sqrt{2} }



In \mathrm{\triangle \mathrm{OBC}, \mathrm{OB}=\mathrm{BC} \operatorname{cosec} 45^{\circ}=(1 / \sqrt{2})^{\prime} \sqrt{2}=1 }
Area of the circle \mathrm{=\pi(\mathrm{OB})^{2}=\pi }.

Hence (C) is the correct answer.

Posted by

manish painkra

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