(C) Let $AB$ be the chord of length $\sqrt{2}$ and $O$ be the centre of the circle.Given that the angle subtended by the chord at the centre is $\angl

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(C) Let $AB$ be the chord of length $\sqrt{2}$ and $O$ be the centre of the circle.Given that the angle subtended by the chord at the centre is $\angle AOB = \frac{\pi}{2} = 90^\circ$.In $\Delta AOB$,$OA = OB = r$ (radius of the circle).By the Pythagorean theorem in $\Delta AOB$:$OA^2 + OB^2 = AB^2$$r^2 + r^2 = (\sqrt{2})^2$$2r^2 = 2$$r^2 = 1$Therefore,the area of the circle is $\pi r^2 = \pi(1) = \pi$.

Class 11 Mathematics 10-1.Circle and System of Circles Geometrical problems regarding circle and its properties

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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
$\frac{\pi}{4}$

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10-1.Circle and System of Circles

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Geometrical problems regarding circle and its properties

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

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