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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
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Verified Answer
The correct answer is:
$\pi$ sq units
Let $\mathrm{AB}$ be the chord of length $\sqrt{2}$. Let $\mathrm{O}$ be the centre of the circle and let $O C$ be the perpendicular from $\mathrm{O}$ on $\mathrm{AB}$.
Then $, \mathrm{AC}=\mathrm{BC}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
In $\Delta \mathrm{OBC},$ we have
$$
\begin{aligned}
\mathrm{OB} &=\mathrm{BC} \operatorname{cosec} 45^{\circ} \\
&=\frac{1}{\sqrt{2}} \times \sqrt{2}=1
\end{aligned}
$$
$\therefore$ Area of the circle $=\pi(\mathrm{OB})^{2}=\pi$ sq units
Then $, \mathrm{AC}=\mathrm{BC}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
In $\Delta \mathrm{OBC},$ we have
$$
\begin{aligned}
\mathrm{OB} &=\mathrm{BC} \operatorname{cosec} 45^{\circ} \\
&=\frac{1}{\sqrt{2}} \times \sqrt{2}=1
\end{aligned}
$$
$\therefore$ Area of the circle $=\pi(\mathrm{OB})^{2}=\pi$ sq units
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