Area of the circle in which a chord of length 2 makes an angle π /...

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 $$\begin{aligned}\mathrm{OB}&=\mathrm{BC} \operatorname{cosec}45^{{}^{\circ}} \\ \text{ and } =\frac{1}{\sqrt{2}} \times \sqrt{2}=1 \end{aligned}$$\therefore$Area of the circle$=\pi($\mathrm{OB}$)^{2}=\pi$sq units <img src="/api/proxy-image?url=https$ and } clean=1 \text{ and } v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FrkS101Yo4ZfgZ7sOrAb33YtaA4es0JxVX1sCGTXbd48.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FrkS101Yo4ZfgZ7sOrAb33YtaA4es0JxVX1sCGTXbd48.original.fullsize.png?alt=media">