Mathematics - Circle Question with Solution | TestHub

Let $\mathrm{AB}$ be the chord of length $\sqrt{2}$. Let $\mathrm{O}$ be the centre of the circle and let $OC$ 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 $$\begin{array}{rl}\mathrm{OB}& =\mathrm{BC}\mathrm{cosec}45^{\circ }\\ & =\frac{1}{\sqrt{2}}\times \sqrt{2}=1\end{array}$$ $\therefore$ Area of the circle $=\pi (\mathrm{OB}{)}^2=\pi$ sq units