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

• A

  $\pi / 2$ sq units
• B

  $2 \pi$ sq units
• Correct answer
  C

  $\pi$ sq units
• D

  $\pi / 4 \mathrm{sq}$ units

Solution

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

Hence, the correct answer is option: 'C'.