question_answer

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)            \[\frac{\pi }{4}\]

Correct Answer: C

Solution :

           Let AB be the chord of length\[\sqrt{2}\], O be centre of the circle and let OC be the perpendicular from O on AB. Then                    \[AC=BC=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\]                    In \[\Delta \ OBC,\ OB=BC\,\text{cosec }{{45}^{o}}=\frac{1}{\sqrt{2}}.\sqrt{2}=1\]                    \[\therefore \]Area of the circle \[=\pi {{(OB)}^{2}}=\pi \].