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 \[\Delta OBC,\] \[OB=BC\,\,\text{cosec 45}{}^\circ \text{=}\frac{1}{\sqrt{2}}.\sqrt{2}=1\] \[\therefore \]   Area of the circle  \[=\pi \,\,{{(OB)}^{2}}=\pi \]