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 \[=\pi {{(OB)}^{2}}=\pi \].You need to login to perform this action.
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