CUET Preparation Today
The area of the smaller portion enclosed by the circle $x^2+y^2=4 $ and the straight line $x+y = 2 $ is : |
$\pi + 2 $ $\pi - 2 $ $2- \pi $ $-\pi + 2 $ |
$\pi - 2 $ |
The correct answer is Option (2) → $\pi - 2 $
So area required = area of quarter circle - area of triangle $=\frac{4\pi}{4}-\frac{1}{2}×2×2$ $=\pi-2$ |
