In the given figure,y is both the radius of t | vedclass

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(B) $1$. Radius of the larger circle $(R)$ = $y$.$2$. Area of the larger circle = $\pi R^{2} = \pi y^{2}$.$3$. Diameter of the smaller circle $(d)$ =

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$\frac{3}{4} \pi y^{2}$

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(B) $1$. Radius of the larger circle $(R)$ = $y$.$2$. Area of the larger circle = $\pi R^{2} = \pi y^{2}$.$3$. Diameter of the smaller circle $(d)$ = $y$,so its radius $(r)$ = $\frac{y}{2}$.$4$. Area of the smaller circle = $\pi r^{2} = \pi (\frac{y}{2})^{2} = \frac{\pi y^{2}}{4}$.$5$. The shaded region is the area of the larger circle minus the area of the smaller circle.$6$. Area of shaded region = $\pi y^{2} - \frac{\pi y^{2}}{4} = \frac{4\pi y^{2} - \pi y^{2}}{4} = \frac{3}{4} \pi y^{2}$.

In the given figure,$y$ is both the radius of the larger circle and the diameter of the smaller circle. Find the area of the shaded region.

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