In given figure we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is $l$. Show that the area of the green region enclosed between the two circles is $\frac{1}{4} \pi l^2$.

Let $R$ = radius of the larger circle, $r$ = radius of the smaller circle.

Since the circles are concentric, the centre is the same point O.

$BC = l$ is a chord of the larger circle tangent to the smaller circle at A.

$OA \perp BC$ (tangent $\perp$ radius), $OA = r$.

Also: $OB^2 = r^2 + \frac{l^2}{4}$ (right triangle OAB) and $OB = R$.

$$\Rightarrow R^2 = r^2 + \frac{l^2}{4} \rightarrow R^2 - r^2 = \frac{l^2}{4}.$$

Green area = Area of annulus = $\pi(R^2 - r^2)$

$$= \pi \times \frac{l^2}{4} = \frac{\pi l^2}{4}.$$

Hence, proved.