AB is a chord of a circle with centre O and radius 4 cm.

Solution:

We know that the area of minor segment of angle θ in a circle of radius r is,

$A=\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} r^{2}$

It is given that the chord AB divides the circle in two segments.

We have $O A=4 \mathrm{~cm}$ and $A B=4 \mathrm{~cm}$. So,

$A L=\frac{A B}{2} \mathrm{~cm}$

$=\frac{4}{2} \mathrm{~cm}$

$=2 \mathrm{~cm}$

Let $\angle A O B=2 \theta$. Then,

$\angle A O L=\angle B O L$

$=\theta$

In $\triangle O L A$, we have

$\sin \theta=\frac{A L}{O A}$

$=\frac{2}{4}$

$=\frac{1}{2}$

$\theta=\sin ^{-1} \frac{1}{2}$

$=30^{\circ}$

Hence, $\angle A O B=60^{\circ}$

Now using the value of r and θ, we will find the area of minor segment

$A=\left\{\frac{\pi \times 60^{\circ}}{360^{\circ}}-\sin \frac{60^{\circ}}{2} \cos \frac{60^{\circ}}{2}\right\} \times 4 \times 4$

$=\left\{\frac{\pi}{6}-\sin 30^{\circ} \cos 30^{\circ}\right\} \times 16$

$=\left\{\frac{16 \times \pi}{6}-\frac{1}{2} \times \frac{\sqrt{3}}{2} \times 16\right\}$

$=\left\{\frac{8 \pi}{3}-4 \sqrt{3}\right\} \mathrm{cm}^{2}$