Step 1: Diameter $=20$ cm, so radius $r=10$ cm and the central angle $\theta=60^{\circ}$.
Step 2: Area of the corresponding sector
$$=\frac{\theta}{360^{\circ}}\times \pi r^{2}=\frac{60}{360}\times 3.14\times 10\times 10$$
$$=\frac{1}{6}\times 314=52.33\text{ cm}^{2}$$
Step 3: Since the two radii are equal and the included angle is $60^{\circ}$, the triangle formed is EQUILATERAL with side 10 cm.
$$\text{Area of triangle}=\frac{\sqrt{3}}{4}a^{2}=\frac{1.73}{4}\times 100=43.25\text{ cm}^{2}$$
Step 4: Area of minor segment $=$ Area of sector $-$ Area of triangle.
$$=52.33-43.25=9.08\text{ cm}^{2}$$
Hence the area of the minor segment is approximately $9.08$ cm$^{2}$.