A chord of a circle of …

Here, we have
r = 15cm, {tex}\theta{/tex} = 60^o
Let OACBO be the given sector and OAB is a triangle. Then

(i). Area of the sector (OACBO)
{tex}=\pi r^{2} \times \frac{\theta}{360}{/tex}
{tex}=\left(3.14 \times 15 \times 15 \times \frac{60}{360}\right) \mathrm{cm}^{2}{/tex}
= 117.75 cm²
(ii) Area of the triangle (AOB)
{tex}=\frac{1}{2} r^{2} \sin \theta{/tex}
{tex}=\frac{1}{2} \times 15 \times 15 \times \frac{\sqrt{3}}{2}{/tex}
{tex}=\frac{15 \times 15 \times 1.73}{4}=97.313 \mathrm{cm}^{2}{/tex}
Now,
Area of MInor segment (ACBA)
= Area of sector (OACBO) - Area of triangle (AOB)
= (117.75 - 97.313) cm²
= 20.437 cm²
and Area of Major segment (ABDA)
= Area of circle - Area of Minor segment
= ({tex}\pi {/tex}r³ - 20.437) cm²
= (3.14 {tex}\times{/tex}15 {tex}\times{/tex} 15 - 20.437) cm²
= (706.5 - 20.437) cm²
= 686.063 cm²