A chord of a circle of …

r = 15 cm, θ  = 60^o
Area of the minor sector =  {tex}\frac\theta{360^\circ}\mathrm{πr}^2\;=\;\frac{\displaystyle60^\circ}{\displaystyle360^\circ}\times3.14\;\times15\times15{/tex} = 117.75 cm²
In {tex}\triangle{/tex}AOB, draw OM {tex}\perp{/tex} AB
In right triangle OMA and OMB,
OA = OB .........Radii of the same circle
OM = OM .........Common side
{tex}\therefore{/tex} {tex}\triangle{/tex}OMA {tex}\cong{/tex} {tex}\triangle{/tex}OMB .........RHS congruence criterion
{tex}\therefore{/tex} AM = BM .......CPCT
{tex}\Rightarrow{/tex} AM = BM = {tex}\frac 12{/tex}AB
{tex}\angle{/tex}AOM = {tex}\angle{/tex}BOM .......CPCT
{tex}\Rightarrow{/tex} {tex}\angle{/tex}AOM = {tex}\angle{/tex}BOM = {tex}\frac 12{/tex}{tex}\angle{/tex}AOB = {tex}\frac 12{/tex} {tex}\times{/tex} 60^o = 30^o
{tex}\therefore{/tex} In right triangle OMA, cos30^o = {tex}\frac {OM}{OA}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{\sqrt3}2{/tex}= {tex}\frac {OM}{15}{/tex}
{tex}\Rightarrow{/tex} OM =  {tex}\frac{15\sqrt3}2{/tex}cm
sin30^o  = {tex}\frac {AM}{OA}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac 12{/tex}= {tex}\frac {AM}{15}{/tex}
{tex}\Rightarrow{/tex} AM =  {tex}\frac{15}2{/tex}cm
{tex}\Rightarrow{/tex} AB = 15 cm
{tex}\therefore{/tex} Area of {tex}\triangle{/tex}AOB = {tex}\frac 12{/tex} {tex}\times{/tex} AB {tex}\times{/tex} OM
= {tex}\frac 12{/tex} {tex}\times{/tex} 15 {tex}\times{/tex} {tex}\frac{15\sqrt3}2{/tex} = {tex}\frac{225\sqrt3}4{/tex}
= {tex}\frac {225 × 1.73}4{/tex} = 97.3125 cm²
{tex}\therefore{/tex} Area of the corresponding minor segment of the circle = Area of minor sector - Area of {tex}\triangle{/tex}AOB
= 117.75 - 97.3125 = 20.4375 cm²
and, area of the corresponding major segment of the circle = {tex}\pi{/tex}r² - area of the corresponding minor segment of the circle
= 3.14 {tex}\times{/tex} 15 {tex}\times{/tex} 15 - 20.4375
= 706.5 - 20.4375 = 686.0625 cm²