Answer:
\[r=10\text{ }cm,\text{ }i=60{}^\circ \] Area of minor segment \[=\frac{\theta }{360}\times \pi \,\,{{r}^{2}}-\frac{1}{2}{{r}^{2}}\,\,\sin \,\,i\] \[=\frac{60}{360}\times 3.14\times 10\times 10-\frac{1}{2}\times 10\times 10\,\,\sin \,\,60{}^\circ \] \[=\frac{1}{6}\times 3.14\times 100-\frac{1}{2}\times 100\times \frac{\sqrt{3}}{2}\] \[=\frac{314}{6}-\frac{100}{4}\times 1.73\] \[=\frac{314}{6}-\frac{173}{4}=\frac{628-519}{12}=\frac{109}{12}c{{m}^{2}}\] Area of major segment = Area of circle\[\]Area of minor segment \[=\pi {{r}^{2}}-\frac{109}{12}c{{m}^{2}}=3.14\times 10\times 10-\frac{109}{12}\] \[=314-\frac{109}{12}=\frac{3768-109}{12}=\frac{3659}{12}c{{m}^{2}}\]
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