question_answer

A chord PQ of a circle of radius 10 cm subtends an angle of \[60{}^\circ \] at the centre of circle. Find the area of major and minor segments of the circle.

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

\[=\pi {{r}^{2}}-\frac{109}{12}c{{m}^{2}}=3.14\times 10\times 10-\frac{109}{12}\]

\[=314-\frac{109}{12}=\frac{3768-109}{12}=\frac{3659}{12}c{{m}^{2}}\]