question_answer

Find the area of the minor segment of a circle of radius 14 cm, when its central angle is \[60{}^\circ \]. Also find the area of the corresponding major segment.  [Use \[\pi =\frac{22}{7}\]]

Answer:

Let ACB be the given arc subtending an angle of \[60{}^\circ \] at the centre.

Here, \[r=14\text{ }cm\] and\[\theta =60{}^\circ \].

Area of the minor segment ACBA = (Area of the sector OACBO)\[-\](Area of \[\Delta \,OAB\])

\[=\frac{\pi {{r}^{2}}\theta }{360{}^\circ }-\frac{1}{2}{{r}^{2}}\sin \theta \]

\[=\frac{22}{7}\times 14\times 14\times \frac{60{}^\circ }{360{}^\circ }-\frac{1}{2}\times 14\times 14\times \sin 60{}^\circ \]

\[=\frac{308}{3}-7\times 14\times \frac{\sqrt{3}}{2}\]

\[=\frac{308}{3}-49\sqrt{3}\]

\[=17.89\,c{{m}^{2}}\,\]

Area of the major segment BDAB = Area of circle\[-\]Area of minor segment ACBA

\[=\pi {{r}^{2}}-17.89\]

\[=\frac{22}{7}\times 14\times 14-17.89\]

\[=616-17.89\]

\[=598.11\approx 598\,c{{m}^{2}}\]