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 \[=\pi {{r}^{2}}-17.89\] \[=\frac{22}{7}\times 14\times 14-17.89\] \[=616-17.89\] \[=598.11\approx 598\,c{{m}^{2}}\]
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