A) \[91\text{ }c{{m}^{2}}\]
B) \[95\,c{{m}^{2}}\]
C) \[97\,c{{m}^{2}}\]
D) \[88\,c{{m}^{2}}~\]
Correct Answer: A
Solution :
Area of \[\Delta \,ABC=\frac{1}{2}\times AB\times BC\] \[=\frac{1}{2}\times 14\times 24=168c{{m}^{2}}\] Area of sector centred at \[A=\frac{\angle A}{{{360}^{o}}}\times \pi {{(7)}^{2}}c{{m}^{2}}\] Area of sector centred at \[B=\frac{\angle B}{{{360}^{o}}}\times \pi {{(7)}^{2}}c{{m}^{2}}\] Area of sector centred at \[C=\frac{\angle C}{{{360}^{o}}}\times \pi {{(7)}^{2}}c{{m}^{2}}\] Total Area of all the three sectors \[=\pi \frac{{{(7)}^{2}}}{{{360}^{o}}}(\angle A+\angle B+\angle C)\] \[=\frac{22}{7}\times \frac{7\times 7}{{{360}^{o}}}\times 180=77c{{m}^{2}}\] Area of shaded region = Area of \[\Delta ABC-\] - Area of three sectors \[=(168-77)c{{m}^{2}}=91c{{m}^{2}}\]
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