A) \[30720.5\,c{{m}^{2}}\]
B) \[31021.42\,c{{m}^{2}}\]
C) \[30391.7\,c{{m}^{2}}\]
D) \[30520.61\,c{{m}^{2}}\]
Correct Answer: D
Solution :
In \[\Delta \,\,OAB,\] by angle sum property \[={{60}^{o}}+\angle OAB+\angle OBA={{180}^{o}}\] \[\Rightarrow \] \[{{60}^{o}}+\angle OAB+\angle OBA={{180}^{o}}\] \[\Rightarrow \]\[2\angle OAB={{120}^{o}}\Rightarrow \angle OAB={{60}^{o}}\] \[\Rightarrow \] \[\Delta OAB\] is equilaterai triangle. \[\Rightarrow \] \[r=100\text{ }cm\] Area of major segment = Area of major sector + Area of \[\Delta \,OAB\] \[=\frac{\left( {{360}^{o}}-{{60}^{o}} \right)}{{{360}^{o}}}\times \pi {{r}^{2}}+\frac{\sqrt{3}}{4}{{r}^{2}}\] \[=\frac{300}{360}\times \frac{22}{7}\times {{(100)}^{2}}+\frac{\sqrt{3}}{4}\times {{100}^{2}}\] \[=\frac{5}{6}\times \frac{22}{7}\times {{(100)}^{2}}+\sqrt{3}\times 2500\] \[=26190.48+4330.13=30520.61\text{ }c{{m}^{2}}\]You need to login to perform this action.
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