A) \[304.97\,c{{m}^{2}}\]
B) \[295\,c{{m}^{2}}\]
C) \[310\,c{{m}^{2}}\]
D) \[335\,c{{m}^{2}}\]
Correct Answer: A
Solution :
| [a] We have, \[OA=OB\] (Radii of circle) |
| \[\Rightarrow \,\,\,\angle A=\angle B=60{}^\circ \,\,\,\,\,\,\,\,\,\,\,\,\,\,(\angle AOB=60{}^\circ )\] |
| \[\Rightarrow \,\,\,\Delta AOB\]is equilateral triangle. |
| \[\therefore \,\,\,\,\,\,OA=OB=AB=10cm\] |
| and \[\theta =60{}^\circ \] |
 |
| Area of the minor segment |
| \[=\text{Area of minor sector}-\text{Area of }\Delta \text{AOB}\] |
| \[=\left( \frac{\pi {{r}^{2}}\theta }{360{}^\circ }-\frac{\sqrt{3}}{4}{{r}^{2}} \right)\] |
| \[=\left\{ \left( 3.14\times 10\times 10\times \frac{60{}^\circ }{360{}^\circ } \right)-\left( \frac{\sqrt{3}}{4}\times 10\times 10 \right) \right\}\] |
| \[=\left( \frac{157}{3}-25\times \sqrt{3} \right)=\left( \frac{157}{3}\,\,\frac{433}{10} \right)\] |
| \[=\frac{(1570-1299)}{30}=\frac{271}{30}=9.03c{{m}^{2}}\] |
| Area of major segment |
| \[=\pi {{r}^{2}}-\] Area of minor segment |
| \[=[(\pi \times 10\times 10)-9.03]\,c{{m}^{2}}\] |
| \[=\{(3.14\times 100)-9.03\}c{{m}^{2}}\] |
| \[=(314-9.03)c{{m}^{2}}=304.97c{{m}^{2}}\] |
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