| A copper wire is bent in the form of an equilateral triangle and has an area \[121\sqrt{3}\,\,c{{m}^{2}}.\]If the same wire is bent into the form of a circle, then the area \[(in\,\,c{{m}^{2}})\] enclosed by the wire is [SSC (CGL) 2011] |
A) 364.5
B) 346.5
C) 693.5
D) 639.5
Correct Answer: B
Solution :
| Let a be the side of equilateral triangle. |
| \[\therefore \] \[\frac{\sqrt{3}}{4}{{a}^{2}}=121\sqrt{3}\] |
| \[\Rightarrow \] \[{{a}^{2}}=121\times 4\] |
| \[\Rightarrow \] \[a=11\times 2=22\,\,cm\] |
| \[\therefore \]Perimeter of triangle \[=3\times 22=66\,\,cm\] |
| \[\Rightarrow \]Circumference of circle = Perimeter of equilateral triangle |
| \[\Rightarrow \] \[2\pi r=66\] |
| \[\Rightarrow \] \[r=\frac{66\times 7}{2\times 22}=\frac{21}{2}=10.5\] |
| Hence, area of circle \[=\pi \,\,({{r}^{2}})\] |
| \[=\frac{22}{7}\times {{(10.5)}^{2}}=346.5\,\,c{{m}^{2}}\] |
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