A) \[160\sqrt{2}\,m\]
B) \[260\sqrt{2}\,m\]
C) \[360\sqrt{2}\,m\]
D) \[460\sqrt{2}\,m\]
Correct Answer: C
Solution :
| [c] Let the radius of the field be r. |
| Then, \[\frac{\pi {{r}^{2}}}{2}=15400\] |
| \[\frac{1}{2}\times \frac{22}{7}\times {{r}^{2}}=15400\] |
| \[{{r}^{2}}=15400\times 2\times \frac{7}{22}=9800\] |
| \[r=70\sqrt{2}\,m\] |
| Thus, perimeter of the field |
| \[=\pi r+2r\] |
| \[=\frac{22}{7}\times 70\sqrt{2}+2\times 70\times \sqrt{2}\] |
| \[=220\sqrt{2}+140\sqrt{2}\] |
| \[=\sqrt{2}(220+140)=360\,\sqrt{2}m\] |
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