A) 20 to 22%
B) 25 to 27%
C) 30 to 32%
D) 48 to 50%
Correct Answer: C
Solution :
Let D be the middle point of AB. Draw a circular arc DE. Then the horse can graze over the portion ADEA. Let \[AB=x.\]Then \[AD=\frac{x}{2}\]. \[\therefore \]Area of portion \[ADEA=\frac{\pi }{6}.{{\left( \frac{x}{2} \right)}^{2}}=\frac{\pi {{x}^{2}}}{24}\] and total area of the field \[=\frac{{{x}^{2}}\sqrt{3}}{4}\] Required \[%=\frac{\frac{\pi {{x}^{2}}}{24}}{\frac{{{x}^{2}}\sqrt{3}}{4}}\times 100\] \[=\frac{\pi }{6\sqrt{3}}\times 100\] \[=30.2%\] (approx.)You need to login to perform this action.
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