A cow is tied to a corner (vertex) of a regular hexagonal fenced area of side a meters by a rope of length 5a/2 meters in a grass field. (The cow cannot graze inside the fenced area.) What is the maximum possible area of the grass field to which the cow has access to graze?

5 π a^{2}

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\frac{5}{2} π a^{2}

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6 π a^{2}

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3 π a^{2}

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Solution

The correct option is A $5 π a^{2}$

$A_{1} = A_{5} = \frac{60}{360} π {(\frac{a}{2})}^{2} = \frac{π a^{2}}{24}, A_{2} = A_{4} = \frac{60}{360} π {(\frac{3 a}{2})}^{2} = \frac{3 a^{2} π}{8}$
$A_{3} = \frac{240}{360} π {(\frac{5 a}{2})}^{2} = \frac{25 a^{2} π}{6}$
Area that can be grazed $= 2 (\frac{π a^{2}}{24}) + 2 (\frac{3 a^{2} π}{8}) + \frac{25 a^{2} π}{6} = 5 π a^{2}$

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A cow is tied to a corner (vertex) of a regular hexagonal fenced area of side a meters by a rope of length 5a/2 meters in a grass field. (The cow cannot graze inside the fenced area.) What is the maximum possible area of the grass field to which the cow has access to graze?

The correct option is A 5πa2A1=A5=60360π(a2)2=πa224,A2=A4=60360π(3a2)2=3a2π8A3=240360π(5a2)2=25a2π6Area that can be grazed =2(πa224)+2(3a2π8)+25a2π6=5πa2