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(B) The interior angle of a regular hexagon is $120^\circ$. The exterior angle is $360^\circ - 120^\circ = 240^\circ$. The cow can graze in a sector o

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$\frac{5}{2} \pi a^2$

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(B) The interior angle of a regular hexagon is $120^\circ$. The exterior angle is $360^\circ - 120^\circ = 240^\circ$. The cow can graze in a sector of radius $R = \frac{5a}{2}$ and angle $240^\circ$. As the rope wraps around the vertices of the hexagon,the radius decreases by $a$ at each turn. $1$. Sector $1$: Radius $R_1 = \frac{5a}{2}$,Angle $	heta_1 = 240^\circ = \frac{2\pi}{3}$ radians. Area $A_1 = \frac{1}{2} R_1^2 	heta_1 = \frac{1}{2} (\frac{5a}{2})^2 (\frac{2\pi}{3}) = \frac{25\pi a^2}{12}$. $2$. Sector $2$: Radius $R_2 = \frac{5a}{2} - a = \frac{3a}{2}$,Angle $	heta_2 = 60^\circ = \frac{\pi}{3}$ radians. Area $A_2 = \frac{1}{2} R_2^2 	heta_2 = \frac{1}{2} (\frac{3a}{2})^2 (\frac{\pi}{3}) = \frac{3\pi a^2}{8}$. $3$. Sector $3$: Radius $R_3 = \frac{3a}{2} - a = \frac{a}{2}$,Angle $	heta_3 = 60^\circ = \frac{\pi}{3}$ radians. Area $A_3 = \frac{1}{2} R_3^2 	heta_3 = \frac{1}{2} (\frac{a}{2})^2 (\frac{\pi}{3}) = \frac{\pi a^2}{24}$. Total Area $= A_1 + A_2 + A_3 = \pi a^2 (\frac{25}{12} + \frac{3}{8} + \frac{1}{24}) = \pi a^2 (\frac{50 + 9 + 1}{24}) = \pi a^2 (\frac{60}{24}) = \frac{5}{2} \pi a^2$.

$A$ cow is tied to a corner (vertex) of a regular hexagonal fenced area of side $a \ m$ by a rope of length $\frac{5a}{2} \ m$ 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?

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