A calf is tied with a rope of length 6 ,m at | vedclass

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Question

(C) Let the calf be tied at the corner $A$ of the square lawn.The increase in area is the difference between the areas of two sectors with a central a

Vedclass · Accepted Answer

$75.625$

Vedclass · Answer

(C) Let the calf be tied at the corner $A$ of the square lawn.The increase in area is the difference between the areas of two sectors with a central angle of $90^{\circ}$ each and radii $R = 11.5 \,m$ $(6 \,m + 5.5 \,m)$ and $r = 6 \,m$.Required increase in area $= 	ext{Area of sector with radius } R - 	ext{Area of sector with radius } r$$= \frac{90^{\circ}}{360^{\circ}} 	imes \pi 	imes R^2 - \frac{90^{\circ}}{360^{\circ}} 	imes \pi 	imes r^2$$= \frac{1}{4} 	imes \pi 	imes (R^2 - r^2)$$= \frac{1}{4} 	imes \frac{22}{7} 	imes (11.5^2 - 6^2)$$= \frac{1}{4} 	imes \frac{22}{7} 	imes (11.5 + 6) 	imes (11.5 - 6)$$= \frac{1}{4} 	imes \frac{22}{7} 	imes 17.5 	imes 5.5$$= \frac{1}{4} 	imes \frac{22}{7} 	imes \frac{35}{2} 	imes \frac{11}{2}$$= \frac{11 	imes 5 	imes 11}{8} = \frac{605}{8} = 75.625 \,m^2$.

$A$ calf is tied with a rope of length $6 \,m$ at the corner of a square grassy lawn of side $20 \,m$. If the length of the rope is increased by $5.5 \,m$,find the increase in area of the grassy lawn in which the calf can graze. (in $m^2$)

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