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5. 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 5.5m, find the increase in the area of the grassy lawn in which the calf can graze.

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To solve this problem, we first find the area the calf can graze with a rope of length 6 m and then find the area it can graze with a rope of length 11.5 m. The difference between these areas will give us the increase in the area of the grassy lawn the calf can graze.
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To solve this problem, we first find the area the calf can graze with a rope of length 6 m and then find the area it can graze with a rope of length 11.5 m. The difference between these areas will give us the increase in the area of the grassy lawn the calf can graze.
$1.&space;**Area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;6&space;m:**&space;When&space;the&space;calf&space;is&space;tied&space;with&space;a&space;rope&space;of&space;length&space;6&space;m,&space;it&space;can&space;graze&space;within&space;a&space;circle&space;whose&space;radius&space;is&space;6&space;m.&space;This&space;forms&space;a&space;circular&space;grazing&space;area.&space;The&space;area&space;of&space;the&space;circular&space;grazing&space;area&space;is&space;given&space;by&space;$$&space;\pi&space;r^2&space;$$,&space;where&space;$$&space;r&space;$$&space;is&space;the&space;radius&space;of&space;the&space;circle&space;(6&space;m).&space;Therefore,&space;the&space;area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;6&space;m&space;is&space;$$&space;\pi&space;\times&space;6^2&space;=&space;36\pi&space;$$&space;square&space;meters.&space;2.&space;**Area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;11.5&space;m:**&space;When&space;the&space;length&space;of&space;the&space;rope&space;is&space;increased&space;to&space;11.5&space;m,&space;the&space;radius&space;of&space;the&space;circular&space;grazing&space;area&space;also&space;increases&space;to&space;11.5&space;m.&space;The&space;area&space;of&space;the&space;circular&space;grazing&space;area&space;with&space;a&space;radius&space;of&space;11.5&space;m&space;is&space;$$&space;\pi&space;\times&space;11.5^2&space;$$&space;square&space;meters.&space;Therefore,&space;the&space;area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;11.5&space;m&space;is&space;$$&space;\pi&space;\times&space;11.5^2&space;$$&space;square&space;meters.$

$Now,&space;let's&space;find&space;the&space;difference&space;between&space;the&space;areas&space;the&space;calf&space;can&space;graze&space;with&space;these&space;two&space;rope&space;lengths:&space;$&space;\text{Increase&space;in&space;area}&space;=&space;(\pi&space;\times&space;11.5^2)&space;-&space;(\pi&space;\times&space;6^2)&space;$&space;$&space;\text{Increase&space;in&space;area}&space;=&space;\pi&space;\times&space;(11.5^2&space;-&space;6^2)&space;$&space;$&space;\text{Increase&space;in&space;area}&space;=&space;\pi&space;\times&space;(132.25&space;-&space;36)&space;$&space;$&space;\text{Increase&space;in&space;area}&space;=&space;\pi&space;\times&space;96.25&space;$&space;Using&space;$$&space;\pi&space;\approx&space;\frac{22}{7}&space;$$:&space;$&space;\text{Increase&space;in&space;area}&space;\approx&space;\frac{22}{7}&space;\times&space;96.25&space;$&space;$&space;\text{Increase&space;in&space;area}&space;\approx&space;303.75&space;\,&space;\text{square&space;meters}&space;$&space;So,&space;the&space;increase&space;in&space;the&space;area&space;of&space;the&space;grassy&space;lawn&space;in&space;which&space;the&space;calf&space;can&space;graze&space;is&space;approximately&space;$$&space;303.75&space;\,&space;\text{square&space;meters}&space;$$.$

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