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The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? Assume that the cow is able to move on all sides with equal ease. Use \(\pi = \frac{22}{7}\) in your calculations.
A. 1594 sq m
B. 1694 sq m
C. 1696 sq m
D. 1756 sq.m
E. 1896 sq.m
A. 1594 sq m
B. 1694 sq m
C. 1696 sq m
D. 1756 sq.m
E. 1896 sq.m
14 Mar 2016, 11:28
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Length of rope will act as the radius of circular region which the cow can graze
Additional ground which the cow will be able to graze =\(\pi\) * ( 30^2 - 19^2)
= 22/7 (900 - 361)
=22/7 * 539
=22 * 77
= 1694 sq m
Answer B
Additional ground which the cow will be able to graze =\(\pi\) * ( 30^2 - 19^2)
= 22/7 (900 - 361)
=22/7 * 539
=22 * 77
= 1694 sq m
Answer B
14 Mar 2016, 11:49
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The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? Assume that the cow is able to move on all sides with equal ease. Use \(\pi = \frac{22}{7}\) in your calculations.
A. 1594 sq m
B. 1694 sq m
C. 1696 sq m
D. 1756 sq.m
E. 1896 sq.m
A. 1594 sq m
B. 1694 sq m
C. 1696 sq m
D. 1756 sq.m
E. 1896 sq.m
Since cow can only eat around the area in circular motion, considering it is tied in field, the extra area will be of shape of donut which is available due to extra length.
\(\frac{22}{7} * (30^2 - 19^2)\) => \(\frac{22}{7}*(49)(11)\) => 1694
