John has a rectangular shaped field whose length is 62.5 yards and width is 45.3
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The radius r used in below equations would equal 2in because it is half the given diameter of 4in.
The volume of the cone would be
The volume of a cylinder with the same dimensions would be
The cylinder is 24 - 8 = 16 cubic inches greater.
The volume of the cone would be
The volume of a cylinder with the same dimensions would be
The cylinder is 24 - 8 = 16 cubic inches greater.