SOLUTION: A cylindrical rod of length "h" is melted and cast into a cone of base radius twice that one of the cylinder. What is the height of the cone??.. please reply... THANKYOU VERY MUCH
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Question 923589: A cylindrical rod of length "h" is melted and cast into a cone of base radius twice that one of the cylinder. What is the height of the cone??.. please reply... THANKYOU VERY MUCH Found 2 solutions by ewatrrr, Theo:Answer by ewatrrr(24785) (Show Source):
the length of the cylinder is equal to the height of the cylinder.
the volume of the cylinder is equal to pi * r^2 * h
the volume of the cone is equal to 1/3 * pi * (2r)^2 * x
x is the height of the cone.
since their volumes are equal, you get:
pi * r^2 * h = 1/3 * pi * (2r)^2 * x
simplify to get:
pi * r^2 * h = 1/3 * pi * 4r^2 * x
simplify further to get:
pi * r^2 * h = 4/3 * pi * r^2 * x
divide both sides of this equation by pi * r^2 and you get:
h = 4/3 * x
solve for x and you get:
x = 3/4 * h
the height of the cone will be 3/4 * the height of the cylinder and the radius of the base of the cone will be 2 times the radius of the base of the cylinder and the volume of the cone will be equal to the volume of the cylinder.
to test, give some values to r and h and see if the volumes are the same.
let r = 3 and h = 8 for the cylinder.
then 2r = 6 and 3/4 * h = 6 and you get:
r = 6 and h = 6 for the cone.
the volume of the cylinder will be pi * r^2 * h = pi * 3^2 * 8 = 72 * pi.
the volume of the cone will be 1/3 * pi * r^2 * h = 1/3 * pi * 6^2 * 6 = = 72 * pi.
the volumes are the same so the formula looks good.
