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 923588: 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 Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! volume(V) of the cylinder is pi*r^2*h
volume(V) of the cone is (1/3)*pi*(2r)^2*hc where hc is height of cone
volumes are the same so set equations equal to each other
pi*r^2*h = (1/3)*pi*(2r)^2*hc
pi*r^2*h = (1/3)*pi*4r^2*hc
cancel what we can from both sides of =
h = (1/3)*4*hc
multiply both sides of = by 3
3h = 4*hc
hc = 3h/4
therefore the height of the cone is 3/4 of the height of the cylinder
