SOLUTION: If the height of cone is decreased by 64% then by what percent should the radius be raised to keep its volume same?
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Question 760227: If the height of cone is decreased by 64% then by what percent should the radius be raised to keep its volume same? Answer by kmadison(20) (Show Source):
You can put this solution on YOUR website! Let r be the original radius, R the new radius, h the original height and H the new decreased height. Then we have:
1.) volume of original cone = (1/3)pi*(r^2)*h
2.) volume of cylinder with decreased height = (1/3)pi*(r^2)*H (100-64 gives us the remaining height) H = .36h
For the volume not to change we must have:
(1/3)pi*(r^2)*h = (1/3)pi*(r^2)*H or (1/3)pi*(r^2)*0.36*h
Divide both sides by (1/3)*pi
(r^2)*h = 0.36*(R^2)*h
Divide both sides by h (we assume the height is not zero)
r^2 = 0.36(R^2)
Take the square root of both sides:
r = 0.6R or r = -0.6R (the negative sign only tells the direction of the cone)
Now divide by 0.6 or 3/5 as a fraction:
5/3r = R = 1 2/3r
So the radius needs to be increased by 2/3 to maintain the volume of the cone.
