SOLUTION: A hemisphere and a right circular cone on equal bases are of equal height. find the ratio of their volumes.
Algebra.Com
Question 1151079: A hemisphere and a right circular cone on equal bases are of equal height. find the ratio of their volumes.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Vs = Volume of sphere
Vs = (4/3)*pi*r^3
A hemisphere is half a sphere.
Vh = Volume of hemisphere
Vh = (1/2)*Vs
Vh = (1/2)*(4/3)*pi*r^3
Vh = (4/6)*pi*r^3
Vh = (2/3)*pi*r^3
We're told that both the hemisphere and cone have the same height.
The height of the hemisphere is the radius r, so for the cone, h = r.
Vc = Volume of cone
Vc = (1/3)*pi*r^2*h
Vc = (1/3)*pi*r^2*r ... plug in h = r
Vc = (1/3)*pi*r^3
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Divide the hemisphere volume over the cone volume.
The pi terms cancel.
The r^3 terms cancel.
Flip the second fraction and multiply.
The ratio of the hemisphere volume to the cone volume is 2:1.
This means the hemisphere has twice the volume of the cone.
Put another way,
Vh = 2*Vc
which can be rearranged to
Vc = (1/2)*Vh
This only works if the cone and hemisphere share the same circular base, and also have the same height (h = r).
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side note:
If we start with Vh = 2*Vc and plug in Vc = (1/3)*pi*r^3, then we get,
Vh = 2*Vc
Vh = 2*(1/3)*pi*r^3
Vh = (2/3)*pi*r^3
Or we could start with Vc = (1/2)*Vh and plug in Vh = (2/3)*pi*r^3 to get,
Vc = (1/2)*Vh
Vc = (1/2)*(2/3)*pi*r^3
Vc = (1/3)*pi*r^3
This helps confirm our answer.
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