SOLUTION: The largest sphere that will fit is placed inside a cone with base radius 𝑟 and height ℎ. What proportion of the cone is filled by the sphere?

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Question 1205736: The largest sphere that will fit is placed inside a cone with base radius 𝑟
and height ℎ.
What proportion of the cone is filled by the sphere?

Found 2 solutions by Edwin McCravy, mccravyedwin:
Answer by Edwin McCravy(20062) About Me  (Show Source):
You can put this solution on YOUR website!

First we look at the mid cross section.
Let the radius of the circle (which is also the radius of the sphere),
be "a".



Then the ratio of the volume of the sphere to the volume of the cone,
which is the answer you're looking for, will be

ratio%22%22=%22%22%28expr%284%2F3%29pi%2Aa%5E3%29%2F%28expr%281%2F3%29pi%2Ar%5E2%2Ah%29%22%22=%22%224a%5E3%2F%28r%5E2%2Ah%29

The hard part is finding the radius of the sphere "a", in terms
of "r" and "h".

By similar right triangles OPQ and TSQ,

OP%2FOQ=+TS%2FTQ

SQ = h
SO = a
OQ = SQ-SO = h-a
OP = a
TS = r
TQ=sqrt%28TS%5E2%2BSQ%5E2%29%22%22=%22%22sqrt%28r%5E2%2Bh%5E2%29

Substituting,

a%5E%22%22%2F%28h-a%5E%22%22%29%22%22=%22%22r%5E%22%22%2Fsqrt%28r%5E2%2Bh%5E2%29 

Cross-multiplying:

a%2Asqrt%28r%5E2%2Bh%5E2%29%22%22=%22%22r%28h-a%29  

a%2Asqrt%28r%5E2%2Bh%5E2%29%22%22=%22%22rh-ra 

a%2Asqrt%28r%5E2%2Bh%5E2%29%2Bra%22%22=%22%22rh

a%28sqrt%28r%5E2%2Bh%5E2%29%2Br%29%22%22=%22%22rh

a%22%22=%22%22%28rh%29%2F%28sqrt%28r%5E2%2Bh%5E2%29%2Br%29
 
Now we substitute that horrible thing in: 

ratio%22%22=%22%224a%5E3%2F%28r%5E2%2Ah%29

ratio%22%22=%22%224%28%28rh%29%2F%28sqrt%28r%5E2%2Bh%5E2%29%2Br%29%29%5E3%2F%28r%5E2%2Ah%29%22%22=%22%22expr%284%2F%28r%5E2%2Ah%29%29+%0D%0A%28%28rh%29%2F%28sqrt%28r%5E2%2Bh%5E2%29%2Br%29%29%5E3%22%22=%22%22expr%284%2F%28r%5E2%2Ah%29%29+%28r%5E3h%5E3%29%2F%28sqrt%28r%5E2%2Bh%5E2%29%2Br%29%5E3%22%22=%22%22%284r%2Ah%5E2%29%2F%28sqrt%28r%5E2%2Bh%5E2%29%2Br%29%5E3

Edwin

Answer by mccravyedwin(408) About Me  (Show Source):
You can put this solution on YOUR website!
Now the above problem is correct!

Edwin