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Math Help - Perplexing Mensuration Question - sphere in a cone

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Perplexing Mensuration Question - sphere in a cone

    Hello everyone.

    This maths question is somewhat beyond my ken. One part of it talks about similarity while the other talks about mensuration. Still, I could not find an interrelationship between the clues given.

    The image is at Mensuration Question | Flickr

    What I know intuitively is that there is a Pythagoras' Theorem involved in finding the radius of the sphere. However, VD, which is 8cm, is solely a part of the hypotenuse. If I have the value of the hypotenuse, then I can use the Pythagoras' Theorem to find the radius of the cone.

    Also, since triangle VDO is similar to triangle VCA, then it would be VD=VC, VO=VA and OD=CA.

    Can anyone give me hints? Help is greatly appreciated!
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  2. #2
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    Hello!

    Here is a hint for you!

    Let the radius of the sphere be r.
    Since VD = 8cm, by the pythagoras theorem, VO is  \sqrt{64+r^2} . Since OC is the radius, it has also length r.

    Now VC = VO + OC = 12 cm. Substituting, you'll get  r + \sqrt{64 + r^2} = 12

    Move the r over, square both sides, and solve accordingly.

    I did this quickly, so you may want to check the working for the equations.
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  3. #3
    Junior Member PythagorasNeophyte's Avatar
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    Hello Gusbob!

    Thank you for your reply.

    So here's how it would be done:

    r + \sqrt{64 + r^2} = 12

    r + 64 + r^2 = 144

    r + r^2 = 80

    Oh no... it seems that I have to solve two unknowns. The answer may be wrong.
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  4. #4
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    You made a mistake squaring: you need to square  r + \sqrt{64+r^2} as a whole, or move the r over with the 12.

    Here is how I'll do it.

     r + \sqrt{64 + r^2} = 12
     \sqrt{64+r^2} = 12 -r
     64 + r^2 = 144 - 24r + r^2
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  5. #5
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    Quote Originally Posted by PythagorasNeophyte View Post
    Hello Gusbob!

    Thank you for your reply.

    So here's how it would be done:

    r + \sqrt{64 + r^2} = 12

    r + 64 + r^2 = 144

    r + r^2 = 80

    Oh no... it seems that I have to solve two unknowns. The answer may be wrong.
    Sorry ... but what you've done is nearly a crime.

    As Gusbob had told you, you have to move the linear r to the RHS first:

    \sqrt{64+r^2}=12-r

    Now square both sides:

    64+r^2=144-24r+r^2

    Now solve for r.
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  6. #6
    Junior Member PythagorasNeophyte's Avatar
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    Oh, I see. It seems that I had flopped my algebra.

    64 + r^2 = 144 - 24r + r^2

    64-144 + r^2 - r^2 +24r = 0

    -80 +24r = 0

    24r = 80

    r = 3\left(\frac{1}{3}\right)

    Now that I had found the radius of the sphere, I must find the volume of it:

    \dfrac{4}{3}\pi r^3

    = \dfrac{4}{3}\pi (\dfrac{10}{3)}^3

    = \dfrac{4}{3}\pi \dfrac{100}{9}

    = \dfrac{400}{27}\pi

    Can anyone help me?
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  7. #7
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    Quote Originally Posted by PythagorasNeophyte View Post
    Oh, I see. It seems that I had flopped my algebra.

    ....
    Now that I had found the radius of the sphere, I must find the volume of it:

    \dfrac{4}{3}\pi r^3

    = \dfrac{4}{3}\pi \left(\dfrac{10}{3}\right)^3

    = \dfrac{4}{3}\pi \dfrac{100}{9} <--- You have squared the value of the radius

    ...
    The correct calculation:

    = \dfrac{4}{3}\pi \cdot \dfrac{1000}{27} = \dfrac{4000}{81} \pi
    Last edited by earboth; September 28th 2010 at 03:41 AM. Reason: typo
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  8. #8
    Junior Member PythagorasNeophyte's Avatar
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    I am very sorry, earboth.

    So now I would have to find the radius of the cone.

    I discern that triangle VDO is similar to triangle VCA.

    So:

    \dfrac{CA}{3\left(\frac{1}{3}\right)} = \dfrac{12}{8] (corr. sides of similar triangles)

    CA = 5cm

    Volume of the cone is \dfrac{1}{3}\pi r^2h

    = \dfrac{!}{3}\pi5^2(!2)

    = 100\pi

    So the volume of cone that is not occupied by the sphere is:

    100\pi - \dfrac{4000}{81} \pi

    Am I on the right track?
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  9. #9
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    Quote Originally Posted by PythagorasNeophyte View Post
    I am very sorry, earboth. <--- why?

    So now I would have to find the radius of the cone.

    I discern that triangle VDO is similar to triangle VCA.

    So:

    \dfrac{CA}{3\left(\frac{1}{3}\right)} = \dfrac{12}{8} (corr. sides of similar triangles)

    CA = 5cm <<<<<<

    Volume of the cone is \dfrac{1}{3}\pi r^2h

    = \dfrac{1}{3}\pi5^2(12)

    = 100\pi <<<<<<

    So the volume of cone that is not occupied by the sphere is:

    100\pi - \dfrac{4000}{81} \pi

    Am I on the right track?
    Perfect!

    If you simplify your result you'll see that the sphere nearly takes half of the volume of the cone.

    (Btw: I've corrected your LaTeX)
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