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Math Help - right circular cylinder is inscribed in a right circular cone

  1. #1
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    right circular cylinder is inscribed in a right circular cone

    A right circular cylinder is inscribed in a right circular cone of altitude h and radius of base x, as shown in the figure. Find the radius of the cylinder if its lateral area is equal to the lateral area of the small cone which surmounts the cylinder.



    This is my effort and i don't know if my effort is really an effort.

    First i solved for the lateral area of the cone:
    A= xhpi(sqrt of (x^2 + h^2))

    Second i solved for the lateral area of the right cylinder:
    A = 2pirh

    And lastly i equated them and getting the result of the radius since the lateral area of the cone is equal to the lateral area of the right cylinder:
    xhpi(sqrt of (x^2 + h^2)) = 2pirh
    r = x(sqrt of (x^2 + h^2))/2

    THE CORRECT ANSWER stated on the book is:
    (2xh)/(2h+(sqrt of (x^2 + h^2)))

    I want to know where in my solution is wrong..Thanks!
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  2. #2
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    I think you are using the wrong formulas. Can you please tell me which is which, because I don't know what you mean by all the variables. In the drawing you are given r and h are the radius and height of the big cone.
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  3. #3
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    Quote Originally Posted by cutiemike1 View Post
    A right circular cylinder is inscribed in a right circular cone of altitude h and radius of base x, as shown in the figure. Find the radius of the cylinder if its lateral area is equal to the lateral area of the small cone which surmounts the cylinder.

    ...Thanks!
    1. I've modified your sketch a little bit. (see attachment)

    2. As far as I understand the question you have 2 constant values (r, h) and you are looking for the radius x of the cylinder which satisfies the given conditions(?).

    If so:

    3. Use proportions:

    \frac tx = \frac hr~\implies~t = \frac{h x}r

    4. Lateral surface of the cylinder: A_{cyl} = 2\cdot \pi \cdot x \cdot h

    Lateral surface of the small cone on top: A_{topcon} = \pi\cdot x \cdot \sqrt{x^2+t^2}

    Both areas are equal:

    2\cdot \pi \cdot x \cdot h = \pi\cdot x \cdot \sqrt{x^2+t^2}

    5. This can be simplified to:

    x^2+t^2 = 4h^2. Substitute t from 3.:

    x^2+\left(\frac{h x}r\right)^2 = 4h^2 and solve for x.
    Attached Thumbnails Attached Thumbnails right circular cylinder is inscribed in a right circular cone-cyl_in_cone.png  
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    OH! I AM REALLY SORRY!!! I messed up my question.

    Ok.. h refers to the height of the cone and r is supposed to be x as the base radius of the cone..

    @earboth.. where and how did you get x^2 + t^2 = 4h^2??
    Last edited by cutiemike1; August 22nd 2010 at 01:34 AM.
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  5. #5
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    I wrote

    Quote Originally Posted by earboth View Post
    ...
    Both areas are equal:

    2\cdot \pi \cdot x \cdot h = \pi\cdot x \cdot \sqrt{x^2+t^2}

    5. This can be simplified to:

    x^2+t^2 = 4h^2.
    Quote Originally Posted by cutiemike1 View Post
    OH! I AM REALLY SORRY!!! I messed up my question.

    Ok.. h refers to the height of the cone and r is supposed to be x as the base radius of the cone..

    @earboth.. where and how did you get x^2 + t^2 = 4h^2??
    I'll show you the steps of transformation:

    2\cdot \pi \cdot x \cdot h = \pi\cdot x \cdot \sqrt{x^2+t^2}

    This equation implies that x = 0 which isn't very plausible with your question. So divide the equation through by \pi x which yields:

    2\cdot h =  \sqrt{x^2+t^2}

    To get rid of the square-root square both sides of the equation:

    4h^2 = x^2+t^2
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  6. #6
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    got it!...but it is different from the answer written in the book? how's that?
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  7. #7
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    Quote Originally Posted by cutiemike1 View Post
    got it!...but it is different from the answer written in the book? how's that?
    Well ... that's easy to explain:

    Quote Originally Posted by cutiemike1 View Post
    OH! I AM REALLY SORRY!!! I messed up my question.

    ...
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  8. #8
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    oh sorry sir..it really is my bad.
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  9. #9
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    so whats the final answer to this?
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