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Thread: A question I had in an exam.

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    Default A question I had in an exam.

    This is a question which I had in an exam yesterday, which I don't think anyone answered.

    There is a hemisphere and a cone.
    Both shapes have base radius x.
    The shapes both have the same surface area.
    Work out the height of the cone (h) in terms of x.

    Any idea on how to do it?
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    well i have been out of math for a few years now... but does h=4x

    pie(x)(s)+pie(x)^2=4pie(x)^2 (surface area of cone = surface area of sphere)

    pie(x)(s) = 3pie(x)^2 (move the pie(x)^2 over)

    s=3x (divide the pie(x) out...?)

    h^2 = 3x^2 +x^2 (Pythagorean theorem)

    h^2 = 4x^2

    h=4x

    no clue whats so ever if this is correct

    edit: LOL i totally did Pythagorean theorem wrong XD . ehh i should get into a math class
    Last edited by x[Warrior]x3500; 06-07-2011 at 03:13 PM.

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    Quote Originally Posted by x[Warrior]x3500 View Post
    well i have been out of math for a few years now... but does h=4x
    How would you get that?
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    I don't care if you're black, white, straight, bisexual, gay, lesbian, short, tall, fat, skinny, rich or poor. If you're nice to me, I'll be nice to you. Simple as that.

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    from seeing this question i can see we can work out h through Pythagoras therom as long as we have r and s. Radius (r) = x so we just need to find s.


    Start with the equations for surface area:
    surface area of a cone = pi x (s + x)
    surface area of a sphere = 4 pi x^2
    surface area of a circle = pi x^2
    area of a hemisphere = 1/2 4 pi x^2 + pi x^2
    = 3 pi x^2

    make the equations = each other:
    pi x (s + x) = 3 pi x^2
    rearrange to make s the subject
    s + x = 3 pi x^2 / pi x
    pi and an x cancel
    s + x = 3 x
    s = 2x

    using Pythagoras:

    s^2 = x^2 + h^2
    substitute s in:
    4x^2 = x^2 + h^2
    rearrange to make h the subject:
    sqr( 4 x^2 - x^2 ) = h
    h = sqr( 3 x^2)

    hope that made sence

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    Quote Originally Posted by x[Warrior]x3500 View Post
    well i have been out of math for a few years now... but does h=4x

    pie(x)(s)+pie(x)^2=4pie(x)^2 (surface area of cone = surface area of sphere)

    pie(x)(s) = 3pie(x)^2 (move the pie(x)^2 over)

    s=3x (divide the pie(x) out...?)

    h^2 = 3x^2 +x^2 (Pythagorean theorem)

    h^2 = 4x^2

    h=4x

    no clue whats so ever if this is correct
    right idea but "There is a hemisphere and a cone"
    not a sphere :P

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    I feel like a noob, lol, we just started Algebra. Anyways, what kind of math are you guys doing, seems kind of crazy....

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    Quote Originally Posted by BazzBarrett View Post
    *SNIP*
    hope that made sence
    Yep, that makes perfect sense. Now I've seen it, it's actually fairly easy. Thanks

    Quote Originally Posted by Huthaifah View Post
    I feel like a noob, lol, we just started Algebra. Anyways, what kind of math are you guys doing, seems kind of crazy....
    EdExcel GCSE Modular (UK).
    <3

    Quote Originally Posted by Eminem
    I don't care if you're black, white, straight, bisexual, gay, lesbian, short, tall, fat, skinny, rich or poor. If you're nice to me, I'll be nice to you. Simple as that.

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