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    A frustrum is cut from a cone. The radius of the bottom circle is double the radius of the top circle, and the height of the frustrum is h.
    Write down the volume of the Frustrum in terms of r and h.

    So I worked out that the height of the entire cone is 2h. Then when I work out the volumes of each cone I can’t seem to simplify when I subtract to work out the volume of the frustrum. Anyone got any ideas?
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    What's r?
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    (Original post by IrrationalRoot)
    What's r?
    r is r which is why it wants the answer in terms of r and h
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    (Original post by Y11_Maths)
    A frustrum is cut from a cone. The radius of the bottom circle is double the radius of the top circle, and the height of the frustrum is h.
    Write down the volume of the Frustrum in terms of r and h.

    So I worked out that the height of the entire cone is 2h. Then when I work out the volumes of each cone I can’t seem to simplify when I subtract to work out the volume of the frustrum. Anyone got any ideas?
    I'm not sure what you're expecting as the answer other than simplify, that's literally the only thing you can do, so do it correctly.
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    (Original post by Y11_Maths)
    r is r which is why it wants the answer in terms of r and h
    No, I mean what is r besides being just some random letter? You've forgotten to define it. Do you mean the radius of the top circle?
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    (Original post by RDKGames)
    I'm not sure what you're expecting as the answer other than simplify, that's literally the only thing you can do, so do it correctly.
    I don’t know how to simplify it though. I’ve seen the answer on the mark scheme and it was a fraction with pi on the bottom but I don’t know how it got to there
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    (Original post by IrrationalRoot)
    No, I mean what is r besides being just some random letter? You've forgotten to define it. Do you mean the radius of the top circle?
    That’s the question quoted exactly. r is the radius of the top circle and 2r is the radius of the bottom
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    (Original post by Y11_Maths)
    I don’t know how to simplify it though. I’ve seen the answer on the mark scheme and it was a fraction with pi on the bottom but I don’t know how it got to there
    Sounds wrong. What is it?
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    (Original post by RDKGames)
    Sounds wrong. What is it?
    I can’t remember it exactly since it was in class, all I know is that is was a fraction and there were some numbers with pi in the numerator. I’m left with the volumes of the big and little cone and I don’t know how to simplify it
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    (Original post by Y11_Maths)
    I can’t remember it exactly since it was in class, all I know is that is was a fraction and there were some numbers with pi in the numerator. I’m left with the volumes of the big and little cone and I don’t know how to simplify it
    Oh ok, I thought you meant pi in the denominator.

    Not sure what you expect me to say... what did you end up with?
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    (Original post by Y11_Maths)
    That’s the question quoted exactly. r is the radius of the top circle and 2r is the radius of the bottom
    If you quoted the question exactly, then it's a very bad question for not even telling you what it means by 'r'.

    Anyway I'll leave this to RDK, but I should point out that if you've corrected the volume correctly, all you can to to simplify it is take out common factors, and if you don't get the same result as the mark scheme then the mark scheme must be wrong.
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    No I did mean denominator that was a typo sorry. I will send you a picture
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    My teacher wasn’t sure if the mark scheme was correct anyways so it could be wrong
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    Name:  E48B1194-5200-4CC6-8C53-525624A09B87.jpg.jpeg
Views: 7
Size:  28.7 KB
    I don’t know what to do??
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    (Original post by Y11_Maths)
    I don’t know what to do??
    Looks fine apart from the last line where you lose \pi.

    The final answer is \dfrac{7\pi}{3}r^2 h so there should not be \pi in the denominator.
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    (Original post by RDKGames)
    Looks fine apart from the last line where you lose \pi.

    The final answer is \dfrac{7\pi}{3}r^2 h so there should not be \pi in the denominator.
    Thank you. Please can you explain how you got to this conclusion? Thanks
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    (Original post by Y11_Maths)
    Thank you. Please can you explain how you got to this conclusion? Thanks
    Take your second to last line and factor everything that is common out
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    (Original post by Y11_Maths)
    A frustrum is cut from a cone. The radius of the bottom circle is double the radius of the top circle, and the height of the frustrum is h.
    Write down the volume of the Frustrum in terms of r and h.

    So I worked out that the height of the entire cone is 2h. Then when I work out the volumes of each cone I can’t seem to simplify when I subtract to work out the volume of the frustrum. Anyone got any ideas?
    Shouldn’t the total height be 3/2h because the smaller cone’s height should be half of the larger cone
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    (Original post by DipdyeSpork)
    Shouldn’t the total height be 3/2h because the smaller cone’s height should be half of the larger cone
    Let a be the height of the smaller cone. Let H be the height of the larger cone. Then H=a+h. We have linear scale factor of \frac{1}{2} from larger to smaller since the ratio of the radii is 2:1 and thus we have H=2a. Substitute into the previous eq. and we have 2a=a+h. hence a=h and hence the height of the smaller cone is h, meaning the height of the larger cone is H=2h

    :qed:
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    (Original post by Y11_Maths)
    Thank you. Please can you explain how you got to this conclusion? Thanks
    Yay I finally got it thank you so much!
 
 
 
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