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Shape/Circle Questions

  1. Aug 17, 2009 #1
    1. The problem statement, all variables and given/known data
    page 9 question 12
    http://www.physics.ox.ac.uk/admissions/undergraduate/admissions_test/tests06.pdf [Broken]

    page 3 question 12
    http://www.physics.ox.ac.uk/admissions/undergraduate/admissions_test/tests08.pdf [Broken]

    3. The attempt at a solution

    I really REALLY struggle with these types of question and honestly, i can't see any way to do it (in my knowledge) I think maybe it relies on certain assumptions about circles i'm not sure about but the annoying thing is I think they should be easy :(....

    thanks for any help / points in the right direction, they're so annoying :/

    also bear in mind a calculator isn't allowed :(
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 17, 2009 #2
    Some observations on the first problem. It looks like the centers of the circle lie on the squares diagonal. The distance from the center of the large circle to the center of the small circle is known. That distance becomes the hypotenuse of a triangle that is similar to the triangle that formed by the diagonal and two sides of the square. It is aglebra from there on out.
     
  4. Aug 18, 2009 #3
    Some observations on the second problem. You will need to find the length of AO and the height of triangle ABC. Set all the equal lengths to 1 unit. Make a triangle AOC. What are the interior angles of AOC? You should be able to find the lenght AO and the distance from O to the base of the triangle AOC if you know the side relationships of a 30-60-90 triangle.
     
  5. Aug 18, 2009 #4
    Ah that's great thanks! I got the 2nd one however still having a bit of trouble with the first;

    i've made the triangle with 3r diagonally and (4.5)^0.5 r on each side and like you said i know it's similiar so the edge of the square must be some product of (4.5)^0.5 however I've no idea how to go about working that out >.< as soon as I know that i've got the answer....

    Thanks for the help though, just need to get this :P
     
  6. Aug 18, 2009 #5
    Also i got a REALLY nasty answer for the second, I know it's eight marks but I got;

    ( [tex]\frac{13\pi}{12\sqrt{0.75}}[/tex] + [tex]\frac{4\pi}{3}[/tex] ) : 1...um...lol which is around 8 : 1, do you reckon that's about right? I know it's meant to be a hard test, but that's ridiculous lol
     
    Last edited: Aug 18, 2009
  7. Aug 18, 2009 #6
    On the first problem, notice that you can draw an extra square with its upper left corner coinciding with the upper left corner and its lower right corner in the center of the big circle. The lengths of the sides of this square should be easy to find, as well as its diagonal. From there you can calculate the length of the diagonal of the big square and from there its sides and area.

    On the second one I got a different answer, [tex] (4/3 + 4/\sqrt{3})\pi[/tex]. Can you show what your approach was?
     
  8. Aug 18, 2009 #7
    that's great, genius! :P
     
  9. Aug 19, 2009 #8
    For the first problem, with the square and the two circles of radii R and 2R, I have got an answer that I'm not sure about. I worked it out myself and I think I did was kanato is explaining. Can I show you my result? I think it is [tex]\frac{10\pi}{9(1 + \sqrt{2} )^2}[/tex]? But I'm not sure.
     
    Last edited: Aug 19, 2009
  10. Aug 19, 2009 #9
    Here are hints for the first problem:
    n6ul4g.png

    If d is the diagonal of the big square then [tex]d=r\sqrt{2}+r+\frac{r}{2}+\frac{r\sqrt{2}}{2}[/tex]

    And if the side of the square is a then [tex]a=\frac{d}{\sqrt{2}}[/tex]

    Regards.
     
  11. Aug 19, 2009 #10
    Here are hints for the second one:

    21nqtkk.png

    The area of the ring is [itex]P=(R^2-r^2)\pi[/itex]

    [tex]r=\frac{2}{3}h=\frac{a\sqrt{3}}{3}[/tex]

    and

    [tex]a=r\sqrt{3}[/tex]

    so

    [tex]R=a+r=r(1+\sqrt{3})[/tex]

    Now [itex]P=r^2\pi[(1+\sqrt{3})^2-1][/itex]

    And the area of the spherical segment is [itex]S=\frac{R^2*\pi*120}{360}[/itex]

    Finally the area of ADEC is [itex]P_{ADEC}=P-S=(R^2-r^2)\pi-\frac{R^2*\pi*120}{360}+\frac{r^2*\pi*120}{360}[/itex]

    [tex]P_{ABC}=\frac{a^2\sqrt{3}}{4}[/tex]

    Regards.
     
    Last edited: Aug 20, 2009
  12. Aug 19, 2009 #11
    okay this is what i've got (i'm pretty sure about it)

    [tex]\frac{2\pi}{3}[/tex] (1 + [tex]\frac{2}{\sqrt{3}}[/tex]) : [tex]\frac{\sqrt{3}}{4}[/tex]
     
  13. Aug 19, 2009 #12
    I came up with:

    [tex]\frac{8\pi\sqrt{3}(5+2\sqrt{3})}{27}[/tex]

    Regards.
     
    Last edited: Aug 20, 2009
  14. Aug 19, 2009 #13
    mkay...i don't think that's it, firstly it's supposed to be a ratio, and secondly i've done it a couple of times and so has another guy (who's good at maths) and we both my answer... not sure though xO
     
  15. Aug 20, 2009 #14
    Yep, its ratio:

    [tex]\frac{P_{ADEC}}{P_{ABC}}=\frac{\frac{2r^2\pi(5+2\sqrt{3})}{3}}{\frac{3r^2\sqrt{3}}{4}}[/tex]

    When you divide both of them (since it is ratio) you come up with the final result.

    Do you have some results so we can compare it?

    Regards.
     
    Last edited: Aug 20, 2009
  16. Aug 20, 2009 #15
    Hmm i'll write out the working and post it a bit later, this WILL be sorted! :P
     
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