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Math Help - Square inscribed in triangle

  1. #1
    Member realintegerz's Avatar
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    Square inscribed in triangle

    A square is inscribed in a right triangle whose short sides are in the ratio of 1:2. What is the length of the side of the square in terms of the length of the shortest side of the circumscribed triangle?

    I tried assuming that the shortest sides were 1 and 2, which makes the hypotenuse sq. root 5

    Then I solved for the area of the whole triangle and I got 1

    So then I made 1 = (1/2)(1-w)(2-w)

    But for some reason I got w = 3 or 0 which doesn't work out...
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  2. #2
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    i may be making up some BS but does this look correct/true?

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  3. #3
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    Quote Originally Posted by realintegerz View Post
    A square is inscribed in a right triangle whose short sides are in the ratio of 1:2. What is the length of the side of the square in terms of the length of the shortest side of the circumscribed triangle?

    I tried assuming that the shortest sides were 1 and 2, which makes the hypotenuse sq. root 5

    Then I solved for the area of the whole triangle and I got 1

    So then I made 1 = (1/2)(1-w)(2-w)

    But for some reason I got w = 3 or 0 which doesn't work out...

    I will try to explain my solution without a picture... I somehow cannot upload a picture from my school...

    Draw a right triangle, B = 90 degrees AB = 1 BC = 2
    Draw inside the triangle a square label the square EBFG with E on AB en F on BC and G on AC

    EB = BF = x
    Then AE = 1 - x and FC = 2 - x

    You've got two similar triangles AEG and GFC there is a ratio between the two triangles and because of that

    AE : GF = EG : FC

    but also AE x FC = GF x EG and
    (1 - x)(2 - x ) = x*x
     2 - 3x + x^2 = x^2
     2 - 3x = 0
    and x = 2/3

    so we get the ratio 2/3 : 1 : 2
    [in my school we have to multiply by 3 ..... 2 : 3 : 6]
    Last edited by Shoreline; December 3rd 2008 at 02:38 AM.
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