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Why is is this answer incorrect? [Geometry]

  1. Oct 7, 2004 #1
    Why is this answer incorrect? [Geometry]

    ---edited
    I see where I went wrong, but with my current level of mathematical knowledge, I am at a dead end:


    I do not know how do draw this question in ASCII, so I will describe it. There is an equilateral triangle with a square inscribed within it. The square has sides of 6 (cm), it asks how long each side of the triangle is. (NOTE: the top of the square fits perfectly to form another triangle, and the two sides of the square form 2 right triangles)

    Now, in the top triangle, since it is as well equilateral, and the base (the top of the square) is 6, so the other sides must be 6.

    -let y represent the hypotenuse of the two right triangles formed by the square and the larger triangle.
    -let n represent the base of the two right triangles (the side other than that formed by the 6 cm side of the square)
    -let x represent the side lengths of the larger equilateral triangle which the square is in.
    (it would help to draw this while following along)

    AXIOMS:
    - [tex] x = 6 + y[/tex]
    - [tex] n = ( x - 6 ) / 2 [/tex]
    therefore [tex] n = [ ( y + 6 ) - 6 ] / 2
    = y / 2 [/tex]
    Now, with this all settled....

    [tex]y^2 = 6^2 + n^2 [/tex]

    [tex]y^2 = 36 + (y/2)^2 [/tex]

    [tex]y^2 = 36 + (y^2 /4) [/tex]

    [tex]y = \sqrt{36+(y^2/4)}[/tex]

    [tex]y = \sqrt{ \frac {36} {1} + \frac {y^2} {4} }[/tex]

    [tex]y = \sqrt{ \frac {144 + y^2} {4} }[/tex]

    [tex]y = \frac { \sqrt{144+y^2} } {2} [/tex]

    Is it possible to derive a value of [tex]y[/tex] from this?
     
    Last edited: Oct 7, 2004
  2. jcsd
  3. Oct 7, 2004 #2
    Go back up to [tex]y^2=36+\frac{y^2}{4}[/tex] retry from here.
     
  4. Oct 7, 2004 #3

    Tide

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    Homework Helper

    Yes, you can find y from that equation but I think a simpler way would be to recognize that if you take the two triangle are 30-60-right triangles (i.e. if you put them together then they form a equilateral triangle!). The altitude of the triangles is the same as the side of the square. Pythagoras will give you the side of the triangles in short order.
     
  5. Oct 8, 2004 #4
    Thanks; I always miss out on the simple details in math. :frown:
     
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