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Math Help - Square

  1. #1
    Junior Member Singular's Avatar
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    Square

    A B C D is a square

    Find the Area of triangle AYX

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  2. #2
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    Hello, Singular!

    I found a solution, but I hope there is a better way . . .


    ABCD is a square.
    Find the area of triangle AYX.
    Code:
               q    X    s-q
        C * - - - - * - - - - * D
          |        :::*       |
          |       *:::::*  4  | s-p
          |  5   :::::::::*   |
          |     *:::::::::::* |
        s |    :::::::::::::::* Y
          |   *:::::::::::*   |
          |  :::::::::*       | p
          | *:::::*      3    |
          |:::*               |
        A * - - - - - - - - - * B
                    s

    Let s = side of the square.
    . . We see that: .s .= .∆ + 12 .[1]

    Let YB = p, then DY = s - p
    Let CX = q, then XD = s - q

    From right triangle YBA: .ps = 3 . . ps = 6---[2]
    From right triangle XCA: .qs = 5 . . qs = 10 .[3]
    From right triangle XDY: .(s - p)(s - q) = 4 . . s - ps - qs + pq .= .8 .[4]

    Substitute [2] and [3] into [4]: .s - 6 - 10 + pq .= .8 . . s + pq .= .24 .[5]


    Multiply [2] and [3]: .pqs = 60 . . pq = 60/s

    Substitute into [5]: .s + 60/s .= .24 . . s^4 - 24s + 60 .= .0
    . . . . . . . . . - . . . - . . . - . . . . . .____________
    . . . . . . . . . . . . . . . . . . . 24 √24 - 4(1)(60) . . . . . . . . .__
    Quadratic Formula: . s .= .------------------------- .= .12 2√21
    . . . . . . . . . . . . . . . . . . . . . . . . 2(1)

    . . . . . . . . . . . . - . . . - . . . . . . . .__
    Equate to [1]: . ∆ + 12 .= .12 2√21
    . . . . . . . . . . . . . . . . .__
    . . Therefore: . .= .2√21

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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    I've got a doubt, what does mean 3, 4 and 5?, sides of the triangle?

    In that case it'd be a right triangle...
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    I've got a doubt, what does mean 3, 4 and 5?, sides of the triangle?

    In that case it'd be a right triangle...
    Look at the units in the original post. The units are in cm^2, which designates an area. So I'd say that the area of the triangle is the area of the square (whatever it is) minus 3 cm^2 + 4 cm^2 + 5 cm^2.

    Looking at it, Soroban likely has the solution, unless there is some kind of clever shortcut.

    -Dan
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  5. #5
    Math Engineering Student
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    Jejeje

    Thanks, first time that I see a problem with that notation.
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  6. #6
    Junior Member Singular's Avatar
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    Thanks SOROBAN
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