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Thread: Integrated Geometry and Algebra

  1. #1
    Member BobBali's Avatar
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    Cool Integrated Geometry and Algebra

    Hi all,

    In diag belo I have to find length of red triangle.

    1). Equation x^2 + y^ = 9 Do i make y^2 subject here? And place in y^2 below:

    (12- x^2)^2 - (9- y^2)^2 = c^2 ?
    Attached Thumbnails Attached Thumbnails Integrated Geometry and Algebra-img_20170630_221952.png  
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  2. #2
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    Re: Integrated Geometry and Algebra

    ?? The red portion is a rectangle, not a triangle.

    You started out the way I would but you have only two equations in three unknowns, x, y, and c. You need a third, independent, equation.
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  3. #3
    Member BobBali's Avatar
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    Re: Integrated Geometry and Algebra

    Yes, sorry for the error. It's length of red rectangle.. How do I obtain 3rd equation?
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  4. #4
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    Re: Integrated Geometry and Algebra

    The three equations:
    $x^2+y^2=3^2$
    $(12-x)^2+(9-y)^2=z^2$
    $xy+(12-x)(9-y)+3z = 12\cdot 9$

    The last equation comes from adding up areas.

    This gives $z \approx. 12.2553$
    Last edited by SlipEternal; Jul 5th 2017 at 06:58 AM.
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  5. #5
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    Re: Integrated Geometry and Algebra

    You can also use similar triangles:
    $\dfrac{y}{12-x} = \dfrac{x}{9-y} = \dfrac{3}{z}$
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