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Math Help - OMG

  1. #1
    Junior Member Nadeshiko's Avatar
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    OMG

    Iduj wears a very funny hat in the shape of a triangle OMG. The altitude from point M is drawn to point A on side OG. If OM is 13 inches, MG is 20 inches, and GA is 16 inches, find the area, in inches squared, of Idujís triangular hat.

    *Hints or steps on how to get the answer is appreciated.*
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  2. #2
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    Re: OMG

    You could use Pythagorean theorem to calculate the height of the triangle OMG. You will need to calculate AO, Can yo do it now. the are of the triangle is height*base/2. What is the base of the the triangle OMG?
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  3. #3
    Junior Member Nadeshiko's Avatar
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    Re: OMG

    Quote Originally Posted by votan View Post
    You could use Pythagorean theorem to calculate the height of the triangle OMG. You will need to calculate AO, Can yo do it now. the are of the triangle is height*base/2. What is the base of the the triangle OMG?
    Is the base 16 inches? Height*16/2=A?
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  4. #4
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    Re: OMG

    Hello, Nadeshiko!

    Did you make a sketch?


    Iduj wears a very funny hat in the shape of a triangle OMG.
    The altitude from point M is drawn to point A on side OG.
    If OM is 13 inches, MG is 20 inches, and GA is 16 inches,
    find the area, in inches squared, of Idujís triangular hat.

    Code:
                  M
                  *
                 *| *
                * |   *
            13 *  |     *  20
              *   |       *
             *    |         *
            *     |           *
           *      |             *
        O * * * * * * * * * * * * * G
                  A       16
    In \Delta M\!AG: \;M\!A^2 + 16^2 \:=\:20^2 \quad\Ruightarrow\quad M\!A^2 + 256 \:=\:400 \quad\Rightarrow\quad M\!A^2 \,=\,144

    . . Hence: . M\!A \,=\,12


    In \Delta M\!AO: \:OA^2 + 12^2 \:=\:13^2 \quad\Rightarrow\quad OA^2 + 144 \:=\:169 \quad\Rightarrow\quad OA^2 \,=\,25

    . . Hence: . OA \,=\,5


    Therefore, triangle OMG has:
    . . base OG \,=\,5+16\,=\,21
    . . height M\!A \,=\,12

    Got it?
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  5. #5
    Junior Member Nadeshiko's Avatar
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    Re: OMG

    Quote Originally Posted by Soroban View Post
    Hello, Nadeshiko!

    Did you make a sketch?



    Code:
                  M
                  *
                 *| *
                * |   *
            13 *  |     *  20
              *   |       *
             *    |         *
            *     |           *
           *      |             *
        O * * * * * * * * * * * * * G
                  A       16
    In \Delta M\!AG: \;M\!A^2 + 16^2 \:=\:20^2 \quad\Ruightarrow\quad M\!A^2 + 256 \:=\:400 \quad\Rightarrow\quad M\!A^2 \,=\,144

    . . Hence: . M\!A \,=\,12


    In \Delta M\!AO: \:OA^2 + 12^2 \:=\:13^2 \quad\Rightarrow\quad OA^2 + 144 \:=\:169 \quad\Rightarrow\quad OA^2 \,=\,25

    . . Hence: . OA \,=\,5


    Therefore, triangle OMG has:
    . . base OG \,=\,5+16\,=\,21
    . . height M\!A \,=\,12

    Got it?
    Yes. I got it! I made a big mistake. I forgot to add the A!!! I' m sorry for causing sooo much trouble and thanks for the help!
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  6. #6
    Forum Admin topsquark's Avatar
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    Re: OMG

    You could also "cheap out" and use Heron's formula.

    -Dan
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  7. #7
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    Re: OMG

    Quote Originally Posted by topsquark View Post
    You could also "cheap out" and use Heron's formula.

    -Dan
    Hello Dan,
    Heron's formula gives area with 3 sides of triangle.

    A quick answer using a scientific calculator

    sin (arc cosine 16/20)=0.6
    MA = 0.6*20=12
    sin (arc cosine 12/13=0.3846
    OA =0.3846*13=5
    area =1/2*12*21
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