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Math Help - Finding the area of a triangle

  1. #1
    Junior Member lanierms's Avatar
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    Finding the area of a triangle

    a, b, c are the lengths of the sides of a triangle.

    When they satisfy the equations : a^2+b^2+c^2=6,           a^4+b^4+c^4=14, find the area of the triangle.

    I got to finding the value of a^2b^2+b^2c^2+c^2a^2 , but I can't progress any further.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Another version of Heron's formula is:

    Area = \dfrac14 \sqrt{(a^2 + b^2 + c^2)^2 - 2 (a^4 + b^4 + c^4)}
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  3. #3
    Junior Member lanierms's Avatar
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    Isn't there another way of solving it? I haven't learned Heron's formula yet.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Oh? I thought you did, since the a^2b^2 + b^2c^2 + a^2c^2 is in one of the Heron's formula for the area of a triangle..

    Heron's formula - Wikipedia, the free encyclopedia

    A = \sqrt{\dfrac{2(a^2b^2 + b^2c^2 + a^2c^2) - (a^4 + b^4 + c^4)}{16}}
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  5. #5
    Junior Member lanierms's Avatar
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    Can't we just make it end up as a right triangle or a regular triangle (without using Heron's formula) so it would actually be possible to get the area?

    Something like a=b=c, or a^2+b^2=c^2.

    Anyway, very useful. I didn't know Heron's formula could change like that.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Sorry, but form what is given, I can't find of any way...
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