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    a gardener encloses a triangular plot of land by using an existing fence 16m long for one side, and fencing of combined length 20m for the other 2. if the total area is 24sqrt3m^2 find the lengths of the sides of the triangle.
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    let a = 16
    =======

    then b + c = 20

    b = 20 - c
    =======

    s = ½(a + b + c)
    s = ½(16 + 20)
    s = 18
    =====

    A = √(s(s-a)(s-b)(s-c)
    A² = s(s-a)(s-b)(s-c)
    24²*3 = 18(18 - 16)(18 - 20 + c)(18 - c)
    96 = 2(c - 2)(18 - c)
    48 = 18c - c² - 36 + 2c
    c² - 20c + 84 = 0
    (c - 14)(c - 6) = 0
    c = 14, c = 6
    :. b = 6, b = 14

    a = 16, b = 14, c = 6
    ================
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    (Original post by Fermat)
    let a = 16
    =======

    then b + c = 20

    b = 20 - c
    =======

    s = ½(a + b + c)
    s = ½(16 + 20)
    s = 18
    =====

    A = √(s(s-a)(s-b)(s-c)
    A² = s(s-a)(s-b)(s-c)
    24²*3 = 18(18 - 16)(18 - 20 + c)(18 - c)
    96 = 2(c - 2)(18 - c)
    48 = 18c - c² - 36 + 2c
    c² - 20c + 84 = 0
    (c - 14)(c - 6) = 0
    c = 14, c = 6
    :. b = 6, b = 14

    a = 16, b = 14, c = 6
    ================
    damn. i got upto thw quadratic and then though that it couldn't be solved. what was i thinking...
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    The lengths of the triangle are printed in blue. I have checked the solution and the area does come to 24 as required.

    You should see two pdf files. One with the calculation and one with the check.

    If you don't see the pdf's let me know.

    steve2005
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    (Original post by Fermat)
    let a = 16
    =======

    then b + c = 20

    b = 20 - c
    =======

    s = ½(a + b + c)
    s = ½(16 + 20)
    s = 18
    =====

    A = ?(s(s-a)(s-b)(s-c)
    A² = s(s-a)(s-b)(s-c)
    24²*3 = 18(18 - 16)(18 - 20 + c)(18 - c)
    96 = 2(c - 2)(18 - c)
    48 = 18c - c² - 36 + 2c
    c² - 20c + 84 = 0
    (c - 14)(c - 6) = 0
    c = 14, c = 6
    :. b = 6, b = 14

    a = 16, b = 14, c = 6
    ================
    I disagree with your solution. If the sides are 16, 14 and 6 then I calculate the area as 41.59. The required area was 24. I can't see where you have gone wrong!!!

    steve2005
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    (Original post by chewwy)
    damn. i got upto thw quadratic and then though that it couldn't be solved. what was i thinking...
    I don't think you should have multiplied the A squared by 3.

    A²= 576

    steve2005
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    (Original post by steve2005)
    I don't think you should have multiplied the A squared by 3.

    A²= 576

    steve2005
    The area given by chewwy in his original post is 24√3 m², rather than 24 m².

    (Original post by chewwy)
    ... if the total area is 24sqrt3m^2 find the lengths of the sides of the triangle.
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    go Fermat:cool:
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    (Original post by Fermat)
    The area given by chewwy in his original post is 24?3 m², rather than 24 m².
    OK, my mistake.
 
 
 
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