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Math Help - Triangle Area - Correct Procedure ???

  1. #1
    Newbie Monocerotis's Avatar
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    Triangle Area - Correct Procedure ???

    The sum of the base and the height of a triangle is 21cm. Determine the maximum area of the triangle, in square centimeters.
    Solution
    Area of a triangle: (b)(h)
    b: base measurement of the triangle, h: height measurement of the triangle.
    b + h = 21 || b = 21 – h


    21h – h^2 = 0

    h^2 – 21h = 0

    h^2 – 21h +110.25 – 110.25 = 0

    h^2 – 21h +110.25 = 110.25

    (h-10.5)^2 = 110.25

    √(h-10.5)^2 = √110.25

    h-10.5 = 10.5

    h = 10.5 10.5 , h = 10.5 + 10.5, h = 21 || h = 10.5 – 10.5, h = 0

    h = 10.5 + 10.5, h = 21

    Divide 21 by 2

    b = 21 – (h), b = 21 – (10.5), b= 10.5

    h = 10.5, b = 10.5

    Now plug in these values to find the area of the triangle.
    Therefore, the area of the triangle, is 55.125 cm2
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by Monocerotis View Post
    The sum of the base and the height of a triangle is 21cm. Determine the maximum area of the triangle, in square centimeters.
    Solution
    Area of a triangle: (b)(h)
    b: base measurement of the triangle, h: height measurement of the triangle.
    b + h = 21 || b = 21 h
    you sure took the long road to get there ...

    A = \frac{1}{2}(21-h)h

    A = \frac{1}{2}(21h - h^2)

    max value at h = \frac{-b}{2a} = \frac{-21}{-2} = 10.5

    A_{max} = \frac{1}{2}(10.5)^2
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  3. #3
    Newbie Monocerotis's Avatar
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    Quote Originally Posted by skeeter View Post
    you sure took the long road to get there ...

    A = \frac{1}{2}(21-h)h

    A = \frac{1}{2}(21h - h^2)

    max value at h = \frac{-b}{2a} = \frac{-21}{-2} = 10.5

    A_{max} = \frac{1}{2}(10.5)^2
    lol yeah looks like I took the long road, thanks for checking it over I'll also write down your method for future problems.
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