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Math Help - Nice Qu but i need help pleeeease

  1. #1
    Newbie
    Joined
    Apr 2009
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    Nice Qu but i need help pleeeease

    hi guys ,

    please , i need ur supporting ,,,

    how i can solve this qu

    about A i solved as the below :
    supposing that L1 = perimeter of square , L2 = circumference of circle
    L = L1 + L2
    L - 4s = L2
    s = (L-L2) / 4
    area of square = (L - L2)/4)^2
    area of circle = pi ( L2 / 2pi )^2

    ( (L - L2)/4)^2 + pi ( L2 / 2pi )^2 ,

    Is This right ???

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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Hello, Learner guy!

    I'll get you started . . .


    A circle and a square are to be constructed
    from a piece of wire of length L.

    (a) Give an expression for the total area of the circle and the square.
    I think you have too many variables.
    I would do it like this . . .


    Let x = length of wire for the circle.
    Then L-x = length of wire for the square.

    The circumference of a circle is: . 2\pi r \:=\:C
    . . So we have: . 2\pi r \:=\:x\quad\Rightarrow\quad r \:=\:\frac{x}{2\pi}

    The area of a circle is: . A \:=\:\pi r^2
    . . So we have: . A_c \:=\:\pi\left(\frac{x}{2\pi}\right)^2 \:=\:\frac{x^2}{4\pi}


    The perimeter of a square is: . 4s \:=\:P
    . . So we have: . 4s \:=\:L-x \quad\Rightarrow\quad s \:=\:\frac{L-x}{4}
    The area of a square is: . A \:=\:s^2

    . . So we have: . A_s \;=\;\left(\frac{L-x}{4}\right)^2


    Therefore, the total area is: . A \;=\;\frac{1}{4\pi}x^2 + \frac{1}{16}(L-x)^2

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