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Math Help - sum of areas

  1. #1
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    sum of areas

    A piece of wire 18 m long is cut into two pieces, the length of the first piece being x m. The first piece is bent into a circle, and the other is bent into a rectangle with length twice the width. Give an expression for the total area A enclosed in the two shapes in terms of x.
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  2. #2
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    start with the circle its perimetre is X then X=2pi*R then R=X/2pi

    Area of the circle is = pi*R^2 = (pi * X^2 )/(4pi^2) = X^2/(4pi)

    now the rectangle : we have 18-X perimetre of the rectangle its perimetre = 2L+2W and we have L=2W => 18-X=6W => W=3-X/6 and L = 2W = 6-X/3

    => Area of the rectangle is = W*L = ( 3-X/6 ) ( 6-X/3 )

    Total Area = ( 3-X/6 ) ( 6-X/3 ) + X^2/(4pi)
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  3. #3
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    T.A = 18 - X -X +X^2/18 + X^2/4pi
    T.A = (1/18+1/4pi)X^2 - 2X + 18
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  4. #4
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    Quote Originally Posted by samtheman17 View Post
    A piece of wire 18 m long is cut into two pieces, the length of the first piece being x m. The first piece is bent into a circle, and the other is bent into a rectangle with length twice the width. Give an expression for the total area A enclosed in the two shapes in terms of x.
    For circle:
    circumference x = 2 \pi r
    r = \frac{x}{2\pi}
    area of circle = (\pi)(\frac{x}{2\pi})^2 = \frac{x^2}{4\pi}
    For rectangle:
    perimeter = 18 - x
    sum of a length and a width = \frac{18 - x}{2}
    length = (\frac{2}{3})(\frac{18 - x}{2})
    width = (\frac{1}{3})(\frac{18 - x}{2})
    area = (\frac{2}{3})(\frac{18 - x}{2})(\frac{1}{3})(\frac{18 - x}{2})
    = \frac{(18 - x)^2}{18}
    = \frac{324 - 36x + x^2}{18}
    = \frac{x^2}{18} -2x + 18
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