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Math Help - Semi-Circle Rectangle Area, perimeter

  1. #1
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    Semi-Circle Rectangle Area, perimeter

    A window is in the shape of a semi-circle on top of a rectangle. The height of the rectangle is twice its width.

    a) Give an expression for A, the area of the window, in terms of w, its width.
    b) Give an expression for P, the perimeter of the window, in terms of w, its width.
    c) Give an expression for A, in terms of P only.
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  2. #2
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    Quote Originally Posted by goliath View Post
    A window is in the shape of a semi-circle on top of a rectangle. The height of the rectangle is twice its width.

    a) Give an expression for A, the area of the window, in terms of w, its width.
    b) Give an expression for P, the perimeter of the window, in terms of w, its width.
    c) Give an expression for A, in terms of P only.
    I'll show you how you can do a):

    The area of a rectangle is calculated by A_{rect}=length \cdot width

    The area of a semi-circle is calcvulated by A_{sem-circ}=\frac12 \cdot \pi \cdot (radius)^2

    ... and now it's your turn.

    With your question you know:
    length = 2 \cdot width
    radius = \frac12 \cdot width

    Therefore the area of the window is:

    A = 2 \cdot width \cdot width + \frac12 \cdot \pi \cdot \left(\frac12 width\right)^2 = (width)^2 \left(2 + \frac18 \pi\right)
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  3. #3
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    Quote Originally Posted by goliath View Post
    b) Give an expression for P, the perimeter of the window, in terms of w, its width.
    Do you "see" that P = height*2 + width + half-circle-circumference,
    which is 5w + pi(w)/2 ?
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  4. #4
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    K guys, I got those 2, I need help with part C
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  5. #5
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    Quote Originally Posted by goliath View Post
    ...

    c) Give an expression for A, in terms of P only.
    1. If you need further information or assistance ask publicly at MHF.

    2. It would be much easier for us to help you if you could show some work so we can see where you have difficulties. Then we have a chance to offer adaquate help.
    ---------------------------------------------------------------------------------------

    3. From Wilmer's post you learned that:

    P=w\left(5+\frac12 \pi\right) ~\implies~w=\dfrac{P}{5+\frac12 \pi}

    Plug the term for w into the equation of A=w^2\left(2+\frac18 \pi\right)

    You'll get:

    A=\left(\dfrac{P}{5+\frac12 \pi} \right)^2\left(2+\frac18 \pi\right)

    After moving around some stuff you'll get:

    A=\dfrac{\pi + 16}{2(\pi +10)^2} \cdot P^2
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  6. #6
    Senior Member nikhil's Avatar
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    Lightbulb

    area of window A=2w^2+(pi/2)(w/2)^2..........eq 1
    perimeter.........p=5w+pi(w/2)
    or w(5+pi/2)=p
    or w=p/(5+pi/2)
    put this value in eq 1
    A interm of P will be obtained
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  7. #7
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    K got it! Awesome guys, thanks!
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