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Math Help - Designing a Rug With Ellipses and Circles

  1. #1
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    Designing a Rug With Ellipses and Circles

    Louise uses an ellipse (see attachment) and two circles to design a rug. The diagram shows her design, drawn on a number plane.

    a. Calculate the area of the shaded region.

    How would I do so?

    Note: The width of the ellipse is 5 and the radius of the small circle is 3.
    Attached Thumbnails Attached Thumbnails Designing a Rug With Ellipses and Circles-area-rug.jpg  
    Last edited by JadeKiara; July 5th 2009 at 10:45 PM.
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  2. #2
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    Quote Originally Posted by JadeKiara View Post
    Louise uses an ellipse (see attachment) and two circles to design a rug. The diagram shows her design, drawn on a number plane.

    a. Calculate the area of the shaded region.

    How would I do so?

    Note: The width of the ellipse is 5 and the radius of the small circle is 3.
    the width of the ellipse is 5 the radius is also 5.
    area of ellipse = pi*a*b = pi*5*3
    now we only have to deal with one half of the ellipse and larger circle.
    so 0.5*pi*5*3
    the area of the semicircle would be 0.5*pi*5^2
    now area of semicircle minus area of half the ellipse and you get the area of the shaded region
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    Thank you for your above post. A good explanation makes Mathematics seem so much easier.

    b. Louise is binding the circumference of the smaller circle with wool. She uses 80 g of wool for every metre. How many grams of wool will she use?

    Correct me if I am wrong.

    2*pi*3 = 18.85 m
    18.85*80 = 1508 g

    Therefore, she will use 1508 g of wool.
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  4. #4
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    Quote Originally Posted by JadeKiara View Post
    Thank you for your above post. A good explanation makes Mathematics seem so much easier.

    b. Louise is binding the circumference of the larger circle with wool. She uses 80 g of wool for every metre. How many grams of wool will she use?

    Correct me if I am wrong.

    2*pi*3 = 18.85 m
    18.85*80 = 1508 g

    Therefore, she will use 1508 g of wool.
    if she is binding the larger circle, it would be 2*pi*5 instead of 3, but you have the concept correct
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