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Thread: Find the area of the shaded region

  1. #1
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    Find the area of the shaded region

    What is the area of the shaded region in this shape ؟
    رفع الصور
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  2. #2
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    Re: Find the area of the shaded region

    Quote Originally Posted by Mhmh96 View Post
    What is the area of the shaded region in this shape ؟
    رفع الصور
    Are the shaded areas made from intersecting quarter circles?
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  3. #3
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    Re: Find the area of the shaded region

    Hello, Mhmh96!

    You should have supplied more information.
    Even if your diagram were symmetric, we're not sure what the figure is.

    I'll assume that "Prove It" was correct . . . These are quarter-circles.

    Code:
                  * * *           * * *
              *           *   *           *
            *               *               *
           *               *:*               *
                           :::
          *               *:::*               *
          *               *:::*               *
          *               *:::*               *
                 .*.*.*.   :::   .*.*.*.
           *  *:::::::::::**:**:::::::::::*  *
            *:::::::::::::::*:::::::::::::::*
           *  *:::::::::::**:**:::::::::::*  *
                  * * *    :::    * * *
          *               *:::*               *
          *               *:::*               *
          *               *:::*               *
                           :::
           *               *:*               *
            *               *               *
              *           *   *           *
                  * * *           * * *
    Your diagram says the length of two "petals" is 4 units.
    Then the radius of the circles is \sqrt{2}.

    I'll let you work out the details.

    I get an answer of 4(\pi - 2)
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  4. #4
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    Re: Find the area of the shaded region

    Yes quarter-circles,but how did you find the radius ?
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  5. #5
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    Re: Find the area of the shaded region

    Hello, Mhmh96!

    How did you find the radius?

    Code:
                  * * *
              *           *
            *               *
           *                 *
                              
          *         C         *
          *         o         *
          *    r  *   *  r    *
                *.*.*.*.*
           *  *:::::::::::*  *
          A o - - - - - - - o B
           *  *:::: 2 ::::*
                  * * *           
          *                   *
          *                   *
          *                   *
    
           *                 *
            *               *
              *           *
                  * * *
    Chord AB is 2 units long.

    Triangle ABC is an isosceles right triangle.

    Hence, its legs (the radius) is: \frac{2}{\sqrt{2}} \,=\,\sqrt{2}
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