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Math Help - Finding the Area (parameter)

  1. #1
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    Finding the Area (parameter)

    Given is a square, ABCD, the lenght of a side is equal to "a". A and D are centers of circles, BD and AC are arcs of circles and they intersect each other in K. Find the area of brightened figure on the picture

    Please Help!

    Finding the Area (parameter)-1111.png
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  2. #2
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    Re: Finding the Area (parameter)

    Hello, Telo!

    Given square ABCD with side a.
    A and D are centers of quarter-circles.
    The arcs intersect at K.
    Find the area of shaded region.

    Code:
        B *-*-------*-* C
          |    K*:::::|
          |   */..*:::|
          | *./.....*:|
          |*./.......*| a
          |./.........|
          */..........*
        A *-----------* D
                a
    Draw chord AK.

    The area of quarter-circle AKCD is:- \tfrac{1}{4}\pi a^2

    The area of sector AKD is:- \tfrac{1}{6}\pi a^2

    The area of segment AK is:- \tfrac{1}{6}\pi a^2 - \tfrac{\sqrt{3}}{4}a^2


    \text{Shaded region} \:=\:\text{(quarter-circle)} - \text{(sector + segment)}

    . . . . . . . . . . =\;\tfrac{1}{4}\pi a^2 - \left(\tfrac{1}{6}\pi a^2 + \left[\tfrac{1}{6}\pi a^2 - \tfrac{\sqrt{3}}{4}a^2\right]\right)

    . . . . . . . . . . =\;\left(\tfrac{\sqrt{3}}{4} - \tfrac{\pi}{12}\right)a^2
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  3. #3
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    Re: Finding the Area (parameter)

    Quote Originally Posted by Telo View Post
    Given is a square, ABCD, the lenght of a side is equal to "a". A and D are centers of circles, BD and AC are arcs of circles and they intersect each other in K. Find the area of brightened figure on the picture

    Please Help!

    Click image for larger version. 

Name:	1111.png 
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    area of square = a^2 area of quarter circle ADB = area of quarter circle ADC = 1/4 ( pi x a^2)
    Area of square ABCD - area of quarter circle ADB - area of quarter circle AD= areaBKC
    Now I am sure you can compute the area of the shaded region.
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  4. #4
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    Re: Finding the Area (parameter)

    Hi,
    Soroban beat me to the same solution. I was busy constructing the attached drawing; I enjoy drawing pictures. I'll post my version of his solution anyway.

    Finding the Area (parameter)-mhfgeometry32.png
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