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# help

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In the diagram, ABCD is a square with a side length of 20 with four inscribed quarter circles as shown.  Find the area of the yellow region. Dec 20, 2019

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See the following image : Let I  be the center of the square = (10,10)

Let one of the quarter circles  have the equation x^2 + y^2  = 400

And let another of the quarter circles have the equation of (x - 20)^2 + y^2  =400

The x coordinate of the  intersection of these circles  = 10

So  the x coordinate of E   = 10

So...using the equation of the circle we can find the y  coordinate as

10^2 + y^2  =400

100 + y^2 = 400

y^2 = 300

y = 10√3

So  IE = the distance between I and E =   10√3 - 10   = 10 (√3 -1)

And by symmetry....this equals IF

So triangle  EIF is right and its area =   (1/2)(IE)(IF) = (1/2) [ 10 (√3 -1)]^2  =

50 (3 - 2√3 + 1)  = [200 - 100√3]units^2      (1)

And the area between sector AEF  and triangle AEF  =

(1/2)*(20^2)(pi/6)  -  (1/2)(20^2)(sin ( EAF) )  =

(1/2) (20^2)(pi/6)  - (1/2)(20^2)(1/2)  =

200 [ pi/6 - 1/2 ]  units^2   =

(100/3)pi - 100   units^2        (2)

So....by symmetry.....the yellow area  =  4 [ (1) + (2)  ] =

4 [ ( 200 -100√3 + (100/3)pi - 100]  units^2  =

4 [ 100 - 100√3 + (100/3)pi ] units^2  =

400 [ 1 - √3 + pi /3 ] units^2 ≈ 126.06 units^2   Dec 21, 2019