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.
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