The area of the whole square is 36
the area of tone quadrant = 0.25πr2=0.25∗π∗(2√3)2=3πcm2
Let the bottom of this diagram be the x- axis.
and (0,0) is at the bottom left corner then the equation of the bottom left quadrant is
x2+y2=12y2=12−x2y=√12−x2
The area of the blue/ green overlap is
A=2∫2√33√12−x2dx=2[0.5x√12−x2+6∗asin(x2√3)]2√33Wolfram|Alpha=2[0.5(2√3)√0+6∗asin(1)]−2[1.5√3+6∗asin(32√3)]=2[6∗π2]−2[1.5√3+6∗asin(√32)]=6π−2[1.5√3+6∗π3)]=6π−2[1.5√3+2π)]=6π−3√3−4π=2π−3√3
So the desired area in the middle is
middlearea=36−[π∗(2√3)2−4∗(2π−3√3)]middlearea=36−[12π−8π+12√3)]middlearea=36−[4π+12√3)]middlearea=36−12√3−4πcm2middlearea≈2.65cm2
Thanks, Melody...
Here's another way without using Calculus
Positioning the circles as Melody did :
The intersection of two circles at the left side of the figure occurs at A = (sqrt(3) , 3)
Similarly, the intersection of the two circles at the bottom occurs at ( 3, sqrt (3) )
So the distance between these two points is AD = sqrt [ 24 - 12 sqrt (3) ] cm
Therefore...using symmetry....we can construct a square AFGD of this side with an area of [ 24 - 12sqrt(3)] cm^2 (1)
Looking at the segment AE... the slope of this segment = 3/sqrt (3) = sqrt (3)
Looking at the slope of DE....the slope of this segment is sqrt (3) / 3 = 1 / sqrt (3)
So arctan (sqrt (3) ) - arctan (1/sqrt(3) ) = 60° - 30° = angle AED = 30°
So the area of sector AED is (1/2)(12)(pi/6) = pi cm^2
And the area of triangle AED = (1/2)(12)sin(30) = 3 cm^2
So....the area between the sector and the triangle is [pi - 3] cm^2
And using symmetry....4 of these areas = [4pi -12] cm^2 (2)
So....the shaded area is (1) - (2) =
[ 24 - 12sqrt (3)] - [ 4pi - 12 ] cm^2 = [ 36 - 12sqrt (3) - 4pi] cm^2 (exact) ≈ 2.65 cm^2 (rounded)