+118680 The area of the whole square is 36
the area of tone quadrant = \(0.25 \pi r^2= 0.25*\pi*(2\sqrt3)^2=3\pi\;\;cm^2\)
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
\(x^2+y^2=12\\ y^2=12-x^2\\ y=\sqrt{12-x^2}\)
The area of the blue/ green overlap is
\(A=\displaystyle 2\int_3^{2\sqrt3}\;\;\sqrt{12-x^2}\;\;dx\\ =2\left[ 0.5x\sqrt{12-x^2}+6 * asin(\frac{x}{2\sqrt3} ) \right]_3^{2\sqrt3} \qquad \text{Wolfram|Alpha}\\ =2\left[ 0.5(2\sqrt3)\sqrt{0}+6 * asin(1 ) \right]-2\left[ 1.5\sqrt{3}+6 * asin(\frac{3}{2\sqrt3} ) \right]\\ =2\left[ 6 * \frac{\pi}{2} \right]-2\left[ 1.5\sqrt{3}+6 * asin(\frac{\sqrt3}{2} ) \right]\\ =6\pi-2\left[ 1.5\sqrt{3}+6 * \frac{\pi}{3} ) \right]\\ =6\pi-2\left[ 1.5\sqrt{3}+2\pi ) \right]\\ =6\pi-3\sqrt{3}-4\pi \\ =2\pi-3\sqrt3\)
So the desired area in the middle is
\(middle \;area =36-[\pi*(2\sqrt3)^2-4*(2\pi-3\sqrt3)]\\ middle \;area =36-[12\pi-8\pi+12\sqrt3)]\\ middle \;area =36-[4\pi+12\sqrt3)]\\ middle \;area =36-12\sqrt3-4\pi \;\;cm^2\\ middle \;area \approx 2.65 cm^2\\\)
+129898 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)
