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The vertices of a square are the centers of four circles as shown above. Given each side of the square is 6cm and the radius of each circle is $$2\sqrt3$$cm, find the area in square centimeters of the shaded region.

Feb 1, 2019
edited by MaxWong  Feb 1, 2019

#1
+104404
+3

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

$$middle \;area =36-[\pi*(2\sqrt3)^2-4*(2\pi-3\sqrt3)]\\ middle \;area =36-[12\pi-8\pi+12\sqrt3)]\\ middle \;area =36-[4\pi+12\sqrt3)]\\ middle \;area =36-12\sqrt3-4\pi \;\;cm^2\\ middle \;area \approx 2.65 cm^2\\$$

Feb 1, 2019
#2
+103915
+3

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)

Feb 1, 2019
edited by CPhill  Feb 1, 2019
edited by CPhill  Feb 1, 2019
#3
+104404
+2

Very nice Chris I like your method better. :)

Melody  Feb 1, 2019
#4
+103915
+2

THX....!!!!

[ It does avoid that nasty integral...LOL !!!   ]

CPhill  Feb 2, 2019