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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
avatar+102763 
+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
avatar+102372 
+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)

 

 

cool cool cool

 Feb 1, 2019
edited by CPhill  Feb 1, 2019
edited by CPhill  Feb 1, 2019
 #3
avatar+102763 
+2

Very nice Chris I like your method better. :)

Melody  Feb 1, 2019
 #4
avatar+102372 
+2

THX....!!!!

 

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

 

 

cool cool cool

CPhill  Feb 2, 2019

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