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Seven identical circles are arranged like the diagram below so that each one is tangent to one another. A hexagon is then drawn around the circles. The radius of each circle is 3. Find the area of the space inside of the hexagon but outside of the circles.

 

 May 2, 2020
 #1
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The only hard part here is trying to determine the side length of the hexagon

 

The vertex angle of a hexagon  =  (2/3)(180) = 120°

 

Looking at the top right circle...if we draw a segment  connnecting the center of this circle with  the top right vertex of the hexagon....this segment will bisect the vertex angle 

 

Next from the center of ths circle draw a radius perpendicular to the top edge of the hexagon....this distance = 3

 

So....we  have a 30- 60 -90  right triangle...call  the distance  from the top right vertex of the hexagon to the point where the radial line meets the top edge   =  x

 

And we can find this distance as  

 

3/sin 60  =  x/sin 30

 

x =  3 sin (30)/ sin (60)

x = 3 (1/2)/ [√ 3/2]  

x = 3(1/2) * (2/√ 3]  

x = √ 3

 

So  using symmetry.....the side length of the hexagon  = x + 3 + 3 + x  = 2( 3 + x) =  2 (3 + √ 3)  =  6 + √ 12

 

And the area of the hexagon  =  3√ 3 side^2 / 2  =    3√ 3 [ 6 + √ 12]^2 / 2  =  

 

3√ 3  [ 36 + 12√ 12 + 12]  / 2  =

 

3√ 3 [ 48 + 12√ 12] / 2   =

 

3√ 3 [ 24 + 6√ 12]  =

 

72√ 3 + 18 √ 36  =

 

72√ 3  + 108

 

The area of the circles is easy  =   7pi 3^2  =  63 pi

 

Then the area  inside the hexagon but outside the circles  =

 

72√ 3   + 108  -  63 pi   ≈  34.8  units^2

 

 

cool cool cool

 May 3, 2020

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