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Six equal circles are placed uniformly such that each touch two other

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Six equal circles are placed uniformly such that each touch two other  [#permalink]

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New post 09 Jul 2018, 07:26
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Question Stats:

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As shown in the figure below, Six equal circles are placed uniformly such that each touch two other circles and that an equal circle touching all 6 circles can be placed between the circles.

Image

Find the area enclosed within the circles if the radius of each circle is 1 unit.

A) \(4\sqrt{3} - 2\pi\)
B) \(6\sqrt{3} - 2\pi\)
C) \(6\sqrt{3} - \pi\)
D) \(4\sqrt{3} - \pi\)
E) None of these


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Capture 6 circles.PNG
Capture 6 circles.PNG [ 26.76 KiB | Viewed 551 times ]

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Six equal circles are placed uniformly such that each touch two other  [#permalink]

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New post 09 Jul 2018, 08:12
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Princ wrote:
As shown in the figure below, Six equal circles are placed uniformly such that each touch two other circles and that an equal circle touching all 6 circles can be placed between the circles.
Attachment:
The attachment Capture 6 circles.PNG is no longer available

Find the area enclosed within the circles if the radius of each circle is 1 unit.

A) \(4\sqrt{3} - 2\pi\)
B) \(6\sqrt{3} - 2\pi\)
C) \(6\sqrt{3} - \pi\)
D) \(4\sqrt{3} - \pi\)
E) None of these



The radius of inside circle too is 1..
So look at the attached figure.
Join all the 6 circles center, and we have a hexagon of side 1+1=2.
Join a side with the centre of this hexagon and we have an equilateral triangle of side 2..
The area of each equilateral triangle is √3/4 *2^2=√3 and each sector inside the vertice of triangle = area of circle /6, since each angle is 60°...
Area of circle = π*1^3=π
Area of each sector therefore is π/6..
Therefore area other than the two sector in triangle =√3-2*π/6
The hexagon consists of 6 triangle , so the area of central portion = 6(√3-π/3)=6√3-2π

B
Attachments

PicsArt_07-09-09.31.40.png
PicsArt_07-09-09.31.40.png [ 43.18 KiB | Viewed 459 times ]


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Six equal circles are placed uniformly such that each touch two other  [#permalink]

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New post 09 Jul 2018, 08:47
Hi
The radius of circle is 1
The Side of regular hexagon is 2.
Now, the hexagon can be divide into 6 equilateral triangle of side 2. So area of hexagon = 6* \(\sqrt{3}\)\(side^2\)/4

Hence Area of hexagon = 3*\(\sqrt{3}\)*\(side^2\)/2 , side = 2
= 6\(\sqrt{3}\)

As each angle of hexagon = 120 deg, Area of each sector= Area of circle /3 = \(\frac{pi}{3}\)
Area of 6 sectors = 2pi

Hence the area between circles = Area of hexagon - area of sectors = 6\(\sqrt{3}\) - 2 pi

Answer =B


(For Science Nerds: This question is an example of the highest-density lattice arrangement of circles, the hexagonal packing arrangement, in which the centers of the circles are arranged in a hexagonal lattice (staggered rows, like a honeycomb), and each circle is surrounded by 6 other circles :angel: ...)
Attachments

Circular packing....jpg
Circular packing....jpg [ 24 KiB | Viewed 444 times ]


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Six equal circles are placed uniformly such that each touch two other &nbs [#permalink] 09 Jul 2018, 08:47
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