Hexagon Problem

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by anuprajan5 » Mon Oct 22, 2012 7:38 am
Ritika,

There are 2 parts to this problems.

First - the area of the hexagon. SInce the perimeter is 36, each side is 6 and the hexagon is a combination of 6 equilateral triangles.

Since the area of 1 equilateral triangle is root 3/4 * a^2

The total area of the hexagon equals the area of 6 equilateral triangles - ie root 3/4 * 6^2 = 54root3


Area of circles - If the circles are tangent to each other then the radii of each circle is equal ie: 3

Area of 1 circle - pi r^2 ie 9*pi

Area of 6 circles - 54 pi. But the area of circle within the hexagon is only one third the area (because the angle subtended by one angle of a hexagon is 120 degrees)

Therefore the area of 6 circles within the hexagon is 18pi.

if the inner circle is also tangent, then the whole are is 9 pi.

Shaded are equals Area of hexagon - Area of circles ie; 54root 3 - 27pi


Image
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Anup

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by GMATGuruNY » Mon Oct 22, 2012 8:08 am
Image

Shaded region = hexagon - circle areas.

Hexagon:
The drawing shows 6 equilateral triangles, each with side 6.
In an equilateral triangle, A = (s²√3)/4:
(6²)(√3)/4 = 9√3.
Since the hexagon is made up of 6 of these triangles:
A = 6(9√3) = 54√3.

Circle areas:
Since r=3, the area of each circle = π(3²) = 9π.
The central angle of each red sector = (angle of equilateral triangle) + (angle of equilateral triangle) = 60+60 = 120 degrees.
Since 120/360 = 1/3, each red sector constitutes 1/3 of a circle.
Thus, the 6 red sectors -- along with the circle with center O -- constitute 3 entire circles:
3(9Ï€) = 27Ï€.

Shaded region = 54√3 - 27π.

The correct answer is E.
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by ritika_bsg » Tue Oct 23, 2012 9:36 am
Thanks a lot ! :)