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65) What is the area of the hexagon?

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by sanju09 » Wed May 12, 2010 4:37 am
ern5231 wrote:Perimeter of hexagon is 36. What is the area of the shadowed region?
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The figure maintains that it's a regular hexagon of side 6. The shadowed region is hexagon region minus six one-thirds of circle of radius 3 with one more circle of radius 3, which is in the middle

= 6 × √3/4 × (6) ^2 - {6 × 1/3 × π × (3) ^2 + π × (3) ^2}

= 54 √3 - 27 π

= [spoiler]27 (2 √3 - π)

please supply choices else switch to the suitable forum[/spoiler]
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by kstv » Wed May 12, 2010 6:03 am
Area of the equilateral triangle with vertex at the centre of the inner circle and base as the side of the hexagon
= 3*5 = 15 , height = h, h² = 6²-3² or h = 5
Area of the three sector inside the triangle = semicircle as each angle = 60° = pi*9 /2
Area of one shaded area = 15 - 9/2 pi
Area of 6 shaded are = 6*3(5-1.5pi)
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by clock60 » Wed May 12, 2010 2:45 pm
ern5231 wrote:Perimeter of hexagon is 36. What is the area of the shadowed region?
Image
not too sure in my solution but anyway
the propety of hexagon is that its side equals the radius of circumscribed circle, it means that
perimeter of the hexagon=6*R=36, R=6
and the Radius of any given circle ( i hope that they are equal)=3
let us find the area of the sector of circle
120/360*pi*3^2=3*pi
and we have 6 sectors so total sum will be 6*3*pi
we have 1 whole circle with area=pi*3^2=9*pi
so total area of not bolded sum inside the hexagon=18*pi+9*pi=27*pi

the area of hexagon=3/2*(3^1/2)*R^2=3/2*(3^1/2)*6^2=54*3^1/2



so bolded area=area of hexagon-not bolded area=54*3^1/2-27*pi=27(2*3^1/2-pi)
wow the same ans with that of sanju09
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by harshavardhanc » Wed May 12, 2010 3:10 pm
ern5231 wrote:Perimeter of hexagon is 36. What is the area of the shadowed region?
Image
required area = area of hexagon - area of 3 circles with r as side/2

area of hexagon = 6 * 6^2 / 4 * V3

area of 3 circles = 3 * pi * 3^2

= 6 * 6^2 / 4 * V3 - 3 * pi * 3^2

= 3^3 ( 2 V3 - pi )
Regards,
Harsha
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