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regular hexagon

by rahul.s » Sun Feb 14, 2010 2:53 am
Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?

(A) 108 - 18pi
(B) 54√3 - 9pi
(C) 54√3 - 18pi
(D) 108 - 27pi
(E) 54√3 - 27pi

OA: E
Source: MGMAT
Level: 700 - 800
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by ajith » Sun Feb 14, 2010 3:11 am
rahul.s wrote:Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?

(A) 108 - 18pi
(B) 54√3 - 9pi
(C) 54√3 - 18pi
(D) 108 - 27pi
(E) 54√3 - 27pi

OA: E
Source: MGMAT
Level: 700 - 800
Radius of the circles A B C D E F and O = 3cm
Area of the circles A B C D E F inside the hexagon = 6*(120/360)* pi*3*3 = 18pi
Area of the circle O inside the hexagon = pi*3*3 = 9 pi
Total Area of the hexagon = 3*sqrt(3)/2*6*6 = 54 √3
Area of the shaded region (inside the hexagon but outside the circles) = 54 √3- (18pi+9pi) = 54 √3- 27pi
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by DeepthiRajan » Sun Feb 14, 2010 2:46 pm
ajith wrote:
rahul.s wrote:Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?

(A) 108 - 18pi
(B) 54√3 - 9pi
(C) 54√3 - 18pi
(D) 108 - 27pi
(E) 54√3 - 27pi

OA: E
Source: MGMAT
Level: 700 - 800
Radius of the circles A B C D E F and O = 3cm
Area of the circles A B C D E F inside the hexagon = 6*(120/360)* pi*3*3 = 18pi
Area of the circle O inside the hexagon = pi*3*3 = 9 pi
Total Area of the hexagon = 3*sqrt(3)/2*6*6 = 54 √3
Area of the shaded region (inside the hexagon but outside the circles) = 54 √3- (18pi+9pi) = 54 √3- 27pi

Ajith,

Could you please explain why angle=120 was used while calculating the area of the circles.

Also, it would be helpful if you could elaborate on the calculation of area of hexagon.

Thanks in advance.
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by ajith » Sun Feb 14, 2010 9:38 pm
DeepthiRajan wrote:
Ajith,

Could you please explain why angle=120 was used while calculating the area of the circles.

Also, it would be helpful if you could elaborate on the calculation of area of hexagon.

Thanks in advance.
The internal angle of a regular polygon with "n" sides can be calculated using:

(n-2) × 180° / n

for hexagon n=6 so, internal angle = 180*4/6 =120

In this case the angle subtended at the center of the circle is the same as the internal angle of the regular hexagon

Hence 120 degree is used in the area calculation.


Area of a regular hexagon is best remembered as a standard formula = 3*sqrt(3)/2*a^2 [where a is the side]

[it is 6*area of an isosceles triangle with side a]


Hope it helps
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by thephoenix » Sun Feb 14, 2010 10:27 pm
area of hexagon:-6* area if an eq trianglewith side 6=54v3

angle of hexagon=(6-2)*180/6=120

area of a sector with radius=3 and angle=120 is 120/360(pi) *(3)^2=3 pi for 6 circles=18 pi

area of a circle=9pi

reqd area=54v3-18 pi-9pi=54v3-27pi
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by bhumika.k.shah » Mon Feb 15, 2010 2:47 am
ajith wrote:
DeepthiRajan wrote:
Ajith,

Could you please explain why angle=120 was used while calculating the area of the circles.

Also, it would be helpful if you could elaborate on the calculation of area of hexagon.

Thanks in advance.
The internal angle of a regular polygon with "n" sides can be calculated using:

(n-2) × 180° / n

for hexagon n=6 so, internal angle = 180*4/6 =120

In this case the angle subtended at the center of the circle is the same as the internal angle of the regular hexagon

Hence 120 degree is used in the area calculation.


Area of a regular hexagon is best remembered as a standard formula = 3*sqrt(3)/2*a^2 [where a is the side]

[it is 6*area of an isosceles triangle with side a]


Hope it helps
Ajith , could you elaborate a l'l more on how u got the area of the hexagon. i am not quite getting the 6 equilaterla triangles thingy ...:(
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by ajith » Mon Feb 15, 2010 3:33 am
bhumika.k.shah wrote: Ajith , could you elaborate a l'l more on how u got the area of the hexagon. i am not quite getting the 6 equilaterla triangles thingy ...:(
Image

That is illustrated well in this page

Source: https://homepage.mac.com/terhorab/iblog/ ... onArea.pdf
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by Atrain13gm » Sat Oct 09, 2010 6:37 pm
I understand the formula for figuring out the area of the hexagon and getting 54 sqrt3, but why do I get 108 if I try calculating the area of the hexagon as two trapezoids?

Trapezoid Area=
1/2 * (6) * (12+6) = 54

54*2 = 108 area for the hexagon
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