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
regular hexagon
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- ajith
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Radius of the circles A B C D E F and O = 3cmrahul.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
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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ajith wrote:Radius of the circles A B C D E F and O = 3cmrahul.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
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.
- ajith
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The internal angle of a regular polygon with "n" sides can be calculated using: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.
(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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- thephoenix
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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
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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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 ...ajith wrote:The internal angle of a regular polygon with "n" sides can be calculated using: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.
(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
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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 ...
That is illustrated well in this page
Source: https://homepage.mac.com/terhorab/iblog/ ... onArea.pdf
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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
Trapezoid Area=
1/2 * (6) * (12+6) = 54
54*2 = 108 area for the hexagon