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Regular hexagon ABCDEF has a perimeter of 36. O is the cente

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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente [#permalink]

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New post 03 Nov 2014, 02:11
russ9 wrote:
Bunuel wrote:
mirzohidjon 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. \(108-18\pi\)
B. \(54\sqrt{3}-9\pi\)
C. \(54\sqrt{3}-18\pi\)
D. \(108-27\pi\)
E. \(54sqrt{3}-27\pi\)


Guys,
I did not understand why we can not form two trapezoids of this hexagon, and solve through it. The official solution says, that we have to form 6 equiaterial triangles in the hexagon and form through it. Forming trapezoids, I got 108 as a square of the hexagon.
Please. elaborate


You can do this by calculating the area of the trapezoid as well.

First of all the angle = 120 degrees. Non shaded area will consist of the areas of 6*120 degrees sectors plus one whole circle = 3 whole circles = \(3\pi{r^2}=27\pi\), (as \(r=3\), half of the side of the hexagon).

Now if we want to proceed with the trapezoid areas. We would have two trapezoids. Bases would be \(a=6\) (the smallest base) and \(b=12\) (the largest base). \(Area=\frac{a+b}{2}*h\). The height of the trapezoid can be calculated as the side in 30-60-90 right triangle. Height would be the opposite side to the 60 degrees angle (half of 120 degrees). In 30-60-90 right triangle the ratio of the sides is \(1/\sqrt{3}/2\), the height would correspond with \sqrt{3} and the hypotenuse (which on the other hand would be the side of the hexagon=6) would correspond with 2. so we'll have \(2/\sqrt{3}=\frac{6}{h}\) --> \(h=3\sqrt{3}\). Area of the trapezoid would be \(Area=\frac{a+b}{2}*h=\frac{6+12}{2}*3\sqrt{3}=27\sqrt{3}\). As we have two trapezoids the whole area would be \(2*27\sqrt{3}=54\sqrt{3}\).

Area of the shaded region \(54\sqrt{3}-27\pi\).

Answer: E.



Hi Bunuel,

Two questions:

1) Regarding your method above, it's still a little blurry as to how you calculated the area of the trapezoid, namely the height. Can you please elaborate a little?

2) The way I went about calculating the area of the trapezoid was 2 traingles plus a rectangle. To find the area of the triangle, I used pyt theorem -- Square Root of 72 was the hypotenuse of the triangle and the length of the rectangle. So I had (2*1/2*6*6) + (Root 72 * 6) = 36 + 36root 2. Why is that wrong? Doesn't that encapsulate the total hexagon?

Thanks!


1. Check the diagram below:
Attachment:
Untitled.png
Untitled.png [ 25.77 KiB | Viewed 1122 times ]

2. Cannot follow you. How did you get that the hypotenuse, which is the side of the hexagon, equals to \(\sqrt{72}\)?
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente [#permalink]

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New post 01 Aug 2016, 10:15
gmatjon wrote:
Attachment:
The attachment 1.JPG is no longer available
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. \(108-18\pi\)
B. \(54\sqrt{3}-9\pi\)
C. \(54\sqrt{3}-18\pi\)
D. \(108-27\pi\)
E. \(54sqrt{3}-27\pi\)


Please check the explanation in the attached file
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente [#permalink]

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New post 21 Oct 2016, 23:13
Hello ,

Just adding this as an alternative approach which i think would be pretty easy to get through :

Consider :

Regular Hexagon : ABCDEF

thus Equilateral triangle : ABO (See attached image )

we need to calculate the shaded portion so what we can do is Calculate the area of equilateral triangle ABO and deduct the area of smaller quadrants within that triangle (that are formed by extending the radius of all the triangle's) : so here we will deduct quadrant made from Circle A , Circle B and Circle O from the bigger equlilateral triangle and hence we can conclude the remaining area !

Now : as regular hexagon (ab=6) thus (a+b / 2) =3 the radius of all the circle

Area of ABO (bigger equilateral triangle ) = [[square_root]3 /4 ]s*s so ([square_root]3 / 4) *6 *6 = 9 [square_root]3

Area of 1 smaller quadrant : pie * r^2 / 6 ( because if we extend all the radius of adjacent circles to O six equal quadrants would be formed because ABCDEF is a regular hexagon )
thus : pie *3^2 /6 = 3pie /2

now as 3 quadrants the total area would be (3pie /2) * 3 = 9 pie /2

no subtract 3 quadrant area from the total area of equilateral triangle to get the area of shaded portion :

subtract : 9 [square_root]3 - 9pie /2 = [18 [square_root]3 - 9 pie /2 ] multiply this into 6 to get the area of all the shaded portions thus

54 [square_root]3 - 27 pie .... Option d

Sorry for pie ! :-D

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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente [#permalink]

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New post 22 Oct 2016, 01:39
While this Q can be solved in multiple ways one of which includes utilizing knowledge of trigonometry, the most intuitive way is visualizing the diagram and solving for it.

(1) Area of a sector = Pi *R^2 (Arc angle /360)
= Pi *3^3 (120/360)
= 3*Pi

Area of 6 sectors = 6 * ( 3*Pi) = 18*Pi

(2) Area of innermost circle = Pi * R^2 = Pi * 3^2 = 9* Pi

(3) Area of hexagon = 6 * (Area of equilateral triangles)
= 6 * (Sq Rt 3/4 * (side)^2 )
= 6 * (Sq Rt 3/4 * 36 )
= 6 * (9 Sq Rt 3)
= 54 Sq Rt 3

Therefore , area of shaded part = Area of hexagon - (area of 6 sectors + Area of innermost circle)
= 54 Sq Rt 3 - (18*Pi +9* Pi )
= 54 Sq Rt 3 - 27* Pi

Option E
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Regular hexagon ABCDEF has a perimeter of 36. O is the cente [#permalink]

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New post 13 May 2018, 04:20
Bunuel

Can you please tell me why the radius of all circles is assumed to be 30 ?

For circles A, B, C, D, E, and F to have centers on the vertices of the hexagon and to be tangent to one another, the circles must be the same size. Their radii must be equal to half of the side of the hexagon, 3. For circle O to be tangent to the other six circles, it too must have a radius of 3.- I cannot understand this point
Regular hexagon ABCDEF has a perimeter of 36. O is the cente   [#permalink] 13 May 2018, 04:20

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