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

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

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09 Dec 2007, 19:36
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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. $$54sqrt{3}-27\pi$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: regular-hexagon-abcdef-has-a-perimeter-of-36-o-is-the-cente-89544.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Sep 2014, 19:30, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the [#permalink]

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09 Dec 2007, 20:08
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This is MGMAT Question =)

1) The side of a hex = 6 => r of the circle = 3.
2) Area of a hex = 6 * area of triangles with the side =6:
S = 1/2 height * base. The height divide triangle (with the sides = 6) to 2 triangles with the sides 6, 3*sqrt3 and 3 (as this is 1:sqrt3:2 triangles)
S of the triangle = 1/2 * (3*sqrt 3) * 6 = 9 * sqrt 3

So the area of hex = 6 * 9 * sqrt 3 = 54 * sqrt 3

The area of shaded region = 54 * sqrt 3 - 3 (Pi*3^2) = 54 * sqrt 3 - 27*Pi
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the [#permalink]

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28 Aug 2014, 22:26
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kizito2001 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)?

My approach:
Divide the hexagon into 6 equilateral triangles:

1. Area of hexagon. = 6 * area of a triangle (As all triangles are equilateral with same side).
= $$6 * \sqrt{3} /4 * 6^2$$
= $$54\sqrt{3}$$

2. Area of orange part in the diagram:
$$pie*r^2* 120/360$$
$$pie*9* 1/3$$
$$3* pie$$

Total of 6 areas like this + the central circle = $$18 * pie + 9 * pie = 27 * pie$$

Now area of black shaded region = area point 1 - area point 2.
= $$54\sqrt{3} - 27 * pie$$

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

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02 Sep 2014, 09:56
My approach was as follows :

1. Area of hexagon : 6 * \sqrt{3} / 4 * 3 * 3

2. Now calculate the area of 6 sectors of circles of angle 120 degree using sector formula : 120/360 * 3.14 * 3 * 3

3. Area of step 2 + Area of another circle : 3.14 * 3 * 3

4. Final Solution : Area of hexagon(step 1) - (Area of sectors+Area of 7th circle)(Step 3)
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the [#permalink]

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02 Sep 2014, 19:33
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kizito2001 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. $$54sqrt{3}-27\pi$$

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

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 $$54\sqrt{3}-27\pi$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: regular-hexagon-abcdef-has-a-perimeter-of-36-o-is-the-cente-89544.html
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the   [#permalink] 02 Sep 2014, 19:33
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