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

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

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05 Aug 2006, 13:51
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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)? (see attached file)

a. 108-18pi
b. 54sqrt(3)-9pi
c. 54sqrt(3)-18pi
d. 108-27pi
e. 54sqrt(3)-27pi
Attachments

Hex.doc [72.5 KiB]

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09 Aug 2006, 13:41
E

hexagon:
if I don't know that the area of a regualr hexagon is (edge length)^2*(3/2)*sqrt(3)... (and why should I?)

...then I can figure it's made up of six equilateral triangles with sides of length 6.

by the pythagorean theorem, the height of each triangle is 3*sqrt(3)
(note each 60-60-60 triangle is 2 30-60-90 triangles and these have the old "one, two, radical three" ratio of sides we all loved in grade school, here the "two" is the length of the eqilateral triangle's side)

so the area is (1/2)*6*3*sqrt(3).

the area of the hexagon is then 6*(1/2)*6*3*sqrt(3) = 54*sqrt(3)

circles, all of radius 3, area 9pi:

the vertex angle of a regular hexagon is 120degrees.

so (120/360) or one third of each of six circles, plus the whole of the one circle in the middle of the hexagon, lies within the perimeter of the hexagon. three circles' area, all told.

inside the hexagon but outside the circle:

Ahex - 3*Acircle = 54*sqrt(3)- 3*9pi
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09 Aug 2006, 13:50
Yup E.

6*9(sqrt3) - 9pi - 18pi

equals

54sqrt(3)-27pi
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09 Aug 2006, 13:52
E it is!

Area of Hexagon - 3*(Area of Circle of radius 3)
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Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

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09 Aug 2006, 13:57
E

Area of circles A,B, C, D E and F that is inside the hexagon = 6*PI * 9/3 = 18 PI

Area of inner circle = 9 PI

Area of hexagon = 6 * 1/2 * 6 * SQRT(27) = 54 * SQRT(3)

Area of shaded region = 54 * SQRT(3) - 18PI - 9PI = 54 SQRT(3) - 27PI
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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09 Aug 2006, 16:46
sorry for the delay....

but you are all winners!!!!!! Keep up the great work!!!!

E is the OA
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11 Aug 2006, 04:23
Side of hexagon = 6
Area of hexangons = 6*Sqr3/4 * 36 = 54sqr3 (Hexagon has 6 equilateral triangles)

Area of circle in the center = 9pi

Area of sector of the circle = 120/360 * 9pi = 3pi
Area of the six sectors = 6*3pi = 18pi

Area of shaded = 54sqr3 - (18pi+9Pi)
= 54sqr3 - 27pi
11 Aug 2006, 04:23
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