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Ruben
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Post subject: Regula Hexagon  Posted: Tue Nov 13, 2007 10:29 pm |
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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)?
108 – 18
54 rt3 – 9
54 rt3– 18
108 – 27
54 rt3 – 27
If I use the Pitag. theo. to solve it instead of the 30-60-90 rule, I get a trianbgle that is 6 of hypot. and 6/2 that would give me a third side of 5, right?
Am I missing something?
Thanks
Ruben
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RonPurewal
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Post subject: Re: Regula Hexagon Posted: Wed Nov 14, 2007 5:07 am |
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| ManhattanGMAT Staff |
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Ruben wrote: If I use the Pitag. theo. to solve it instead of the 30-60-90 rule, I get a trianbgle that is 6 of hypot. and 6/2 that would give me a third side of 5, right?
Am I missing something?
I'm not sure exactly what you're trying to say here, but: If you draw a right triangle (by erecting a side from the center, perpendicular to one of the sides of the hexagon), then the short leg of the triangle is 6/2 = 3, and the hypotenuse is 6. So the third side is root(6^2 - 3^2), which is root(36 - 9) = root(27) = 3*root(3).
What I think is happening here is that you're subtracting 36 - 9 and getting 25. It's a common mistake, right along with the likes of 16 - 9 = 5, and other such digit-switching errors.
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RonPurewal
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Post subject: Posted: Wed Nov 14, 2007 5:07 am |
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Posts: 7146
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By the way, I'm quite impressed with your ability to embed images within posts.
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Ruben
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Post subject: thanks Posted: Mon Nov 19, 2007 1:46 pm |
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Thank you so much!
I guess when you study too long even a easy subtraction becomes an obstacle;)
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guest612
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Post subject: ok Posted: Sun Apr 27, 2008 5:50 pm |
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got it. totally get the explanation provided and how to solve this question as explained in the CAT.
However, can you please show me the error of my ways and tell me what is wrong with calculating the area of the hexagon using two trapezoids (resulting in the incorrect answer, D,) rather than six equalitaral triangles? This is more for future problems going forward so I don't actually make this error on the exam.
Thanks!
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RonPurewal
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Post subject: Re: ok Posted: Wed Apr 30, 2008 5:02 am |
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| ManhattanGMAT Staff |
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guest612 wrote: got it. totally get the explanation provided and how to solve this question as explained in the CAT.
However, can you please show me the error of my ways and tell me what is wrong with calculating the area of the hexagon using two trapezoids (resulting in the incorrect answer, D,) rather than six equalitaral triangles? This is more for future problems going forward so I don't actually make this error on the exam.
Thanks!
my mind-reading is a little bit off today, so, unfortunately, i can't tell you what's wrong with your solution unless you post it.
using two trapezoids should still give you the correct answer of 54√3 for the hexagon's area, though:
each trapezoid has height 3√3 (see my post above - this is almost certainly where you made your mistake)
each trapezoid has base1 = 6 and base2 = 12
so area of each trapezoid = (1/2)(3√3)(6 + 12) = 27√3
so hexagon = 2(27√3) = 54√3
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chris111
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Post subject: why isn't the hexagon's area 81? Posted: Tue Jan 06, 2009 9:50 pm |
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This is a regular hexagon... why can't you slice off triangle BCD (and then split it into two pieces, one above the other), and then re-attach them on the other side to form a square? Each side of the square is then equal to three radiuses, and since r=3, it is simple to get the area of it as 81.
What am I missing here?
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Guest
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Post subject: Posted: Fri Jan 09, 2009 9:32 am |
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In my opinion the Ans should be: 54 rt3- 27 pi
There are basically 6 Equilateral Traingles with side 6 which form a regular hexagon.
From This we deduct the area of the central circle with Radius 3 and 6* 1/3 circles of radius 3.
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chris111
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Post subject: Re: why isn't the hexagon's area 81? Posted: Mon Jan 12, 2009 6:22 pm |
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chris111 wrote: This is a regular hexagon... why can't you slice off triangle BCD (and then split it into two pieces, one above the other), and then re-attach them on the other side to form a square? Each side of the square is then equal to three radiuses, and since r=3, it is simple to get the area of it as 81.
What am I missing here?
Still waiting on an answer here.... doesn't that make sense?
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MICHAEL_SHAUNN
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Post subject: DARK SPACE... Posted: Fri Jan 16, 2009 2:37 pm |
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HI CHRIS,
HERE IS THE EXPLANATION.HOPE U WILL FIND IT HELPFUL.
THE AREA SUBSTENDED BY EACH CIRCLE AT THE ORIGIN O IS 120 DEGREES.THUS,THE THE AREA ENCLOSED WITHIN THE HEXAGON CORRESPONDING TO ALL THE CIRCLES EXCEPT THE ONE WITH ORIGIN AT O IS EQUAL TO THE AREA OF TWO SUCH CIRCLES WHICH IS 18pi.ADD TO IT THE AREA OF THE CIRCLE AT THE ORIGIN O GIVING 27pi.
NOW THE AREA OF THE HEXAGON IS EQUAL TO THE AREA OF SIX TIMES THE EQUILATERAL TRIALGLE(join origin O to each vertex of the hexagon,the area substended by each side of the hexagon at origin O is 60 degrees,the other two sides are equal and thus by the property of isosceles triangle the other two angles are also equal and equal to 60 degress and thus the HEXAGON has been represented by six equilateral triangles of equal sides).SUBTRACT FROM AREA OF THE HEXAGON(54*sqrt3) THE VALUE OF 27pi TO GET THE ANSWER.IT'S SAME AS GIVEN BY THE GUEST.
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MICHAEL_SHAUNN
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Post subject: A DARK SPACE.. Posted: Fri Jan 16, 2009 2:43 pm |
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A STATEMENT WENT WRONG WHICH IS "THE AREA SUBSTENDED BY EACH CIRCLE AT THE ORIGIN O IS 120 DEGREES".
ITS ACTUALLY LIKE THIS: "THE AREA OF EACH CIRCLE ENCLOSED BY THE HEXAGON EXCEPT THE ONE AT ORIGIN O SUBSTENDS AN ANGLE OF 120 DEGREES AT THEIR RESPECTIVE CENTRES."
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JonathanSchneider
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Post subject: Posted: Wed Jan 28, 2009 12:29 pm |
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Chris, you are right that you can cut that triangle BCD into two pieces, "move them" (in your mind) over to the left, and form a complete square. HOWEVER, the side length of that square is NOT equal to 3 times the radius of the circle. Note the line segment (again, in your mind) from B to D. It includes two radii (one in the top circle, one in the bottom) and a smaller segment in between. For the total to be euqal to 3 times the radius, that middle section would need to also be the length of the radius. This is where your approach breaks down. That segment can be seen to be the hypotenuse of an isosceles triangle in Circle C, where the two legs are each radii. Note that the angle at C is 120 degrees, not the 60 that would be required to make the hypotenuse equal to the length of either base. Thus, that middle segment is > the radius. Thus, the segment from B to D is greater than 3 times the radius.
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rgaddam
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Post subject: Re: Regula Hexagon Posted: Sun Feb 06, 2011 2:27 am |
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I need a quick clarification. I know I am interpreting the question wrong because I cannot get the correct answer. Initially when I worked on this question, I calculated area of 6 outer circles and subtracted the inner circle area from it. Does this method not give the area of shaded region?
So, it is (6 pi r^2 - pi r^2 = 45pi, r=3). What am I doing wrong. Please help.
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jnelson0612
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Post subject: Re: Regula Hexagon Posted: Sun Feb 06, 2011 4:34 pm |
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rgaddam wrote: I need a quick clarification. I know I am interpreting the question wrong because I cannot get the correct answer. Initially when I worked on this question, I calculated area of 6 outer circles and subtracted the inner circle area from it. Does this method not give the area of shaded region?
So, it is (6 pi r^2 - pi r^2 = 45pi, r=3). What am I doing wrong. Please help. Be careful here. You only want the shaded area *inside* the hexagon. If you are calculating the area of the six exterior circles you are including area that is not actually inside the hexagon. Thus, you will not be able to obtain the correct answer. Please read through the explanations given here again as to how to proceed, and if you are still stuck let us know. Thanks,
_________________
Jamie Nelson
ManhattanGMAT Instructor
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