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Ratio of Circle's Area to Hexagon's Area

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Ratio of Circle's Area to Hexagon's Area [#permalink] New post 26 Jul 2003, 11:21
Need some help with this one... I can't seem to get the same answer as the one suggested by the material.

Anyways, here it is...

"In the figure above, a circle is inscribed in a hexagon in which all the sides are equal. If r is the radius of the circle, what is the ratio of the area of the circle to the area of the hexagon?"

(The answer I get is: pi / (6 root 3), but this is not correct. The actual answer is (pi * root 3)/ 6).

Does anyone care to give this one a try? Maybe it's something really silly that I'm overlooking.
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 [#permalink] New post 26 Jul 2003, 12:07
Should be broken up into equilateral. triangles.

But, radius doesn't extend all the way to the side, don't know where did you get this stuff from?
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 [#permalink] New post 26 Jul 2003, 19:17
This question came from some of the PR practice stuff. In breaking the hexagon into 6 equilateral triangles, I get an area for each triangle to be r root 3 * r (= r^2 root 3), which is then multiplied by 6, for the number of triangles..... But PR says this is not the correct answer???


Curly05 wrote:
Should be broken up into equilateral. triangles.

But, radius doesn't extend all the way to the side, don't know where did you get this stuff from?
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 [#permalink] New post 27 Jul 2003, 06:50
If u see it carefully , u will realize that radius of the circle is equal to height of the equliateral triangle. So side of the traingle equals

r / sin60 = r/ SQRT3 / 2 = 2r / SQRT3

So area of the regualr hexgon = 6 * 1/2 * 2r /SQRT3 * r
= (6/SQRT3) r^2

Hence, Ratio = pi r^2 / (6/sqrt3) r^2

= pi (SQRT3) /6
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Curly Here [#permalink] New post 27 Jul 2003, 08:00
Curly Here :yes

Radius should be = sq rt 3
-------- -
2
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 [#permalink] New post 29 Jul 2003, 12:56
the area of the hexagon is

6/ root 3 rr

area of circle pi r r

so ratio circle to hex = pi root 3 /6
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Baronharsh [#permalink] New post 29 Jul 2003, 13:13
Welcome on board, buddy!

I totally disagree with you and Brainless. Brainless is some hot chick supposedly. I hate it the girls here don't want to get hit on :wink:

The radius is sqrt 3
and the base is 2.
The area is 1/2 * 2 sq rt 3
----------------
2
= sq rt 3
six hexagon area's is 6 sq rt 3
Anyone want to disprove me?? 8-)

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Re: Baronharsh [#permalink] New post 29 Jul 2003, 15:59
Curly05 wrote:
Welcome on board, buddy!

I totally disagree with you and Brainless. Brainless is some hot chick supposedly. I hate it the girls here don't want to get hit on :wink:

The radius is sqrt 3
and the base is 2.
The area is 1/2 * 2 sq rt 3
----------------
2
= sq rt 3
six hexagon area's is 6 sq rt 3
Anyone want to disprove me?? 8-)


Give it another shot , Curly

Better brush up your trigonometry basics instead of wasting time hating girls !! You may need 'em sometime ( trigonometry basics , not girls )

:bebe
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Re: Brainless [#permalink] New post 29 Jul 2003, 16:13
Curly05 wrote:
You don't have to know trigonometry for GMAT. How did you get what you get? So you are a girl? I don't hate girls? What is your nationality?


Hey, from the beginning I have defended your right to ask questions -- any questions -- and make whatever comments so deemed necessary in the interest of all of us learning. You can confirm this with Stolyar.

But now you are simply being obnoxious to no one's benefit and everyone's irritation.

Please cut it out.

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MBA, Anderson School of Management, UCLA, Class of 1993

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Private [#permalink] New post 29 Jul 2003, 17:05
I realize that Akami, sorry :yikes

Please answer the question that I have asked.

:-D
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 [#permalink] New post 29 Jul 2003, 19:28
To err is human ... My apologies too !!!
:oops:
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Brainless [#permalink] New post 01 Aug 2003, 08:18
Hey Brainless Beauty,

Can you show me my error?

I'm dumber than a babe :oops:
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Re: Brainless [#permalink] New post 01 Aug 2003, 18:36
Curly05 wrote:
Hey Brainless Beauty,

Can you show me my error?

I'm dumber than a babe :oops:


In your earlier message, you wrote

The radius is sqrt 3
and the base is 2.
The area is 1/2 * 2 sq rt 3
----------------
2
= sq rt 3
six hexagon area's is 6 sq rt 3
Anyone want to disprove me??
----------------------------------------------------------------------
Actually, radius (r) is the height of the equilateral traingle.
Lets say , each side of this equilateral triangle is x
So, r/x = Sin60 = SQRT3 / 2
==> x = 2r / SQRT3
Area of traingle is 1/2 * 2r/SQRT3 * r = r^2 / SQRT3
Hence, hexagon area = 6 x r^2 /SQRT3

Hope this helps you !!
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Re: Brainless   [#permalink] 01 Aug 2003, 18:36
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