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Ratio of Circle's Area to Hexagon's Area [#permalink]
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.

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?

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

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?? _________________

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

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??

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 )

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. _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

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