|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 23 Jun 2003
Posts: 16
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Ratio of Circle's Area to Hexagon's Area [#permalink]
 26 Jul 2003, 12:21
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.
|
|
|
|
|
|
|
Eternal Intern
Joined: 07 Jun 2003
Posts: 484
Location: Lone Star State
Followers: 1
Kudos [?]:
3
[0], given: 0
|
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?
|
|
|
|
|
|
Intern
Joined: 23 Jun 2003
Posts: 16
Followers: 0
Kudos [?]:
0
[0], given: 0
|
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?
|
|
|
|
|
|
Manager
Joined: 25 Jun 2003
Posts: 100
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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
_________________
Brainless
|
|
|
|
|
|
Eternal Intern
Joined: 07 Jun 2003
Posts: 484
Location: Lone Star State
Followers: 1
Kudos [?]:
3
[0], given: 0
|
Curly Here 
Radius should be = sq rt 3
-------- -
2
|
|
|
|
|
|
Intern
Joined: 27 Apr 2003
Posts: 38
Followers: 0
Kudos [?]:
0
[0], given: 0
|
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
|
|
|
|
|
|
Eternal Intern
Joined: 07 Jun 2003
Posts: 484
Location: Lone Star State
Followers: 1
Kudos [?]:
3
[0], given: 0
|
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 
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?? 
_________________
Ride em cowboy
|
|
|
|
|
|
Manager
Joined: 25 Jun 2003
Posts: 100
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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 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 )
_________________
Brainless
|
|
|
|
|
|
GMAT Instructor
Joined: 07 Jul 2003
Posts: 773
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 5
Kudos [?]:
9
[0], given: 0
|
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.
_________________
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
|
|
|
|
|
|
Eternal Intern
Joined: 07 Jun 2003
Posts: 484
Location: Lone Star State
Followers: 1
Kudos [?]:
3
[0], given: 0
|
I realize that Akami, sorry 
Please answer the question that I have asked.
_________________
Ride em cowboy
|
|
|
|
|
|
Manager
Joined: 25 Jun 2003
Posts: 100
Followers: 1
Kudos [?]:
0
[0], given: 0
|
To err is human ... My apologies too !!!
_________________
Brainless
|
|
|
|
|
|
Eternal Intern
Joined: 07 Jun 2003
Posts: 484
Location: Lone Star State
Followers: 1
Kudos [?]:
3
[0], given: 0
|
Hey Brainless Beauty,
Can you show me my error?
I'm dumber than a babe
_________________
Ride em cowboy
|
|
|
|
|
|
Manager
Joined: 25 Jun 2003
Posts: 100
Followers: 1
Kudos [?]:
0
[0], given: 0
|
Curly05 wrote: Hey Brainless Beauty, Can you show me my error? I'm dumber than a babe 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 !!
_________________
Brainless
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
