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In the figure below, if the radius of circle O is r and tria

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In the figure below, if the radius of circle O is r and tria [#permalink] New post 31 Mar 2012, 02:10
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Question Stats:

74% (02:31) correct 26% (01:22) wrong based on 87 sessions
In the figure below, if the radius of circle O is r and triangle ABC is equilateral, what is the length of AB, in terms of r?
Attachment:
Triangle.GIF
Triangle.GIF [ 3.47 KiB | Viewed 2823 times ]


(A) r \(\sqrt{2}\)
(B) r \(\sqrt{3}\)
(C) 2r \(\sqrt{3}\)
(D) \(\frac{3}{2}\) r
(E) 2r
[Reveal] Spoiler: OA

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Re: Inscribed circle in an Equilateral triangle [#permalink] New post 31 Mar 2012, 04:01
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enigma123 wrote:
In the fi…gure below, if the radius of circle O is r and triangle ABC is equilateral, what is the length of AB, in terms of r?

(A) r \(\sqrt{2}\)
(B) r \(\sqrt{3}\)
(C) 2r \(\sqrt{3}\)
(D) \(\frac{3}{2}\) r
(E) 2r

Any idea how to solve?


There are many ways to solve this question including using the direct formula saying that the radius of the inscribed circle in an equilateral triangle is \(r=a*\frac{\sqrt{3}}{6}\).

One can also do the following, consider the diagram below:
Attachment:
Circle.gif
Circle.gif [ 6.02 KiB | Viewed 2808 times ]
Angle DCO is 60/2=30 degrees, hence triangle DOC is 30°-60°-90° right triangle (angle DOC is 60°). Next, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\) and the leg opposite 30° (OD=r) corresponds with \(1\) and the leg opposite 60° (DC) corresponds with \(\sqrt{3}\), so \(\frac{r}{DC}=\frac{1}{\sqrt{3}}\) --> \(DC=r\sqrt{3}\). Now, since DC=AC/2 then \(AC=2DC=2r\sqrt{3}\).

Answer: C.
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Re: Inscribed circle in an Equilateral triangle [#permalink] New post 31 Mar 2012, 03:37
Ok so as in the diagram.
the ratio of green line to red line will be 2:1.
Pink line will be perpendicular to triangle edge.
therefore

(2a)^2 = (x/2)^2 + a^2

Upon solving you will get x = 2(3)^1/2 a

Hope it helps.
Attachments

Untitled.png
Untitled.png [ 5.14 KiB | Viewed 2796 times ]


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Director
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Status: Finally Done. Admitted in Kellogg for 2015 intake
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Kudos [?]: 1376 [0], given: 217

Re: In the fi…gure below, if the radius of circle O is r and [#permalink] New post 31 Mar 2012, 04:11
Bunuel - Thanks. But how can (angle DOC is 60°)?
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MGMAT 1 --> 530
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MGMAT 3 ---> 610
GMAT ==> 730

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Re: In the fi…gure below, if the radius of circle O is r and [#permalink] New post 31 Mar 2012, 04:17
Expert's post
Director
Director
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Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
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Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V40
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Kudos [?]: 1376 [0], given: 217

Re: In the fi…gure below, if the radius of circle O is r and [#permalink] New post 31 Mar 2012, 16:00
Sorry Bunuel - what I meant to ask was - How come OC will bisect the angle DCO?
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GMAT ==> 730

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Re: In the fi…gure below, if the radius of circle O is r and [#permalink] New post 31 Mar 2012, 16:16
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Re: In the figure below, if the radius of circle O is r and tria [#permalink] New post 06 Sep 2013, 02:48
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Re: In the figure below, if the radius of circle O is r and tria [#permalink] New post 10 Sep 2013, 21:59
Hi Bunuel, can we say that in equilateral triangle median = angle bisector = altitude = perpendicular bisector?
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Re: In the figure below, if the radius of circle O is r and tria [#permalink] New post 11 Sep 2013, 00:45
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Re: In the figure below, if the radius of circle O is r and tria [#permalink] New post 07 Jan 2014, 01:57
Hi Bunnel,

How we solve this problem by using the direct formula for inradius of equilateral triangle
Re: In the figure below, if the radius of circle O is r and tria   [#permalink] 07 Jan 2014, 01:57
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In the figure below, if the radius of circle O is r and tria

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