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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]  31 Mar 2012, 02:10
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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 [ 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]  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 [ 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 $$AC=2DC=2r\sqrt{3}$$.

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Re: Inscribed circle in an Equilateral triangle [#permalink]  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 [ 5.14 KiB | Viewed 2796 times ]

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Director
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]  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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Kudos [?]: 50152 [0], given: 7529

Re: In the figure below, if the radius of circle O is r and [#permalink]  31 Mar 2012, 04:17
Expert's post
enigma123 wrote:
Bunuel - Thanks. But how can (angle DOC is 60°)?

DOC is 30°-60°-90° right triangle. Angle D=90°, angle DCO=30° and angle DOC=60°.
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Status: Finally Done. Admitted in Kellogg for 2015 intake
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Re: In the figure below, if the radius of circle O is r and [#permalink]  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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Kudos [?]: 50152 [0], given: 7529

Re: In the figure below, if the radius of circle O is r and [#permalink]  31 Mar 2012, 16:16
Expert's post
enigma123 wrote:
Sorry Bunuel - what I meant to ask was - How come OC will bisect the angle DCO?

Ask yourself, why should any angle from ACO and BCO be greater than another?
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Re: In the figure below, if the radius of circle O is r and tria [#permalink]  06 Sep 2013, 02:48
Expert's post
Bumping for review and further discussion.
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Re: In the figure below, if the radius of circle O is r and tria [#permalink]  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]  11 Sep 2013, 00:45
Expert's post
b2bt wrote:
Hi Bunuel, can we say that in equilateral triangle median = angle bisector = altitude = perpendicular bisector?

Yes, that's true. For more check here: math-triangles-87197.html
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Re: In the figure below, if the radius of circle O is r and tria [#permalink]  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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