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# In the figure below, if the radius of circle O is r and

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In the figure below, if the radius of circle O is r and [#permalink]  31 Mar 2012, 03:10
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83% (02:08) correct 16% (01:07) wrong based on 1 sessions
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?
[Reveal] Spoiler: OA

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Re: Inscribed circle in an Equilateral triangle [#permalink]  31 Mar 2012, 04:37
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.
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Re: Inscribed circle in an Equilateral triangle [#permalink]  31 Mar 2012, 05:01
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 772 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: In the figure below, if the radius of circle O is r and [#permalink]  31 Mar 2012, 05:11
Bunuel - Thanks. But how can (angle DOC is 60°)?
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Re: In the figure below, if the radius of circle O is r and [#permalink]  31 Mar 2012, 05:17
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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Re: In the figure below, if the radius of circle O is r and [#permalink]  31 Mar 2012, 17:00
Sorry Bunuel - what I meant to ask was - How come OC will bisect the angle DCO?
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Re: In the figure below, if the radius of circle O is r and [#permalink]  31 Mar 2012, 17:16
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 figure below, if the radius of circle O is r and   [#permalink] 31 Mar 2012, 17:16
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