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# Vertex A of equilateral triangle ABC is the center of the circle above

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Math Expert
Joined: 02 Sep 2009
Posts: 46319
Vertex A of equilateral triangle ABC is the center of the circle above [#permalink]

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11 Sep 2016, 05:23
00:00

Difficulty:

5% (low)

Question Stats:

90% (00:31) correct 10% (00:50) wrong based on 56 sessions

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Vertex A of equilateral triangle ABC is the center of the circle above. If the side of the triangle is 2, what is the radius of the circle?

A. 1/(2√3)
B. 1/2
C. (√3)/2
D. 1
E. √3

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Re: Vertex A of equilateral triangle ABC is the center of the circle above [#permalink]

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11 Sep 2016, 06:37
A radius drawn to the point of tangency is always perpendicular to the tangent, thus line AD is the height of triangle ABC.
height of an equilateral triangle is also the median, BD =DC = 1
AB^2 = AD^2 + DB ^2
=> AD^2 = AB^2 - DB^2
= 4 -1

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Re: Vertex A of equilateral triangle ABC is the center of the circle above [#permalink]

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16 Sep 2016, 07:25
side BC is a tangent to the circle. Line joining the point of contact of this tangent with the circle and the center of the circle is the perpendicular bisector of the triangle.
perpendicular bisector of equilateral triangle = $$\sqrt{3}/2$$ * side

= $$\sqrt{3}/2$$ * 2
= $$\sqrt{3}$$
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Re: Vertex A of equilateral triangle ABC is the center of the circle above [#permalink]

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25 May 2017, 19:36
Since the triangle is equilateral, we can easily use one of its properties namely by drawing a perpendicular line from the the center of the circle to the opposite vertex and create a 90, 60, and 30 Triangle with sides 2, $$\sqrt{3}$$ and 1.

That will easily bring us to the option: E
Re: Vertex A of equilateral triangle ABC is the center of the circle above   [#permalink] 25 May 2017, 19:36
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