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

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V
Joined: 02 Sep 2009
Posts: 60555
Vertex A of equilateral triangle ABC is the center of the circle above  [#permalink]

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New post 11 Sep 2016, 05:23
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

95% (00:53) correct 5% (01:41) wrong based on 62 sessions

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Status: Greatness begins beyond your comfort zone
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Re: Vertex A of equilateral triangle ABC is the center of the circle above  [#permalink]

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New post 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
=> AD = (3)^(1/2)

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

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New post 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

Radius = perpendicular bisector
= \(\sqrt{3}/2\) * 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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New post 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
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Re: Vertex A of equilateral triangle ABC is the center of the circle above   [#permalink] 25 May 2017, 19:36
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Vertex A of equilateral triangle ABC is the center of the circle above

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