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In the figure above, an equilateral triangle is inscribed in

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In the figure above, an equilateral triangle is inscribed in [#permalink] New post 12 Aug 2018, 10:12
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Question Stats:

57% (00:42) correct 42% (02:19) wrong based on 7 sessions
Attachment:
circle.jpg
circle.jpg [ 20.08 KiB | Viewed 398 times ]


In the figure above, an equilateral triangle is inscribed in a circle. If the arc bounded by adjacent corners of the triangle is between \(4\pi\) and \(6\pi\) long, which of the following could be the diameter of the circle?

(A) 6.5
(B) 9
(C) 11.9
(D) 15
(E) 23.5
[Reveal] Spoiler: OA

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Re: In the figure above, an equilateral triangle is inscribed in [#permalink] New post 18 Aug 2018, 20:09
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Carcass wrote:
Attachment:
circle.jpg


In the figure above, an equilateral triangle is inscribed in a circle. If the arc bounded by adjacent corners of the triangle is between \(4\pi\) and \(6\pi\) long, which of the following could be the diameter of the circle?

(A) 6.5
(B) 9
(C) 11.9
(D) 15
(E) 23.5


Here we need to consider the both the arc length i.e. \(4\pi\) and \(6\pi\)

Since all the angles of equilateral triangle = 60°

Therefore
\(\frac{{central angle}}{360}= \frac{{Arc length}}{circumference}\)

\(\frac{120}{360}= \frac{{4\pi}}{{2\pi*radius}}\)

or \(radius = 6\)

or \(diameter = 12\)

Now if we consider arc length as \(6\pi\) then,

\(\frac{{central angle}}{360} = \frac{{6\pi}}{{2\pi*radius}}\)

\(\frac{120}{360}= \frac{{6\pi}}{{2\pi*radius}}\)

or \(radius = 9\)

or \(diameter = 18\)

So only Option D satisfy
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Re: In the figure above, an equilateral triangle is inscribed in   [#permalink] 18 Aug 2018, 20:09
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In the figure above, an equilateral triangle is inscribed in

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