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# If equilateral triangle MNP is inscribed in circle O with

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If equilateral triangle MNP is inscribed in circle O with  [#permalink]

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31 Mar 2012, 15:46
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15% (low)

Question Stats:

78% (01:27) correct 22% (01:39) wrong based on 144 sessions

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If equilateral triangle MNP is inscribed in circle O with radius of 6, what is the length of minor arc MN?

(A) 2 pi
(B) 4 pi
(C) 6 pi
(D) 8 pi
(E) 12 pi

How come the answer is B guys?

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Joined: 02 Sep 2009
Posts: 64144
Re: Length of Arc MN  [#permalink]

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31 Mar 2012, 15:52
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1
enigma123 wrote:
If equilateral triangle MNP is inscribed in circle O with radius of 6, what is the length of minor arc MN?
(A) 2 pi
(B) 4 pi
(C) 6 pi
(D) 8 pi
(E) 12 pi

How come the answer is B guys?

Since triangle MNP is equilateral each of the 3 minor arcs (MN, NP, PM) will equal to 1/3rd of the circumference, which is $$2\pi{r}=12\pi$$. Hence, the length of minor arc MN is $$\frac{12\pi}{3}=4\pi$$.

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Re: Length of Arc MN  [#permalink]

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01 Apr 2012, 14:18
Bunuel wrote:
enigma123 wrote:
If equilateral triangle MNP is inscribed in circle O with radius of 6, what is the length of minor arc MN?
(A) 2 pi
(B) 4 pi
(C) 6 pi
(D) 8 pi
(E) 12 pi

How come the answer is B guys?

Since triangle MNP is equilateral each of the 3 minor arcs (MN, NP, PM) will equal to 1/3rd of the circumference, which is $$2\pi{r}=12\pi$$. Hence, the length of minor arc MN is $$\frac{12\pi}{3}=4\pi$$.

I understand the answer, but was initially thrown off by calculating that each triangle angles is 60 degrees and circumference is 12 pi so 60/360 * 12pi = 2pi. What's wrong with this reasoning? Thank you.
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Joined: 02 Sep 2009
Posts: 64144
Re: Length of Arc MN  [#permalink]

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01 Apr 2012, 15:12
1
bohdan01 wrote:
Bunuel wrote:
enigma123 wrote:
If equilateral triangle MNP is inscribed in circle O with radius of 6, what is the length of minor arc MN?
(A) 2 pi
(B) 4 pi
(C) 6 pi
(D) 8 pi
(E) 12 pi

How come the answer is B guys?

Since triangle MNP is equilateral each of the 3 minor arcs (MN, NP, PM) will equal to 1/3rd of the circumference, which is $$2\pi{r}=12\pi$$. Hence, the length of minor arc MN is $$\frac{12\pi}{3}=4\pi$$.

I understand the answer, but was initially thrown off by calculating that each triangle angles is 60 degrees and circumference is 12 pi so 60/360 * 12pi = 2pi. What's wrong with this reasoning? Thank you.

The angle in the formula you are using should be a central angle and 60 degrees angle is inscribed angle.

Now, according to the central angle theorem the measure of inscribed angle is always half the measure of the central angle.. So, the corresponding central angles would be 2*60=120 degrees and if we put this value in your formula we'll get 120/360*12pi=4pi, which is a correct answer.

For more on this subject check Circles chapter of Math Book: math-circles-87957.html

Hope it helps..
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Re: If equilateral triangle MNP is inscribed in circle O with  [#permalink]

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08 Jul 2016, 04:14
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Re: If equilateral triangle MNP is inscribed in circle O with   [#permalink] 08 Jul 2016, 04:14