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M12-15

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M12-15  [#permalink]

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New post 15 Sep 2014, 23:46
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A regular hexagon is inscribed in a circle. If the radius of the circle is 1, what is the length of the side of the hexagon? (A hexagon is a six-sided polygon).

A. \(\frac{1}{\sqrt{2}}\)
B. \(2\sqrt{3}\)
C. \(1\)
D. \(\sqrt{2}\)
E. \(\sqrt{3}\)

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Re M12-15  [#permalink]

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New post 15 Sep 2014, 23:46
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Official Solution:

A regular hexagon is inscribed in a circle. If the radius of the circle is 1, what is the length of the side of the hexagon? (A hexagon is a six-sided polygon).

A. \(\frac{1}{\sqrt{2}}\)
B. \(2\sqrt{3}\)
C. \(1\)
D. \(\sqrt{2}\)
E. \(\sqrt{3}\)

Connect the center of the circle with each of the hexagon's vertices to get six triangles. In each triangle, the angle at the center is 60 degrees \((\frac{360}{6})\). The two other angles are also 60 degrees each. Thus, all six triangles are equilateral.

Answer: C
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M12-15  [#permalink]

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New post 14 Dec 2015, 12:56
could you please elaborate further on how you got the answer?
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Re: M12-15  [#permalink]

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Re: M12-15  [#permalink]

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New post 10 Jan 2016, 10:47
Could you please explain why the other 2 angles are also 60?

Bunuel wrote:
rhio wrote:
could you please elaborate further on how you got the answer?


Please be a little bit more specific. Thank you.
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Re: M12-15  [#permalink]

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New post 10 Jan 2016, 12:31
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mestrec wrote:
Could you please explain why the other 2 angles are also 60?

Bunuel wrote:
rhio wrote:
could you please elaborate further on how you got the answer?


Please be a little bit more specific. Thank you.


Check here: a-regular-hexagon-is-inscribed-in-a-circle-if-the-radius-of-the-circl-75199.html
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New to the Math Forum?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M12-15  [#permalink]

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New post 15 Jan 2016, 08:52
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mestrec wrote:
Could you please explain why the other 2 angles are also 60?

Bunuel wrote:
rhio wrote:
could you please elaborate further on how you got the answer?


Please be a little bit more specific. Thank you.


Hi,
hexagon as shown is inscribed in the circle..
each angle of hexagon is 120...
when you join opposite vertices, these are also the diagonals and bisect the angles 120..

this is the reason why all three angles are 60 and teh side of hexa too equals radius =1..
hope it helps
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Re: M12-15  [#permalink]

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New post 25 Mar 2018, 20:20
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Bunuel We really needed a picture with this question. could not anticipate anything without a pic. :(
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Re: M12-15 &nbs [#permalink] 25 Mar 2018, 20:20
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