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# If the side of the regular hexagon above is 2, what is the circumferen

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Math Expert
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If the side of the regular hexagon above is 2, what is the circumferen  [#permalink]

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28 Sep 2016, 02:51
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55% (hard)

Question Stats:

65% (02:06) correct 35% (02:25) wrong based on 130 sessions

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If the side of the regular hexagon above is 2, what is the circumference of the inscribed circle?

A. 2∏√3

B. 3∏

C. 4∏/√3

D. 2∏/√3

E. ∏√3

Attachment:

T6150.png [ 4.96 KiB | Viewed 2327 times ]

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Re: If the side of the regular hexagon above is 2, what is the circumferen  [#permalink]

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28 Sep 2016, 03:47
2
Bunuel wrote:

If the side of the regular hexagon above is 2, what is the circumference of the inscribed circle?

A. 2∏√3

B. 3∏

C. 4∏/√3

D. 2∏/√3

E. ∏√3

Attachment:
The attachment T6150.png is no longer available

Refer the diagram which I mentioned below:

Now AC will be 1 because circle is touching the midpoints of the circle.

Then we get kite ACBO.

We have AC = 1 and CB = 1, now we need to find BO which is radius

from triplets we get 30:60:90 = 1:√3:2

Now CBO is right angled triangle, then we get BO as √3.

r = √3...circumference is 2∏r is 2∏√3.

IMO option A.
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Hexa.PNG [ 9.9 KiB | Viewed 1999 times ]

Math Expert
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Posts: 7984
Re: If the side of the regular hexagon above is 2, what is the circumferen  [#permalink]

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28 Sep 2016, 04:06
Bunuel wrote:

If the side of the regular hexagon above is 2, what is the circumference of the inscribed circle?

A. 2∏√3

B. 3∏

C. 4∏/√3

D. 2∏/√3

E. ∏√3

Attachment:
T6150.png

Hi,
Each side is 2..
If each vertex is joined at centre, side will be 2 as it becomes 6*equilateral triangle.

Now the height of each equilateral triangle is nothing but the radius of circle.
Ht = √3/2 *2=√3....
So radius=√3 and CIRCUMFERENCE = 2*π*√3..
A
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If the side of the regular hexagon above is 2, what is the circumferen  [#permalink]

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28 Sep 2016, 05:37
I solved it the following way:

On joining the vertices from center of hexagon we got a equilateral triangle with side as 2. On bisecting this triangle we get a right angle triangle with base as 1 and hypotenuse as 2, from these two we can calculate the height for the triangle (Which is the radius of circle in questions) which comes out to be √3. So the circumference will be 2∏√3.

So, option A. Pls let me know if this approach is correct.

Bunuel - can you pls tell if this approach is correct? TIA
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Re: If the side of the regular hexagon above is 2, what is the circumferen  [#permalink]

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01 Mar 2017, 06:57
The side is 2. \
The height, which is the radius will be √3, if we connect each vertices to the center of the circle and each triangle will be equilateral.
Therefore the circumference will be 2∏r= 2∏√3
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Re: If the side of the regular hexagon above is 2, what is the circumferen  [#permalink]

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01 Mar 2017, 13:18
Bunuel wrote:

If the side of the regular hexagon above is 2, what is the circumference of the inscribed circle?

A. 2∏√3

B. 3∏

C. 4∏/√3

D. 2∏/√3

E. ∏√3

Attachment:
The attachment T6150.png is no longer available

Attachments

hexagon.png [ 33.51 KiB | Viewed 1641 times ]

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Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2815
Re: If the side of the regular hexagon above is 2, what is the circumferen  [#permalink]

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02 Mar 2017, 18:29
3
Bunuel wrote:

If the side of the regular hexagon above is 2, what is the circumference of the inscribed circle?

A. 2∏√3

B. 3∏

C. 4∏/√3

D. 2∏/√3

E. ∏√3

We can break up the hexagon into 6 equilateral triangles in which each side is equal to 2.

If we “cut” one of the equilateral triangles in half, we have a 30-60-90 right triangle. The ratio of the sides of a 30-60-90 triangle are x : x√3 : 2x.

We see that the side opposite the 30-degree angle = 1, and thus, the side opposite the 60-degree angle, which also represents the radius of the circle, is √3. We now can determine the circumference of the circle.

Circumference = 2Πr = 2Π(√3) = 2Π√3

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Re: If the side of the regular hexagon above is 2, what is the circumferen  [#permalink]

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04 Apr 2018, 09:43
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Re: If the side of the regular hexagon above is 2, what is the circumferen   [#permalink] 04 Apr 2018, 09:43
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