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# A circle is inscribed within a regular hexagon in such a way that the

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Math Expert
Joined: 02 Sep 2009
Posts: 50627
A circle is inscribed within a regular hexagon in such a way that the  [#permalink]

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24 Apr 2015, 01:52
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Difficulty:

45% (medium)

Question Stats:

69% (02:35) correct 31% (02:20) wrong based on 133 sessions

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A circle is inscribed within a regular hexagon in such a way that the circle touches all sides of the hexagon at exactly one point per side. Another circle is drawn to connect all the vertices of the hexagon. Expressed as a fraction, what is the ratio of the area of the smaller circle to the area of the larger circle?

A. √(2/3)
B. (√2)/3
C. (√3)/2
D. (√3)/4
E. 3/4

Kudos for a correct solution.

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Posts: 50627
Re: A circle is inscribed within a regular hexagon in such a way that the  [#permalink]

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27 Apr 2015, 01:08
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Bunuel wrote:
A circle is inscribed within a regular hexagon in such a way that the circle touches all sides of the hexagon at exactly one point per side. Another circle is drawn to connect all the vertices of the hexagon. Expressed as a fraction, what is the ratio of the area of the smaller circle to the area of the larger circle?

A. √(2/3)
B. (√2)/3
C. (√3)/2
D. (√3)/4
E. 3/4

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

The first thing to recognize is that a regular hexagon has 6 equal sides and 6 equal internal angles of 120°, since the angles inside the hexagon must add up to (n – 2)×180 = 720°, or 120° per angle. This is a very symmetrical figure. The circle will touch each side exactly in the middle of the side, by symmetry, like so:
Attachment:

hex-1.jpg [ 3.58 KiB | Viewed 4202 times ]

The outer circle will touch each vertex:
Attachment:

hex-2.jpg [ 5.23 KiB | Viewed 4206 times ]

Now, to compare the areas of the two circles, we should compare their radii. The obvious place to draw radii is from the points of contact with the circles:
Attachment:

hex-3.jpg [ 6 KiB | Viewed 4204 times ]

The triangle that’s been created is a 30-60-90 triangle. At point B, the radius is perpendicular to the side of the hexagon (which is tangent to the circle). The 120° angle of the hexagon is equally split by the longer radius, again by symmetry, creating a 60° angle at point A.

The ratios of the sides of a 30-60-90 triangle are 1: √3 : 2, with 2 as the longest side (the hypotenuse). The longer radius is the “2” side, while the shorter radius is the “√3” side.

Since the areas are proportional to the square of the radius (by A = πr^2), we know that the smaller area to the larger area would be $$\sqrt{(3)^2} : 2^2$$, or 3 : 4. Expressed as a fraction, this ratio is 3/4.

Note that C is a trap answer: it expresses the ratios of the radii themselves.

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Joined: 16 Mar 2015
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Re: A circle is inscribed within a regular hexagon in such a way that the  [#permalink]

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24 Apr 2015, 10:11
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given x the radius of the circle external to hexagon. The area of the external circle is X^2pi

Each angle of the hexagon is 120° . The hexagon is composed of 6 equilateral triangles. Each hight is the radius of the inner circle. The height is also the hight of a right triangle with angles 60°,90°, 30° . So, given the formula, the height/radius is (x/2)SQR3.

The area of the inner circle is ((x/2)SQR3)^2Pi = (¾)x^2Pi
The area of the external circle is x^2Pi

Hence the ratio is ((¾)x^2Pi ) / x^2Pi = ¾ Answer E.
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Joined: 28 Feb 2014
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Location: United States
Concentration: Strategy, General Management
Re: A circle is inscribed within a regular hexagon in such a way that the  [#permalink]

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24 Apr 2015, 14:03
1
A hexagon has 6 sides.
360/6=60

We know that there are 6 even central angles, 60 degrees each.
Splicing a 60 degree central angle in half, we have a 30-60-90 triangle.
The hypotenuse of that triangle (the radius of the large circle) is in the ratio of 2, and the height of the triangle is in the ratio of sqrt(3).

sqrt(3)^2 *pi / 2^2 * pi
= 3/4

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Re: A circle is inscribed within a regular hexagon in such a way that the  [#permalink]

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07 Apr 2018, 05:12
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Re: A circle is inscribed within a regular hexagon in such a way that the &nbs [#permalink] 07 Apr 2018, 05:12
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