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# In the figure above, a regular hexagon EFGHIJ is inscribed in the rect

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Joined: 14 Sep 2015
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GMAT 1: 700 Q45 V40
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In the figure above, a regular hexagon EFGHIJ is inscribed in the rect  [#permalink]

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Updated on: 25 May 2017, 14:35
4
00:00

Difficulty:

75% (hard)

Question Stats:

54% (03:25) correct 46% (02:39) wrong based on 55 sessions

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In the figure above, a regular hexagon EFGHIJ is inscribed in the rectangle ABCD. If the ratio of the magnitudes of the area and the perimeter of the hexagon is $$\frac{√3}{2}$$, what is the ratio of the magnitudes of the area and the perimeter of the rectangle ABCD?

A. $$2\sqrt{3} - 4$$

B. $$4\sqrt{3} - 6$$

C. $$2\sqrt{3} + 4$$

D. $$4\sqrt{3} + 6$$

E. None of the above

Attachment:

XFWWxun.png [ 4.75 KiB | Viewed 2442 times ]

Originally posted by niteshwaghray on 25 May 2017, 13:28.
Last edited by Bunuel on 25 May 2017, 14:35, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 58434
Re: In the figure above, a regular hexagon EFGHIJ is inscribed in the rect  [#permalink]

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25 May 2017, 15:13
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niteshwaghray wrote:

In the figure above, a regular hexagon EFGHIJ is inscribed in the rectangle ABCD. If the ratio of the magnitudes of the area and the perimeter of the hexagon is $$\frac{√3}{2}$$, what is the ratio of the magnitudes of the area and the perimeter of the rectangle ABCD?

A. $$2\sqrt{3} - 4$$

B. $$4\sqrt{3} - 6$$

C. $$2\sqrt{3} + 4$$

D. $$4\sqrt{3} + 6$$

E. None of the above

Attachment:
The attachment XFWWxun.png is no longer available

Some redundant info in the question...

Check the image below:

Each interior angle of a regular polygon is given by: $$\frac{180(n−2)}{n}$$, where n is the number of sides. Thus, each interior angel in the regular hexagon = 120°. This will make triangles at the corner 30°-60°-90°. The sides are always in the ratio 1:√3:2. Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

So, we have that one side of the rectangle is 1 + 2 + 1 = 4 and another is $$\sqrt{3}+\sqrt{3}=2\sqrt{3}$$

Area = $$4*2\sqrt{3}=8\sqrt{3}$$;

Perimeter = $$2*(4+2\sqrt{3})=8+4\sqrt{3}$$.

Ratio = $$\frac{8\sqrt{3}}{8+4\sqrt{3}}=\frac{2\sqrt{3}}{2+\sqrt{3}}$$.

Rationalize by multiplying numerator and denominator by $$2-\sqrt{3}$$:

$$\frac{2\sqrt{3}(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=\frac{2\sqrt{3}(2-\sqrt{3})}{4-1}=4\sqrt{3}-6$$.

Attachment:

Hexagon.png [ 6.04 KiB | Viewed 2302 times ]

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Re: In the figure above, a regular hexagon EFGHIJ is inscribed in the rect  [#permalink]

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13 Oct 2018, 03:16
Hello from the GMAT Club BumpBot!

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Re: In the figure above, a regular hexagon EFGHIJ is inscribed in the rect   [#permalink] 13 Oct 2018, 03:16
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