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

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

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New post Updated on: 25 May 2017, 14:35
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A
B
C
D
E

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

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New post 25 May 2017, 15:13
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niteshwaghray wrote:
Image

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:

Image

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 \(2-\sqrt{3}\):

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

Answer: B.


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

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New post 13 Oct 2018, 03:16
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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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