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Math Expert V
Joined: 02 Sep 2009
Posts: 58320
The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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26 00:00

Difficulty:   95% (hard)

Question Stats: 37% (03:11) correct 63% (02:50) wrong based on 103 sessions

HideShow timer Statistics The octagon in the diagram above is regular: all of its sides are of equal length, and all of its angles are of equal measure. If the octagon’s perimeter is 8 inches, and every other vertex of the octagon is connected to create a square as shown above, what is the area of the square?

A. 2
B. 2√2
C. 2 + √2
D. 2√3
E. 2 + √3

Kudos for a correct solution.

Attachment: dg1.png [ 8.44 KiB | Viewed 3251 times ]

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Math Expert V
Joined: 02 Sep 2009
Posts: 58320
Re: The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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Bunuel wrote: The octagon in the diagram above is regular: all of its sides are of equal length, and all of its angles are of equal measure. If the octagon’s perimeter is 8 inches, and every other vertex of the octagon is connected to create a square as shown above, what is the area of the square?

A. 2
B. 2√2
C. 2 + √2
D. 2√3
E. 2 + √3

Kudos for a correct solution.

Attachment:
The attachment dg1.png is no longer available

MANHATTAN GMAT OFFICIAL SOLUTION:

To find the area of the square, we need the length of one of the square’s sides. That length can be found if we construct a right triangle outside of the existing diagram, with one side of the octagon as hypotenuse, like this: There are several ways to determine that the small triangle thus created is a 45º-45º-90º triangle. This fact can be deduced from the symmetry of the figure; alternatively, the interior angles of the octagon are 135º each, so that the two marked angles are 180º – 135º = 45º.

The perimeter of the octagon is 8 inches, so each side of the octagon is 1 inch long. Using the ratios for a 45º-45º-90º triangle, the length of each of the red sides in the diagram above is $$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ inches.

We now have both legs of a larger right triangle: the shorter (red) leg is $$\frac{\sqrt{2}}{2}$$ inch, and the longer (red + black) leg is $$(1 + \frac{\sqrt{2}}{2})$$ inches. Let s stand for the third side of the larger triangle and the side of the square. Use the Pythagorean theorem to find it:

$$s^2= (\frac{\sqrt{2}}{2})^2 + (1 + \frac{\sqrt{2}}{2})^2$$

$$s^2= \frac{1}{2} + (1 + \sqrt{2} + \frac{1}{2})$$

$$s^2 = 2+\sqrt{2}$$

Conveniently, the area of the square is s^2, so there is actually no need to find the value of s itself; the expression $$s^2 = 2 + \sqrt{2}$$ represents the area of the square.

The correct answer is (C).

Attachment: dg2.png [ 10.64 KiB | Viewed 2869 times ]

Attachment: dg3.png [ 11.79 KiB | Viewed 2865 times ]

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General Discussion
Manager  Joined: 17 Mar 2015
Posts: 115
The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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1
The value of an angle in an octagon is 180*(8-2)/8 = 180*6/8 = 180*3/4 = 135 degrees.
Sides are equal to 1 coz its regular octagon.
Cosine theorem helps us find the side of the square X which is $$\sqrt{1 + 1 - 2*1*1*cos(135)} = \sqrt{2 +2^\frac{1}{2}} = X$$
The area ends up being $$X^2$$ which corresponds to C
Director  Joined: 07 Aug 2011
Posts: 502
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27 Re: The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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2
2
Bunuel wrote: The octagon in the diagram above is regular: all of its sides are of equal length, and all of its angles are of equal measure. If the octagon’s perimeter is 8 inches, and every other vertex of the octagon is connected to create a square as shown above, what is the area of the square?

A. 2
B. 2√2
C. 2 + √2
D. 2√3
E. 2 + √3

Kudos for a correct solution.

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The attachment dg1.png is no longer available

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Manager  Joined: 01 Jan 2015
Posts: 55
Re: The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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angle = 135
on solving answer = c ; 2+root2
Manager  G
Joined: 01 Sep 2016
Posts: 189
GMAT 1: 690 Q49 V35 Re: The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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Another way to deal with this question is to construct a circle around the octagon, and assume that the side of 1 unit is equal to the 1/8 th arc of the circle. This assumption and subsequent solving will help you to eliminate choice A, B and E. You will be left with Choice C and D . I would go with C as at it contains root 2 in it.( high chances of this question being solved by 45, 45 and 90. But still that is not a clear cut approach. But since this is a vehry difficult question, we are now at higher odds of getting the right answer.
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Joined: 02 Apr 2014
Posts: 468
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34 GPA: 3.5
Re: The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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1
one more method:
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Non-Human User Joined: 09 Sep 2013
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Re: The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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_________________ Re: The octagon in the diagram above is regular: all of its sides are of e   [#permalink] 25 Jun 2019, 21:22
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