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# The octagon in the diagram above is regular: all of its sides are of e

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Math Expert
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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30 Apr 2015, 04:14
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Difficulty:

95% (hard)

Question Stats:

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

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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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Re: The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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04 May 2015, 03:40
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.

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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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30 Apr 2015, 12:17
1
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
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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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30 Apr 2015, 23:38
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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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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01 May 2015, 13:43
angle = 135
on solving answer = c ; 2+root2
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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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23 Sep 2017, 11:18
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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Re: The octagon in the diagram above is regular: all of its sides are of e  [#permalink]

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23 Jan 2018, 12:51
1
one more method:
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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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25 Jun 2019, 21:22
Hello from the GMAT Club BumpBot!

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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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