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

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

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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


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

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New post 30 Apr 2015, 12:17
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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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New post 30 Apr 2015, 23:38
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Bunuel wrote:
Image
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.


[Reveal] Spoiler:
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The attachment dg1.png is no longer available

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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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New post 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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Bunuel wrote:
Image
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.


[Reveal] Spoiler:
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:
Image
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.
Image

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).

[Reveal] Spoiler:
Attachment:
dg2.png
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Attachment:
dg3.png
dg3.png [ 11.79 KiB | Viewed 1717 times ]

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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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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New post 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] 23 Sep 2017, 11:18
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The octagon in the diagram above is regular: all of its sides are of e

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