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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:

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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Re: The octagon in the diagram above is regular: all of its sides are of e [#permalink]
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]
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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one more method:
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Re: The octagon in the diagram above is regular: all of its sides are of e [#permalink]
Can be done, very easily, using cosine formula.
Directly fetches you area of square

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Re: The octagon in the diagram above is regular: all of its sides are of e [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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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