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A square, whose side is 2 meters, has its corners cut away

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A square, whose side is 2 meters, has its corners cut away  [#permalink]

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New post 12 Oct 2018, 08:14
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Difficulty:

  55% (hard)

Question Stats:

53% (01:37) correct 47% (01:33) wrong based on 28 sessions

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A square, whose side is 2 meters, has its corners cut away so as to form an octagon with all sides equal. Then the length of each side of the octagon in meters is.

a) \(\frac{\sqrt{2}}{\sqrt{2}+1}\)
b) \(\frac{2}{\sqrt{2}+1}\)
c) \(\frac{\sqrt{2}}{\sqrt{2}-1}\)
d) \(\frac{2}{\sqrt{2}-1}\)
e) None of these

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A square, whose side is 2 meters, has its corners cut away  [#permalink]

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New post 12 Oct 2018, 10:18
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Afc0892 wrote:
A square, whose side is 2 meters, has its corners cut away so as to form an octagon with all sides equal. Then the length of each side of the octagon in meters is.

a) \(\frac{\sqrt{2}}{\sqrt{2}+1}\)
b) \(\frac{2}{\sqrt{2}+1}\)
c) \(\frac{\sqrt{2}}{\sqrt{2}-1}\)
d) \(\frac{2}{\sqrt{2}-1}\)
e) None of these


In the below square ABCD, we need to find the value of EF or GE (since both are equal)

In triangle AGE, AG=AE=x. Since it's an isosceles right angled triangle, GE=\(\sqrt{2}x\) .(Sides are in the ratio 1:1:\(\sqrt{2}\) )

EF= 2 - 2x (Since the side of square is 2)

Since EF=GE (Given Octagon with equal sides)

2-2x=\(\sqrt{2}x\)

On simplifying the above equation, we get x=\(\frac{\sqrt{2}}{\sqrt{2}+1}\)

We need to find the value of \(\sqrt{2}x\), on simplification we get \(\frac{2}{\sqrt{2}+1}\)

Hence B.

Cheers!
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A square, whose side is 2 meters, has its corners cut away   [#permalink] 12 Oct 2018, 10:18
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