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ABCD is a square of side 4 inch. If each corner of square is cut ident

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ABCD is a square of side 4 inch. If each corner of square is cut ident  [#permalink]

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20 Sep 2018, 21:38
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33% (02:20) correct 67% (02:23) wrong based on 24 sessions

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ABCD is a square of side 4 inch. If each corner of square is cut identically so that resultant structure becomes regular eight sided polygon. What is length of side of octagon?

A) $$4/ (2-√2)$$
B) $$2√2/ (√2+1)$$
C) $$4(√2-1)$$
D) $$4(√2+1)$$
E) $$4/ (√2-1)$$

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Re: ABCD is a square of side 4 inch. If each corner of square is cut ident  [#permalink]

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21 Sep 2018, 00:35
ABCD is a square of side 4 inch. If each corner of square is cut identically so that resultant structure becomes regular eight sided polygon. What is length of side of octagon?

A) 4/(2−√2)4/(2−√2)
B) 2√2/(√2+1)2√2/(√2+1)
C) 4(√2−1)4(√2−1)
D) 4(√2+1)4(√2+1)
E) 4/(√2−1)

GMATinsight I am not able to solve the question completely, below are my two cents :

Given each side of square is 4, so and it is cut equally to an octagon so lets assume it to be cut equally on each by x so now each side of the square will be ( 4-2x) , perimeter of the square shape with 8 points will be (4-2x) * 4: (16-8x) and we can deduce that each side of the octagon would be (16-8x)*1/8 : (2-x).

Each side of the octagon would be (2-x).

I am not able to solve further than this..not even sure whether my aforementioned approach is correct or not....
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ABCD is a square of side 4 inch. If each corner of square is cut ident  [#permalink]

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21 Sep 2018, 01:01
1
archish3113 wrote:
ABCD is a square of side 4 inch. If each corner of square is cut identically so that resultant structure becomes regular eight sided polygon. What is length of side of octagon?

A) 4/(2−√2)4/(2−√2)
B) 2√2/(√2+1)2√2/(√2+1)
C) 4(√2−1)4(√2−1)
D) 4(√2+1)4(√2+1)
E) 4/(√2−1)

GMATinsight I am not able to solve the question completely, below are my two cents :

Given each side of square is 4, so and it is cut equally to an octagon so lets assume it to be cut equally on each by x so now each side of the square will be ( 4-2x) , perimeter of the square shape with 8 points will be (4-2x) * 4: (16-8x) and we can deduce that each side of the octagon would be (16-8x)*1/8 : (2-x).

Each side of the octagon would be (2-x).

I am not able to solve further than this..not even sure whether my aforementioned approach is correct or not....

Refer the figure attached for how the octagon will be obtained by chopping of fthe corners identically

Here Side of square, S = 4
and Side of Octagon, a = ?

Now, we see that each corner cut off is a right triangle with sides $$a√2$$ as hypotenuse of the triangle being a

i.e. $$\frac{a}{√2}+a+\frac{a}{√2} = 4$$

i.e. $$a+\frac{2a}{√2} = a+a√2 = 4$$

i.e. $$a = 4/(√2+1) = 4(√2-1)$$

Hope this helps!!!
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Re: ABCD is a square of side 4 inch. If each corner of square is cut ident  [#permalink]

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21 Sep 2018, 08:47
GMATinsight wrote:
ABCD is a square of side 4 inch. If each corner of square is cut identically so that resultant structure becomes regular eight sided polygon. What is length of side of octagon?

A) $$4/ (2-√2)$$
B) $$2√2/ (√2+1)$$
C) $$4(√2-1)$$
D) $$4(√2+1)$$
E) $$4/ (√2-1)$$

Hi GMATinsight!

First of all, congratulations for the beautiful question. It´s exactly GMAT-like at its best: clearly written and easily-solved with proper understanding/tools!

$$? = x$$

$$\left\{ \begin{gathered} 2L + x = 4 \hfill \\ x = L\sqrt 2 \,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,\sqrt 2 } \,\,\,\,2L = x\sqrt 2 \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,x\sqrt 2 + x = 4$$

$$\,x\left( {\sqrt 2 + 1} \right) = \,\,4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = x = \frac{4}{{\sqrt 2 + 1}} = \frac{{4\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} = 4\left( {\sqrt 2 - 1} \right)$$

The correct answer is therefore (C).

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: ABCD is a square of side 4 inch. If each corner of square is cut ident &nbs [#permalink] 21 Sep 2018, 08:47
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