Problem Solving

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) [m]4/ (2-√2)[/m]
B) [m]2√2/ (√2+1)[/m]
C) [m]4(√2-1)[/m]
D) [m]4(√2+1)[/m]
E) [m]4/ (√2-1)[/m]

www.GMATinsight.com

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)

www.GMATinsight.com

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!

(All measures are in inches.)


\[? = x\]
\[\left\{ \begin{gathered}
2L + x = 4 \hfill \\
x = L\sqrt 2 \,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,\sqrt 2 } \,\,\,\,2L = x\sqrt 2 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,x\sqrt 2 + x = 4\]
\[\,x\left( {\sqrt 2 + 1} \right) = \,\,4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = x = \frac{4}{{\sqrt 2 + 1}} = \frac{{4\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} = 4\left( {\sqrt 2 - 1} \right)\]

The correct answer is therefore [spoiler]__(C)______[/spoiler] .


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

Regards,
Fabio.