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What is the area of a regular octagon with side-length 2?

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by Max@Math Revolution » Mon Nov 12, 2018 12:02 am

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[Math Revolution GMAT math practice question]

What is the area of a regular octagon with side-length 2?

A. 4 + 4 √2.
B. 8 + 8 √2.
C. 16 + 16 √2.
D. 16
E. 32
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by fskilnik@GMATH » Mon Nov 12, 2018 7:00 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

What is the area of a regular octagon with side-length 2?

A. 4 + 4 √2.
B. 8 + 8 √2.
C. 16 + 16 √2.
D. 16
E. 32
$$? = {S_{{\rm{tot}}}} = 2 \cdot {S_{ABCD}} + {S_{ADEH}}$$

Image

$$\left. \matrix{
{S_{ABCD}}\,\,\, = \,\,\,2\left( {{{x \cdot x} \over 2}} \right) + 2x = {x^2} + 2x\,\,\,\, \hfill \cr
{S_{ADEH}} = 2\left( {2x + 2} \right) = 4x + 4 \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2\left( {{x^2} + 2x} \right) + 4x + 4 = 2\left( {{x^2} + 4x + 2} \right)$$

$$L,L,L\sqrt 2 \,\,\,{\rm{shortcut}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,2 = x \cdot \sqrt 2 \,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = \sqrt 2 $$

$$? = 2\left( {2 + 4\sqrt 2 + 2} \right) = 8 + 8\sqrt 2 \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( B \right)$$


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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Max@Math Revolution » Wed Nov 14, 2018 12:21 am
=>

Image

The above figure shows that the octagon can be divided up into one square of side-length 2, 4 rectangles of side-lengths 2 and √2, and 4 right-angled isosceles triangles with equal sides of length √2. Thus its area is the sum of the areas of the square (2*2), 4 rectangles (2*√2) and 4 triangles ((1/2) (√2)( √2)).
The area of the octagon is 2^2 + 4*2 √2 + 4*1 = 8 + 8 √2.

Therefore, the answer is B.
Answer: B
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