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In the regular octagon above, the area of rectangle BCDE

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In the regular octagon above, the area of rectangle BCDE  [#permalink]

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In the regular octagon above, the area of rectangle BCDE is what fraction of the area of rectangle ACDF?


A) \(\frac{1}{\sqrt{2}}\)

B) \(\frac{1}{2*\sqrt{2})}\)

C) \(\frac{1}{(2+ \sqrt{2})}\)

D) \(\frac{2}{(1+ \sqrt{2})}\)

E) \(\frac{4}{(2+ \sqrt{2})}\)


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In the regular octagon above, the area of rectangle BCDE  [#permalink]

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New post 30 Jun 2018, 11:14
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1
gmatbusters wrote:
In the regular octagon above, the area of rectangle BCDE is what fraction of the area of rectangle ACDF?


A) \(\frac{1}{\sqrt{2}}\)

B) \(\frac{1}{2*\sqrt{2})}\)

C) \(\frac{1}{(2+ \sqrt{2})}\)

D) \(\frac{2}{(1+ \sqrt{2})}\)

E) \(\frac{4}{(2+ \sqrt{2})}\)

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gmatbusters - I highly doubt this question is sub-600! ;)

Side lengths and interior angles
Regular octagon: all sides and all interior angles are equal

Each interior angle = sum of interior angles divided by the number of sides

Sum of interior angles = \((n-2)180=(6*180)=1,080\)
8 sides: \(\frac{1,080}{8}=135°\) per angle
Internal angles each = 135° (see ∠S)

Sides of a regular octagon have equal lengths.
Let each equal side = \(x\)

Other properties
• ACDF and BCDE are rectangles. Perpendicular sides.
Thus there are right angles at every lettered point and at intersections

• Vertices are divided: the 90° angles divide the octagon's vertex of 135°
Thus there are \(90°\) and \((135° - 90°)=45°\) angles (see vertex A)

• The 4 identical triangles have angle measures 45-45-90, thus
All are isosceles right triangles with opposite side lengths in ratio \(a : a : a\sqrt{2}\)
See, e.g., ∆ BCP and ∆ DEQ (right side of the diagram)

• For the four identical triangles:
Hypotenuse = \(x\)
Legs each* = \(\frac{x}{\sqrt{2}}\)
And \(\frac{x}{\sqrt{2}}\) = width of small rectangle (shared side)

Area of rectangle BCDE
\(CD=L = x\)
\(BC=W =\frac{x}{\sqrt{2}}\)
\(A= (L*W)=(x*\frac{x}{\sqrt{2}})=\frac{x^2}{\sqrt{2}}\)


Area of rectangle ACDF (= a square between rectangles, see middle figure of diagram)
\(CD=W=x\)
\(AC=L=x+\frac{x}{\sqrt{2}}+\frac{x}{\sqrt{2}}\)
\(AC=L=x+\frac{2x}{\sqrt{2}}\)
Simplify. LCD: Multiply \(x\) by \(\frac{\sqrt{2}}{\sqrt{2}}\), then
\(L=\frac{x\sqrt{2}}{\sqrt{2}}+\frac{2x}{\sqrt{2}}\)
\(L=\frac{x\sqrt{2}+2x}{\sqrt{2}}\)
\(A=(\frac{x\sqrt{2}+2x}{\sqrt{2}})*x=\frac{x^2\sqrt{2}+2x^2}{\sqrt{2}}\)


Area of BCDE as a fraction of area of ACDF?

Note: Fraction is Big Rectangle/Small Rectangle
Format issues. Flip the fraction at the end.

\(\frac{ACDF}{BCDE}=\)

\(\frac{\frac{x^2 \sqrt{2}+2x^2}{\sqrt{2}}}{(\frac{x^2}{\sqrt{2}})}=\)

\(\frac{x^2\sqrt{2}+2x^2}{\sqrt{2}}*\frac{\sqrt{2}}{x^2}=\)

\(\frac{x^2\sqrt{2}+2x^2}{x^2}\)
Divide all terms by \(x^2\)
Area of ACDF divided by area of BCDE =
\(\frac{\sqrt{2}+2}{1}\)

\(\frac{ACDF}{BCDE}=\frac{\sqrt{2}+2}{1}\)

Flip the fraction. \(\frac{BCDE}{ACDF}\)
\(=\frac{1}{\sqrt{2}+2}\)

Answer C

*In the 45-45-90 triangles, side lengths opposite those angles correspond with \(a : a : a\sqrt{2}\).
The hypotenuse, side length \(x\), corresponds with \(a\sqrt{2}\)
\(x=a\sqrt{2}\)
\(\frac{x}{\sqrt{2}}=a\)
Each leg corresponds with \(a\). The length of each leg = \(\frac{x}{\sqrt{2}}\)

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Re: In the regular octagon above, the area of rectangle BCDE  [#permalink]

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New post 30 Jun 2018, 18:03
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We can see that the rectangle in the middle is actually a square, with the length of the sides being equal to the side of the octagon. The sides of the octagon (and square in the middle) are greater than the height of the rectangle BCDE, since it is the hypotenuse of the triangle formed in the corners of the octagon.
So, the area of the square in the middle is greater than the area of rectangle BCDE, but less than twice as large. So, the area of ACDF is between 3 to 4 times the area of BCDE, or in other words, BCDE is between \(\frac{1}{3}\) and \(\frac{1}{4}\)of ACDF. Only (C) matches these criteria.
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Re: In the regular octagon above, the area of rectangle BCDE &nbs [#permalink] 30 Jun 2018, 18:03
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