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 anglesRegular 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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