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