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The area of a regular octagon [#permalink]
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This problem asks you to compare the area of an octagon to the area of a square. First, the simpler Quantity B is calculated using the formula for the area of a square:

Area = side^2 = \(8^2 = 64\)

Next, calculate Quantity A.

The easiest way to calculate the area of a regular octagon only knowing the length of a side is to sketch out the octagon and divide into regions whose areas are easy to calculate.

The octagon is divided into,(1) a square with area 4 x 4, (2) 4 rectangles each with an area of 4 x A, and (3) 4 triangles each with an area \(\frac{1}{2}A^2\).

From these pieces we can derive an area formula as follows:

Area of Octagon \(= 4^2 + 4 * \frac{1}{2} A^2 + 4 * 4 * A\)

A can be calculated using the Pythagorean Theorem:

\(4^2 = A^2 + A^2\)
\(16 = 2A^2\)
\(4 = A\) \(\sqrt{2}\)
\(A= \dfrac{4}{\sqrt{2}}=2 \sqrt{2}\)

Now substitute value for A into the original Area Formula:

Area of Octagon \(= 4^2 + 2 · (2\sqrt{2})^2 + 4 · 4 · 2\sqrt{2}\)

\(16 + 16 + 32\sqrt{2} = 77.25\)

77.25 is greater than 64, therefore Quantity A is greater than Quantity B. The correct answer choice is A.
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