Bunuel wrote:

The octagon in the diagram above is regular: all of its sides are of equal length, and all of its angles are of equal measure. If the octagon’s perimeter is 8 inches, and every other vertex of the octagon is connected to create a square as shown above, what is the area of the square?

A. 2

B. 2√2

C. 2 + √2

D. 2√3

E. 2 + √3

Kudos for a correct solution.Attachment:

The attachment **dg1.png** is no longer available

MANHATTAN GMAT OFFICIAL SOLUTION:To find the area of the square, we need the length of one of the square’s sides. That length can be found if we construct a right triangle outside of the existing diagram, with one side of the octagon as hypotenuse, like this:

There are several ways to determine that the small triangle thus created is a 45º-45º-90º triangle. This fact can be deduced from the symmetry of the figure; alternatively, the interior angles of the octagon are 135º each, so that the two marked angles are 180º – 135º = 45º.

The perimeter of the octagon is 8 inches, so each side of the octagon is 1 inch long. Using the ratios for a 45º-45º-90º triangle, the length of each of the red sides in the diagram above is \(\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\) inches.

We now have both legs of a larger right triangle: the shorter (red) leg is \(\frac{\sqrt{2}}{2}\) inch, and the longer (red + black) leg is \((1 + \frac{\sqrt{2}}{2})\) inches.

Let s stand for the third side of the larger triangle and the side of the square. Use the Pythagorean theorem to find it:

\(s^2= (\frac{\sqrt{2}}{2})^2 + (1 + \frac{\sqrt{2}}{2})^2\)

\(s^2= \frac{1}{2} + (1 + \sqrt{2} + \frac{1}{2})\)

\(s^2 = 2+\sqrt{2}\)

Conveniently, the area of the square is s^2, so there is actually no need to find the value of s itself; the expression \(s^2 = 2 + \sqrt{2}\) represents the area of the square.

The correct answer is (C).Attachment:

dg2.png [ 10.64 KiB | Viewed 2429 times ]
Attachment:

dg3.png [ 11.79 KiB | Viewed 2425 times ]
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