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30 Apr 2015, 04:14
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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:
dg1.png [ 8.44 KiB | Viewed 12007 times ]
04 May 2015, 03:40
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Bunuel
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 11512 times ]
Attachment:
dg3.png [ 11.79 KiB | Viewed 11444 times ]
General Discussion
30 Apr 2015, 12:17
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The value of an angle in an octagon is 180*(8-2)/8 = 180*6/8 = 180*3/4 = 135 degrees.
Sides are equal to 1 coz its regular octagon.
Cosine theorem helps us find the side of the square X which is \(\sqrt{1 + 1 - 2*1*1*cos(135)} = \sqrt{2 +2^\frac{1}{2}} = X\)
The area ends up being \(X^2\) which corresponds to C
Sides are equal to 1 coz its regular octagon.
Cosine theorem helps us find the side of the square X which is \(\sqrt{1 + 1 - 2*1*1*cos(135)} = \sqrt{2 +2^\frac{1}{2}} = X\)
The area ends up being \(X^2\) which corresponds to C
