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21 Aug 2014, 00:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
67% (03:37) correct
33%
(03:26)
wrong
based on 159
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In the figure below, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of shaded region DEG
square.png [ 7.06 KiB | Viewed 6014 times ]
A: \(64\sqrt{2} - 32 - 16\pi\)
B: \(8[4\sqrt{2} - (\pi + 2)]\)
C: \(64 - 32\pi\)
D: \(64 - 16\pi\)
E: \(64\sqrt{2} - 8\pi\)
Attachment:
square.png [ 7.06 KiB | Viewed 6014 times ]
A: \(64\sqrt{2} - 32 - 16\pi\)
B: \(8[4\sqrt{2} - (\pi + 2)]\)
C: \(64 - 32\pi\)
D: \(64 - 16\pi\)
E: \(64\sqrt{2} - 8\pi\)
21 Aug 2014, 13:10
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PareshGmat
Area of the square ABCD = 64
Area of sector CBD = \(\frac{\pi * 8^2}{4} = 16\pi\)
Attachment:
1.jpg [ 11.31 KiB | Viewed 5921 times ]
For square GHCI, GC is the diagonal. Therefore, \(GH = GI = 4\sqrt{2}\) which makes \(FG = GE = 8 - 4\sqrt{2} = 4(2 - \sqrt{2})\)
Now, as the diagonal AC divides both the square ABCD and the sector CBD into equal halves, Area of region DAG = Area of region BAG
Also, the same diagonal divides the small square AEGF into two congruent triangles. So, Area of region DEG = Area of region BFG
Now, Area of region DAG + Area of region BAG = Area of the square ABCD - Area of sector CBD - Area of square AEGF
Area of region DAG = \(\frac{64 - 16\pi - [4(2 - \sqrt{2})]^2}{2} = 32 - 8\pi - \frac{16(4 + 2 - 4\sqrt{2})}{2} = 8(4 - \pi - 6 + 4\sqrt{2}) = 8 [4\sqrt{2} - (\pi + 2)]\)
Hence the answer is B.
Hope it's clear.
21 Aug 2014, 13:33
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PareshGmat
Dear PareshGmat,
That's a great question. I think it is very much like something the GMAT could ask on one of the harder questions.
Here's a solution.
If we take the big square ABCD (Area = 64) and subtract both the quarter circle and the little square AFGE, then divide the remaining area in half, we will get the shaded area. Very elegant.
Of course, big square ABCD = 64.
Quarter circle = \(16\pi\)
What about the little square?
Attachment:
square, circular arc, smaller square.JPG [ 20.01 KiB | Viewed 5860 times ]
Of course, GC is just the radius of the circle: GC = 8
Well, \(AG = AC - GC = 8sqrt(2) - 8 = 8(sqrt(2) - 1)\)
That's the diagonal of the square AFGE.
As a shortcut, we can find the area of any square by squaring the diagonal, and then dividing by two.
Area of AFGE = \(\frac{(AG)^2}{2}\)
Area of AFGE = \(\frac{(8(sqrt(2) - 1))^2}{2}\)
Area of AFGE = \(32(sqrt(2) - 1)^2\)
Area of AFGE = \(32(2 - 2sqrt(2) + 1)\)
Area of AFGE = \(96 - 64sqrt(2)\)
Twice shaded region = (Square ABCD) - (quarter circle) - (Square AFGE)
Twice shaded region = \(64 - 16\pi - (96 - 64sqrt(2))\)
Twice shaded region = \(64sqrt(2) - 32 - 16\pi\)
Divide by 2
Shaded region = \(32sqrt(2) - 16 - 8\pi\)
Factor out an 8
Shaded region = \(8(4sqrt(2) - 2 - \pi)\)
Shaded region = \(8(4sqrt(2) - (2 + \pi))\)
Answer = (B)
Great problem!
Mike
