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24 Jun 2021, 06:00
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D
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Difficulty:
95%
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Question Stats:
27% (03:52) correct
73%
(02:24)
wrong
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In the figure below, the two small circles are identical. What is the radius (in cm) of each small circle, if the side of the square is 2 cm?
A. \(2 - \sqrt{2} - \sqrt{2\sqrt{1 + \sqrt{2}}}\)
B. \(2 + \sqrt{2} + 2\sqrt{1 + \sqrt{2}}\)
C. \(2 - \sqrt{2} + 2\sqrt{1 + \sqrt{2}}\)
D. \(2 + \sqrt{2} - 2\sqrt{1 + \sqrt{2}}\)
E. \(2 + \sqrt{2} - 2\sqrt{\sqrt{2} - 1}\)
Attachment:
2021-06-16_18-23-39.png [ 17.8 KiB | Viewed 1301 times ]
25 Jun 2021, 11:33
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Bunuel
(This is not a GMAT question so feel free to skip if you're focusing on GMAT)
The diagonal is \(2\sqrt{2}\) so the radius of the big circle is \(\sqrt{2}\). Next, we need to find the radius of the small circle, there are many ways to do this but it starts with setting the radius as x and finding a relation around it. In the graph below, I use the right triangle ABC and the Pythagorean theorem to find the relation:
\((\sqrt{2} + x)^2 = (2 - x)^2 + x^2\)
\(x^2 + 2\sqrt{2}x + 2 = 2x^2 - 4x + 4\)
\(x^2 + (-2\sqrt{2} - 4)x + 2 = 0\)
Then we can use the quadratic formula to solve for a solution:
\(b^2 - 4ac = 8 + 16\sqrt{2} + 16 - 4*2 = 16\sqrt{2} + 16\)
Note we will have two positive solutions, we want the smaller one as in the graph x should be quite small. Thus we take the solution that subtracts the sqrt of the determinant.
\(x = \frac{2\sqrt{2} + 4 - \sqrt{16\sqrt{2} + 16}}{2}\) (we can take 16 out and it becomes 4, divided by 2 it becomes 2)
\(x = \sqrt{2} + 2 - 2\sqrt{\sqrt{2} + 1}\)
Ans: D
Attachments
graph.png [ 95.69 KiB | Viewed 1119 times ]
