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21 Nov 2017, 23:18
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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:
45%
(medium)
Question Stats:
67% (02:19) correct
33%
(02:15)
wrong
based on 85
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In the figure, the circles with centers A and B are tangent to each other at C, and are tangents to the lines k and m at D and F respectively. If the radius of each circle is √2 and k is parallel to m, what is the length of EF?
(A) 2√2 + 1
(B) 2√2 + 2
(C) 2√2 + 3
(D) 3√2
(E) 3√2 + 1
Attachment:
2017-11-21_1032_003.png [ 11.11 KiB | Viewed 3007 times ]
22 Nov 2017, 05:41
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Answer is B.
EF Is summation of FB+AD+ the middle portion. Let the intersection of A produced to intersect EF be G.
Hence finding BG solves the problem.
From Triangle ABG, we can prove that the triangle is isosceles (although this is quite clear from the viewpoint of symmetry).
So Applying Pythagorean Theorem to ABG we get AG=BG= 2 (as BC+AC is known which is Sqrt2).
Hence the answer is 2*sqrt2 + 2.
Hence option B.
Sent from my iPhone using GMAT Club Forum
EF Is summation of FB+AD+ the middle portion. Let the intersection of A produced to intersect EF be G.
Hence finding BG solves the problem.
From Triangle ABG, we can prove that the triangle is isosceles (although this is quite clear from the viewpoint of symmetry).
So Applying Pythagorean Theorem to ABG we get AG=BG= 2 (as BC+AC is known which is Sqrt2).
Hence the answer is 2*sqrt2 + 2.
Hence option B.
Sent from my iPhone using GMAT Club Forum
27 Nov 2017, 12:26
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Bunuel
We see that EF consists of the radius of the top circle, a leg of the triangle, and the radius of the bottom circle.
We see that the two radii combined are √2 + √2 = 2√2.
We also see that BA = 2√2 and that the triangle is a 45-45-90 right triangle, so each leg is 2.
Thus, EF = 2√2 + 2.
Answer: B
