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In the figure, the circles with centers A and B are tangent to each

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In the figure, the circles with centers A and B are tangent to each [#permalink]

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New post 21 Nov 2017, 22:18
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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


[Reveal] Spoiler:
Attachment:
2017-11-21_1032_003.png
2017-11-21_1032_003.png [ 11.11 KiB | Viewed 343 times ]
[Reveal] Spoiler: OA

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Re: In the figure, the circles with centers A and B are tangent to each [#permalink]

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New post 22 Nov 2017, 04:41
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.


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Re: In the figure, the circles with centers A and B are tangent to each [#permalink]

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New post 27 Nov 2017, 11:26
Bunuel wrote:
Image
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


[Reveal] Spoiler:
Attachment:
2017-11-21_1032_003.png


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
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In the figure, the circles with centers A and B are tangent to each [#permalink]

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New post 27 Nov 2017, 13:51
JeffTargetTestPrepw wrote:
Bunuel wrote:
Image
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


[Reveal] Spoiler:
Attachment:
2017-11-21_1032_003.png


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

JeffTargetTestPrep , or anyone who can explain, I have a question about this problem: How do we know the triangle is a right isosceles triangle?

The sides are equal right up to the point they cross each circumference (because radii are equal). How do we know the distance of each leg's small segment outside each circle -- from where the leg crosses each circumference to where it intercepts line EF -- is equal?

I have been doodling around with external tangent lines to see whether or not that might help to prove that the legs of the right triangle are equal in length. So far, nothing.

And if it is isosceles because its two smaller angles = 45°, I don't understand that, either.

The two answers assert that the sides are equal. Would someone please explain why?

Kudos [?]: 407 [0], given: 647

In the figure, the circles with centers A and B are tangent to each   [#permalink] 27 Nov 2017, 13:51
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