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

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Bunuel
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In the figure, the circles with centers A and B are tangent to each [#permalink] New post   21 Nov 2017, 23:18
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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


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


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Re: In the figure, the circles with centers A and B are tangent to each [#permalink] New post   27 Nov 2017, 12:26
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Bunuel wrote:

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


Show 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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Re: In the figure, the circles with centers A and B are tangent to each [#permalink] New post   07 Apr 2019, 10:28
Bunuel wrote:

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


Show Spoiler
Attachment:
2017-11-21_1032_003.png



Bunuel , could you help to understand how the triangle is isoscelece

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Re: In the figure, the circles with centers A and B are tangent to each [#permalink] New post   16 May 2019, 02:31
m1033512

Draw a tangent as shown in the figure.
In triangle CPA
Angle APC=45
Angle ACP=90
Therefore Angle CAP=45

Similarly In triangle BCP, Angle CBP = 45
Hence, Triangle BAP is a 45-45-90 triangle.
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Re: In the figure, the circles with centers A and B are tangent to each [#permalink] New post   16 May 2019, 09:41
moropa2018 wrote:
m1033512

Draw a tangent as shown in the figure.
In triangle CPA
Angle APC=45
Angle ACP=90
Therefore Angle CAP=45

Similarly In triangle BCP, Angle CBP = 45
Hence, Triangle BAP is a 45-45-90 triangle.



Hi, a small query, why is angle APC 45, is there any theorem related to tangent, i am missing out on.
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Re: In the figure, the circles with centers A and B are tangent to each [#permalink] New post   17 May 2019, 07:48
moropa2018 wrote:
m1033512

Draw a tangent as shown in the figure.
In triangle CPA
Angle APC=45
Angle ACP=90
Therefore Angle CAP=45

Similarly In triangle BCP, Angle CBP = 45
Hence, Triangle BAP is a 45-45-90 triangle.



How did you find the angle is 45?
Can you please help?

If circle is rotated slightly upwards then it will give different triangle. Not isosceles. And i have heard that we can not go by figure to judge the lengths of lines because we can not measure. If we have been given specific lengths then only we can find relations of lines.

Am I missing something here?

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Re: In the figure, the circles with centers A and B are tangent to each [#permalink] New post   15 Jul 2020, 04:45
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Re: In the figure, the circles with centers A and B are tangent to each [#permalink]
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