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In the figure above, A is the center of the circle, DF is 5, and EF is

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In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink]

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In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

A. 5
B. 7
C. 8
D. 10
E. 15

[Reveal] Spoiler:
Attachment:
ABCDEF.PNG
ABCDEF.PNG [ 13.71 KiB | Viewed 860 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 118840 [0], given: 12007

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Re: In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink]

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New post 09 Mar 2017, 04:56
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let AF be x
EA = AB = AD = AF +5 =x +5

Using pythagoras theorem in triangle EAF

25^2 = (x+5)^2 +x^2 or
2x^2 +10x - 600 = 0
or x^2 +5x -300 = 0
or x =15 or -20
since x cannot be negative, x has to be 15

using similar triangle property in triangle EBC and EAF, we can say that
EC/EB = EA/EF

or EC/40 = 20/25
or EC =32

EC = EF +FC = 32
or 25+FC =32
or FC = 7

option B

Kudos [?]: 52 [2] , given: 15

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Kudos [?]: 12 [0], given: 40

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Re: In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink]

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New post 16 Apr 2017, 01:22
ByjusGMATapp wrote:
let AF be x
EA = AB = AD = AF +5 =x +5

Using pythagoras theorem in triangle EAF

25^2 = (x+5)^2 +x^2 or
2x^2 +10x - 600 = 0
or x^2 +5x -300 = 0
or x =15 or -20
since x cannot be negative, x has to be 15

using similar triangle property in triangle EBC and EAF, we can say that
EC/EB = EA/EF

or EC/40 = 20/25
or EC =32

EC = EF +FC = 32
or 25+FC =32
or FC = 7

option B


Thanks for your brief explanation. can you please shed some light on the statement "using similar triangle property in EBC and EAF".
what is this property and when we can use this.

Kudos [?]: 12 [0], given: 40

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Kudos [?]: 3184 [2] , given: 170

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Re: In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink]

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New post 17 Apr 2017, 11:21
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Hi All,

When dealing with complex-looking questions, it's important to remember that GMAT prompts are built around patterns. When you're dealing with right triangles, there's a pretty good chance that the triangle will be based on an established triangle 'pattern' (such as a 30/60/90, 45/45/90, 3/4/5 or 5/12/13). While not every right triangle question will involve one of those patterns, many of them will. This is meant to say that you can use those patterns 'against' the prompt and potentially answer the question that's asked without doing any complex math.

That having been established, it's interesting that right triangle AEF has a hypotenuse of 25... that might be a 3/4/5 "times 5"... meaning that there's a pretty good chance that the two legs of that triangle are 15 and 20. If that's actually the case, then we know that the radius of the circle is either 15 or 20.

Next, triangle BCE is a RIGHT TRIANGLE (any triangle that has a hypotenuse = diameter and has all 3 vertices on the circumference of the circle IS a right triangle). Thus, with a likely radius of 15 or 20, the likely diameter would be 30 or 40.

From the picture, it certainly appears that FC is greater than FD, so FC is probably greater than 5... meaning that EC is a little greater than 30. Since EC is a 'leg' of the large right triangle, and that leg is greater than 30, then the hypotenuse couldn't be 30... it would have to be 40.

Based on all of those assumptions, here's what we have...
1) The radius of the circle is 20 and the diameter is 40
2) Triangle AEF has sides of 20, __ and 25, so side AF would have to be 15 (this is confirmed since AD is ALSO a radius = 20)
3) Big triangle BCE has a hypotenuse of 40, so that's likely a 3/4/5 "times 8"... meaning that the 3 sides would be 24/32/40. CE cannot be 24 (we already deduced that the total length is a little more than 30), so CE would have to be 32. In that case, segments FC = 32 - 25 = 7.

Notice that EVERY number that I've discussed "fits" the information in the prompt, so this MUST be the answer to the question.

Final Answer:
[Reveal] Spoiler:
B


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Kudos [?]: 3184 [2] , given: 170

Re: In the figure above, A is the center of the circle, DF is 5, and EF is   [#permalink] 17 Apr 2017, 11:21
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In the figure above, A is the center of the circle, DF is 5, and EF is

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