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Math Expert V
Joined: 02 Sep 2009
Posts: 55266
In the figure above, A is the center of the circle, DF is 5, and EF is  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 64% (02:56) correct 36% (03:20) wrong based on 156 sessions

### HideShow timer Statistics 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

Attachment: ABCDEF.PNG [ 13.71 KiB | Viewed 2780 times ]

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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In the figure above, A is the center of the circle, DF is 5, and EF is  [#permalink]

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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:

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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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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
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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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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.
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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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Top Contributor
Bunuel wrote: 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

Attachment:
ABCDEF.PNG

Let x = the length of AF This means AD = x + 5 = radius of the circle.

This is convenient, because AE is also a radius of the circle.
So, AE must have length x + 5 At this point, we can focus on the RIGHT TRIANGLE below: When we apply the Pythagorean Theorem we get: (x + 5 )² + x² = 25²
Expand to get: x² + 10x + 25 + x² = 625
Simplify: 2x² + 10x + 25 = 625
Set equal to zero: 2x² + 10x - 600 = 0
Divide both sides by 2 to get: x² + 5x - 300 = 0
Factor: (x + 20)(x - 15) = 0
So, EITHER x = -20 or x - 15
Since x cannot be negative, we know that x = 15

So, let's add this to our diagram. Also, notice that I added some symbols to represent the 3 angles in ∆EAF

At this point we might recognize that ∆EAF and that ∆ECB are similar triangles
Here's why:
Since EB is the diameter of the circle, we know that ∠C is 90°
Also, both triangles have ∠B
If ∆EAF and ∆ECB share two angles, then the 3rd angles must also be equal.
So, the two triangles must be similar. Let y = the length of CF
So, side EC has length 25 + y

If two triangles are similar, the ratio of their corresponding sides must be equal.
We can write: 25/40 = 20/(25 + y)
Cross multiply to get: 25(25 + y) = (40)(20)
Expand: 625 + 25y = 800
Then: 25y = 175
So y = 7

In other words, CF = 7

Answer: B

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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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Top Contributor
Bunuel wrote: 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

Attachment:
ABCDEF.PNG

As you can see from my solution above, this question is a time-killer (even if you answer it correctly!!)
So, if you're running short on time, you can reduce the correct answers to two options, and make your best guess
How so?

IMPORTANT: the diagrams in GMAT problem solving questions are DRAWN TO SCALE unless stated otherwise.
For more on this, see the video below. So, if FD has length 5, how long do you think CF is?
To me, CF looks a little longer than FD (which has length 5). So, I'd say the length of CF is either 7 or 8.
So, the correct answer is either B or C.
Make your best guess and move on!

DOWNSIDE: You just guessed
UPSIDE: You have a 50% chance of guessing correctly, AND you just saved a ton of time.

RELATED VIDEO FROM OUR COURSE

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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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Bunuel wrote: 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

Brent´s solution is full of VERY important GMAT properties/details.

On the other hand, the "3k, 4k, 5k" shortcut (GMAT´s scope) and a very-easy Geometry property (out-of-GMAT´s scope) do the job in 15 seconds:

$$? = CF = x$$

$$\Delta AFE\,\, = \left( {R - 5,R,25} \right)\,\,\mathop = \limits^{{\text{works}}\,!} \,\,\left( {3 \cdot 5,4 \cdot 5,5 \cdot 5} \right)\,\,\,\,\, \Rightarrow \,\,\,R = 20$$

$$\underbrace {\left[ {R + (R - 5)} \right]}_{GF}\,\,\, \cdot \,\,\,\underbrace {\,5\,}_{FD}\,\, = \underbrace {\,25\,}_{EF} \cdot \underbrace {\,x\,}_{FC}\,\,\,\,\,\mathop \Rightarrow \limits^{R\, = \,20} \,\,\,\,\,\,? = x = 7$$

Regards,
Fabio.
Attachments 04Oct18_9q.gif [ 31.63 KiB | Viewed 375 times ]

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net In the figure above, A is the center of the circle, DF is 5, and EF is   [#permalink] 04 Oct 2018, 11:51
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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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