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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] New post   09 Mar 2017, 02:13
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A
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E

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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

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ABCDEF.PNG
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BrentGMATPrepNow
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In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] New post   Updated on: 14 Dec 2021, 16:10
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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

Show Spoiler
Attachment:
ABCDEF.PNG


APPROACH #1: Apply geometric properties
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

------------------------------------------------------------------------------------------------------

APPROACH #2: Visually estimate the length of CF
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 try to reduce the correct answer to two options, and make your best guess. To do so we'll use the following property:

The diagrams in GMAT problem solving questions are DRAWN TO SCALE unless stated otherwise.



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.

Answer: B

RELATED VIDEOS FROM OUR COURSE



Originally posted by BrentGMATPrepNow on 04 Oct 2018, 09:37.
Last edited by BrentGMATPrepNow on 14 Dec 2021, 16:10, edited 1 time in total.
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Re: In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] 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:
Show Spoiler
B


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Re: In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] 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
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Re: In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] 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.
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In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] New post   Updated on: 14 Dec 2021, 16:11
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Sorry, I posted a duplicate response.
See my post above.

Originally posted by BrentGMATPrepNow on 04 Oct 2018, 09:48.
Last edited by BrentGMATPrepNow on 14 Dec 2021, 16:11, edited 1 time in total.
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In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] New post   04 Oct 2018, 11:51
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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.
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Re: In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] New post   06 Sep 2020, 02:17
BrentGMATPrepNow wrote:
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

Show Spoiler
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


Hi BrentGMATPrepNow

Thanks for the explanation. While I understood why the 2 triangles are similar (because we have 2 similar angles 90 and E) I did not understand how you arrived at the ratios (highlighted) especially the LHS. RHS if I am not wrong is EA/EC (the 90 degree angles).
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In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] New post   06 Sep 2020, 20:02
once we get past the hard part of figuring out the Pythagoras Theorem for Triangle EAF, and you determine that the Radius = 20

Triangle FAE is SIMILIAR to Triangle BCE

Triangle FAE's Side Lengths are in Multiples of a 3-4-5 Triangle: 15 = 3 * (5) --- 20 = 4 * (5) --- 25 = 5 * (5)

Since Triangle BCE is Similar, it must therefore ALSO be a Multiple of a 3-4-5 Triangle.

We know that the Hypotenuse of this 3-4-5 Triangle BCE = 40

40 / 5 = 8 is the Multiple of the 3-4-5 Triangle BCE

Thus, the other 2 Legs (CE and CB) must = 24 and 32

Since EF is already 25 in Length, the Side CE must be the Leg of 32 Length.

CF is the Difference between: 32 - 25 = 7

A.C. -B-
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In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] New post   08 Sep 2020, 05:29
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Pritishd wrote:
BrentGMATPrepNow wrote:
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

Show Spoiler
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


Hi BrentGMATPrepNow

Thanks for the explanation. While I understood why the 2 triangles are similar (because we have 2 similar angles 90 and E) I did not understand how you arrived at the ratios (highlighted) especially the LHS. RHS if I am not wrong is EA/EC (the 90 degree angles).


In the highlighted area we are comparing corresponding sides
The hypotenuse of the red triangle is 25, and the hypotenuse of the blue triangle is 40.
So we have the ratio 25/40

In the red triangle, the side between the right angle and the star has length 20. In the blue triangle, the side between the right angle and the star has length 25+y.
So we have the ratio 20/(25 + y)

In similar triangles, the ratios of corresponding sides are equal
So we can write: 25/40 = 20/(25 + y)
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Re: In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] New post   08 Sep 2020, 07:51
BrentGMATPrepNow wrote:
Pritishd wrote:
BrentGMATPrepNow wrote:
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

Show Spoiler
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


Hi BrentGMATPrepNow

Thanks for the explanation. While I understood why the 2 triangles are similar (because we have 2 similar angles 90 and E) I did not understand how you arrived at the ratios (highlighted) especially the LHS. RHS if I am not wrong is EA/EC (the 90 degree angles).


In the highlighted area we are comparing corresponding sides
The hypotenuse of the red triangle is 25, and the hypotenuse of the blue triangle is 40.
So we have the ratio 25/40

In the red triangle, the side between the right angle and the star has length 20. In the blue triangle, the side between the right angle and the star has length 25+y.
So we have the ratio 20/(25 + y)

In similar triangles, the ratios of corresponding sides are equal
So we can write: 25/40 = 20/(25 + y)


Hi BrentGMATPrepNow

Thank you very much for the explanation. My doubt is cleared and +1 to you

Warm Regards,
Pritishd
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In the figure above, A is the center of the circle, DF is 5, and EF is [#permalink] New post   08 Apr 2024, 07:36
Knowing common pythagorean triples would let you answer this in seconds
3,4,5 being the most common triplet
3*5, 4*5, 5*5
15,20,25 - seems like the triplet here
therefore AF = 15, DF = 5 ie. AE = AD = 20
hence EB = 40 (diameter)
3*8,4*8,5*8 (since 8*5 = 40)
24,32,40
Since EF is already 25, therefore EC HAS TO BE 32
so CF = 7­
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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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