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

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Wed Oct 03, 2018 10:09 pm

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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

OA B

Source: Veritas Prep

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Source: Beat The GMAT — Problem Solving | Wed Oct 03, 2018 10:09 pm

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 04, 2018 8:36 am

BTGmoderatorDC 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

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 )Â² + xÂ² = 25Â²
Expand to get: xÂ² + 10x + 25 + xÂ² = 625
Simplify: 2xÂ² + 10x + 25 = 625
Set equal to zero: 2xÂ² + 10x - 600 = 0
Divide both sides by 2 to get: xÂ² + 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 âˆ†EAF

At this point we might recognize that âˆ†EAF and that âˆ†ECB are similar triangles
Here's why:
Since EB is the diameter of the circle, we know that âˆ C is 90Â°
Also, both triangles have âˆ B
If âˆ†EAF and âˆ†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

Brent Hanneson - Creator of GMATPrepNow.com

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by Brent@GMATPrepNow » Thu Oct 04, 2018 8:39 am

BTGmoderatorDC 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

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.

Brent Hanneson - Creator of GMATPrepNow.com

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GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 04, 2018 8:45 am

BTGmoderatorDC 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

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.

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.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

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

by fskilnik@GMATH » Thu Oct 04, 2018 10:39 am

BTGmoderatorDC 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

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

On the other hand, the "3k, 4k, 5k" shortcut (GMATÂ´s scope) and a very-easy Geometry property (out-of-GMATÂ´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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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swerve: Legendary Member; Posts: 2535; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Thu Oct 04, 2018 11:30 am

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

Hence, B is the correct answer. Regards!

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deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Sat Oct 06, 2018 12:52 am

Let AF =X
Length = EA = AB = AD = AF, 15 = X + 15
Using Pythagoras theorem in triangle EAF
$$25^2=\ \left(x\ +\ 15^2\right)+x^2$$
$$625=\ \left(x\ +\ 5\right)+\left(x+5\right)\ +\ x^2$$
$$625=\ x^2+5x\ +\ 5x\ +\ 25\ +\ x^2$$
$$2x^2\ +\ 10x\ -\ 600\ =\ 0$$
Dividing through with 2
$$\frac{2x^2}{2}+\ \frac{10x}{2}-\ \frac{600}{2}=0$$
$$x^2+5x\ -\ 300\ =\ 0$$
Factorizing the equation.
(x - 15) (x + 20)= 0
x - 15 = 0 or x + 20 = 0
x = 15 or x = -20
The length AF cannot be negative, hence x = 15
For triangle EBC and EAF they are similar or congruent triangle in which their corresponding sides are all in the same proportion.
Hence, $$\frac{EC}{EB}\ =\ \frac{EA}{EF}$$
$$\frac{EC}{40}\ =\frac{20}{25}$$
$$EC=\ \frac{\left(20\ \cdot\ 40\right)}{25}=32$$
EC = EF + FC = 32
25 +FC = 32
FC = 32 - 25
FC = 7

Option B is CORRECT.

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