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

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

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
Image
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
Image

At this point, we can focus on the RIGHT TRIANGLE below:
Image
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.
Image
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.
Image

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

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.
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by Brent@GMATPrepNow » Thu Oct 04, 2018 8:45 am
BTGmoderatorDC wrote:Image

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.
Image

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
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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:Image

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:
$$? = CF = x$$
$$\Delta AFE\,\, = \left( {R - 5,R,25} \right)\,\,\mathop = \limits^{{\text{works}}\,!} \,\,\left( {3 \cdot 5,4 \cdot 5,5 \cdot 5} \right)\,\,\,\,\, \Rightarrow \,\,\,R = 20$$
$$\underbrace {\left[ {R + (R - 5)} \right]}_{GF}\,\,\, \cdot \,\,\,\underbrace {\,5\,}_{FD}\,\, = \underbrace {\,25\,}_{EF} \cdot \underbrace {\,x\,}_{FC}\,\,\,\,\,\mathop \Rightarrow \limits^{R\, = \,20} \,\,\,\,\,\,? = x = 7$$
Image

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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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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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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