In the figure above, A is the center of the circle, DF is 5,
This topic has expert replies
-
- Moderator
- Posts: 7187
- Joined: Thu Sep 07, 2017 4:43 pm
- Followed by:23 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
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
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
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.
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
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
- fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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!
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!
-
- Legendary Member
- Posts: 2214
- Joined: Fri Mar 02, 2018 2:22 pm
- Followed by:5 members
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.
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.