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09 Mar 2017, 02:13
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
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Question Stats:
58% (03:08) correct
42%
(03:17)
wrong
based on 277
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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
Attachment:
ABCDEF.PNG [ 13.71 KiB | Viewed 12993 times ]
Updated on: 14 Dec 2021, 16:10
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.
Last edited by BrentGMATPrepNow on 14 Dec 2021, 16:10, edited 1 time in total.
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Bunuel
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
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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:
GMAT assassins aren't born, they're made,
Rich
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:
GMAT assassins aren't born, they're made,
Rich
General Discussion
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
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
