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D, E and F are points on the sides of triangle ABC above, such that qu
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Updated on: 12 Sep 2016, 04:40
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D, E and F are points on the sides of triangle ABC above, such that quadrilateral AEFD is a rectangle. If DC=1, and CF=4, what is the value of AD? (1) 3EB = AB (2) FB = 2 Attachment:
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Originally posted by mvictor on 03 Oct 2014, 15:05.
Last edited by Bunuel on 12 Sep 2016, 04:40, edited 2 times in total.
Renamed the topic, edited the question and added the OA.




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Re: D, E and F are points on the sides of triangle ABC above, such that qu
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04 Oct 2014, 13:41
mvictor wrote: D, E and F are points on the sides of triangle ABC above, such that quadrilateral AEFD is a rectangle. If DC=1, and CF=4, what is the value of AD? (1) 3EB = AB (2) FB = 2 It is very easy to solve if you mention two similar triangles. Triangle CDF is similar to FEB since angle D=angle E=90 degrees angle DFC=angle EBF since DFAB. Because of similarity the corresponding sides of triangles must be proportional. Our task is to find such ratio. (1) Sufficient. Here we know that proportional coefficient is 2, since DF=AE=ABEB=3EBEB=2EB. Each side in triangle CDF must be twice the corresponding side of triangle FEB. Therefore, AD=FE=1/2 (2) Sufficietnt CF=4, FB=2, hence proportional coefficient is 2. Each side in triangle CDF must be twice the corresponding side of triangle FEB. Therefore, AD=FE=1/2. The correct answer is D.
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Re: D, E and F are points on the sides of triangle ABC above, such that qu
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03 Oct 2014, 15:09
since <FDA=90 we can conclude that <CDF=90 knowing CD & CF, we can find DF
in order to find AD, we need to know AE & EB (1) we can find AE & AB since DF = AE. but we cannot find FB (2) does not tell us anything about AB
1+2 can solve for DA answer C
am I right?



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Re: D, E and F are points on the sides of triangle ABC above, such that qu
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04 Oct 2014, 01:02
Given : DC=1 and CF=4 AEFD is a rectangle.
DF=(4^21^2)^1/2 DF=(161)^1/2 DF=15^1/2 Therefore AE=15^1/2... AEFD is a rectangle
Statement 1 : 3EB = AB Therefore, AE=AB*2/3 AB=3AE/2 AB=(3*15^1/2)/2 Therefore EB=(15^1/2)/2
Let AD be x. Therefore FE=x using Pythagoras,
AC^2+AB^2=BC^2 (1+x)^2+[(3*15^1/2)/2]^2=(4+FB)^2 ..............1
Length of FB can be derived in terms of x using Pythagoras. FB^2=[(15^1/2)/2]^2 +x^2 So equation 1 can be solved to get value of x.
statement is sufficient.
statement 2 : FB = 2
BC= CF+FB BC=4+2 BC=6
Let AD be x, therefore FE=x (AEFD is a rectangle)
using Pythagoras, AC^2+AB^2=BC^2 (1+x)^2+[(15^1/2)/2+EB]^2= 6^2 .............. 1
EB can be derived in terms of x using Pythagoras EB^2= FB^2FE^2 EB^2= 2^2x^2 Value of EB^2 can be put in equation 1 and value of x can be obtained.
Statement is sufficient.
Ans  D



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Re: D, E and F are points on the sides of triangle ABC above, such that qu
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17 May 2015, 18:44
With data sufficiency questions, particularly geometry ones, you always want to make as many inferences as you can upfront. We can see that triangle CDF is a right triangle, because using the supplementary angle rule we know that angle CDF is 90 degrees. When they give us that leg DC = 1 and hypotenuse CF = 4 then, we can use the formula a^2 + b^2 = c^2 to see that leg DF = sqrt15. Because AEFD is a rectangle, AE must also be sqrt15. Statement 1 tells us that 3EB = AB, and therefore we know that EB = AB/2. Using the supplementary angle rule again we see that angle FEB is a 90 degree angle, and therefore triangle FEB is also a right triangle. Statement 2 gives us the hypotenuse of this triangle, and because we already have the information for leg EB (its sqrt15/2), we know that we can find the other leg EF. Again using the fact that AEFD is a rectangle, we know that EF will equal AD, and therefore with both statements together we can find AD, yielding answer choice C. We of course would not do the actual math, because all we want to prove is sufficiency. I hope this helps!
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Re: D, E and F are points on the sides of triangle ABC above, such that qu
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28 May 2015, 18:51
So the OA is C or D ?. I Got C...
Regards, Aj



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Re: D, E and F are points on the sides of triangle ABC above, such that qu
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29 May 2015, 01:26
AjChakravarthy wrote: So the OA is C or D ?. I Got C...
Regards, Aj Dear AjChakravarthyThe correct answer is D. Here's an easy solution for this question. Let the length of AD be x. First of all, let's represent the given information on the diagram. By applying Pythagoras Theorem on right Triangle CDF, we can see that DF = \(\sqrt{15}\). Since opposite sides of a rectangle are equal, AE = \(\sqrt{15}\). Analyzing Statement 13EB = ABThis means, EB = \(\frac{AB}{3}\) Now, AB = AE + EB So, AB = \(\sqrt{15} +\frac{AB}{3}\). Upon solving this equation, we get \(AB = \frac{3√15}{2}\) Now, in right triangle CAB, \(tanC = \frac{AB}{AC} = \frac{3√15}{2(1+x)}\) . . . (1) In right triangle CDF, \(tanC = \frac{DF}{CD} =\frac{√15}{1}\) . . . (2) Upon equating (1) and (2) and solving for x, we get x = 1/2 So, St. 1 alone is sufficient to determine a unique value of x. Analyzing Statement 2FB = 2This means, BC = 4 + 2 = 6 in right triangle CAB, \(cosC = \frac{AC}{BC} = \frac{(1+x)}{6}\) . . . (1) In right triangle CDF, \(cosC = \frac{CD}{CF} =\frac{1}{4}\) . . . (2) Upon equating (1) and (2) and solving for x, we get x = 1/2 So, St. 2 alone is sufficient to determine a unique value of x. Hope this helped! Best Regards Japinder
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Re: D, E and F are points on the sides of triangle ABC above, such that qu
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13 Jul 2016, 07:19
My answer with THALES
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