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Updated on: 29 Jun 2018, 11:05
Originally posted by CAMANISHPARMAR on 29 Jun 2018, 11:00.
Last edited by Bunuel on 29 Jun 2018, 11:05, edited 1 time in total.
Last edited by Bunuel on 29 Jun 2018, 11:05, edited 1 time in total.
Formatted.
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B
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Difficulty:
75%
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Question Stats:
50% (02:34) correct
50%
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In the figure above, the measure of angle EAB in triangle ABE is 90 degrees, and BCDE is a square. What is the length of AB?
(1) The length of AE is 12, and the ratio of the area of triangle ABE to the area of square BCDE is \(\frac{6}{25}\).
(2) The perimeter of square BCDE is 80 and the ratio of the length of AE to the length of AB is 3 to 4.
Attachment:
ZZ1.jpg [ 12.5 KiB | Viewed 7236 times ]
29 Jun 2018, 12:20
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CAMANISHPARMAR
Question stem:- AB=?
Statement1:-
Given \(\frac{0.5*AE *AB}{EB^2}=\frac{6}{25}\)
Let AB=x unit , hence \(x^2+12^2=EB^2\)
Substituting , we have, \(\frac{0.5*12*x}{x^2+12^2}=\frac{6}{25}\)
Or,\(x^2-25x+12^2=0\)
Since we have two positive roots which implies more than one value of AB. So, st1 is insufficient.
Statement2:-
Given,\(\frac{AE}{AB}=\frac{3}{4}\) & \(BE=\frac{perimeter}{4}=\frac{80}{4}=20\)
Now, we can definitely obtain the length of AB by using Pythagoras property in the right angled triangle AEB.(\((3k)^2+(4k)^2=20^2\), k can be calculated , so AB=4k)
Therefore, st2 is sufficient.
Ans. (B)
29 Jun 2018, 12:41
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CAMANISHPARMAR
Always look for SPECIAL TRIANGLES.
Statement 1:
Since the length of AE is divisible by 3 and 4, test whether it's possible that ABE is a multiple of a 3:4:5 triangle.
Case 1: AE=12, AB=16 and BE=20, with the result that AE:AB:BE = 3:4:5
In this case:
\(\frac{ABE}{BCDE} = \frac{(0.5 * 12 *16)}{20^2} = \frac{96}{400} = \frac{6}{25}\).
Case 2: AE=12, AB=9 and BE=15. with the result that AB:AE:BE = 3:4:5
In this case:
\(\frac{ABE}{BCDE} = \frac{(0.5 * 12 * 9)}{15^2} = \frac{54}{225} = \frac{6}{25}\).
Statement 1 is satisfied by both cases.
Since AB can be different values, INSUFFICIENT.
Statement 2:
Statement 2 is satisfied only by Case 1.
Thus, AB=16.
SUFFICIENT.
