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)
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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.


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Hi,

How did u arrive at case 2, AB=9 unit?

Case-I is clearly understood(3x:4x:5x with x=4).
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Because AE=12 is a multiple of both 3 and 4, I tested whether it could constitute the SMALLEST side of a 3:4:5 triangle and whether it could constitute the MIDDLE side of a 3:4:5 triangle.

In Case 1, AE=12 constitutes the SMALLEST side of a 3:4:5 triangle:
AE --> 3*4 = 12
AB --> 4*4 = 16
BE --> 5*4 = 20

In Case 2, AE=12 constitutes the MIDDLE side of a 3:4:5 triangle:
AB = 3*3 = 9
AE = 4*3 = 12
BE = 5*3 = 15

As shown in my earlier solution, both cases satisfy Statement 1.
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Triangle ABE is right angled \(\triangle\) and its hypotenuse BE is also the side of square ABCD..

(1) The length of AE is 12, and the ratio of the area of triangle ABE to the area of square BCDE is \(\frac{6}{25}=6x/25x\).
area of \(\triangle {AEB}\)= \(\frac{1}{2}12*AB=6x.....AB=x\)
Area of square = \(25x=BE^2=AB^2+AE^2=12^2+x^2\)......
\(25x=x^2+144.......................x^2-25x+144=0.....................(x-9)(x-16)=0\)
so x= AB can be 16 or 9
insuff

(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.
so BE = \(\frac{80}{4}=20\)....
ratio of AE:AB = 3:4 means it is 3:4:5
so AE:AB:BE=3x:4x:5x where 5x=20....x=4
so AB=4*4=16
suff

B
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Hi chetan2u,

I think it is: AB=\(\sqrt{(BE^2-AE^2)}\)

And the ratio becomes an equation with one unknown variable.
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Thanks...
typo.. Dont know what I was thinking
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x=AB
12x/2 : 144+x^2 = 6 : 25
since Hypotenuse = sqrt(144+x^2)

6x / (6x+144+x^2)= 6/31 should cancel 6s
6x=36/31x+864/31+6/31x^2
186x=36x+864+6x^2
0=6x^2-150x+864
0=x^2-25x+144
0=(x-9)(x-16)
x=9,16 Insufficient

(2) square side=20
AE:20=3:4
AE=15
x=remaining side of right triangle side
Sufficient

Everything is correct but you are wrong in the coloured portion..
AE:AB=3:4
So it's 3:4:5 triangle and sides are AE:AB:EB=AE:AB:20 =3:4:5
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