GMAT CLUB OLYMPICS: In the image above, if

Question

https://gmatclub.com/forum/download/file.php?id=62913 In the image above, if AE = \sqrt{17} , BE = \sqrt{5} , and CE = \sqrt{13} then what is the area of square ABCD? A. 2 B. 3 C. 4 D. 5 E. 6 Untitled3.png

Kinshook · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=62913 Asked: In the image above, if AE = \sqrt{17} , BE = \sqrt{5} , and CE = \sqrt{13} then what is the area of square ABCD? Screenshot 2021-08-17 at 9.02.16 PM.png z^2 + (x+y)^2 = 17 z^2 + y^2 = 5 (z+x)^2 + y^2 = 13 (x+2y)x = x^2 + 2xy = 17-5 = 12 x(x+2z) = x^2 + 2xz= 13-5 = 8 2x(y-z) = 12-8 = 4 Area of the square ABCD = x^2 If x = 2; y = 2; z = 1; All 3 equations are satisfied. z^2 + (x+y)^2 = 1^2 + (2+2)^2 = 1 + 16 = 17 z^2 + y^2 = 1^2 + 2^2 = 1+ 4 = 5 (z+x)^2 + y^2 = (1+2)^2 + 2^2 = 9 + 4 = 13 IMO C

Bunuel · Answer

Official Solution:

https://gmatclub.com/tests-beta/resources/import/img/M37-30.png 
 
In the image above, if  AE = \sqrt{17} ,  BE = \sqrt{5} , and  CE = \sqrt{13}  then what is the area of square ABCD?

A.  2 
B.  3 
C.  4 
D.  5 
E.  6

Consider the image below:
 
  https://gmatclub.com/tests-beta/resources/import/img/M37-30-2.png 
 
Apply Pythagorean theorem to the three triangles shown in the image:
 
 a^2+b^2=(\sqrt{5})^2 ;
 
 (x+a)^2+b^2=(\sqrt{13})^2 . Expand:  x^2+2ax+a^2+b^2=13 . Substitute  a^2+b^2=5 :  x^2+2ax+5=13 . So,  a=\frac{8-x^2}{2x} ;
 
 (x+b)^2+a^2=(\sqrt{17})^2 . Expand:  x^2+2bx+b^2+a^2=17 . Substitute  a^2+b^2=5 :  x^2+2bx+5=17 . So,  b=\frac{12-x^2}{2x} .
 
Substitute  a=\frac{8-x^2}{2x}  and  b=\frac{12-x^2}{2x}  into  a^2+b^2=5 
 
 (\frac{8-x^2}{2x})^2+ (\frac{12-x^2}{2x})^2=5 ; 
 
 \frac{(8-x^2)^2}{4x^2}+ \frac{(12-x^2)^2}{4x^2}=5 
 
At this point it's better to plug-in the options. We are asked to fins the area of the square, which would be  x^2 , so plug-in options for  x^2  and check which gives the right answer.
 
Option C (4) works:  \frac{(8-4)^2}{4*4}+ \frac{(12-4)^2}{4*4}=5

Answer: C

Archit3110 · Answer

for ∆ ABE and ∆ BCE
AB/ BE = BE / BC
AB * BC = BE^2
x^2 = 5
x= √5
area of square ; 5
option D

neutronstar · Answer

We know the property of triangle that sum of 2 sides will always be more than the other side. Hence, if we convert the sides of both triangles into near integers such as √17 is little more than 4, √13 is little more than 3 and √5 is little more than 2.
We also know that ABCD is a square - i.e. all sides must be equal.
Logically, we need to find a solution (say 'x') such that 
A - i) x + 2 > 4, ii) x + 4 > 2, iii) 4 + 2 > x in 1st triangle ABE
and 
B - i) x + 2 > 3 ii) x + 3 > 2, iii) 2 + 3 > x in 2nd triangle CBE

Hence, 
Option A is eliminated because if we consider x to be 2, it does not satisfy the equation Ai)
Option D is eliminated because if we consider x to be 5, it does not satisfy Biii)
Option E is eliminated because if we consider x to be 6, it does not satisfy Aiii) and Biii)
Out of option B & C I guessed Option C

cs15811581 · Answer

We can utilise the property: Any side of a rectangle will be more than the difference of the other two sides and less than the sum of other two sides.

Take triangle ABE
AE= \sqrt{17}=4.1 
BE= \sqrt{5}=2.2 
So AB is between 4.1-2.2 and 4.1+2.2, that is 1.9 and 6.3

Take triangle CBE
CE= \sqrt{15}=3.6 
BE= \sqrt{5}=2.2 
So AB is between 3.6-2.2 and 3.6+2.2, that is 1.6 and 5.8

5 means each side is  \sqrt{5} , making each ABE and CBE as isosceles. This case is not possible as then both AE and BE would be equal.

Only possibility is 4, as each side is 2.

C

HarshaBujji · Answer

Please refer to the attachment for a better understanding. Let's assume each side of sq is s and cos x = s^2-8/2s\sqrt{5} Similarly, we can apply the cosine theorem for EBA; cos(270-x) = s^2 + 5 -17/2s\sqrt{5} => +/- sin x = s^2-12/2s\sqrt{5} As we know (sin x)^2 + (cos x)^2 =1 ; If we slove the equation we will end up s^4 -30*s^2+104 =0 ; By solving this we will get s^2 as 4 or 26; But 26 won't satisfy the given conditions, Hence S^2, which is area of the square, is 4; IMO C

rajatchopra1994 · Answer

Explanation:

Extended lines AD to Point G, AB to Point I & BC to Point H, GE is Parallel line to AI, EF||HC , DC extends to F as shown in the attached figure.
As ABCD is square, Let side of square is a
AB = BC = CD = DA = GH = IF = a
HE = CF = BI = b
GA = HB = EI = c

in Right angle Triangle BEI, 
BE^2 = BI^2 + IE^2
(√5)^2 = b^2 + c^2
b^2 + c^2 = 5 ---- Eqn 1

in right angle triangle EHC,
CH^2 + HE^2 = EC^2 
(a+c)^2 + b^2 = (√13)^2 
a^2 + c^2 + 2ac + b^2 = 13
Putting value of b^2 + c^2 from Eqn 1,
a^2 + 2ac = 8
c = (8-a^2)/(2a) --- Eqn 2

Similarly in right angle triangle AEI
AI^2 + IE^2 = AE^2
(a+b)^2 + c^2 =(√17)^2
a^2 + b^2 + 2ab + c^2 = 17
Putting value of b^2 + c^2 from Eqn 1,
a^2 + 2ab = 12
b = (12 - a^2)/2a -- Eqn 3

Putting the value of c & b in Eqn 1,
b^2 + c^2 = 5 
{(8-a^2)/(2a)}^2 + {(12 - a^2)/(2a)}^2 = 5
64 + (a^4) - 16a^2 + 144 + a^4 - 24a^2 = 5*4a^2
2a^4 - 40a^2 + 208 = 20a^2
2(a^2)^2 - 60a^2 + 208 = 0

b = -60
a = 2
c = 208
the roots of the quadratic equation = {−b ± √(b^2−4ac)} / 2a 
a^2 (which is area of Square ABCD) = 60 ± √(3600 − 4x2x208) / 2x2
= 60 ± √(3600 − 1664) / 4
= 60 ± 44 / 4
= 26 ; 4

4 is in option

IMO-C

GMATinsight · Answer

Answer: Option C

Step-by-Step Video solution by  GMATinsight

https://www.youtube.com/watch?v=gf5-ywmdMKc

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