In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what

Question

https://gmatclub.com/forum/download/file.php?id=37916 In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what is the length of a side of the square? (A) 16 (B) 6√5 (C) 4√5 (D) 8 (E) 2√6 2017-10-04_1125.png

Nikkb · Answer

In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what is the length of a side of the square?

Let side of Square = 2a = DC = BC 
=> BE = a

Given DE = 8√5

In triangle DEC =>  DE^2 = DC^2 + CE^2  =>  (8√5)^2 = (2a)^2 +a^2  =>  64*5 = 5a^2  =>  a^2 = 64  => a=8

Answer: D

raki99 · Answer

In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what is the length of a side of the square?

Let side of Square = 2a = DC = BC
=> BE = a

Given DE = 8√5

In triangle DEC => DE2=DC2+CE2DE2=DC2+CE2 => (8√5)2=(2a)2+a2(8√5)2=(2a)2+a2 => 64∗5=5a264∗5=5a2 => a2=64a2=64 => a=8

Answer: D

worship · Answer

Yes, but BE is a half of a side, so we should double it to 16, right?
Isn't correct answer A?

Luckisnoexcuse · Answer

let the side of square be x
 x^2 + (x^2)/4 = (8\sqrt{5})^2

x^2 = 16^2
x=16
A

Shobhit7 · Answer

Let DC=2x
Therefore, CE=x

So, in Triangle CDE, 
Using Pythagoras: DE=x√5
So, x√5=8√5 , x=8
Hence, each side of square=2x=16.

Ans A

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generis · Answer

Careful if you use  x  and  2x  to solve. I found out the hard way. The length of a side in that case is not  x . It is  2  x

As the midpoint of a side, E marks one-half the length of a side. So CE = half a side.

The square's bottom side, CD, and CE = legs of a right triangle CDE whose hypotenuse is length 8√5.

Let CE = x and CD = 2x

x^2 + (2x)^2 = (8√5)^2 
 5x^2 = (64)(5) 
 x^2 = 64 
 x = 8

Side CD (or any side) is  2 x.

Side of square = (2 * 8) = 16

Answer A

mj12345 · Answer

Same solution, but a is 1/2 the side of the square. Side = 2a = 16.

ScottTargetTestPrep · Answer

We can let DC = n and EC = (1/2)n. Using the Pythagorean theorem, we can determine n:

n^2 + (n/2)^2 = (8√5)^2

n^2 + (n^2/4) = 320

4n^2 + n^2 = 320 x 4

5n^2 = 320 x 4

n^2 = 64 x 4

n = 8 x 2 = 16

Answer: A

souvonik2k · Answer

Let each side of the square be x.
CD=x, CE=x/2 (Since E is midpoint of the side)
In right angled triangle CDE, by Pythagoras theorem
 x^2+(x/2)^2 =(8 \sqrt{5}) ^2=64*5
 \frac{5x^2}{4}  = 64*5
x^2 = 64 * 4
x=8*2 = 16
Answer A.

pra1785 · Answer

You got that right. a=16 but the question is asking for side of the square. Since side of the square will be 2a, the answer will be 16 (A)

bumpbot · Answer

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In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what

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In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what

Prep Toolkit

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