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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 [#permalink]

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05 Oct 2017, 00:36
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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

[Reveal] Spoiler:
Attachment:

2017-10-04_1125.png [ 4.66 KiB | Viewed 964 times ]
[Reveal] Spoiler: OA

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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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05 Oct 2017, 01:54
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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

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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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05 Oct 2017, 03:35
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

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

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05 Oct 2017, 04:03
Nikkb wrote:
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

Yes, but BE is a half of a side, so we should double it to 16, right?

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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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05 Oct 2017, 11:03
Bunuel wrote:

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

[Reveal] Spoiler:
Attachment:
2017-10-04_1125.png

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

x^2 = 16^2
x=16
A
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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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05 Oct 2017, 21:36
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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In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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06 Oct 2017, 13:46
Bunuel wrote:

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

[Reveal] Spoiler:
Attachment:
2017-10-04_1125.png

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 2x.

Side of square = (2 * 8) = 16

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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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06 Oct 2017, 14:53
Nikkb wrote:
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

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

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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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09 Oct 2017, 15:44
Bunuel wrote:

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

[Reveal] Spoiler:
Attachment:
2017-10-04_1125.png

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

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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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10 Oct 2017, 06:58
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
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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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11 Oct 2017, 10:38
Nikkb wrote:
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

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)

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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what   [#permalink] 11 Oct 2017, 10:38
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