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Square ABCD is the base of the cube while square EFGH is the [#permalink]

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27 Nov 2007, 00:24

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Square ABCD is the base of the cube while square EFGH is the cube's top facet such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

A. \(\frac{1}{\sqrt{2}}\) B. 1 C. \(\sqrt{2}\) D. \(\sqrt{3}\) E. \(2\sqrt{3}\)

Square ABCD is the base of the cube while square EFGH is the cube's top face such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

1/sqrt2 1 sqrt2 sqrt3 2sqrt3

Please explain your answer.

distance from mid point of AB to AD = sqrt [(1/sqrt2)^2+(1/sqrt2)^2] = 1

the distance between the midpoint of edge AB and the midpoint of edge EH = sqrt [1^2+(sqrt2)^2] = sqrt3.

Square ABCD is the base of the cube while square EFGH is the cube's top face such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

1/sqrt2 1 sqrt2 sqrt3 2sqrt3

Please explain your answer.

sqrt 3.

let midpoint of EH = M
let midpoint of AB = N
Drop perpendicular from M to side AD on F
AN = (sqrt 2)/2
AF = (sqrt 2)/2
FN = [(sqrt 2)/2]^2 + [(sqrt 2)/2]^2 = 1
MN^2 = 1^2 + (sqrt 2)^2 = 3
MN = sqrt 3

Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

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29 May 2014, 05:46

I used deluxe pythag to solve....

We know that the area of the square is 2, therefore the side = sqrt2 or 2^1/2. We know the midpoints are (2^1/2)/2.

So in reality, we are just finding the main diagonal of a rectangular solid with lengths (2^1/2)/2, (2^1/2)/2, and (2^1/2); apply deluxe pythag theorum.

Find x - Main diagonal ((2^1/2)/2)^2 + ((2^1/2)/2)^2 + (2^1/2)^2 = x^2 (2/2)+(2/2)+2=x^2 1/2+1/2+2=x^2 3=x^2 3^1/2=x

Square ABCD is the base of the cube while square EFGH is the cube's top facet such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

A. \(\frac{1}{\sqrt{2}}\) B. 1 C. \(\sqrt{2}\) D. \(\sqrt{3}\) E. \(2\sqrt{3}\)

Look at the diagram below:

Attachment:

Cube.png [ 14.44 KiB | Viewed 3818 times ]

Notice that Z is the midpoint of AD. We need to find the length of line segment XY.

Now, since the area of ABCD is 2 then each edge of the cube equals to \(\sqrt{2}\).

Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

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16 Jun 2014, 06:10

Why angle Z is the right angle?

Bunuel wrote:

Square ABCD is the base of the cube while square EFGH is the cube's top facet such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

A. \(\frac{1}{\sqrt{2}}\) B. 1 C. \(\sqrt{2}\) D. \(\sqrt{3}\) E. \(2\sqrt{3}\)

Look at the diagram below:

Attachment:

Cube.png

Notice that Z is the midpoint of AD. We need to find the length of line segment XY.

Now, since the area of ABCD is 2 then each edge of the cube equals to \(\sqrt{2}\).

Square ABCD is the base of the cube while square EFGH is the cube's top facet such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

A. \(\frac{1}{\sqrt{2}}\) B. 1 C. \(\sqrt{2}\) D. \(\sqrt{3}\) E. \(2\sqrt{3}\)

Look at the diagram below:

Attachment:

Cube.png

Notice that Z is the midpoint of AD. We need to find the length of line segment XY.

Now, since the area of ABCD is 2 then each edge of the cube equals to \(\sqrt{2}\).

Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

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28 Jun 2015, 11:13

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

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01 Jul 2015, 05:52

Bunuel wrote:

amar13 wrote:

Why angle Z is the right angle?

Bunuel wrote:

Square ABCD is the base of the cube while square EFGH is the cube's top facet such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

A. \(\frac{1}{\sqrt{2}}\) B. 1 C. \(\sqrt{2}\) D. \(\sqrt{3}\) E. \(2\sqrt{3}\)

Look at the diagram below:

Attachment:

Cube.png

Notice that Z is the midpoint of AD. We need to find the length of line segment XY.

Now, since the area of ABCD is 2 then each edge of the cube equals to \(\sqrt{2}\).

Square ABCD is the base of the cube while square EFGH is the [#permalink]

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14 Oct 2015, 10:09

Bunuel wrote:

Square ABCD is the base of the cube while square EFGH is the cube's top facet such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

A. \(\frac{1}{\sqrt{2}}\) B. 1 C. \(\sqrt{2}\) D. \(\sqrt{3}\) E. \(2\sqrt{3}\)

Look at the diagram below:

Attachment:

Cube.png

Notice that Z is the midpoint of AD. We need to find the length of line segment XY.

Now, since the area of ABCD is 2 then each edge of the cube equals to \(\sqrt{2}\).

Can u elaborate a bit more on how the line is perpendicular?

Planes ABCD and ADHE are perpendicular to each other. As such, any such lines drawn on these 2 mutually perpendicular planes will be perpendicular to each other.
_________________

Square ABCD is the base of the cube while square EFGH is the cube's top facet such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

A. \(\frac{1}{\sqrt{2}}\) B. 1 C. \(\sqrt{2}\) D. \(\sqrt{3}\) E. \(2\sqrt{3}\)

Look at the diagram below:

Attachment:

Cube.png

Notice that Z is the midpoint of AD. We need to find the length of line segment XY.

Now, since the area of ABCD is 2 then each edge of the cube equals to \(\sqrt{2}\).

abcd looks like rombos and it is not like a square then how <xaz= 90?

ABCD can not be a hombus as it a face of a cube. All 6 faces in a cube are squares. The 1st few words of the question itself mention that ABCD is a square face of the cube.

Additionally, you are basing your observation on a 3D figure drawn on a 2D surface. This is going to lead to a bit of distortion and hence the square 'looks like' a rhombus. In reality, all 6 faces of a cube are square, by definition.
_________________

Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

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22 Oct 2016, 05:41

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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