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Square ABCD is the base of the cube while square EFGH is the [#permalink]
26 Nov 2007, 23:24
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63% (03:17) correct
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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]
29 May 2014, 04:46
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
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
29 May 2014, 05:45
1
This post received KUDOS
Expert's post
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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. \(2\sqrt{3}\)
Look at the diagram below:
Attachment:
Cube.png [ 14.44 KiB | Viewed 2485 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]
16 Jun 2014, 05:10
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 \(\sqrt{2}\).
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
16 Jun 2014, 05:42
Expert's post
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 \(\sqrt{2}\).
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
28 Jun 2015, 10:13
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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
01 Jul 2015, 04:52
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]
14 Oct 2015, 09:09
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 \(\sqrt{2}\).
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
14 Oct 2015, 09:46
jayanthjanardhan wrote:
Bunuel,
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 [#permalink]
14 Oct 2015, 09:48
anik19890 wrote:
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 \(\sqrt{2}\).
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. _________________
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