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27 Nov 2007, 00:24
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D
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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. \(2\sqrt{3}\)
m14 q23
A. \(\frac{1}{\sqrt{2}}\)
B. 1
C. \(\sqrt{2}\)
D. \(\sqrt{3}\)
E. \(2\sqrt{3}\)
m14 q23
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29 May 2014, 06:45
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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:
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 \(\sqrt{2}\).
\(XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1\);
\(XY=\sqrt{XZ^2+YZ^2}=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}\).
Answer: D.
Cube.png [ 14.44 KiB | Viewed 21396 times ]
A. \(\frac{1}{\sqrt{2}}\)
B. 1
C. \(\sqrt{2}\)
D. \(\sqrt{3}\)
E. \(2\sqrt{3}\)
Look at the diagram below:
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 \(\sqrt{2}\).
\(XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1\);
\(XY=\sqrt{XZ^2+YZ^2}=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}\).
Answer: D.
Attachment:
Cube.png [ 14.44 KiB | Viewed 21396 times ]
General Discussion
27 Nov 2007, 00:33
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bmwhype2
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.
D.
