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If the box shown is a cube, then the difference in length

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If the box shown is a cube, then the difference in length [#permalink] New post 12 Feb 2012, 20:50
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76% (01:00) correct 24% (01:35) wrong based on 232 sessions
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Cube.PNG [ 2.46 KiB | Viewed 4978 times ]
If the box shown is a cube, then the difference in length between line segment BC and line segment AB is approximately what fraction of the distance from A to C?

A. 10%
B. 20%
C. 30%
D. 40%
E. 50%
[Reveal] Spoiler: OA

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Re: Cube [#permalink] New post 12 Feb 2012, 21:38
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enigma123 wrote:
If the box shown is a cube, then the difference in length between line segment BC and line segment AB is approximately what fraction of the distance from A to C?

A)10%
B)20%
C)30%
D)40%
E)50%

Any idea how to solve this guys?


AC is the edge of the cube. Let's say its length is 'a'.
AB is just the diagonal of a face of the cube i.e. the diagonal of the square whose each side is of length 'a'. Using pythagorean theorem, we know that AB =\(\sqrt{2}a\)

Now think of the two dimensional triangle ABC (it is right angled at A)
AC = a and AB = \(\sqrt{2}a\)
Again using pythagorean theorem, \(BC^2 = a^2 + (\sqrt{2}a)^2\)
\(BC = \sqrt{3}a\)

So, \((BC - AB)/AC * 100 = (\sqrt{3} - \sqrt{2}) * 100 = (1.732 - 1.414) * 100 = apprx 30%\)

By the way, you would probably be given the value of root 3.
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Re: If the box shown is a cube, then the difference in length [#permalink] New post 13 Feb 2012, 03:52
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If the box shown is a cube, then the difference in length between line segment BC and line segment AB is approximately what fraction of the distance from A to C?
A. 10%
B. 20%
C. 30%
D. 40%
E. 50%
Attachment:
Cube.PNG
Cube.PNG [ 2.46 KiB | Viewed 3870 times ]


AC is the edge (the side) of a cube, suppose it equals to 1;
AB is the diagonal of a face, hence is equals to \(\sqrt{2}\), (either from 45-45-90 triangle properties or form Pythagorean theorem);
BC is the diagonal of the cube itself and is equal to \(\sqrt{1^2+1^2+1^2}=\sqrt{3}\);

Ratio: \(\frac{BC-AB}{AC}=\frac{\sqrt{3}-\sqrt{2}}{1}\approx{1.7-1.4}={0.3}\).

Answer: C.
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Re: If the box shown is a cube, then the difference in length [#permalink] New post 23 Feb 2012, 06:09
looking at the figure CB is the longest diagonal of cube and we know longest diagonal of cube is
CB = X*rt3 ( assumed side of cube asX)
for AB we will use pythogoras theorem we will get Xrt2 ( X is the side of cube)
and AC is X
we have to find X*rt3 - Xrt2 divded by X will give us rt3 - rt2 that equal to close to 0.30 or 30%
answer is C
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Re: If the box shown is a cube, then the difference in length [#permalink] New post 20 Jun 2013, 05:39
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Re: If the box shown is a cube, then the difference in length [#permalink] New post 21 Jul 2014, 06:08
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Re: If the box shown is a cube, then the difference in length [#permalink] New post 19 Jan 2015, 00:43
Since it is a cube, we know that the diagonal of the cube(CB) is s(sqrt(3)) and the diagonal of the square (AB) is s(sqrt(3))
with s being the side (which is the same in both cases because it is a cube.

What we are essentially looking for is (s(sqrt(3))-s(sqrt(2)))/s=x/100
we can divide out the s from the left side to get sqrt(3)-sqrt(2)=x/100
sqrt(3)=1.73 sqrt(2)=1.41 (I recommend memorizing the square roots of 1-10 as they come up on the test and are very difficult to figure out without a calculator)
thus .32=x/100 ------> x=32 and the closest option is C (30%)
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If the box shown is a cube, then the difference in length [#permalink] New post 20 Jan 2015, 20:59
Let AC = 1, then

AB \(= \sqrt{1+1} = \sqrt{2}\)

BC \(= \sqrt{1+1+1} = \sqrt{3}\)

Fraction \(= \frac{\sqrt{3} - \sqrt{2}}{1} * 100 = (1.732 - 1.414) * 100 = 32%\)

Answer = C
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If the box shown is a cube, then the difference in length   [#permalink] 20 Jan 2015, 20:59
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