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If the box shown is a cube, then the difference in length [#permalink]

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12 Feb 2012, 21:50

6

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A

B

C

D

E

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25% (medium)

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76% (02:07) correct
24% (01:36) wrong based on 423 sessions

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Attachment:

Cube.PNG [ 2.46 KiB | Viewed 9155 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?

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, \(BC^2 = a^2 + (\sqrt{2}a)^2\) \(BC = \sqrt{3}a\)

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 [ 2.46 KiB | Viewed 6733 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 \(\sqrt{1^2+1^2+1^2}=\sqrt{3}\);

Re: If the box shown is a cube, then the difference in length [#permalink]

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23 Feb 2012, 07:09

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

Re: If the box shown is a cube, then the difference in length [#permalink]

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21 Jul 2014, 07:08

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Re: If the box shown is a cube, then the difference in length [#permalink]

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19 Jan 2015, 01:43

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]

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17 Jul 2016, 00:45

enigma123 wrote:

Attachment:

Cube.PNG

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%

enigma123 wrote:

Attachment:

Cube.PNG

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%

AC=side of the cube = x AB= hypotenuse of the base (which is a square with side x)= \(x\sqrt{2}\) BC=deluxe hypotenuse or diagonal of the cube =\(x\sqrt{3}\) BC-AB= \(x\sqrt{3}\) - \(x\sqrt{2}\) = x(1.73-1.41)=0.32x \(\frac{(BC-AB)}{AC}=\frac{0.30x}{x} ==>0.32==>32\)%

ANSWER IS C
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