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

Question

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%

KarishmaB · Accepted Answer

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.

Bunuel · Answer

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% Cube.PNG 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.

pbull78 · Answer

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

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE Theory on Geometry: Triangles Polygons Circles Coordinate geometry 3-D Geometries All DS Geometry to practice: search.php?search_id=tag&tag_id=32 All PS Geometry to practice: search.php?search_id=tag&tag_id=53

elizaanne · Answer

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%)

PareshGmat · Answer

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

LogicGuru1 · Answer

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

devctg · Answer

If we simplify the question it is about to find value of x in below equation:
AC*x = (BC - AB)

suppose the side of cube is 1.

we know that value of diagonal of cube is √3 of any side
and diagonal of nay square part of cube is √2 of any side
then =>

1*x = 1(√3 - √2)
x ≈ 0.30
x ≈ 30% Ans

bumpbot · Answer

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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

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