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The greatest possible (straight line) distance, in inches, between any

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The greatest possible (straight line) distance, in inches, between any  [#permalink]

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New post 07 Oct 2016, 02:33
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The greatest possible (straight line) distance, in inches, between any two points on a certain cube is 10. If the cube is modified so that its length is doubled and its width is doubled while its height remains unchanged, then what is the greatest possible (straight line) distance, in inches, between any two points on the modified box?

A. 10√2
B. 10√3
C. 20
D. 30
E. 30√3

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The greatest possible (straight line) distance, in inches, between any  [#permalink]

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New post 07 Oct 2016, 05:35
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The greatest distance(straight line) between two points in a cube is its diagonal.
Here the length of the digaonal is 10.
Therefore, Side*\(\sqrt{3} = 10\)
Side = \(\frac{10}{\sqrt{3}}\)

Since the length and breadth doubles, and height remains the same.
The new box is a cuboid with length = breadth = \(\frac{20}{\sqrt{3}}\) and height is \(\frac{10}{\sqrt{3}}\)

The straight line greatest distance is the diagonal.
Diagonal = \(\sqrt{length^2 + breadth^2 + height^2} = \sqrt{300} = 10*\sqrt{3}\)(Option B)
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Re: The greatest possible (straight line) distance, in inches, between any  [#permalink]

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New post 19 Feb 2018, 17:55
pushpitkc wrote:
The greatest distance(straight line) between two points in a cube is its diagonal.
Here the length of the digaonal is 10.
Therefore, Side*\(\sqrt{3} = 10\)
Side = \(\frac{10}{\sqrt{3}}\)

Since the length and breadth doubles, and height remains the same.
The new box is a cuboid with length = breadth = \(\frac{20}{\sqrt{3}}\) and height is \(\frac{10}{\sqrt{3}}\)

The straight line greatest distance is the diagonal.
Diagonal = \(\sqrt{length^2 + breadth^2 + height^2} = \sqrt{300} = 10*\sqrt{3}\)(Option B)


Can you please explain how you got the square root of 300? I am so lost at that point?
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Re: The greatest possible (straight line) distance, in inches, between any  [#permalink]

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New post 19 Feb 2018, 19:32
300 is 10*10*3 or \(3 * 10^2\)

The square root of 300 is 10*sqft(3) as 10 comes outside the square

Hope that helps!
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The greatest possible (straight line) distance, in inches, between any  [#permalink]

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New post 19 Feb 2018, 19:35
pushpitkc wrote:
300 is 10*10*3 or \(3 * 10^2\)

The square root of 300 is 10*sqft(3) as 10 comes outside the square

Hope that helps!


But if the new box is a cuboid with length = breadth = 20/√3 and height is 10/√3 then where are the 10*10*3 coming from? Shountld you be multiplying the two 20/√3 with the 10/√3 ?

Sorry but for some reason this is literally going over my head....
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Re: The greatest possible (straight line) distance, in inches, between any  [#permalink]

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New post 19 Feb 2018, 20:15
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aanjumz92 wrote:
pushpitkc wrote:
300 is 10*10*3 or \(3 * 10^2\)

The square root of 300 is 10*sqft(3) as 10 comes outside the square

Hope that helps!


But if the new box is a cuboid with length = breadth = 20/√3 and height is 10/√3 then where are the 10*10*3 coming from? Shountld you be multiplying the two 20/√3 with the 10/√3 ?

Sorry but for some reason this is literally going over my head....


Diagonal = \(\sqrt{length^2 + breadth^2 + height^2} = \sqrt{(\frac{20}{sqroot3})^2 + (\frac{20}{sqroot3})^2 + (\frac{10}{sqroot3})^2}= \sqrt{(\frac{400}{3}) + (\frac{400}{3}) + (\frac{100}{3})}= \sqrt{\frac{900}{3} =}
\sqrt{300} = 10*\sqrt{3}\)(Option B)
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The greatest possible (straight line) distance, in inches, between any  [#permalink]

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New post 19 Feb 2018, 22:54
Bunuel wrote:
The greatest possible (straight line) distance, in inches, between any two points on a certain cube is 10. If the cube is modified so that its length is doubled and its width is doubled while its height remains unchanged, then what is the greatest possible (straight line) distance, in inches, between any two points on the modified box?

A. 10√2
B. 10√3
C. 20
D. 30
E. 30√3

Attachment:
Unknown.png
Unknown.png [ 1.69 KiB | Viewed 374 times ]

The greatest possible distance between any two points on a cube is the diagonal (D) which is also the hypotenuse of the triangle made by the the shorter diagonal (d) and a. The shorter diagonal (d) is the hypotenuse of sides a and a.
So \(d^2 = a^2 + a^2\)
and \(D^2 = d^2 + a^2 = 3a^2\)
\(D = \sqrt{3}a = 10\)

If length and width are doubled, New \(d^2 = (2a)^2 + (2a)^2 = 8a^2\)
New \(D^2 = 8a^2 + a^2 = 9a^2\)
Then New \(D = 3a = 10\sqrt{3}\)

Answer (B)
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