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If the space diagonal of cube C is 5 inches long, what is the length,

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If the space diagonal of cube C is 5 inches long, what is the length,  [#permalink]

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New post 05 Oct 2016, 03:26
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Re: If the space diagonal of cube C is 5 inches long, what is the length,  [#permalink]

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New post 05 Oct 2016, 04:36
Space diagonal of cube = a*sqrt(3)
Base diaganol of cube = a*sqrt(2)

a*sqrt(3)=5

a=5/sqrt(3)

a*sqrt(2) = 5*sqrt(2/3)


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If the space diagonal of cube C is 5 inches long, what is the length,  [#permalink]

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New post 06 Oct 2016, 06:57
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Bunuel wrote:
If the space diagonal of cube C is 5 inches long, what is the length, in inches, of the diagonal of the base of cube C?

A. 5/(√6)
B. 5·√(2/3)
C. 5·√(3/2)
D. 5·√3
E. 5·√6


When it comes to 3-D diagonals, we can use this formula
Image


In cube C, let's say that each side has side length x
So, the diagonal = √(x² + x² + x²) = √(3x²)
Here, we're told that the diagonal has length 5, so we can write:
5 = √(3x²)
Square both sides to get: 25 = 3x²
Divide both sides by 3 to get: 25/3 = x²
Square root both sides: √(25/3) = x
Or.... (√25)/(√3) = x
Simplify: 5/(√3) = x
We now know that each side in the cube has length 5/(√3)

Our goal is to find the diagonal of the BASE of cube C
The diagonal will be the hypotenuse of a right triangle in which each leg has length 5/(√3)
Using Pythagoras, we can say: diagonal² = [5/(√3)]² + [5/(√3)
So, diagonal² = 25/3 + 25/3
Simplify: diagonal² = 50/3
So, diagonal = √(50/3)
= √[(25)(2/3)]
= [√25][√(2/3)]
= 5√(2/3)
Answer:

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If the space diagonal of cube C is 5 inches long, what is the length,  [#permalink]

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New post 24 Mar 2018, 09:48
Bunuel wrote:
If the space diagonal of cube C is 5 inches long, what is the length, in inches, of the diagonal of the base of cube C?

A. 5/(√6)
B. 5·√(2/3)
C. 5·√(3/2)
D. 5·√3
E. 5·√6

A rectangular prism's space diagonal, \(d\), is calculated with a variation of the Pythagorean theorem:
\(L^2 + W^2 + H^2 = d^2\)

A cube, as a rectangular prism with equal sides, has a space diagonal of length \(3s^2\)
\(L = W = H\)
\(s^2 + s^2 + s^2 = d^2\)
\(3s^2 = d^2\)


Find the cube's side length:
\(3s^2 = d^2\)
\(3s^2 = 5^2\)
\(3s^2 = 25\)
\(s^2 = \frac{25}{3}\)
\(s = \sqrt{\frac{25}{3}}\)


Leave RHS in that form for side length, \(s\)
A cube's face diagonal, \(d\) is that of a square:
\(d = s\sqrt{2}\)
\(d = \sqrt{\frac{25}{3}}*\sqrt{2}\)
\(d= \sqrt{\frac{25*2}{3}}\)
\(d=\frac{\sqrt{25}*\sqrt{2}}{\sqrt{3}}=(5*\frac{\sqrt{2}}{\sqrt{3}})\)
\(d = 5*\sqrt{\frac{2}{3}}\)


Answer B

*If you cannot recall the formula for a square's diagonal, derive it. A square's face diagonal is the hypotenuse of a right triangle. Use Pythagorean theorem. \(d\) = hypotenuse.
\(s^2 +s^2 = d^2\)
\(2s^2 = d^2\)
\(\sqrt{2s^2} = \sqrt{d^2}\)
\(\sqrt{2}\sqrt{s^2}=d\)
\(\sqrt{2}*s=d\)
\(s\sqrt{2}=d\)

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If the space diagonal of cube C is 5 inches long, what is the length,  [#permalink]

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New post 14 Jul 2019, 17:54
Hello GMATPrepNow!

What am I doing wrong?

\(\frac{5}{√3} = x\)

So

\(\frac{5}{√3}*\frac{√3}{√3}= x\)

One side of the square base \((45 - 45 - 90) (1:1:√2)\)

\(\frac{5√3}{3} =x\)

So

\(\frac{5√3}{3}*√2=\) diagonal of the base

Diagonal of the base:

\(\frac{5√6}{3}\)
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Re: If the space diagonal of cube C is 5 inches long, what is the length,  [#permalink]

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New post 15 Jul 2019, 07:06
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jfranciscocuencag wrote:
Hello GMATPrepNow!

What am I doing wrong?

\(\frac{5}{√3} = x\)

So

\(\frac{5}{√3}*\frac{√3}{√3}= x\)

One side of the square base \((45 - 45 - 90) (1:1:√2)\)

\(\frac{5√3}{3} =x\)

So

\(\frac{5√3}{3}*√2=\) diagonal of the base

Diagonal of the base:

\(\frac{5√6}{3}\)


Your solution is perfect.
\(\frac{5√6}{3}\) is EQUIVALENT TO 5√(2/3)
In fact, your solution (with no square roots in the denominator) is the preferred format

To see how this is, take: 5√(2/3)
Rewrite as: 5√2/√3
Multiply top and bottom by √3 to get: 5√6/3.....voila!!

Cheers,
Brent
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Re: If the space diagonal of cube C is 5 inches long, what is the length,   [#permalink] 15 Jul 2019, 07:06
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