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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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05 Oct 2016, 02:26
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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

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

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06 Oct 2016, 05:57
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Top Contributor
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

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)

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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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05 Oct 2016, 03:36
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)

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

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24 Mar 2018, 08:48
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:
$$d = s\sqrt{2}$$
$$d = \sqrt{\frac{25}{3}}*\sqrt{2}$$
$$d= \sqrt{\frac{25*2}{3}}$$
$$d=\frac{\sqrt{25}*\sqrt{2}}{\sqrt{3}}=(5*\frac{\sqrt{2}}{\sqrt{3}})$$
$$d = 5*\sqrt{\frac{2}{3}}$$

*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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14 Jul 2019, 16:54
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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Joined: 11 Sep 2015
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GMAT 1: 770 Q49 V46
Re: If the space diagonal of cube C is 5 inches long, what is the length,  [#permalink]

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15 Jul 2019, 06:06
Top Contributor
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:

$$\frac{5√6}{3}$$

$$\frac{5√6}{3}$$ 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]

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24 May 2020, 04:29
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

Recall that if s is the side length of a cube, then its space diagonal is s√3, and a diagonal of one of its faces is s√2.

Since the space diagonal of the cube is 5, we have:

s√3 = 5

s = 5/√3

Therefore, the diagonal of its base is 5/√3 * √2 = 5√2 / √3 = 5√(2/3).

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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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25 May 2020, 07:07
Option B
By Pythagoras theorem,
( Diagnol of Cube )^2 = Diagnol of Base^2 + Side^2.
25 = (root(2)a)^2 + a^2
25 = 3 a^2
a = 5/root(3).

Diagnol of Base = root(2) * side.
= 5 ( root(2) / Root(3) ).
Re: If the space diagonal of cube C is 5 inches long, what is the length,   [#permalink] 25 May 2020, 07:07