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I encountered this problem while practicing for a mathematics competition.

A cube has a diagonal length of 10. What is the surface area of the cube? No Calculators Allowed.

(Emphasis mine)

I'm not even sure where to start with this, so I scribbled down some numbers and solved for a square instead of a cube. Presumably, you can calculate the diagonal of a cube using the Pythagoras Theorem somehow, though I'm not sure how.

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    $\begingroup$ Is the space-diagonal meant ? In this case, the sides of the cube have length $\frac{10}{\sqrt{3}}$ and the rest is easy. $\endgroup$
    – Peter
    Commented Oct 24, 2017 at 20:35
  • $\begingroup$ A cube with side-length $1$ has surface area $6$ and diagonal length what exactly? Well, you could put one vertex at $(0,0,0)$ and the opposite one at $(1,1,1)$. Now, what factor would you have to scale this by to get a diagonal of length $10$? $\endgroup$
    – Angina Seng
    Commented Oct 24, 2017 at 20:36
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    $\begingroup$ The surface area is $2 d^2$ $\endgroup$
    – orangeskid
    Commented Oct 24, 2017 at 20:51
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    $\begingroup$ Don't say you don't know where to start—your simpler case was a very good thing to try. And you pinpointed the critical information: Does Pythagoras hold in 3D? Once you are satisfied that it does, you are on your way. $\endgroup$
    – Matthew Leingang
    Commented Oct 25, 2017 at 15:37

4 Answers 4

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It is just the pythagorean theorem: enter image description here

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This problem has a simple solution: $s=2d^2$ where $s$ is the surface area and $d$ is the spacial diagonal.

Explanation:

I discovered through Google (while writing the question) that the side length of a cube can be calculated with $d = a\sqrt{3}$ where $d$ is the diagonal of the cube and $a$ is the length of a single side. Nonetheless, I think the question still stands as a valid question and could be useful to future users, so I'll answer it myself.

We know that the surface area is equal to $6a^2$, so isolating a: $$d=a\sqrt{3} \\ d^2=3a^2 \\ \frac{d^2}{3}=a^2 \\ a=\sqrt{\frac{d^2}{3}} \\ a=\frac{d\sqrt{3}}{3}$$ This means that surface area is equal to $6\left(\frac{d\sqrt{3}}{3}\right)^2$ Expanding to simplify further: $$s=6\left(\frac{d\sqrt{3}}{3}\right)\left(\frac{d\sqrt{3}}{3}\right) \\ s=6\left(\frac{3d^2}{9}\right) \\ s = \frac{6d^2}{3} = 2d^2$$

The surface area can be calculated by $2d^2$

Lets test this: Going the long way: $$10=a\sqrt{3} \\ 100=3a^2 \\ \frac{100}{3}=a^2 \\ a=\sqrt{\frac{100}{3}} \\ a = \frac{10}{\sqrt{3}} \\ a=\frac{10\sqrt{3}}{3}$$ This means the side length of the cube is $\frac{10\sqrt{3}}{3}$. If we square this, we get the area of a single face is $\left(\frac{10\sqrt{3}}{3}\right)^2$

$$s=\left(\frac{10\sqrt{3}}{3}\right)^2 \\ s=\left(\frac{10\sqrt{3}}{3}\right)\left(\frac{10\sqrt{3}}{3}\right) \\ s=\frac{100\sqrt{9}}{9} \\ s=\frac{300}{9} \\ s=\frac{100}{3} $$

This means the surface area of a single face is $\frac{100}{3}$, So we multiply by six to get the total surface area: $\frac{600}{3} = 200$

And going the short way gives us $2(10)^2 = 200$. Both ways give us the same solution.

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  • $\begingroup$ I think you have some problems with powers. From $10=a\sqrt{3}$ you would need to square note cube to get rid of the square root. So you don't have a cubic root of 3 in this problem $\endgroup$
    – Andrei
    Commented Oct 24, 2017 at 20:40
  • $\begingroup$ @Andrei Yes, typo, I solved it correctly in the second part. I'll fix that now, thanks for the heads up! $\endgroup$
    – Travis
    Commented Oct 24, 2017 at 20:41
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    $\begingroup$ Too bad you had to find the relationship $d=a\sqrt3$ via an Internet search instead of deriving it yourself. It’s a straightforward application of the Pythagorean theorem. $\endgroup$
    – amd
    Commented Oct 25, 2017 at 0:06
  • $\begingroup$ From $d^2 = 3a^2$, you can directly deduce that $s = 6a^2 = 2d^2$. $\endgroup$
    – Thern
    Commented Oct 25, 2017 at 16:10
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Well. Let $a$ be the side-length of a cube. Then the diagonal of a face is $a \sqrt 2$. Now, again, using Pythagoras, the diagonal of the cube is $$ d= \sqrt{\left(a \sqrt2 \right)^2 + a^2} = a \sqrt3.$$

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  • $\begingroup$ Yes, that's exactly the information I was missing, unfortunately I discovered it shortly before your answer. +1 $\endgroup$
    – Travis
    Commented Oct 24, 2017 at 20:40
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Without using square roots:

$$10^2 = a^2 + a^2 + a^2 = 3a^2 = 100 \\ $$ $$SA = 6a^2 = 2(3a^2) = 2×100 = 200$$

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