11
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 Oct 24 '17 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 Oct 24 '17 at 20:36
  • 1
    $\begingroup$ The surface area is $2 d^2$ $\endgroup$ – orangeskid Oct 24 '17 at 20:51
  • 1
    $\begingroup$ 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 Oct 25 '17 at 15:37
40
$\begingroup$

It is just the pythagorean theorem: enter image description here

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Oct 24 '17 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 Oct 24 '17 at 20:41
  • 2
    $\begingroup$ 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 Oct 25 '17 at 0:06
  • $\begingroup$ From $d^2 = 3a^2$, you can directly deduce that $s = 6a^2 = 2d^2$. $\endgroup$ – Thern Oct 25 '17 at 16:10
5
$\begingroup$

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 $$ d= \sqrt{\left(a \sqrt2 \right)^2 + a^2} = a \sqrt3.$$

$\endgroup$
  • $\begingroup$ Yes, that's exactly the information I was missing, unfortunately I discovered it shortly before your answer. +1 $\endgroup$ – Travis Oct 24 '17 at 20:40
1
$\begingroup$

Without using square roots:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.