There are two cubes. The points $P$ and $Q$ are both on the small cube: $P$ is a point in the centre of one of the faces, and $Q$ is a corner on the opposite face.

The second cube has sides of length $|PQ|$. What is the surface area of the large cube divided by the surface area of the smaller cube?

I drew both of them and know that the SA of the large cube is $6{QP}$. My problem is with the small cube. In the small cube, if I would connect points $P$ and $Q$, they would for a right angled triangle and line $PQ$ is the hypotenuse. However what is the purpose of this line, how can we find the SA using this?

  • $\begingroup$ What do you mean by SA ? $\endgroup$ – Jean Marie Jan 18 '17 at 21:56
  • $\begingroup$ @JeanMarie: Surface Area? Just guessing… $\endgroup$ – MvG Jan 18 '17 at 22:04
  • $\begingroup$ @MvG Good guess, ... $\endgroup$ – Jean Marie Jan 18 '17 at 22:12
  • $\begingroup$ I have proposed a real title (consider that a title with "help" is not very informative for us: everybody here asking a question desires some kind of help :) $\endgroup$ – Jean Marie Jan 18 '17 at 22:16
  • $\begingroup$ yes SA is surface area :) $\endgroup$ – exchangehelpforuni Jan 19 '17 at 21:45

Take the cube $[0,1]^3$ as small cube for example (without loss of generality as you may pick your coordinate system any way you want for this task). It has surface area $6$ as its six sides have area $1$ each. Pick e.g. $Q=(0,0,0)$ and $P=(1,\frac12,\frac12)$. Then the distance is $$\lvert P-Q\rvert=\sqrt{1+\frac14+\frac14}=\sqrt{\frac32}$$ So the surface area of the large cube is $6\cdot\frac32=9$ and the quotient between these suface areas is $$\frac96=\frac32$$


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The area' ratio $R$ is the square of the length's ratio $r$. Consider that the cube has sidelength 1.

Let $P'$ be the center of the opposite face of $P$. It is clearly the orthogonal projection of $P$ onto this face. Thus, triangle $PP'Q$ is rectangle in $P'$ with $PP'=1$ and $P'Q=\sqrt{2}/2$ (remember that the diagonal of a square with side 1 is $\sqrt{2}$).

Pythagoras' theorem, applied to triangle $PP'Q$ gives:

$PQ^2=1^2+(\sqrt{2}/2)^2=3/2$. Thus $PQ=\sqrt{3/2}$, giving $r=PQ/1=\sqrt{3/2}$.

Therefore, the ratio of lengths is $r=\sqrt{3/2}/1=\sqrt{3/2}$

Thus the ratio of areas is the square of the previous ratio : $R=r^2=3/2.$

  • $\begingroup$ I completely did not understad what you said. $\endgroup$ – exchangehelpforuni Jan 19 '17 at 21:20
  • $\begingroup$ I know P' is P on the other side of the cube, however what is PP'?? If you do not mind, may you please draw it? $\endgroup$ – exchangehelpforuni Jan 19 '17 at 21:21
  • $\begingroup$ Also, I tried doing it by drawing triangle PP'Q, with the right angle being at P'. Then distance between P and P' is 1 and distance from P' to Q is 1/2 because its in the center, using pythagorean therom, I got PQ to be sqrt5 /2 $\endgroup$ – exchangehelpforuni Jan 19 '17 at 21:24
  • $\begingroup$ P' is in the center of the opposite face, thus in the middle of the diagonal. P'Q is thus the half of the diagonal, i.e., $\sqrt{2}$, giving $\sqrt{2}/2$, not $1/2.$ $\endgroup$ – Jean Marie Jan 19 '17 at 21:44
  • $\begingroup$ i got that sqrt (3/2) is length PQ, however I do not understand what you did after that. The question asks for the surafec area of large cube / the small one. I got that the surface area of the large one is 9, and the small one is 1, so how does that work?? What did u do with the ratio? and where did u get it from $\endgroup$ – exchangehelpforuni Jan 24 '17 at 16:00

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