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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?

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    What do you mean by SA ? – 2017-01-18
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    @JeanMarie: Surface Area? Just guessing… – 2017-01-18
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    @MvG Good guess, ... – 2017-01-18
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    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 :) – 2017-01-18
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    yes SA is surface area :) – 2017-01-19

2 Answers 2

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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 : $R=r^2=3/2.$

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    I completely did not understad what you said. – 2017-01-19
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    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? – 2017-01-19
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    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 – 2017-01-19
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    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 $1/2.$ – 2017-01-19
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    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 – 2017-01-24
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    plz help me solve this question as i dont understand it – 2017-01-24
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    I have added a figure. Wish it helps. – 2017-01-24
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    thank u for the figure, i understand everything about it. however what i dont get is why u sqaured the length while they were asking for the the surface are of the large cube/ the small cube – 2017-01-24
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    please read the comment above – 2017-01-24
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    I need an answer quick plzz.... i have drawn the exact same figure, howevr i do not know what u mean by ratio. how can i do it without the ratio??? – 2017-01-24
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    surface are of large cube=9, small= 1/ – 2017-01-24
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    Have a look at the new figure in my answer. – 2017-01-24
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    thank u. i completely get what ur saying and oind. by question is why is the thing im doing wrong? why is it wrong to simple divide tha areas – 2017-01-24
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    I have added the (red) cube. And I have placed a side of the black cube symbolically inside a side of the black cube: you see that if a side is 1 and the other side PQ=\sqrt{3/2}, the ratio of the areas is the square of the ratio of the sides i.e., $3/2$. – 2017-01-24
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    In my opinion, you were not considering the right cube. – 2017-01-24
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    what do u mean ?????????? – 2017-01-24
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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 $$\frac96=\frac32$$