Determine the exact length of the diagonal of the prism.

Question

The area of the front, top and side of a right rectangular prism are $36 cm^2, 40cm^2$ and $45cm^2$ respectively. Determine the exact length of the diagonal of the prism.

So I know that the diagonal, if the length, width and height were each $a, b, c$, would be $\sqrt{a^2+b^2+c^2}$. Also, $a^2+b^2+c^2 \ge ab+bc+ca$. WLOG, let $ab = 36, bc = 40, ca = 45$. Then $ab+bc+ca = 121$, and the diagonal will be at least $11$. I'm quite inclined to believe this is the answer, but I'm having a problem making sure. For example, I could calculate that $abc = 180\sqrt{2}$, but that doesn't seem to help me... Any help would be really appreciated!

Answer 1

You already write $$ab=36,\qquad bc=40,\qquad ca=45$$

Then, we have $$36\times 45=ab\times ca=bc\times a^2=40a^2\implies a=\frac{9}{\sqrt 2}$$ from which $$b=\frac{36}{a}=4\sqrt 2,\qquad c=\frac{45}{a}=5\sqrt 2$$ follow.

So, the answer is $$\sqrt{a^2+b^2+c^2}=\frac{7\sqrt{10}}{2}$$

Answer 2

We obtain $$a^2b^2c^2=64800,$$ which gives $abc=180\sqrt2$.

Thus, $$a=\frac{abc}{bc}=\frac{180\sqrt2}{40}=4.5\sqrt2,$$ $$b=\frac{180\sqrt2}{45}=4\sqrt2$$ and $$c=\frac{180\sqrt2}{36}=5\sqrt2.$$ Id est, $$\sqrt{a^2+b^2+c^2}=\sqrt{122.5}=\frac{35}{\sqrt{10}}.$$