Commonly used polyhedral dice all have sides of the same shape. However: is this strictly required?
Take a triangular prism. A triangular prism with small height compared to the base edge would almost always fall on one of its bases when thrown. A triangular prism with large height would fall like a "three-sided cylinder", almost always on one of the flat sides.
How would a fair triangular prism that falls on any of its sides with equal probability look?
Lacking symmetry this question cannot be answered purely mathematically. We need some physics: When the die is tossed against the table surface, it bounces and rotates irregularly until it comes to halt in one of finitely many states of equilibrum. The probability distribution governing this final state can be viewed as following Boltzmann statistics so that the die lands on face $j$ with probability $$p_j\sim e^{-E_j/kT}.$$ Here $E_j$ is the potential energy of state $j$, which is proportional to the height of the center of gravity over the table, $k$ is the Boltzmann constant and $T$ the temperature. The only way to make all probabiliteies equal is to esure that the potential energy is the same for all faces (or for an approximate solution make the room real hot).
If the prism has height $h$ and side length $a$ (and is homogenous of course), then the center of gravity is either at height $\frac h2$ or $\frac a{\sqrt 3}$. Therefore, you need $a=\frac{h\sqrt3} 2$.
Note that the assumption about the applicability of Boltzmann may not be justified if the die does not bounce a lot - I suppose that in that case the final position is mostly governed by the size of its "basin of attraction", essentially tha spacial angle covered by each face ... I suggest you get out your fretsaw and make a long series of experiments to verify if such a 5-die is fair.
If you just want a die that has five numbers that occur with equal probability you could either have a pentagonal prism which you have to roll (with the end faces unmarked), or you could have a regular icosahedron with each of the five numbers repeated four times on the faces, which could be tossed in the usual way.
For a triangular prism where we include the end faces, presumably the triangular face would be equilateral, and the length would have to be arrived at by experiment. You couldn't assume, for example, that square faces would make the die fair, or even that it would be fair if the areas of the rectangular faces were equal to the area of the triangular faces.
However, it is reasonable to assume that there must exist a proportion for the rectangular faces that would make the die fair.
I will use the same arguments as presented in my solution to the physical biased coin problem for determining the exact height of the triangular prism. As stated in the comments there, this is effective if the dice loses most of its energy in the first bounce.
Consider the pyramid in the prism with a base coinciding with one of the triangular sides and an apex at the geometric centre, which is obviously the centre of mass too. We know that the solid angle in a right pyramid with a regular n-gonal base of circumradius r and height h is $$2\pi - 2n \tan^{-1}\left(\frac {\tan \left({\pi\over n}\right)}{\sqrt{1 + {r^2 \over h^2}}} \right)$$ Fix r = 1 and n = 3, then the fraction of the sphere subtended by the triangular base for our case is $$\frac12 - \frac3{2\pi} \tan^{-1}\left(\frac {\sqrt3}{\sqrt{1 + {1\over h^2}}} \right)$$ and this must equal 1/5 for the dice to be fair. Solving, we get $h=\sqrt{\frac{3\sqrt5-5}8}$. The rectangular sides have height $2h$ and width $\sqrt3$, so they are shorter than they are wide with an aspect ratio of $$\sqrt{\frac{3(5+3\sqrt5)}{10}}=1.874156\dots$$ The same technique can be used to construct fair d7's as pentagonal prisms, for which the rectangles are taller than they are wide with an aspect ratio of $$2\tan\frac\pi7\sqrt{\frac2{(5-\sqrt5) (5-2\sqrt5-\tan^2\frac\pi7))}}=1.506037\dots$$ And so on and so forth, with d11 as a nonagonal prism, d13 as a hendecagonal prism, etc. (See also my problem of constructing a hendecagon.) However, these are not really practical.
An alternative idea: use multiple rolls of an ordinary cubic die to approximate the fair five-sided one. You add up $n$ rolls and divide by $5$, then the remainder is $0,1,2,3$, or $4$ with probabilities as nearly equal as possible given $6^n$ roll permutations. Thus with $n=3$ you have $44$ permutations with remainder $3$ and $43$ permutations with each of the other remainders.
