# Is there a relationship between altitude and corresponding side length of a triangle?

If so, what is it?

I've seen that $(\text{altitude}) \cdot (\text{side length}) = 2(\text{area})$ but I'm not sure why this is true...

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I am nearly sure you have seen it, in the version "area is equal to $\frac{1}{2}$ times base times height." Let the area be $A$, let the base (side) be $b$ and let the corresponding height (altitude) be $h$. Then the result you are probably familiar with can be written as $$A=\frac{1}{2}bh.$$ Multiply both sides by $2$, and we get the result you mentioned.
Here is a somewhat dishonest proof. Draw a triangle $XYZ$, say with side $XY$ horizontal. Draw the triangle so that the vertex $Z$ is somewhere above the line segment $XY$. Drop a perpendicular from $Z$ to the base $XY$, meeting $XY$ at say $P$ between $X$ and $Y$. Let $XY=b$, and let $ZP=h$.
Now draw the rectangle with base $XY$ and height $h$. This rectangle has area $bh$. Cut off the two parts of the rectangle outside $\triangle XYZ$. These can be reassembled to make a copy of $\triangle XYZ$. So the rectangle has twice the area of the triangle. This says that twice the area of the triangle is equal to $bh$, which is exactly what we wanted.
The above proof doesn't work if $\triangle XYZ$ is very "tilted," so that $Z$ does not lie above the line segment $XY$. It is not very hard to write a proof that deals with that case.