Visual proof for $[\triangle ABC] = rs$?

Question

I'm teaching a high school geometry class and we just got to the part where we proved that the area of a triangle $[\triangle ABC]$ is equal to the inradius $r$ times the semiperimeter $s$.

A student pointed out to me that it's really the same thing as $[\triangle ABC] = \frac{1}{2}rp$ where $p$ is the perimeter. But this representation looks like the regular formula $\frac{1}{2}bh$.

So I was wondering if there's a way to take the original $\triangle ABC$ and construct a new triangle that will have a base of $p$ and a height of $r$ and show that the original and newly constructed triangle have the same area?

Answer 1

It is possible. In fact, in $\Delta ABC$, extending $BC$ to $A'$ and then to $B'$ such that $$ CA'=AC, A'B'=AB $$ one has \begin{eqnarray} S_{\Delta ABC}&=&S_{\Delta OBC}+S_{\Delta OAC}+S_{\Delta OBC}=\frac12ra+\frac12rb+\frac12rc\\ &=&S_{\Delta OBC}+S_{\Delta OCA'}+S_{\Delta OA'B'}=S_{\Delta OBB'}. \end{eqnarray}

Answer 2

In case you're also interested in a rectangle, I made a diagram.

Rotate one in each pair of the congruent right triangles to obtain a rectangle.

Indeed its length is $$ \color{blue}{(s-a)} + \color{red}{(s-b)} + \color{green}{(s-c)} = s$$

and clearly, $$ rs = \triangle$$