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We can prove that $A=rs$ by drawing altitudes from the Incenter $I$ to the sides $AB$ , $BC$ , and $CA$ at points $P$ , $Q$ , and $R$ respectively. Note that $[ABC]=[AIB]+[BIC]+[CIA]=\dfrac{1}{2}(IP)(AB)+\dfrac{1}{2}(IQ)(BC)+\dfrac{1}{2}(IR)(AC)$

Note that $IP=IQ=IR=r$ , so we have $[ABC]=\dfrac{1}{2}r(AB+BC+CA)$

We see that $[ABC]=A$ and $AB+BC+CA=P$ . So $A=\dfrac{1}{2}rP$

However, $\dfrac{1}{2}P=s$ . Thus, $\boxed{A=rs}$ thus proved. $\Box$

@Daniel Liu This is an elegant solution, but I still wonder how a triangle can have an area that is less than that of its incircle, (as $64\pi \gt 43$ ). Also, a triangle with this perimeter value, (as far as I can determine), would have a maximum area of only $5.5600$ to 4 decimal places. Any thoughts?

Nice solution quite an easy problem.By the way do you attend science bowl and if you do what resources do you use to study @Daniel Liu