imc

We know that all side lengths are integers, so we can test Pythagorean triples for all triangles. First, we focus on $△ABP$. The length of $AB$ is $30$, and the possible (small enough) Pythagorean triples $△ABP$ can be are $(3,4,5),(5,12,13),(8,15,17),$ where the length of the longer leg is a factor of $30$. Testing these, we get that only $(8,15,17)$ is a valid solution. Thus, we know that $BP=16$ and $AP=34$. Next, we move on to $△QDA$. The length of $AD$ is $28$, and the small enough triples are $(3,4,5)$ and $(7,24,25)$. Testing again, we get that $(3,4,5)$ is our triple. We get the value of $DQ=21$, and $AQ=35$. We know that $CQ=CD-DQ$ which is $9$, and $CP=BC-BP$ which is $12$. $△CPQ$ is therefore a right triangle with side length ratios $3,4,5$, and the hypotenuse is equal to $15$. $△APQ$ has side lengths $34,35,$ and $15,$ so the perimeter is equal to $34+35+15=\boxed{84}.$