Answer:
[tex]\bold{p=\dfrac{2SA-2B}l}[/tex]
Step-by-step explanation:
[tex]\bold{SA=\frac12lp+B}\\-B\qquad-B\\\\\bold{SA-B=\frac12lp}\\{}\qquad^{\times2}\qquad^{\times2}\\\\\bold{2SA-2B=lp}\\{}\qquad\div l\qquad\div l\\\\\bold{\dfrac{2SA-2B}{l}=p}[/tex]
