Question

Somebody please help me with this problem.

Answer 1

`\bold{p=\frac{2SA-2B}l}`

Explanation:

`\bold{SA=\frac12lp+B}\\-B\qquad-B\\\\\bold{SA-B=\frac12lp}\\{}\qquad^(*2)\qquad^(*2)\\\\\bold{2SA-2B=lp}\\{}\qquad/ l\qquad/ l\\\\\bold{(2SA-2B)/(l)=p}`