[tex]P=\dfrac{1}{2}(a+b)h \ \ \ \ \ | \cdot2\\\\2P=(a+b)h \ \ \ \ \ | :h\\\\\dfrac{2P}{h} =a+b \ \ \ \ \ |-a\\\\ \dfrac{2P}{h}-a =b\\\\\boxed{b= \dfrac{2P}{h}-a}[/tex]
