Answer:
\(b_{2}=\frac{2A}{h}-b_{1}\)
Step-by-step explanation:
Area of Trapezoid is given by:
\(A=\frac{1}{2}(b_{1}+b_{2})h\)
Using distributive property [ \(a(b+c)=ab+ac\) ] and simplifying:
\(A=\frac{1}{2}hb_{1}+\frac{1}{2}hb_{2}\)
Now, doing some algebra and solving for \(b_{2}\) gives us:
\(A-\frac{1}{2}hb_{1}=\frac{1}{2}hb_{2}\\\frac{A-\frac{1}{2}hb_{1}}{\frac{1}{2}h}=b_{2}\\\frac{A}{\frac{h}{2}}-\frac{\frac{1}{2}hb_{1}}{\frac{1}{2}h}=b_{2}\\\frac{2A}{h}-b_{1}=b_{2}\)
This is our answer.
