Question

How do you solve `A = (1/2)h(b_1 + b_2)" for "b_2`?

Answer 1

`b_2=(2A)/h-b_1`

Explanation:

Yes it is the formula for the Area A of the trapezoid

From the given formula

`A=1/2h(b_1+b_2)`

multiply both sides of the equation by 2

`2*A=2*1/2h(b_1+b_2)`

`2*A=cancel2*(1/(cancel(2)))*h(b_1+b_2)`

`2*A=h(b_1+b_2)`

Divide both sides of the equation by h

`(2*A)/h=(h(b_1+b_2))/h`

`(2*A)/h=(cancelh(b_1+b_2))/cancelh`

`(2*A)/h=b_1+b_2`

Subtract `b_1` from both sides of the equation

`(2*A)/h-b_1=b_1+b_2-b_1`

`(2*A)/h-b_1=b_2`

by symmetric property

`b_2=(2*A)/h-b_1`

God bless....I hope the explanation is useful.