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

1 Answer

Answer:

#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.