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

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Answer 1 · 2016-06-07T10:32:48

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

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.

How do you solve A = (1/2)h(b_1 + b_2)" for "b_2A = (1/2)h(b_1 + b_2)" for "b_2?