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

${b}_{2} = \frac{2 A}{h} - {b}_{1}$

#### Explanation:

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

From the given formula

$A = \frac{1}{2} h \left({b}_{1} + {b}_{2}\right)$

multiply both sides of the equation by 2

$2 \cdot A = 2 \cdot \frac{1}{2} h \left({b}_{1} + {b}_{2}\right)$

$2 \cdot A = \cancel{2} \cdot \left(\frac{1}{\cancel{2}}\right) \cdot h \left({b}_{1} + {b}_{2}\right)$

$2 \cdot A = h \left({b}_{1} + {b}_{2}\right)$

Divide both sides of the equation by h

$\frac{2 \cdot A}{h} = \frac{h \left({b}_{1} + {b}_{2}\right)}{h}$

$\frac{2 \cdot A}{h} = \frac{\cancel{h} \left({b}_{1} + {b}_{2}\right)}{\cancel{h}}$

$\frac{2 \cdot A}{h} = {b}_{1} + {b}_{2}$

Subtract ${b}_{1}$ from both sides of the equation

$\frac{2 \cdot A}{h} - {b}_{1} = {b}_{1} + {b}_{2} - {b}_{1}$

$\frac{2 \cdot A}{h} - {b}_{1} = {b}_{2}$

by symmetric property

${b}_{2} = \frac{2 \cdot A}{h} - {b}_{1}$

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