Question

How do you solve $A = (\dfrac{1}{2}) h (b_1 + b_2)$ for $b_2$?

Answer 1

Hint: In the above question, we have to solve $A = (\dfrac{1}{2}) h (b_1 + b_2)$. The given equation is the formula for calculating area A of a trapezoid. After solving this we will get our answer in terms of b2. So let us see how we can solve this problem.

Complete step-by-step answer:
The given problem is $A = (\dfrac{1}{2}) h (b_1 + b_2)$.
On multiplying both the sides of the equation with 2 we get,
 $\Rightarrow 2.A = 2.\dfrac{1}{2}.h\left( {{b_1} + {b_2}} \right)$
 We can cancel 2 because $\dfrac{2}{2} = 1$
 $\Rightarrow 2.A = h\left( {{b_1} + {b_2}} \right)$
On dividing both sides of the equation with h we get,
 $\Rightarrow \dfrac{{2.A}}{h} = \dfrac{{h\left( {{b_1} + {b_2}} \right)}}{h}$
We can cancel h because $\dfrac{h}{h} = 1$
  $\Rightarrow \dfrac{{2.A}}{h} = {b_1} + {b_2}$
After subtracting b1 from both sides of the equation we get,
 $\Rightarrow \dfrac{{2.A}}{h} - {b_1} = {b_1} + {b_2} - {b_1}$
 $\Rightarrow \dfrac{{2.A}}{h} - {b_1} = {b_2}$
By symmetric property
 $\therefore {b_2} = \dfrac{{2.A}}{h} - {b_1}$
Therefore, on solving $A = (\dfrac{1}{2}) h (b_1 + b_2)$ we get ${b_2} = \dfrac{{2.A}}{h} - {b_1}$ .

Note: In the above formula, we used the symmetric property. The symmetric property means if x = y then y = x. Also, you can see in the solution that from the beginning of the solution we tried to separate b2, because of which we first multiplied the equation with 2, then divide the equation with h, and finally subtracted b1 to get our answer.