Question

How do you use the area of a triangle to solve for B where B stands for base of the triangle?

Answer 1

Hint: We first explain the terms in the equation $A={\displaystyle \frac{Bh}{2}}$. We find the particular triangular figure which follows the formula. We find their volume using the base area. From the values we verify the result.

Complete step by step solution:
The formula for area of a triangle is $A={\displaystyle \frac{Bh}{2}}$.
In the equation three variables have been used. These variables have their own meaning for the figure. The variable $A$ stands for the area of the figure. The variable $B$ stands for the length of the base side of the figure. The variable $h$ stands for the height of the figure.
All three variables are just numbers. We can treat them as algebraic variables.
For example: in case of a triangle the base length is $B=4$. Let’s assume the height of the cube is also 4 units.

Then the area is $A={\displaystyle \frac{Bh}{2}}={\displaystyle \frac{4\times 4}{2}}=8$.
Therefore, we can just divide both sides of the equation $A={\displaystyle \frac{Bh}{2}}$ with ${\displaystyle \frac{h}{2}}$.
The equation becomes
$\begin{array}{rl}& A={\displaystyle \frac{Bh}{2}}\\ & \Rightarrow {\displaystyle \frac{2A}{h}}={\displaystyle \frac{Bh}{h}}\times {\displaystyle \frac{2}{h}}=B\end{array}$
Therefore, the solution for $B$ from the equation $A={\displaystyle \frac{Bh}{2}}$ is $B={\displaystyle \frac{2A}{h}}$.
We verify the result from the perspective of the cube. Base was $B=4$. We verify the result with $B={\displaystyle \frac{2\times 8}{4}}={\displaystyle \frac{16}{4}}=4$. Thus verified $B$ is $B={\displaystyle \frac{2A}{h}}$.

Note: We can only solve this equation in every case as the area follows the same formula. In cases where the formula changes the value of $B$ also changes according to the formula. We need to remember that the equation does not necessarily have to be a formula for area all the time.