Question

Given a trapezoid, if you double the height and the bases, what happens to the area?

Answer 1

Step 1: Determine an expression for the new area of the trapezoid.

The area of a trapezoid is $A=\frac{1}{2}h\left(a+b\right)$, where height $h$ is the distance between its two parallel sides, base $a$ represents the length of one of the parallel sides, and base $b$ is the length of the other parallel side. If the height $h$ is doubled the new height is $h\text{'}=2h$. If the bases are doubled the new bases are $a\text{'}=2a$, and $b\text{'}=2b$. The new area $A\text{'}$ will then become $A\text{'}=\frac{1}{2}h\text{'}\left(a\text{'}+b\text{'}\right)$.

Step 2: Determine an expression for the new area of the trapezoid in terms of its old measurements.

Substitute $h\text{'}=2h$, $a\text{'}=2a$, and $b\text{'}=2b$ in the equation $A\text{'}=\frac{1}{2}h\text{'}\left(a\text{'}+b\text{'}\right)$ and find an expression for the new area $A\text{'}$:

$\begin{array}{rcl}A\text{'}& =& \frac{1}{2}h\text{'}\left(a\text{'}+b\text{'}\right)\\ & =& \frac{1}{2}\left(2h\right)\left(2a+2b\right)\\ & =& \frac{1}{2}\left(2h\right)\times 2\left(a+b\right)\\ & =& \frac{1}{2}\times 4\times h\left(a+b\right)\\ & =& 4\times \left(\frac{1}{2}h\left(a+b\right)\right)\end{array}$

Thus, the new area is $\begin{array}{rcl}A\text{'}& =& 4\times \left(\frac{1}{2}h\left(a+b\right)\right)\end{array}$.

Step 3: Determine an expression for the new area of the trapezoid in terms of its old area.

Substitute, $A=\frac{1}{2}h\left(a+b\right)$ in the equation $\begin{array}{rcl}A\text{'}& =& 4\times \left(\frac{1}{2}h\left(a+b\right)\right)\end{array}$.

$\begin{array}{rcl}A\text{'}& =& 4\times \left(\frac{1}{2}h\left(a+b\right)\right)\\ & =& 4\times A\end{array}$

Thus, $\begin{array}{rcl}A\text{'}& =& 4\times A\end{array}$. The new area is four times the original area.

Hence, if we double the height and the bases of a trapezoid, the new area of the trapezoid becomes four times the original area.