Question

Each side of a pentagon is $20$ $\mathrm{inches}$ longer than the previous side. What is the length of the shortest side if the perimeter is $401$ $\mathrm{inches}$?

Answer 1

Step 1: Find the sides of the pentagon.

Let the shortest side of the pentagon be $x$.

Since each side of a pentagon is $20$ $\mathrm{inches}$ longer than the previous side, thus the sides of the pentagon are as follows:

$\left(x+20\right),\left(x+20+20\right),\left(x+20+20+20\right)$ and $\left(x+20+20+20+20\right)$, i.e. $\left(x+20\right),\left(x+40\right),\left(x+60\right)$ and $\left(x+80\right)$.

Step 2: Find the value of $x$.

Since the perimeter of the pentagon is $401$ $\mathrm{inches}$.

It is known that the sum of all the sides of the polygon is the perimeter of the polygon.

Thus,

$\begin{array}{rcl}x+x+20+x+40+x+60+x+80& =& 401\\ & \Rightarrow & 5x+200=401\\ & \Rightarrow & 5x+200-200=401-200\left[\text{Subtract 200 from both sides of the equation}\right]\\ & \Rightarrow & 5x=201\\ & \Rightarrow & \frac{5x}{5}=\frac{201}{5}\left[\text{Divide both sides of the equation by 5}\right]\\ \therefore x& =& 40.2\end{array}$

Hence, the shortest side of the pentagon is $40.2$ $\mathrm{inches}$.