Chapter 14.4, Problem 21PPS

To determine

To calculate: The length and width of NFL and arena football field.

  $\begin{array}{ccc}\text{Measure}& \text{Arena}& \text{NFL}\\ \text{Width}& a-38& n+3\\ \text{Length}& a& 2n+20\end{array}$

Answer to Problem 21PPS

The length and width of Arena football field are 66 and 28 yards respectively. The length and width of NFL football field are 120 and 53 yards respectively.

Explanation of Solution

Given information:

The perimeter of arena football field is 188 yards and the perimeter of an NFL football field is 346 yards.

Formula used:

Perimeter $\left(P\right)$ of rectangle is $P=2\left(l+w\right)$ where length is $\left(l\right)$ and width is $\left(w\right)$ .

Calculation:

Consider the provided information thatperimeter of arena football field is 188 yards and the perimeter of an NFL football field is 346 yards. The dimensions of field are,

  $\begin{array}{ccc}\text{Measure}& \text{Arena}& \text{NFL}\\ \text{Width}& a-38& n+3\\ \text{Length}& a& 2n+20\end{array}$

For arena football field width is $a-38$ and length is a .

Recall that perimeter $\left(P\right)$ of rectangle is $P=2\left(l+w\right)$ where length is $\left(l\right)$ and width is $\left(w\right)$ .

Apply it,

  $\begin{array}{c}188=2\left(a-38+a\right)\\ 188=2\left(2a-38\right)\\ 188=4a-76\\ 4a=264\end{array}$

Simplify it further as,

  $\begin{array}{c}4a=264\\ a=66\end{array}$

Therefore, dimensions of the field are, length is 66 yards and width is $66-38=28$ yards.

For NFL football field width is $n+3$ and length is $2n+20$ .

Recall that perimeter $\left(P\right)$ of rectangle is $P=2\left(l+w\right)$ where length is $\left(l\right)$ and width is $\left(w\right)$ .

  $\begin{array}{c}346=2\left(2n+20+n+3\right)\\ 346=2\left(3n+23\right)\\ 346=6n+46\\ 6n=300\end{array}$

Simplify it further as,

  $\begin{array}{c}6n=300\\ n=50\end{array}$

Therefore, dimensions of the field are, length is $2\left(50\right)+20=120$ yards and width is $50+3=53$ yards.