Question

Aschool constructing a rectangular play area againstan exterior wall of the school building. It uses 120 feetof fencing material to enclose three sides of the playarea.Complete the table by giving the length and area foreach width. (The width is the side perpendicular to thebuilding.)

Answer 1

Given:

The length of the fencing material is c = 120 feet.

The rectangular play is constructed against an exterior wall.

The objective is to find the various length and area of the play for the given width in the table.

The general formula to find the length of the rectangular play is,

$c=l+2w$

If width is of the play is 10 ft with the available 120 feet fencing material,

$\begin{gathered} 120=l+2(10) \\ 120=l+20 \\ l=120-20 \\ l=100\text{ ft.} \end{gathered}$

Then, area of the play with w = 10 ft is,

$\begin{gathered} A=l\times w \\ =100\times10 \\ =1000ft^2 \end{gathered}$

If width is of the play is 30 ft with the available 120 feet fencing material,

$\begin{gathered} 120=l+2(30) \\ 120=l+60 \\ l=120-60 \\ l=60\text{ ft.} \end{gathered}$

Then, area of the play with w = 30 ft is,

$\begin{gathered} A=l\times w \\ =60\times30 \\ =1800ft^2 \end{gathered}$

If width is of the play is 40 ft with the available 120 feet fencing material,

$\begin{gathered} 120=l+2(40) \\ 120=l+80 \\ l=120-80 \\ l=40\text{ ft} \end{gathered}$

Then, area of the play with w = 40 ft is,

$\begin{gathered} A=l\times w \\ =40\times40 \\ =1600ft^2 \end{gathered}$

If width is of the play is w ft with the available 120 feet fencing material,

$\begin{gathered} 120=l+2w \\ l=(120-2w)\text{ ft} \end{gathered}$

Then, area of the play with w ft is,

$\begin{gathered} A=l\times w \\ =(120-2w)\times w \\ =(120w-2w^2)ft^2 \end{gathered}$

Hence, the table values are,