Question

13. The perimeter P (in yards) of a soccer field is represented by the formula P = 2 + 2w,

where is the length (in yards) and w is the width (in yards).
P = 330 yd
a. Solve the formula for w.
b. Find the width of the field.
c. About what percent of the field
is inside the circle?
10.4
10 yd
L=100 yd

Answer 1

`\textsf{a)} \quad w = (P)/(2)- \ell`

b) w = 65 yd

c) 4.83% (2 d.p.)

Explanation:

Perimeter formula

`P=2 \ell+2w`

where:

• 
  `\ell` is the length
• 
  `w` is the width

Part (a)

To solve the formula for w, use arithmetic operations to isolate w:

`\begin{aligned}P & =2 \ell+2w\\P - 2 \ell& =2 \ell+2w -2 \ell\\P - 2 \ell& =2w\\2w & = P - 2 \ell\\(2w)/(2) & = (P)/(2) - (2 \ell)/(2)\\w & = (P)/(2)- \ell\end{aligned}`

Part (b)

Given:

• 
  `P = 330 \sf \: yd`
• 
  `\ell = 100 \sf \:yd`

Substitute the given values into the formula for width found in part (a) to find w:

`\begin{aligned}w & = (P)/(2)- \ell\\\implies w & = (330)/(2)-100\\& = 165-100\\& = 65 \:\: \sf yd\end{aligned}`

Part (c)

Area of a rectangle

`A= \ell \:w`

where:

• 
  `\ell` is the length
• 
  `w` is the width

Therefore, the area of the field is:

`\begin{aligned}A &= \ell \:w\\\implies A & = 100 * 65\\& = 6500\:\: \sf yd^2\end{aligned}`

Area of a circle

`A= \pi r^2`

where:

• 
  `r` is the radius

Therefore, the area of the central circle of the field is:

`\begin{aligned}A & = \pi r^2\\\implies A&=\pi 10^2\\& = 100 \pi \:\: \sf yd^2\end{aligned}`

To find the percent of the field that is inside the circle, divide the found area of the circle by the found area of the field, and multiply by 100:

`\begin{aligned}\sf Percent & = \sf (Area\:of\:circle)/(Area\:of\:field) * 100\\\\ \implies \sf Percent & = (100 \pi)/(6500) * 100\\\\& = (10000\pi)/(6500)\\\\& = (100\pi)/(65)\\\\& = (20\pi)/(13)\\\\& = 4.83\%\:\:\sf (2\:d.p.)\end{aligned}`