Answer:
[tex]\textsf{a)} \quad w = \dfrac{P}{2}- \ell[/tex]
b) w = 65 yd
c) 4.83% (2 d.p.)
Step-by-step explanation:
Perimeter formula
[tex]P=2 \ell+2w[/tex]
where:
- [tex]\ell[/tex] is the length
- [tex]w[/tex] is the width
Part (a)
To solve the formula for w, use arithmetic operations to isolate w:
[tex]\begin{aligned}P & =2 \ell+2w\\P - 2 \ell& =2 \ell+2w -2 \ell\\P - 2 \ell& =2w\\2w & = P - 2 \ell\\\dfrac{2w}{2} & = \dfrac{P}{2} - \dfrac{2 \ell}{2}\\w & = \dfrac{P}{2}- \ell\end{aligned}[/tex]
Part (b)
Given:
- [tex]P = 330 \sf \: yd[/tex]
- [tex]\ell = 100 \sf \:yd[/tex]
Substitute the given values into the formula for width found in part (a) to find w:
[tex]\begin{aligned}w & = \dfrac{P}{2}- \ell\\\implies w & = \dfrac{330}{2}-100\\& = 165-100\\& = 65 \:\: \sf yd\end{aligned}[/tex]
Part (c)
Area of a rectangle
[tex]A= \ell \:w[/tex]
where:
- [tex]\ell[/tex] is the length
- [tex]w[/tex] is the width
Therefore, the area of the field is:
[tex]\begin{aligned}A &= \ell \:w\\\implies A & = 100 \times 65\\& = 6500\:\: \sf yd^2\end{aligned}[/tex]
Area of a circle
[tex]A= \pi r^2[/tex]
where:
- [tex]r[/tex] is the radius
Therefore, the area of the central circle of the field is:
[tex]\begin{aligned}A & = \pi r^2\\\implies A&=\pi 10^2\\& = 100 \pi \:\: \sf yd^2\end{aligned}[/tex]
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:
[tex]\begin{aligned}\sf Percent & = \sf \dfrac{Area\:of\:circle}{Area\:of\:field} \times 100\\\\ \implies \sf Percent & = \dfrac{100 \pi}{6500} \times 100\\\\& = \dfrac{10000\pi}{6500}\\\\& = \dfrac{100\pi}{65}\\\\& = \dfrac{20\pi}{13}\\\\& = 4.83\%\:\:\sf (2\:d.p.)\end{aligned}[/tex]
