Answer
a) $w = \frac{P}{2} - l$
b) $65$ yards
c) $4.83\%$
Work Step by Step
a) The perimeter, $P$, of a rectangle is $$P = 2l + 2w.$$ Subtract $2l$ from both sides $$P - 2l = 2w.$$ Divide both sides by $2$ $$\frac{P}{2} - l = w.$$ Therefore the formula for the width, $w$, is $$w = \frac{P}{2} - l.\tag{1}$$
b) From the figure, $P = 330$, $l =100$. Substitute the values of $P$ and $l$ in equation $(1)$ to determine $w$ $$w = \frac{330}{2} -100=165-100=65.$$ The width of the field is $w=65$ yards.
c) The area, $A$, of the field is $$A = l \cdot w = 100 \cdot 65 = 6500\text{ yd}^2.$$ The area, $a$, of the circle inside the field is $$a = \pi r^2 = \pi 10^2 = 100 \pi \approx 314.16\text{ yd}^2.$$ We calculate the percentage of the circle from the field $$\frac{\text{Area of circle}}{\text{Area of field}}\cdot100\%=\frac{a}{A}\cdot 100\%= \frac{314.16}{6500}\cdot100 \%\approx
4.83\%.$$
