Digging:
$A_d = \frac{1}{2}(3 + \frac{22}{12})(5) = \frac{145}{12} ~ \text{ft.}^2$
$L = 484 ~ \text{ft.}$
$V_d = A_dL = \frac{145}{12}(484)$
$V_d = 5848.33 ~ \text{ft.}^3 = 5848.33 ~ \text{ft.}^3 \left( \dfrac{1 ~ \text{yd.}}{3 ~ \text{ft.}} \right)^3$
$V_d = 216.60 ~ \text{yd.}^3$
Cost of Digging:
$C_d = 5(216.60) = \$1,083.02$ answer
Concrete:
$\dfrac{a}{53} = \dfrac{7}{60}$
$a = 6.65 ~ \text{in.}$
$b = 22 + 2a = 22 + 2(6.183) = 34.37 ~ \text{in.}$
$b = 2.864 ~ \text{ft.}$
$\cos (\frac{1}{2}\theta) = \dfrac{4}{8}$
$\theta = 120^\circ$
$A_1 = \frac{1}{2}r^2(\theta_{rad} - \sin \theta_{deg})$
$A_1 = \dfrac{1}{2}\left(\dfrac{8}{12}\right)^2\left[ 120^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 120^\circ \right]$
$A_1 = 0.273 ~ \text{ft.}^2$
$A_c = \frac{1}{2}(3 + b)(\frac{7}{12}) - A_1 = \frac{1}{2}(3 + 2.864)(\frac{7}{12}) - 0.273$
$A_c = 1.437 ~ \text{ft.}^2$
$V_c = A_cL = 1.437(484) = 695.67 ~ \text{ft.}^3$
$V_c = 695.67 ~ \text{ft.}^3 \left( \dfrac{1 ~ \text{yd.}}{3 ~ \text{ft.}} \right)^3$
$V_c = 25.76 ~ \text{yd.}^3$
Cost of concrete pouring:
$C_c = 9.00(25.76) = \$231.89$ answer