Question

The driveway of the Santiago's new home is to be blinded by semicircular arcs as shown. The distance from X and Y is 100 ft. The concrete driveways is 10ft wide and 1/3 ft thick. Using 3.14 as an approximation for π, how many cubic feet of concrete are needed to build the driveway? (round to nearest cubic ft)

Answer 1

First, we need to compute the volume of the hemicylinder with a diameter of 100 ft (radius 50 ft) and a height of 1/3 ft, as follows:

`\begin{gathered} V_1=\pi\cdot r^2\cdot h\cdot(1)/(2) \\ V_1=3.14\cdot50^2\cdot(1)/(3)\cdot(1)/(2) \\ V_1=1308.33ft^3 \end{gathered}`

Next, we need to compute the volume of the hemicylinder with a diameter of 80 ft (radius 40 ft) and a height of 1/3 ft, as follows:

`\begin{gathered} V_2=\pi\cdot r^2\cdot h\cdot(1)/(2) \\ V_2=3.14\cdot40^2\cdot(1)/(3)\cdot(1)/(2) \\ V_2=837.33ft^3 \end{gathered}`

Finally, the volume of concrete needed to build the driveway is 1308.33 - 837.33 = 471 cubic feet