Chapter 12.3, Problem 18PPS

To determine

To find: the volume of the mailbox

Answer to Problem 18PPS

The volume of the figure is $1233.43\text{in}{\text{.}}^3$ .

Explanation of Solution

Given:

  

Formula used:

The volume V of the cylinder is $V=B\cdot h$

Calculation:

The objective is to find the volume of the figure.

The required volume is the sum of the volume of half cylinder having diameter 6.75 in. and the volume of a rectangular prism.

The volume V of the cylinder is given by the formula

  $V=B\cdot h$

where $B=\pi r^2$ is the area of the circle and h is the height the cylinder.

The volume V of a rectangular prism is given by the formula

  $V=Bh$

whereB is the area of a face and h is the height.

Since radius is half the diameter of a circle, so radius of the circle is $\frac{6.75}{2}$ in. that is 3.375 in..

The area B of the circle is given

  $\begin{array}{c}{B}_2=\pi r^2\\ =\pi {\left(3.375\right)}^2\\ \approx 11.39\pi \\ \approx 11.39\left(3.14\right)\left[Since\pi \approx 3.14\right]\\ \approx 35.76\end{array}$

Put $B\approx 35.76$ and $h=20.25$ in the volume formula to find the volume.

Thus, the volume $V_1$ of the cylinder is

  $\begin{array}{c}{V}_1=B\cdot h\\ =\left(35.76\right)20.25\\ =724.14\end{array}$

Therefore, the volume of the half cylinder is $\frac{724.14}{2}\text{in}{\text{.}}^3$ that is $362.07\text{in}{\text{.}}^3$ .

Next find the volume of the rectangular prism.

From the figure the area B is a rectangle whose length is 6.75 in. and width is 6.375 in.

The area B of the rectangle is given by

  $B=l\cdot w$ wherel is the length and w is the width of the rectangle

  $\begin{array}{c}=\left(6.75\right)6.375\\ \approx 43.03\end{array}$

Put $B\approx 43.03$ and $h=20.25$ in the volume formula to find the volume.

Thus, the volume $V$ is

  $\begin{array}{c}{V}_2=Bh\\ =\left(43.03\right)20.25\\ \approx 871.36\end{array}$

Therefore, the volume of the rectangular prism is $871.36\text{in}{\text{.}}^3$ .

Thus, the required volume V is

  $\begin{array}{c}V=\frac{1}{2}{V}_1+V_2\\ \approx 362.07+871.36\\ =1233.43\end{array}$

Therefore, the volume of the figure is approximately $1233.43\text{in}{\text{.}}^3$ .

Conclusion:

Therefore, the volume of the figure is $1233.43\text{in}{\text{.}}^3$ .