Question

Find the lateral surface area and volume please round up to nearest integer

Answer 1

For this problem we will use the following formula for the surface area of a truncated cone:

$\begin{gathered} SA=\pi(r_1+r_2)\sqrt[]{(r_1-r_2)^2+h^2}+\pi(r^2_1+r^2_2), \\ \text{Where r}_1\text{ is the lower radius, r}_2\text{ is the upper radius, and h is the height.} \end{gathered}$

Substituting:

$\begin{gathered} r_1=\frac{11in}{2}=5.5in, \\ r_2=\frac{14in}{2}=7in, \\ h=21in, \end{gathered}$

we get:

$\begin{gathered} SA=\pi(7in+5.5in)\sqrt[]{(7in-5.5in)^2+(21in)^2}+\pi((5.5in)^2+(7in)^2) \\ =\pi(12.5in)\sqrt[]{2.25in^2+441in^2}+\pi(30.25in^2+49in^2) \\ =\pi(12.5in)(21.05in)+\pi\cdot79.25in^2 \\ =\pi(263.125+79.25)in^2 \\ =\pi(342.375)in^2 \\ \approx1076in^2. \end{gathered}$

Now, to compute the volume we will use the following formula:

$V=\frac{1}{3}\pi(r^2_1+r_1r_2+r^2_2)h\text{.}$

Substituting the given values we get:

$\begin{gathered} V=\frac{1}{3}\pi((5.5in)^2+(5.5in)(7in)+(7in)^2)21in \\ =\frac{1}{3}\pi(30.25in^2+38.5in^2+49in^2)21in \\ =\frac{1}{3}\pi(117.75in^2)21in \\ =824.25\pi in^3 \\ =2589in^3\text{.} \end{gathered}$

Answer: The total surface area is

$1076in^2\text{.}$

The volume is

$2589in^3\text{.}$