Chapter 12.1, Problem 6CE

a.

To determine

To find: the lateral area.

Answer to Problem 6CE

The lateral area is $120{\text{cm}}^2$ .

Explanation of Solution

Given information:

The figure provided in the question,

  

Calculation: to find the lateral area, multiply the perimeter by the height,

  $\begin{array}{l}\text{L}\text{.A}=6\left(4\right)\left(5\right)\text{ [perimeter multiply by height]}\\ \text{L}\text{.A}=120{\text{cm}}^2\text{ [multiply]}\end{array}$

Thus, The lateral area is $120{\text{cm}}^2$ .

b.

To determine

To find: the base area.

Answer to Problem 6CE

The base area is $24\sqrt{3}{\text{cm}}^2$ .

Explanation of Solution

Given information:

The figure provided in the question,

  

Calculation: to find the base area, use the formula and then substitute and simplify,

  $\begin{array}{l}\text{B}=\frac{3\sqrt{3}{s}^2}{2}\\ \text{B}=\frac{3\sqrt{3}{\left(4\right)}^2}{2}\text{ [substitute]}\\ \text{B}=24\sqrt{3}{\text{cm}}^2\text{ [simplify]}\end{array}$

Thus, The base area is $24\sqrt{3}{\text{cm}}^2$ .

c.

To determine

To find: the total area.

Answer to Problem 6CE

The total area is $\text{120+48}\sqrt{3}{\text{cm}}^2$ .

Explanation of Solution

Given information:

The figure provided in the question,

  

Calculation: to find the total area, add the lateral area and double of base area,

  $\begin{array}{l}\text{T}\text{.A}=\text{L}\text{.A+2B}\\ \text{T}\text{.A}=120\text{+2}\left(24\sqrt{3}\right)\text{ [substitute]}\\ \text{T}\text{.A=120+48}\sqrt{3}{\text{cm}}^2\end{array}$

Thus, The total area is $\text{120+48}\sqrt{3}{\text{cm}}^2$ .

d.

To determine

To find: the volume of prism.

Answer to Problem 6CE

The volume of prism is $120\sqrt{3}{\text{cm}}^3$ .

Explanation of Solution

Given information:

The figure provided in the question,

  

Calculation: to find the volume, multiply the height by the base area,

  $\begin{array}{l}v=Bh\\ v=24\sqrt{3}\left(5\right)\text{ [substitute]}\\ v=120\sqrt{3}{\text{cm}}^3\text{ [simplify]}\end{array}$

Thus, The volume of prism is $120\sqrt{3}{\text{cm}}^3$ .