Chapter 12.1, Problem 7PSB

a.

To determine

To calculate: The lateral area and total area of right square prism with dimensions as 6 and 20.

Answer to Problem 7PSB

The lateral area is $480$ and total area is $552$ .

Explanation of Solution

Given information:

A rectangular box with dimensions as 6 and 20.

Formula used:

Area of rectangle: $A=l\times w$

l = length of rectangle

w= width of rectangle Area of square: $A=s^2$

s = side of square

Calculation:

  

Area of rectangle:

  $\begin{array}{l}A(▭)=l\times w\\ A(▭)=20\times 6\\ A(▭)=120\end{array}$

Area of base:

  $\begin{array}{l}A(Base)=A(Square)=s^2\\ A(Base)={\left(6\right)}^2\\ A(Base)=36\end{array}$

Lateral Area = 4 $\times A(▭)$

Lateral Area = $4\times 120$

Lateral Area = $480$

Total Area = Lateral Area + 2 (Area of Base)

Total Area $=480+2\left(36\right)$

Total Area $=480+72$

Total Area $=552$

b.

To determine

To calculate: The lateral area and total area of right triangular prism with triangle dimensions as 3, 5 and 4.

Answer to Problem 7PSB

The lateral area is $120$ and total area is $132$ .

Explanation of Solution

Given information:

A right triangular prism with triangle dimensions as 3, 5 and 4.Rectangle dimensions are 4 and 10.

Formula used:

Area of rectangle: $A=l\times w$

l = length of rectangle

w= width of rectangle Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

  $\begin{array}{l}A({▭}_1)=l\times w=10\times 4=40\\ A({▭}_2)=l\times w=10\times 5=50\\ A({▭}_3)=l\times w=10\times 3=30\end{array}$

Lateral Area:

  $\begin{array}{l}{A}_{Lateral}=A({▭}_1)+A({▭}_2)+A({▭}_3)\\ A_{Lateral}=40+50+30\\ A_{Lateral}=120\end{array}$

h can be calculated by using Pythagorean Triple $(3-4-5)$

h = 4

  $\begin{array}{l}{A}_{Triangle}=\frac{1}{2}\times b\times h\\ A_{Triangle}=\frac{1}{2}\times 3\times 4\\ A_{Triangle}=6\end{array}$

There are two triangles.

  $\begin{array}{l}{A}_{Bases}=2A_{Triangle}\\ A_{Bases}=2\times 6\\ A_{Bases}=12\end{array}$

Total Area = Lateral Area + Area of Bases

Total Area $=120+12$

Total Area $=132$

c.

To determine

To calculate: The lateral area and total area of right isosceles triangular prism with rectangle dimensions as 50 and 13.

Answer to Problem 7PSB

The lateral area is $2500$ and total area is $2620$ .

Explanation of Solution

Given information:

A right isosceles triangular prism with triangle dimensions as 13, 13 and 24.Rectangle dimensions are 50 and 13.

Formula used:

Area of rectangle: $A=l\times w$

l = length of rectangle

w= width of rectangle Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

  $\begin{array}{l}A({▭}_1)=l\times w=50\times 13=650\\ A({▭}_2)=l\times w=50\times 13=650\\ A({▭}_3)=l\times w=50\times 24=1200\end{array}$

Lateral Area:

  $\begin{array}{l}{A}_{Lateral}=A({▭}_1)+A({▭}_2)+A({▭}_3)\\ A_{Lateral}=650+650+1200\\ A_{Lateral}=2500\end{array}$

h can be calculated by using Pythagorean Triple $(5-12-13)$

h = 5

  $\begin{array}{l}{A}_{Triangle}=\frac{1}{2}\times b\times h\\ A_{Triangle}=\frac{1}{2}\times 24\times 5\\ A_{Triangle}=60\end{array}$

There are two triangles.

  $\begin{array}{l}{A}_{Bases}=2A_{Triangle}\\ A_{Bases}=2\times 60\\ A_{Bases}=120\end{array}$

Total Area = Lateral Area + Area of Bases

Total Area $=2500+120$

Total Area $=2620$

d.

To determine

To calculate: The lateral area and total area of regular hexagonal prism with hexagon side as 6.

Answer to Problem 7PSB

The lateral area is $360$ and total area is $547$ .

Explanation of Solution

Given information:

A regular hexagonal prism with hexagon side as 6.Rectangle dimensions are 10 and 6.

Formula used:

Area of rectangle: $A=l\times w$

l = length of rectangle

w= width of rectangle Area of polygon: $A=n\left(\frac{s^2}{4}\sqrt{3}\right)$

n = number of sides of polygon.

s = side of base.

Calculation:

  

  $A(▭)=l\times w=10\times 6=60$

Lateral Area:

  $\begin{array}{l}{A}_{Lateral}=6A(▭)\\ A_{Lateral}=6\times 60\\ A_{Lateral}=360\end{array}$

  $\begin{array}{l}{A}_{Base}=A_{RegularHexagon}=n\left(\frac{s^2}{4}\sqrt{3}\right)\\ A_{RegularHexagon}=6\left(\frac{{\left(6\right)}^2}{4}\sqrt{3}\right)\\ A_{RegularHexagon}=6\left(\frac{36}{4}\sqrt{3}\right)\\ A_{RegularHexagon}=54\sqrt{3}\end{array}$

Total Area = Lateral Area + 2 (Area of Base)

Total Area $=360+108\sqrt{3}$

Total Area $=547$