Chapter 12.1, Problem 6PSB

a.

To determine

To calculate: The total area of cardboard pieces needed to construct open rectangular box.

Answer to Problem 6PSB

The total area of cardboard pieces needed to construct each open rectangular box is

  $236$ .

Explanation of Solution

Given information:

A rectangular box with dimensions as 6, 8 and 7.

Formula used:

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

l = length of rectangle

w= width of rectangle

Calculation:

  

Area of sides is

  $\begin{array}{l}A({▭}_1)=6\times 8=48\\ A({▭}_2)=6\times 7=42\\ A({▭}_3)=8\times 7=56\end{array}$

Total area:

  $\begin{array}{l}A(▭)=2A({▭}_1)+2A({▭}_2)+A({▭}_3)\\ A(▭)=\left(2\times 48\right)+\left(2\times 42\right)+\left(56\right)\\ A(▭)=96+84+56\\ A(▭)=236\end{array}$

b.

To determine

To find: The total area of cardboard pieces needed to construct open triangular box with dimensions as 6 and 8.

Answer to Problem 6PSB

The total area of cardboard pieces needed to construct open triangular box.is $258.31$ .

Explanation of Solution

Given information:

A triangular box comprising of rectangle with dimensions 6, 8 and ${30}^{\circ }{60}^{\circ }{90}^{\circ }$ triangle.

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

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

b = base of triangle

h = height of triangle Area of rectangle: $A=l\times w$

l = length of rectangle

w= width of rectangle

Calculation:

  

  $A_{OneSide}=6\times 8=48$

The base sides of triangle can be calculated from ${30}^{\circ }{60}^{\circ }{90}^{\circ }$ triangle.

Side opposite ${30}^{\circ }$ = 6,

  $\begin{array}{l}\tan {60}^{\circ }=\sqrt{3}\\ \frac{sideoppositeto60^o}{sideadjacentto60^o}=\sqrt{3}\\ \frac{sideoppositeto60^o}{6}=\sqrt{3}\end{array}$

Side opposite to ${60}^{\circ }$ = $6\sqrt{3}$ and

Hypotenuse can be calculated by applying Pythagoras Theorem,

In right angled triangle, we get

  $\begin{array}{l}{\left(Hypotenuse\right)}^2={\left(6\right)}^2+{\left(6\sqrt{3}\right)}^2\\ {\left(Hypotenuse\right)}^2=36+108\\ {\left(Hypotenuse\right)}^2=144\\ Hypotenuse=\sqrt{144}\\ Hypotenuse=12\end{array}$

  $\begin{array}{l}{A}_{△}=\frac{1}{2}\times 6\times 6\sqrt{3}\\ A_{△}=18\sqrt{3}\end{array}$

  $\begin{array}{l}{A}_{Rectangle1}=12\times 8=96\\ A_{Rectangle2}=\left(6\sqrt{3}\right)\times 8=48\sqrt{3}\end{array}$

The total area:

  $\begin{array}{l}{A}_{Total}=A_{OneSide}+A_{△}+A_{Rectangle1}+A_{Rectangle2}\\ A_{Total}=48+18\sqrt{3}+96+48\sqrt{3}\\ A_{Total}=144+66\sqrt{3}\\ A_{Total}=258.31\end{array}$