Chapter 12.5, Problem 15PPS

(a)

To determine

To find: the surface area of the cube.

Answer to Problem 15PPS

The surface area of the cube is $150\text{in}{\text{.}}^2$ .

Explanation of Solution

Given:

  

Formula used:

The surface area S of a square pyramid is S = L + 2B

Calculation:

The objective is to find the surface area of the cube.

The lateral area L of a square pyramid is given by the formula

L = Ph

Where P is the perimeter of the base and h is the height the square pyramid.

The surface area S of a square pyramid is given by the formula

S = L + 2B

Where L is the lateral surface area and B is the area of the base.

On combining the last two formulas, the surface area is given by

  $\begin{array}{c}S=L+2B\\ =Ph+2B\\ =4\left(side\right)side+2{\left(side\right)}^2\\ =4{\left(side\right)}^2+2{\left(side\right)}^2\\ =6{\left(side\right)}^2\end{array}$

Find the surface area of the square box

Put side = 5 in the surface area formula to find the surface area of the cube.

  $\begin{array}{c}S=6{\left(side\right)}^2\\ =6{\left(5\right)}^2\\ =6\cdot 25\\ =150\end{array}$

Therefore, the surface area of the cube is $150\text{in}{\text{.}}^2$ .

Conclusion:

Therefore, the surface area of the cube is $150\text{in}{\text{.}}^2$ .

(b)

To determine

To complete: the given table.

Answer to Problem 15PPS

Scale Factor of Dilation	Original	$\times 2$	$\times 3$
Length of Side	5	$5\times 2=10$	$5\times 3=15$
Surface Area	$150\text{in}{\text{.}}^2$	$600\text{in}{\text{.}}^2$	$1350\text{in}{\text{.}}^2$

Explanation of Solution

Given:

Scale Factor of Dilation	Original	$\times 2$	$\times 3$
Length of Side	■	■	■
Surface Area	■	■	■

Calculation:

Consider the table

Scale Factor of Dilation	Original	$\times 2$	$\times 3$
Length of Side	5	$5\times 2=10$	$5\times 3=15$
Surface Area

The objective is to complete the table.

Put side = 10 in the volume formula to find the surface area of the cube.

  $\begin{array}{c}S=6{\left(side\right)}^2\\ =6{\left(10\right)}^2\\ =6\cdot 100\\ =600\end{array}$

Therefore, the surface area of the cube is $600\text{in}{\text{.}}^2$ .

Put side = 15 in the volume formula to find the surface area of the cube.

  $\begin{array}{c}S=6{\left(side\right)}^2\\ =6{\left(15\right)}^2\\ =6\cdot 225\\ =1350\end{array}$

Therefore, the surface area of the cube is $1350\text{in}{\text{.}}^2$ .

Therefore, the complete table is

Scale Factor of Dilation	Original	$\times 2$	$\times 3$
Length of Side	5	$5\times 2=10$	$5\times 3=15$
Surface Area	$150\text{in}{\text{.}}^2$	$600\text{in}{\text{.}}^2$	$1350\text{in}{\text{.}}^2$

Conclusion:

Therefore, the given table is completed.

(c)

To determine

To compare: the original surface area to the surface area after dilation.

Answer to Problem 15PPS

The new surface area is 4 times the original surface area when the scale factor is 2

The new surface area is 9 times the original surface area when the scale factor is 3

Explanation of Solution

The objective is to compare the original surface area to the surface are after dilation.

When the side of the cube is dilation by a scale factor 2 the new surface area is 4 times the original surface area.

When the side of the cube is dilation by a scale factor 3 the new surface area is 9 times the original surface area.

(d)

To determine

To explain:whether the same relationship exists with rectangular prism or not

Answer to Problem 15PPS

The same relation exist with the surface are of a rectangular prism

Explanation of Solution

Calculation:

The same relation exist with the surface are of a rectangular prism too because a cube is form of rectangle prism whose sides are equal.

The new surface area is the square of the scale factor of the side multiplied to the original surface area.

The original formula for surface area is

  $\begin{array}{c}S=L+2B\\ =Ph+2B\end{array}$

When length and width are double the new surface area becomes

  $\begin{array}{c}{S}_2=L+2B\\ =2\left(2l+2w\right)2h+2\left(2l\cdot 2w\right)\\ =2\cdot 2\left(l+w\right)\cdot 2h+2\cdot 2\cdot 2\left(lw\right)\\ =4\left(l+w\right)2h+2\cdot 4\left(lw\right)\\ =4\cdot 2\left(l+w\right)h+4\cdot 2\left(lw\right)\\ =4\left\{2\left(l+w\right)h+2\left(lw\right)\right\}\\ =4\left(Ph+2B\right)\\ =4S\end{array}$

Conclusion:

Therefore, the same relation exist with the surface are of a rectangular prism