Chapter 12.8, Problem 31STP

To determine

The value of the surface area of the second prism is to be estimated.

Answer to Problem 31STP

The surface area of the second rectangular prism is $13.75f{t}^2$

Explanation of Solution

Given:

  $A_1=220f{t}^2$

Concept Used:

The value of the surface area of the rectangular prism is given by the following expression:

  $A=2\left(wl+hl+hw\right)$

Where,

Width of rectangular prism: $w$

Height of rectangular prism: $h$

Length of rectangular prism: $l$

Calculation:

Consider, a prism is rectangular has the dimensions:

Surface area of first rectangular prism: $A_1$

Width of first rectangular prism: $w$

Height of first rectangular prism: $h$

Length of first rectangular prism: $l$

Surface area of first rectangular prism: $A_2$

Surface area of first rectangular prism is calculated as:

  $\begin{array}{l}{A}_1=2\left(wl+hl+hw\right)\\ 220=2\left(wl+hl+hw\right)\\ 110=\left(wl+hl+hw\right)\mathrm{...}(1)\end{array}$

The second rectangular prism has dimensions as $\frac{1}{4}$ of first rectangular prism. Then the surface area of the second rectangular prism is calculated as:

  $\begin{array}{l}{A}_2=2\left(\left(\frac{1}{4}w\times \frac{1}{4}l\right)+\left(\frac{1}{4}h\times \frac{1}{4}l\right)+\left(\frac{1}{4}h\times \frac{1}{4}w\right)\right)\\ A_2=2\times \frac{1}{16}\left(wl+hl+hw\right)\\ A_2=\frac{1}{8}\left(wl+hl+hw\right)\end{array}$

But $\left(wl+hl+hw\right)=110$ from eq. (1):

So,

  $\begin{array}{l}{A}_2=\frac{1}{8}\left(wl+hl+hw\right)\\ A_2=\frac{1}{8}\times 110\\ A_2=13.75f{t}^2\end{array}$

Conclusion:

Hence, surface area of the second rectangular prism is calculated as $A_2=13.75f{t}^2$