Chapter 12.8, Problem 20PPS

To determine

To find the surface area of the smaller solid.

Answer to Problem 20PPS

  $946.3$ square centimeters

Explanation of Solution

Given:

Two similar solids.

  

Concept used:

The ratio of the surface areas of the larger and smaller solids equals the square of the ratio of the lengths of any of their corresponding edges.

Formula used:

Lateral surface area of a prism $=Ph$ , where $P=$ perimeter of the prism.

Lateral surface area of a pyramid $=\frac{1}{2}Pl$ , where $l=$ slant height of the pyramid

Calculation:

For the given two solids, the lengths of the corresponding edges of the base are $14$ cm and $12$ cm.

First find the surface area of the larger solid.

The larger solid is a composite solid composed of a square-based prism with sides of $14$ cm and a height of $12$ cm and a square-based pyramid with sides of $14$ cm and a slant height of $15$ cm .

Surface area of composite $=$ lateral surface area of the prism $+$ area of its bottom base $+$ lateral surface area of the pyramid.

  $S{A}_{\text{large solid}}=Ph+B+\frac{1}{2}Pl$

  $\begin{array}{l}=\left(4\times 14\right)\left(12\right)+{\left(14\right)}^2+\frac{1}{2}\left(4\times 14\right)\left(15\right)\\ =56\times 12+196+28\times 15\\ =672+196+420\\ =1288\end{array}$

So, the surface area of the large solid is $1288$ square centimeters.

Now, use the equal ratios to find the surface area of the smaller solid.

  $\Rightarrow \frac{S{A}_{\text{large solid}}}{S{A}_{\text{small solid}}}={\left(\frac{14}{12}\right)}^2$

  $\Rightarrow \frac{S{A}_{\text{large solid}}}{S{A}_{\text{small solid}}}={\left(\frac{7}{6}\right)}^2$

  $\Rightarrow \frac{1288}{S{A}_{\text{small solid}}}=\frac{49}{36}$

  $\Rightarrow S{A}_{\text{small solid}}=\frac{1288\times 36}{49}$

  $\Rightarrow S{A}_{\text{small solid}}=\frac{6624}{7}$

  $\Rightarrow S{A}_{\text{small solid}}\approx 946.3$

Conclusion:

Therefore, the surface area of the smaller solid is about $946.3$ square centimeters.