Chapter 12.7, Problem 19PPS

To determine

To find: the lateral and total surface area.

Answer to Problem 19PPS

The lateral area is ${\text{150m}}^2$

Total surface area is $\text{193}{\text{.5m}}^2$

Explanation of Solution

Given:

Consider the figure shown as below:

  

Formulas used:

The lateral area $L$ of a triangular pyramid is $L=\frac{1}{2}Pl$

The surface area $S$ of a triangular pyramid is $S=L+B$

Calculation:

The figure is a triangular pyramid.

The objective is to find the lateral and surface area of the triangular pyramid.

Find the lateral area of the triangular pyramid as follows:

Each side of the triangle is $10m$ .

The perimeter $P$ of the equilateral triangle is $P=3(side)$

Substitute $side=10$ in $P=3(side)$

Thus,

  $\begin{array}{l}P=3.10\\ \text{ =30}\end{array}$

The slant height of the triangular pyramid is $10m$ .

Substitute $\text{P=30}$ and $\text{l=10}$ in $\text{L=}\frac{1}{2}\text{Pl}$

  $\begin{array}{l}\text{L=}\frac{1}{2}\text{Pl}\\ \text{ =}\frac{1}{2}\mathrm{.30.10}\\ \text{ =15}\text{.10}\\ \text{ =150}\end{array}$

Hence, the lateral area of the triangular pyramid is ${\text{150m}}^2$ .

Find the surface area of the triangular pyramid as follows:

The area $B$ of the base is $B=\frac{1}{2}\times b\times a.$

Here, $b$ is the base and $a$ is the height of the triangle.

Substitute $b=10,a=8.7$ in $B=\frac{1}{2}\times b\times a.$

  $\begin{array}{l}B=\frac{1}{2}.10(8.7)\\ \text{ =5(8}\text{.7)}\\ \text{ =43}\text{.5}\end{array}$

Substitute $\text{B=43}\text{.5}$ and $\text{L=150}$ in $\text{S=L+B}$

  $\begin{array}{l}\text{S=L+B}\\ \text{ =150+43}\text{.5}\\ \text{ =193}\text{.5}\end{array}$

Hence, the surface area of the triangular pyramid is $\text{193}{\text{.5m}}^2$

Conclusion:

Therefore, the lateral area is ${\text{150m}}^2$ and the total surface area is $\text{193}{\text{.5m}}^2$