Chapter 12, Problem 1RP

a

To determine

To calculate : The lateral surface area and total area of the regular pyramid.

Answer to Problem 1RP

The lateral surface area and total area of the regular pyramid are $60{\text{ units}}^2$ and $96{\text{ units}}^2$

Explanation of Solution

Given information:

Side of the base square $\left(a\right)=6\text{ units}$

Slant height $\left(l\right)=5\text{ units}$

Formula used:

Lateral area $\left(LA\right)=4\times \left(\frac{1}{2}\times l\times a\right)$

Total area $\left(TA\right)=\left(\text{Lateral Area}\right)+\left(\text{Base Area}\right)=4\times \left(\frac{1}{2}\times l\times a\right)+\left(a^2\right)$

Calculation:

Lateral area of the regular pyramid

  $\begin{array}{c}\left(LA\right)=4\times \left(\frac{1}{2}\times l\times a\right)\\ =4\times \left(\frac{1}{2}\times 5\times 6\right)\\ =4\times \left(5\times 3\right)\\ =60{\left(\text{units}\right)}^2\end{array}$

Total area of the regular pyramid

  $\begin{array}{c}\left(TA\right)=4\times \left(\frac{1}{2}\times l\times a\right)+\left(a^2\right)\\ =60+{\left(6\right)}^2\\ =60+36\\ =96{\left(\text{units}\right)}^2\end{array}$

b

To determine

To calculate : The lateral surface area and total area of the cylinder.

Answer to Problem 1RP

The lateral surface area and total area of the cylinder are $56{\text{π units}}^2$ and $154{\text{π units}}^2$

Explanation of Solution

Given information:

Base radius of the cylinder $\left(r\right)=4\text{ units}$

Height of the cylinder $\left(h\right)=7\text{ units}$

Formula used:

Lateral area $\left(LA\right)=2\text{π}rh$

Total area $\left(TA\right)=\left(2\text{π}rh\right)+\left(\text{2π}{r}^2\right)$

Calculation:

Lateral area of cylinder

  $\begin{array}{c}\left(LA\right)=2\text{π}r\times h\\ =2\text{π}\times \text{4}\times 7\\ =56\text{π }{\left(\text{units}\right)}^2\end{array}$

Total area of cylinder

  $\begin{array}{c}\left(TA\right)=\left(2\text{π}r\times h\right)+\left(\text{2π}{r}^2\right)\\ =56\text{π}+\left(\text{2π}\times 7^2\right)\\ =56\text{π}+98\text{π}\\ =154\text{π }{\left(\text{units}\right)}^2\end{array}$