Chapter 12, Problem 38SGR

To determine

To calculate:To find the lateral and surface area of pyramid and round to the nearest tenth

Answer to Problem 38SGR

Lateral surface area of pyramid is $45c{m}^2$ and total surface area of pyramid is $51c{m}^2$

Explanation of Solution

Given information:Pyramid is

  

Formula Used:

Lateral and surface area of triangular pyramid is given by the below formula:

  

  $Lateralsurfaceare{a}_{Pyramid}=\frac{3}{2}\left(bs\right)$

  $Surfaceare{a}_{Pyramid}=\left(\frac{3}{2}bs\right)+\left(\frac{1}{2}ac\right)$

Calculation:

Pyramid is given as follows

  

Here, dimensions of pyramidas follows:

  $\begin{array}{l}s=6cm\\ a=4cm\\ b=5cm\\ c=3cm\end{array}$

Lateral surface of pyramid is calculated as follows:

  $Lateralsurfaceare{a}_{Pyramid}=\frac{3}{2}\left(bs\right)$

Substituting the values, volume is

  $\begin{array}{l}Lateralsurfaceare{a}_{Pyramid}=\frac{3}{2}\left(5\times 6\right)\\ Lateralsurfaceare{a}_{Pyramid}=\frac{3}{2}\times 30\\ Lateralsurfaceare{a}_{Pyramid}=45c{m}^2\end{array}$

Surface area of pyramidis calculated as follows:

  $Surfaceare{a}_{Pyramid}=\left(\frac{3}{2}bs\right)+\left(\frac{1}{2}ac\right)$

Substituting the values, volume is

  $\begin{array}{l}Surfaceare{a}_{Pyramid}=\frac{3}{2}\left(5\times 6\right)+\frac{1}{2}\left(3\times 4\right)\\ Surfaceare{a}_{Pyramid}=\left(\frac{3}{2}\times 30\right)+\left(\frac{1}{2}\times 12\right)\\ Surfaceare{a}_{Pyramid}=45+6\\ Surfaceare{a}_{Pyramid}=51c{m}^2\end{array}$

Conclusion:

Hence, Lateral surface area of pyramid is $45c{m}^2$ and total surface area of pyramid is $51c{m}^2$