Chapter 12.3, Problem 10PPS

To determine

Find the Lateral area and surface area of the regular pyramid.

Answer to Problem 10PPS

L = $150f{t}^2$

S = $211.9f{t}^2$

Explanation of Solution

Given:

  

Formula Used: The Lateral area L of a regular Pyramid is

  $L=\frac{1}{2}Pl$ , where l is the slant height and P is the perimeter of the base .

The Surface area S of a regular Pyramid is

  $S=\frac{1}{2}Pl+B$ , where l is the slant height, P is the perimeter of the base and B is the area of the base .

Calculation:

The pyramid has a base that is pentagon , the length of each side is 6 ft. and the slant height is 10 ft.

Perimeter of the base = P = 5(length of each side) = 5(6) = 30 ft.

The Lateral area = $L=\frac{1}{2}Pl=\frac{1}{2}(30)(10)=150f{t}^2$

The base is a pentagon .Divide the base into 5 equal triangles , find the area of a triangle and multiply it by 5 to find the area of the base :

  

Find the angle of the triangle = x = $\frac{360°}{10}=36°$

Now,

  $\begin{array}{l}\tan x=\frac{3}{a}\mathrm{..............}[\tan \theta =\frac{opposite}{adjacent}]\\ \Rightarrow \tan 36°=\frac{3}{a}\\ \Rightarrow a=\frac{3}{\tan 36°}\mathrm{.......}[multiply\text{ each side by }\frac{a}{\tan 36°}]\\ \Rightarrow a\approx 4.13ft\end{array}$

So, the area of the triangle that has base 6 ft. and height 4.13 ft. = $\frac{1}{2}(base)(height)=\frac{1}{2}(6)(4.13)=12.39f{t}^2$

Area of the base = B =5(area of each triangle)= $5(12.39)=61.95f{t}^2$

So, the surface area of the pyramid :

  $S=\frac{1}{2}Pl+B\approx 150+61.9\approx 211.9f{t}^2$

Hence ,

L = $150f{t}^2$

S = $211.9f{t}^2$