Chapter 12.4, Problem 54SPR

To determine

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

Answer to Problem 54SPR

L = $472.5i{n}^2$

S = $758.9i{n}^2$

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 base of the pyramid is regular hexagon whose side is 10.5 in. and slant height is 15 in.

Now, the base is a regular hexagon .Divide the base in 6 equal triangles with height of each triangle 6m.

  

Draw the right angled triangle out and find the base:

  

Now, first find the value of x = $\frac{360°}{12}=30°$

Also,

  $\begin{array}{l}\tan x=\frac{5.25}{a}\mathrm{....................}[\tan \theta =\frac{length\text{ of side opposite to }\theta }{\text{length of side adjacent to }\theta }]\\ \Rightarrow \tan 30°=\frac{5.25}{a}\\ \Rightarrow a=\frac{5.25}{\tan 30°}\mathrm{...........}[multiply\text{ each side by }\frac{a}{\tan 30°}]\\ \Rightarrow a=5.25\sqrt{3}\approx \mathrm{9.093.............}[\tan 30°=\frac{1}{\sqrt{3}}]\end{array}$

Perimeter of the base = P = (length of each side )(number of sides ) = $6(10.5)=63$

So, the Lateral area = $L=\frac{1}{2}Pl=\frac{1}{2}(63)(15)=472.5i{n}^2$

The area of one section of the hexagon , which is a triangle with base 10.5 in. and height

9.093 in. is $\frac{1}{2}(base)(height)=\frac{1}{2}(10.5)(9.093)\approx 47.74i{n}^2$

So, the area of the base = B = $6\times 47.74\approx 286.44i{n}^2$ ......[there are 6 equal triangles ]

So, the surface area of the pyramid :

  $S=\frac{1}{2}Pl+B=472.5+286.44\approx 758.9i{n}^2$

Hence ,

L = $472.5i{n}^2$

S = $758.9i{n}^2$