Chapter 12.3, Problem 12PPS

To determine

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

Answer to Problem 12PPS

L = $162m{m}^2$

S = $255.5m{m}^2$

Explanation of Solution

Given:

Hexagonal pyramid with a base edge of 6 mm and a slant height of 9 mm.

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 6 mm and slant height is 9 mm.

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,

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

Perimeter of the base = P = (length of each side )(number of sides ) = $6\cdot 6=36mm$

So, the Lateral area = $L=\frac{1}{2}Pl=\frac{1}{2}(36)(9)=162m{m}^2$

The area of one section of the hexagon , which is a triangle with base 6 mm and height

  $3\sqrt{3}$ mm is $\frac{1}{2}(base)(height)=\frac{1}{2}(6)(3\sqrt{3})=9\sqrt{3}m{m}^2$

So, the area of the base = B = $6\times 9\sqrt{3}=54\sqrt{3}m{m}^2$ ……[there are 6 equal triangles ]

So, the surface area of the pyramid :

  $S=\frac{1}{2}Pl+B=162+54\sqrt{3}\approx 255.5m{m}^2$

Hence ,

L = $162m{m}^2$

S = $255.5m{m}^2$