Chapter 12.3, Problem 3CYU

To determine

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

Answer to Problem 3CYU

L = $207.8m^2$

S = $332.5m^2$

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.As shown in the figure, find the base of the right angled triangle whose one leg is 8 m and hypotenuse is 10 m.

  

Using Pythagoras Theorem :

So, the length of the base b = $\sqrt{{10}^2-8^2}=\sqrt{100-64}=6m$

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,

  $\begin{array}{l}\tan x=\frac{a}{6}\mathrm{....................}[\tan \theta =\frac{length\text{ of side opposite to }\theta }{\text{length of side adjacent to }\theta }]\\ \Rightarrow \tan 30°=\frac{a}{6}\\ \Rightarrow a=6\tan 30°\mathrm{...........}[multiply\text{ each side by 6}]\\ \Rightarrow a=6\cdot \frac{1}{\sqrt{3}}\mathrm{.............}[\tan 30°=\frac{1}{\sqrt{3}}]\\ \Rightarrow a=\frac{6}{\sqrt{3}}\end{array}$

Now, the length of base of each triangle =

  $2\cdot \frac{6}{\sqrt{3}}=\frac{12}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=4\sqrt{3}m$

Hence , each side of the hexagon is $4\sqrt{3}m$

Perimeter of the base = P = (length of each side )(number of sides ) = $6\cdot 4\sqrt{3}=24\sqrt{3}m$

So, the Lateral area = $L=\frac{1}{2}Pl=\frac{1}{2}(24\sqrt{3})(10)=207.8m^2$

The area of one section of the hexagon , which is a triangle with base $4\sqrt{3}m$ and height

6 m is $\frac{1}{2}(base)(height)=\frac{1}{2}(4\sqrt{3})(6)=12\sqrt{3}c{m}^2$

So, the area of the base = B = $6\times 12\sqrt{3}=124.71c{m}^2$ ....[there are 6 equal triangles ]

So, the surface area of the pyramid :

  $S=\frac{1}{2}Pl+B=207.8+124.71=332.5m^2$

Hence ,

L = $207.8m^2$

S = $332.5m^2$