Chapter 12.3, Problem 9PPS

To determine

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

Answer to Problem 9PPS

L = $178.2c{m}^2$

S = $302.9c{m}^2$

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 triangle , the height of the pyramid is 5 cm and the slant height of the pyramid is 7 cm.

The base can be divided into 3 equal triangles.

  

Find the apothem of the base using the Pythagoras theorem :

a = $\sqrt{7^2-5^2}=\sqrt{49-25}=\sqrt{24}=2\sqrt{6}cm$

The base of the pyramid :

  

Find the angle x = $\frac{360°}{6}=60°$

Now, find the side b of the triangle :

  $\begin{array}{l}\tan x=\frac{b}{a}\mathrm{...............}[\tan \theta =\frac{opposite}{adjacent}]\\ \Rightarrow \tan 60°=\frac{b}{2\sqrt{6}}\\ \Rightarrow b=2\sqrt{6}\cdot \tan 60°\mathrm{.....}[multiply\text{ each side by }2\sqrt{6}]\\ \Rightarrow b=2\sqrt{6}\cdot \sqrt{3}\\ \Rightarrow b=6\sqrt{2}\end{array}$

So, the length of the each side of the base = $2\cdot 6\sqrt{2}=12\sqrt{2}$

Perimeter of the base = P = 4(length of each side) = $3\cdot 12\sqrt{2}=36\sqrt{2}$ cm

The Lateral area = $L=\frac{1}{2}Pl=\frac{1}{2}(36\sqrt{2})(7)=126\sqrt{2}\approx 178.2c{m}^2$

Area of the base = B = $3\cdot \frac{1}{2}(12\sqrt{2})(2\sqrt{6})=124.7c{m}^2$

So, the surface area of the pyramid :

  $S=\frac{1}{2}Pl+B=178.2+124.7=302.9c{m}^2$

Hence ,

L = $178.2c{m}^2$

S = $302.9c{m}^2$