Chapter 12.4, Problem 52SPR

To determine

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

Answer to Problem 52SPR

L = $212.1f{t}^2$

S = $255.4f{t}^2$

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 equilateral triangle with each side 10 ft.

The base of the pyramid :

  

Find h by Pythagoras theorem:

h = $\sqrt{{10}^2-5^2}=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}$

The side of the pyramid is :

  

Find l by Pythagoras theorem:

l = $\sqrt{{15}^2-5^2}=\sqrt{225-25}=\sqrt{200}=10\sqrt{2}$

Perimeter of the base = P = 3(length of each side) = $3\cdot 10=30$ cm

The Lateral area = $L=\frac{1}{2}Pl=\frac{1}{2}(30)(10\sqrt{2})=150\sqrt{2}\approx 212.1f{t}^2$

Area of the base = B = $=\frac{1}{2}bh=\frac{1}{2}(10)(5\sqrt{3})=25\sqrt{3}f{t}^2$

So, the surface area of the pyramid :

  $S=\frac{1}{2}Pl+B=150\sqrt{2}+25\sqrt{3}\approx 255.4f{t}^2$

Hence ,

L = $212.1f{t}^2$

S = $255.4f{t}^2$