Chapter 12.3, Problem 11PPS

To determine

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

Answer to Problem 11PPS

L = $966i{n}^2$

S = $1686i{n}^2$

Explanation of Solution

Given:

Square pyramid with an altitude of 12 inches and a slant height of 18 inches.

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 square and slant height 18 inches , altitude 12 inches.

Let x be the measure from the centre of the base to the edge.

Find x by Pythagoras theorem where slant height is the hypotenuse and altitude is a leg :

x = $\sqrt{{18}^2-12}=\sqrt{324-144}=\sqrt{180}$

Side of the base is = s = $2\sqrt{180}$

Perimeter of the base = P = 4(length of each side) = 4(2 $\sqrt{180}$ ) = $8\sqrt{180}in.$

The Lateral area = $L=\frac{1}{2}Pl=\frac{1}{2}(8\sqrt{180})(18)=72\sqrt{180}\approx 966i{n}^2$

Area of the base = B = (length of each side )(length of each side)= $(2\sqrt{180})(2\sqrt{180})=720i{n}^2$

So, the surface area of the pyramid :

  $S=\frac{1}{2}Pl+B\approx 966+720=1686i{n}^2$

Hence ,

L = $966i{n}^2$

S = $1686i{n}^2$