Chapter 12.4, Problem 53SPR

To determine

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

Answer to Problem 53SPR

L = $126c{m}^2$

S = $175c{m}^2$

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 square whose length is 7 cm and the slant height is 9 cm.

Perimeter of the base = P = 4(length of each side) = 4(7) = 28 cm.

The Lateral area = $L=\frac{1}{2}Pl=\frac{1}{2}(28)(9)=126c{m}^2$

Area of the base = B = (length of each side )(length of each side)= $(7)(7)=49c{m}^2$

So, the surface area of the pyramid :

  $S=\frac{1}{2}Pl+B=126+49=175c{m}^2$

Hence ,

L = $126c{m}^2$

S = $175c{m}^2$