Chapter 12.3, Problem 2ACYP

To determine

To find:

The surface area of square pyramid.

Answer to Problem 2ACYP

The surface area of the pyramid is $179{\text{ cm}}^2$ .

Explanation of Solution

Given information:

Square pyramid with a height of $9\text{ in}\text{.}$ and a lateral height of $11\text{ in}\text{.}$ .

Calculation:

The surface area $S$ of a regular pyramid is $S=L+B$ .

  $L$ is denoted as the lateral area.

  $L=\frac{1}{2}Pl$ , where $l$ is the slant height, $P$ is the perimeter of the base and $B$ is the area of the base.

Here, the height of $9\text{ in}\text{.}$ and the slant height is $11\text{ in}\text{.}$ .

First find the base of square using Pythagoras theorem.

  $c^2=a^2+b^2$

Substitute $c=11$ and $a=9$ in the above formula.

  $\begin{array}{c}{11}^2=9^2+b^2\\ {11}^2-9^2=b^2\\ b=\sqrt{{11}^2-9^2}\\ =\sqrt{121-81}\\ =\sqrt{40}\\ =6.32\end{array}$

  $\begin{array}{c}L=\frac{1}{2}Pl\\ =\frac{1}{2}\left[4\left(6.32\right)\right]\left(11\right)\\ =2\cdot 6.32\cdot 11\\ =139.04\end{array}$

The surface area $S$ of a regular pyramid is $S=L+B$ .

  $\begin{array}{c}S=139.04+{\left(6.32\right)}^2\\ =139.04+39.94\\ =178.98\\ \approx 179\end{array}$

Hence, the surface area of the pyramid is $179{\text{ cm}}^2$ .