Chapter 12.3, Problem 23PPS

To determine

To find:

The slant height of the pyramid.

Answer to Problem 23PPS

The slant height is $5\text{ mm}$ .

Explanation of Solution

Given information:

The surface area of a square pyramid is $24$ square millimeters and the base area is $4$ square millimeters.

Calculation:

The surface area $S$ of a regular pyramid is $S=\frac{1}{2}Pl+B$ .

Substitute $S=24$ and $B=4$ in the above formulae.

  $\begin{array}{c}24=\frac{1}{2}Pl+4\\ 24-4=\frac{1}{2}Pl\\ \frac{1}{2}Pl=20\\ Pl=\mathrm{40.............}\left(I\right)\end{array}$

Since the pyramid is a square with side $a$ .

Area of square is $a^2$ .

  $\begin{array}{c}{a}^2=4\\ a=2\end{array}$

So, the side is $2$ millimeters.

Perimeter of square is $4a$ .

So, $P=4a$ .

Substitute $P=4a$ in the equation $\left(I\right)$ to find the slant height.

  $\begin{array}{c}4al=40\\ al=10\\ 2l=10\\ l=5\end{array}$

Hence, the slant height is $5\text{ mm}$ .