Question

Copy and complete the table below for the regular square pyramid.

$\begin{array}{cc}\text{Height},h& \\ \text{Slant\hspace{0.17em}\hspace{0.17em}Height},l& 12\\ \text{Base\hspace{0.17em}\hspace{0.17em}edge}& \\ \text{Lateral\hspace{0.17em}\hspace{0.17em}edge}& 15\end{array}$

Short answer

$\begin{array}{cc}\text{Height},h& 3\sqrt{7}\\ \text{Slant\hspace{0.17em}\hspace{0.17em}Height},l& 12\\ \text{Base\hspace{0.17em}\hspace{0.17em}edge}& 18\\ \text{Lateral\hspace{0.17em}\hspace{0.17em}edge}& 15\end{array}$

Step-by-step solution

Step 1. Given information.

The slant height and lateral edge of a regular square pyramid are 12 and 15 unit respectively.

Step 2. Write the concept.

The base edge is given by:

$\text{Baseedge}=2\sqrt{{\left(l\right)}^2-{\left(h\right)}^2}$

And the lateral edge is given by:

$\text{lateraledge}=\sqrt{{\left(l\right)}^2+{\left(\frac{\text{baseedge}}{2}\right)}^2}$

Step 3. Determine the values.

Substitute all the given values in the formula.

$\begin{array}{c}\text{Lateraledge}=\sqrt{{\left(l\right)}^2+{\left(\frac{\text{baseedge}}{2}\right)}^2}\\ \text{15}=\sqrt{{\left(12\right)}^2+{\left(\frac{\text{baseedge}}{2}\right)}^2}\text{[substitute]}\\ \text{225}=144+{\left(\frac{\text{baseedge}}{2}\right)}^2\text{[square]}\end{array}$

So,

$\begin{array}{c}{\left(\frac{\text{baseedge}}{2}\right)}^2=81\text{[subtract]}\\ \frac{\text{baseedge}}{2}=9\text{\hspace{0.05em}[sqaureroot]}\\ \text{baseedge}=18\text{[multiply]}\end{array}$

And

$\begin{array}{c}\text{Baseedge}=2\sqrt{{\left(l\right)}^2-{\left(h\right)}^2}\\ \text{18}=2\sqrt{{\left(12\right)}^2-{\left(h\right)}^2}\text{[substitute]}\\ 9=\sqrt{144-h^2}\text{[squareanddivide]}\end{array}$

So, the height is:

$\begin{array}{c}144-h^2=81\text{[square]}\\ h=3\sqrt{7}\text{[simplify]}\end{array}$