Question

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

$\begin{array}{cc}\text{Height},h& 24\\ \text{Slant\hspace{0.17em}\hspace{0.17em}Height},l& \\ \text{Base\hspace{0.17em}\hspace{0.17em}edge}& 14\\ \text{Lateral\hspace{0.17em}\hspace{0.17em}edge}& \end{array}$

Short answer

$\begin{array}{cc}\text{Height},h& 24\\ \text{Slant\hspace{0.17em}\hspace{0.17em}Height},l& 25\\ \text{Base\hspace{0.17em}\hspace{0.17em}edge}& 14\\ \text{Lateral\hspace{0.17em}\hspace{0.17em}edge}& \sqrt{674}\end{array}$

Step-by-step solution

Step 1. Given information.

The height and base edge of a regular square pyramid are 24 and 14 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.

$\begin{array}{c}\text{Baseedge}=2\sqrt{{\left(l\right)}^2-{\left(h\right)}^2}\\ \text{14}=2\sqrt{{\left(l\right)}^2-{\left(24\right)}^2}\text{[substitute]}\\ \text{7}=\sqrt{l^2-576}\text{[squareanddivide]}\end{array}$

So,

$\begin{array}{c}{l}^2-576=49\text{[square]}\\ l=25\text{[simplify]}\end{array}$

And

$\begin{array}{c}\text{Lateraledge}=\sqrt{{\left(25\right)}^2+{\left(7\right)}^2}\text{[substitute]}\\ =\sqrt{674}\text{[simplify]}\end{array}$