Q1. - Geometry

Question

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

$\begin{array}{cc}\text{Height},h& 4\\ \text{Slant\hspace{0.17em}\hspace{0.17em}Height},l& 5\\ \text{Base\hspace{0.17em}\hspace{0.17em}edge}& \\ \text{Lateral\hspace{0.17em}\hspace{0.17em}edge}& \end{array}$

Step-by-Step Solution

Answer

$\begin{array}{cc}\text{Height},h& 4\\ \text{Slant\hspace{0.17em}\hspace{0.17em}Height},l& 5\\ \text{Base\hspace{0.17em}\hspace{0.17em}edge}& 6\\ \text{Lateral\hspace{0.17em}\hspace{0.17em}edge}& \sqrt{34}\end{array}$

Step 1. Given information.

The height and slant height of a regular square pyramid are 4 and 5 unit respectively.

Step 2. Write the concept.

The base edge is given by:

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

 

And the lateral edge is given by:

$\text{lateral edge}=\sqrt{{\left(l\right)}^2+{\left(\frac{\text{base edge}}{2}\right)}^2}$

Step 3. Determine the values.

Substitute all the given values in the formula.

$\begin{array}{c}\text{Base edge}=2\sqrt{{\left(5\right)}^2-{\left(4\right)}^2}\text{ [substitute]}\\ =2\sqrt{25-16}\text{ [square]}\\ =2\sqrt{9}\text{ [subtract]}\\ =\text{6 [simplify]}\end{array}$

 

And

$\begin{array}{c}\text{Lateral edge}=\sqrt{{\left(5\right)}^2+{\left(3\right)}^2}\text{ [substitute]}\\ =\sqrt{34}\end{array}$