Q4. - Geometry

Question

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

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

Step-by-Step Solution

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 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:

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

 

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.

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

 

So,

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

 

And

$\begin{array}{c}\text{Base edge}=2\sqrt{{\left(l\right)}^2-{\left(h\right)}^2}\\ \text{18}=2\sqrt{{\left(12\right)}^2-{\left(h\right)}^2}\text{ [substitute]}\\ 9=\sqrt{144-h^2}\text{ [square and divide]}\end{array}$

 

So, the height is:

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