Q6. - Geometry

Question

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

$\begin{array}{cc}\text{Height},h& 6\\ \text{Slant\hspace{0.17em}\hspace{0.17em}Height},l& \\ \text{Base\hspace{0.17em}\hspace{0.17em}edge}& \\ \text{Lateral\hspace{0.17em}\hspace{0.17em}edge}& 10\end{array}$

Step-by-Step Solution

Answer

$\begin{array}{cc}\text{Height},h& 6\\ \text{Slant\hspace{0.17em}\hspace{0.17em}Height},l& 2\sqrt{17}\\ \text{Base\hspace{0.17em}\hspace{0.17em}edge}& 8\sqrt{2}\\ \text{Lateral\hspace{0.17em}\hspace{0.17em}edge}& 10\end{array}$

Step 1. Given information.

The height and lateral edge of a regular square pyramid are 6 and 10 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.

$\begin{array}{c}\text{Lateral edge}=\sqrt{{\left(l\right)}^2+{\left(\frac{\text{base edge}}{2}\right)}^2}\\ \text{10}=\sqrt{{\left(l\right)}^2+{\left(\sqrt{{\left(l\right)}^2-{\left(6\right)}^2}\right)}^2}\text{ [substitute]}\\ 100=l^2+l^2-36\text{ [square]}\end{array}$

 

So, the slant height is:

$\begin{array}{c}2l^2=136\text{ [add]}\\ l^2=68\text{ \hspace{0.05em} [divide]}\\ l=2\sqrt{17}\text{ [simplify]}\end{array}$

 

And

$\begin{array}{c}\text{Base edge}=2\sqrt{{\left(l\right)}^2-{\left(h\right)}^2}\\ =2\sqrt{{\left(2\sqrt{17}\right)}^2-{\left(6\right)}^2}\text{ [substitute]}\\ =2\sqrt{68-36}\text{ [square]}\end{array}$

 

Then,

$\begin{array}{c}\text{Base edge}=2\sqrt{32}\text{ [subtract]}\\ =8\sqrt{2}\text{ [simplify]}\end{array}$