Chapter 6.5, Problem 39WE

a.

To determine

To show that volume of pyramid is $\frac{x^3\sqrt{2}}{6}$ when each edge of pyramid is of length $x$

Explanation of Solution

Given information:

Length of each of the 8 edges of pyramid is $x$

  

We know that the volume of the pyramid is given by,

  $V=\frac{1}{3}\times (\text{base area)}\times \text{(height)}$

Now we determine the area of base.

It can be observed that the base of pyramid is a square with side $x$

Hence the area of the base becomes $x^2$

Now we determine the height of the pyramid.

From the given figure we can observe that EHC form a right angle triangle.

Side EH is height (name the base point of $h$ as H)

Side HC is the base of the right angle triangle EHC of the length half that of diagonal of square ABCD

Diagonal of the square is $\sqrt{2}\times \text{side}=x\sqrt{2}$

Hence side EH is $\frac{x\sqrt{2}}{2}=\frac{x}{\sqrt{2}}$

Side EC is the hypotenuses of the triangle EHC of length $x$

Now using Pythagoras theorem in triangle EHC we get,

  $\begin{array}{l}{x}^2={\left(\frac{x}{\sqrt{2}}\right)}^2+{\left(\text{EH}\right)}^2\\ {\left(\text{EH}\right)}^2=x^2-\frac{x^2}{2}\\ {\left(\text{EH}\right)}^2=\frac{x^2}{2}\\ \text{EH}=\frac{x}{\sqrt{2}}\end{array}$

Hence the height is $\frac{x}{\sqrt{2}}$

Now we put the value of base area and the height in the formula we get,

  $\begin{array}{l}V=\frac{1}{3}\times (\text{base area)}\times \text{(height)}\\ \text{V=}\frac{1}{3}\times x^2\times \frac{x}{\sqrt{2}}\\ V=\frac{x^3}{3\sqrt{2}}\end{array}$

Multiply $\sqrt{2}$ with numerator and denominator we get volume as

  $V=\frac{x^3\sqrt{2}}{6}$

b.

To determine

To find the value of $x$

Explanation of Solution

Given information:

Volume is given 9

Now we equate the value with the formula

  $\begin{array}{c}V=\frac{x^3\sqrt{2}}{6}\\ 9=\frac{x^3\sqrt{2}}{6}\\ x^3\sqrt{2}=54\\ x^3=\frac{54}{\sqrt{2}}\end{array}$

Take cube root both sides we get,

  $x=\sqrt[3]{\frac{54}{\sqrt{2}}}$