Chapter 12.2, Problem 30WE

To determine

To find: The volume of the pyramid in terms of $y$ .

Answer to Problem 30WE

The volume of the pyramid in terms of $y$ is equal to $\frac{3}{2}{y}^3{\text{cm}}^3$ .

Explanation of Solution

Given information:

The base of a pyramid is a regular hexagon with sides $y\text{cm}$ long and the lateral edges are $2y\text{cm}$ long.

Calculation:

Consider the hexagon.

  

Figure (1)

The formula for the volume of pyramid is one third of the product of area of hexagon and height of hexagon.

  $\text{V}=\frac{1}{3}\left(A\right)\left(h\right)$

As the triangle $AOB$ is an equilateral triangle. So, all the sides of triangle are equal.

So, the length of the side $AO$ is equal to $y$ .

Extend the line $AO$ to the opposite vertex to $A$ and name it $C$ .

Use the Pythagoras theorem in $\Delta ABC$ for the height $h$ of pyramid.

  $\begin{array}{c}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ {\left(BC\right)}^2={\left(2y\right)}^2-y^2\\ BC=\sqrt{4y^2-y^2}\\ BC=\sqrt{3}y\end{array}$

So, the height of the hexagon is equal to $\sqrt{3}y$ .

Simplify the area of hexagon with side equal to $y\text{cm}$ .

  $\begin{array}{c}A=\frac{3\sqrt{3}}{2}{a}^2\\ =\frac{3\sqrt{3}}{2}{y}^2\end{array}$

Substitute $\frac{3\sqrt{3}}{2}{y}^2$ for $A$ and $\sqrt{3}y$ for $h$ in the formula for volume of hexagon.

  $\begin{array}{c}\text{V}=\frac{1}{3}\left(A\right)\left(h\right)\\ =\frac{1}{3}\left(\frac{3\sqrt{3}}{2}{y}^2\right)\left(\sqrt{3}y\right)\\ =\frac{3}{2}{y}^3\end{array}$

Therefore, the volume of the pyramid in terms of $y$ is equal to $\frac{3}{2}{y}^3{\text{cm}}^3$ .