Chapter 1.8, Problem 12WE

To determine

To calculate:

Volume of the pyramid for given condition.

Answer to Problem 12WE

Volume of the pyramid for given condition is $\frac{1}{3}\left(x^3-3x^2+2x\right)$

Explanation of Solution

Given information:

Condition: length and width of the base and height are three consecutive odd integers and $x$ is the largest ineger.

  

Calculations:

Let the largest integer be $x$ .

If the length and width of base and height of a rectangular base pyramid are three consecutive integer.

Then the dimensions of rectangular base pyramid are $x,\left(x-1\right)$ and $\left(x-2\right)$ respectively.

The area of the rectangular base is ,

  $B=x\left(x-1\right)\mathrm{...}\left(1\right)$

The volume of pyramid is,

  $\begin{array}{l}V=\frac{1}{3}Bh\\ V=\frac{1}{3}x\left(x-1\right)\left(x-2\right)\end{array}$

From eq. $\left(1\right)$ and above hypothesis.

  $V=\frac{1}{3}x\left(x-1\right)\left(x-2\right)$

By distributive property,

  $V=\frac{1}{3}\left(x^2\left(x-2\right)-x\left(x-2\right)\right)$

  $V=\frac{1}{3}\left(x^2-2x^2-\left(x^2-2x\right)\right)$

By distributive property,

  $V=\frac{1}{3}\left(x^3-3x^2+2x\right)$

Combine like terms

Therefore, the volume of rectangular box is $\frac{1}{3}\left(x^3-3x^2+2x\right)$ .