Chapter 6.8, Problem 3CYU

To determine

To calculate: The dimensions of the solid,

The volume of a triangular pyramid is $210$ cubic inches.

  

Answer to Problem 3CYU

The value of length, width and height of a triangular pyramid is:

  $\begin{array}{c}h=28\text{inches}\\ \text{base=5inches}\\ \text{triangular height=}9\text{inches}\end{array}$ .

Explanation of Solution

Given information:

The dimensions of the pyramid,

The volume of a triangular pyramid is $210$ cubic inches.

Formula used:

For volume of a triangle is,

  $V=\frac{1}{3}Bh$

Where:

  $\begin{array}{c}\text{length=l}\\ \text{base=B}\\ \text{height=h}\\ \text{area=A}\end{array}$ .

Calculation:

Consider the statement as:

The volume of a triangular pyramid is $210$ cubic inches.

The Area of a pyramid is: $A\text{inches}$

Then the base is: $x\text{ inches}$

The height is: $5x+3$

The volume of a triangular pyramid is: $210\text{ inches}$

Firstly find the area of the triangle:

  $\begin{array}{l}A=\frac{1}{2}Bh\\ B=\frac{1}{2}Ah\\ B=\frac{1}{2}\left(2x-1\right)\left(x\right)\end{array}$

Recall that the area of the rectangle:

  $V=\frac{1}{3}Bh$

Apply the values in the formulas:

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

Substitute the value of $V,B,h$

in the volume formula:

  $210=\frac{1}{3}\cdot \frac{1}{2}\left(2x-1\right)\left(x\right)\left(5x+3\right)$

Multiply both sides by $6$ ,

  $\begin{array}{c}210=\frac{1}{3}\cdot \frac{1}{2}\left(2x-1\right)\left(x\right)\left(5x+3\right)\\ 6\left(210\right)=6\left(\frac{1}{3}\cdot \frac{1}{2}\left(2x-1\right)\left(x\right)\left(5x+3\right)\right)\\ 1260=2x^2-x\left(5x+3\right)\\ 1260=10x^3+x^2-3x\end{array}$

The standard form of the equation:

  $\begin{array}{l}10x^3+x^2-3x-1260=0\\ x^2\left(10x+1\right)-3\left(x-420\right)=0\end{array}$

These are the two factors:

  $\begin{array}{l}\left(10x+1\right)\cdot x^2\\ \left(x-420\right)\cdot -3\end{array}$

There is no factorization is applied, factorization is failed here.

Find the roots and zeros take, $x=0$ .

  $\begin{array}{c}10x^3+x^2-3x-1260=0\\ 10\left(0\right)+{\left(0\right)}^2-3\left(0\right)-1260=0\\ -1260=0\end{array}$

Here the value is too grater than zero.

Check for the value of $x=1$ .

  $\begin{array}{c}10x^3+x^2-3x-1260=0\\ 10{\left(1\right)}^3+{\left(1\right)}^2-3\left(1\right)-1260=0\\ 10+1-3-1260=0\\ -1252=0\end{array}$

Here the value is too grater than zero.

Check for the value of $x=2$ ,

  $\begin{array}{c}10x^3+x^2-3x-1260=0\\ 10{\left(2\right)}^2+{\left(2\right)}^2-3\left(2\right)-1260=0\\ 10\left(4\right)+4-6-1260=0\\ 40-2-1260=0\end{array}$

Simplified further:

  $\begin{array}{c}40-2-1260=0\\ -1182=0\end{array}$

Here the value is too grater than zero.

Check for the value of $x=3$ ,

  $\begin{array}{c}10x^3+x^2-3x-1260=0\\ 10{\left(3\right)}^3+{\left(3\right)}^2-3\left(3\right)-1260=0\\ 10\left(27\right)+9-9-1260=0\\ 270-1260=0\end{array}$

Simplified further:

  $\begin{array}{c}270-1260=0\\ -990=0\end{array}$

Here the value is too grater than zero.

Check for the value of $x=4$ ,

  $\begin{array}{c}10x^3+x^2-3x-1260=0\\ 10{\left(4\right)}^3+{\left(4\right)}^2-3\left(4\right)-1260=0\\ 10\left(64\right)+16-12-1260=0\\ 640-1260=0\end{array}$

Simplified further:

  $\begin{array}{c}640-1260=0\\ -616=0\end{array}$

Here the value is too grater than zero.

Check for the value of $x=5$ ,

  $\begin{array}{c}10x^3+x^2-3x-1260=0\\ 10{\left(5\right)}^3+{\left(5\right)}^2-3\left(5\right)-1260=0\\ 10\left(125\right)+25-15-1260=0\\ 1260-1260=0\end{array}$

Simplified further:

  $\begin{array}{c}1260-1260=0\\ 0=0\end{array}$

  $x=5$ , so now substitute the value to dimensions of the triangular pyramid will give the true values of the dimensions.

  $\begin{array}{c}x=5\text{ inches}\\ \text{5}\left(5\right)+3=28\text{ inches}\\ \text{2}\left(5\right)-1=9\text{ inches}\end{array}$

The dimensions of the triangular pyramid:

Thus, the value of length, width and height of a triangular pyramid is:

  $\begin{array}{c}h=28\text{inches}\\ \text{base=5inches}\\ \text{triangular height=}9\text{inches}\end{array}$