The base of a triangular pyramid is a triangle with corners at (8 ,5 ), (6 ,2 ), and (5 ,9 ). If the pyramid has a height of 8 , what is the pyramid's volume?

Mar 31, 2017

The volume is $22 \frac{2}{3}$

Explanation:

The volume is the area of the base multiplied by the height:

$V = \frac{1}{3} A h$

Because we are given 3 points, the area is best computed using a determinant.

Here is a reference for the determinant that will help to compute the area given 3 points:

$A = \pm \frac{1}{2} | \left({x}_{1} , {y}_{1} , 1\right) , \left({x}_{2} , {y}_{2} , 1\right) , \left({x}_{3} , {y}_{3} , 1\right) |$

Substituting in the 3 points:

A = +-1/2 |(8,5,1), (6,2,1), (5,9,1)|

If find it easier to demonstrate the evaluation of a $3 \times 3$ determinant if the first two rows are repeated:

A = +-1/2 | (8,5,1,8,5), (6,2,1,6,2), (5,9,1,5,9) |

Multiply each of the major diagonals and add them:

A = +-1/2 | (color(red)(8),color(green)(5),color(blue)(1),8,5), (6,color(red)(2),color(green)(1),color(blue)(6),2), (5,9,color(red)(1),color(green)(5),color(blue)(9)) | =

$\textcolor{red}{\left(8\right) \left(2\right) \left(1\right)} + \textcolor{g r e e n}{\left(5\right) \left(1\right) \left(5\right)} + \textcolor{b l u e}{\left(1\right) \left(6\right) \left(9\right)} = 95$

Multiply the minor diagonals and subtract them from the sum of the major diagonals:

A = +-1/2 | (8,5,color(red)(1),color(green)(8),color(blue)(5)), (6,color(red)(2),color(green)(1),color(blue)(6),2), (color(red)(5),color(green)(9),color(blue)(1),5,9) | =

$95 - \textcolor{red}{\left(1\right) \left(2\right) \left(5\right)} - \textcolor{g r e e n}{\left(8\right) \left(1\right) \left(9\right)} - \textcolor{b l u e}{\left(5\right) \left(6\right) \left(1\right)} = - 17$

Because the area cannot be negative, we choose the negative value for the $\pm \frac{1}{2}$

$A = - \frac{1}{2} \left(- 17\right)$

$A = \frac{17}{2}$

Using the $h = 8$ we can compute the volume:

$V = \frac{1}{3} \times \frac{17}{2} \times 8$

$V = \frac{68}{3} = 22 \frac{2}{3}$