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?

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

#V = 1/3Ah#

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 = +-1/2|(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)|#

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 #3xx3# 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)) | = #

#color(red)((8)(2)(1)) + color(green)((5)(1)(5))+ color(blue)((1)(6)(9)) = 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 - color(red)((1)(2)(5)) - color(green)((8)(1)(9)) - color(blue)((5)(6)(1)) = -17#

Because the area cannot be negative, we choose the negative value for the #+-1/2#

#A = -1/2(-17)#

#A = 17/2#

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

#V = 1/3xx17/2xx8#

#V = 68/3=22 2/3#