Question

The base of a triangular pyramid is a triangle with corners at `(3 ,2 )`, `(5 ,6 )`, and `(2 ,8 )`. If the pyramid has a height of `9`, what is the pyramid's volume?

Answer 1

21.0 `(units)^3`

Explanation:

This is a two-part problem. First we must use the “distance formula” to find the base length. Any of the “base lines” may be used, as the value of h is related to each of them. We will need the additional “base” lines to calculate h from the Pythagorean Theorem. Then we will use the formula for the volume of a triangular pyramid (different from that of a square-based pyramid) to find the volume.
`V = (1/3)*A*h = (1/6)*b*h*H`

Distance Formula:
b= `sqrt ((x_2-x_1)^2+(y_2-y_1)^2)`
Pyramid Base Sides:
`b_1= qrt ((5-3)^2+(6-2)^2)` ; `b_1` = 4.47
`b_2= sqrt ((2-5)^2+(8-6)^2)` ; `b_2` = 3.61
`b_3= sqrt ((2-3)^2+(8-2)^2)` ; `b_3` = 6.08

`Let x + y = b_1` ; then `y = b_1 – x`
Using the Pythagorean Theorem to find h:
`h^2 + (b_1 – x)^2 = (b_2)^2`
`h^2 + x^2 = (b_3)^2`
Subtracting the second equation from the first we obtain:
`(b_1 – x)^2 - x^2 = (b_2)^2 – (b_3)^2`
`x^2 -2b_1*x - x^2 = (b_2)^2 – (b_3)^2`
`-2*b_1*x = (b_2)^2 – (b_3)^2`
`x = [(b_2)^2 – (b_3)^2]/(-2*b_1)`
`x = [(3.61)^2 – (6.08)^2]/(-2*4.47)` = (13.03 – 36.97)/(-8.94) = 2.68

`y = b_1 – x ; y = 4.47 – 2.68 = 1.79`

`h^2 + y^2 = (b_2)^2` ; `h^2 = (b_2)^2 – (y^2)` ; `h^2 = (3.61)^2 – (1.79)^2` ; `h^2 = 13.03 – 3.20`

`h = 3.14`

NOW we have our “b”, “h” and H for the triangle volume formula:

`V =(1/6)*b*h*H` = `(1/6)*4.47*3.14*9` = 21.0 `(units)^3`