# 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?

Mar 8, 2017

21.0 ${\left(u n i t s\right)}^{3}$

#### 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.
$V = \left(\frac{1}{3}\right) \cdot A \cdot h = \left(\frac{1}{6}\right) \cdot b \cdot h \cdot H$

Distance Formula:
b= $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
Pyramid Base Sides:
${b}_{1} = q r t \left({\left(5 - 3\right)}^{2} + {\left(6 - 2\right)}^{2}\right)$ ; ${b}_{1}$ = 4.47
${b}_{2} = \sqrt{{\left(2 - 5\right)}^{2} + {\left(8 - 6\right)}^{2}}$ ; ${b}_{2}$ = 3.61
${b}_{3} = \sqrt{{\left(2 - 3\right)}^{2} + {\left(8 - 2\right)}^{2}}$ ; ${b}_{3}$ = 6.08

$L e t x + y = {b}_{1}$ ; then y = b_1 – x
Using the Pythagorean Theorem to find h:
h^2 + (b_1 – x)^2 = (b_2)^2
${h}^{2} + {x}^{2} = {\left({b}_{3}\right)}^{2}$
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

${h}^{2} + {y}^{2} = {\left({b}_{2}\right)}^{2}$ ; 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:

$V = \left(\frac{1}{6}\right) \cdot b \cdot h \cdot H$ = $\left(\frac{1}{6}\right) \cdot 4.47 \cdot 3.14 \cdot 9$ = 21.0 ${\left(u n i t s\right)}^{3}$