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?

1 Answer
Mar 8, 2017

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#