Question

The base of a triangular pyramid is a triangle with corners at `(2 ,6 )`, `(5 ,3 )`, and `(8 ,7 )`. If the pyramid has a height of `18`, what is the pyramid's volume?

Answer 1

`V_(pyr)=1/3 (10.5)*18=10.5*6=63 " cubic units"`

Explanation:

Given the vertices of the base triangle of a pyramid and altitude:
`A(2,6), B(5,3) and C(8,7)`
Altitude, `H=18` units
Required: The volume?
This problem can be solved as follows:
`V_(pyr)=1/3 (BA)*H` where `BA=`Base Area and `H=`Altitude
So the strategy is:
a) Use distance formula to determine the length of sides of the triangle: `bar(AB),bar(BC), bar(CA),`
Distance formula of, `P_1(x_1, y_1, z_1), P_2(x_2, y_2, z_2)`
`bar(P_1P_2)= sqrt((x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2 )`
Since the base is on the x-y plane `z_1=z_2=0` Thus,
`bar(AB) = sqrt((2-5)^2 + (6-3)^2)=3sqrt(2)`
`bar(BC)=sqrt((5-8)^2 + (3-7)^2)=5`
`bar(CA)=sqrt((8-2)^2 + (7-6)^2)=sqrt(37)`
b) Now calculate the area. Since you know all sides you can use Heron's formula:
`Area_(Delta) =A_Delta= sqrt(s(s-a)(s-b)(s-c))`
where `a, b, and d` are sides of base triangle and s is half the perimeter `s=1/2 (a+b+c)= 1/2 (5+3sqrt(2)+sqrt(37))`
`A_Delta=sqrt(1/2 (5+3sqrt(2)+sqrt(37))[1/2 (5+3sqrt(2)+sqrt(37)) -5 ] [1/2 (5+3sqrt(2)+sqrt(37)) -3sqrt(2) ][1/2 (5+3sqrt(2)+sqrt(37)) -sqrt(37) ] )~~10.5` `A_Delta=10.5 " square units"`
b) Now calculate the volume. : `V_(pyr)=1/3 (BA)*H`
`V_(pyr)=1/3 (10.5)*18=10.5*6=63 " cubic units"`