Question

The base of a triangular pyramid is a triangle with corners at `(6 ,8 )`, `(2 ,4 )`, and `(7 ,3 )`. If the pyramid has a height of `2`, what is the pyramid's volume?

Answer 1

`V_(Delta_(pyr)) = BA*H = 24*2 = 48`

Explanation:

Given: Base Triangle Vertices: `A(2,4,0), B(7,3,0), C(6,8,0)`
Using Distance formula find the length of the sides of the base triangle, `bar(AB), bar(BC), bar(CA)`:
`bar(AB) = sqrt((2-7)^2 + (4-3)^2 )=sqrt(26)`
`bar(BC) = sqrt((7-6)^2 + (3-8)^2 )=sqrt(26)`
`bar(CA) = sqrt((6-2)^2 + (8-4)^2 )=4sqrt(2)`

Now you can use the Pyramid volume formula to calculate the Volume. `V = 1/3 BA*H`
where `BA`, and `H` are Base Area and Height of the pyramid respectively.
We are going to use Heron Formula to calculate the area of the triangle:
`A_(Delta) = sqrt(p(p-bar(AB))(p-bar(BC))(p-bar(CA))`
Where `p=P/2=(bar(AB)+bar(BC)+bar(CA))/2`
`p= (sqrt(26)+sqrt(26)+4sqrt(2))/2`
`A_(Delta) = sqrt(((sqrt(26)+sqrt(26)+4sqrt(2))/2)((sqrt(26)+sqrt(26)+4sqrt(2))/2-sqrt(26))((sqrt(26)+sqrt(26)+4sqrt(2))/2-sqrt(26))((sqrt(26)+sqrt(26)+4sqrt(2))/2-4sqrt(2))`
`A_Delta = sqrt(((sqrt(26)+sqrt(26)+4sqrt(2)))/2(4sqrt(2))(4sqrt(2))((sqrt(26)+sqrt(26)-4sqrt(2))/2 )`
`A_Delta = sqrt( (2sqrt(26)+4sqrt(2) )(2sqrt(26)-4sqrt(2))(4*2) )`
`A_Delta = sqrt( (4*26-16*2 )(4*2) ) = 24`
and the Volume is:
`V_(Delta_(pyr)) = BA*H = 24*2 = 48`