Question

The base of a triangular pyramid is a triangle with corners at `(4 ,2 )`, `(3 ,6 )`, and `(7 ,5 )`. If the pyramid has a height of `8`, what is the pyramid's volume?

Answer 1

`20`

Explanation:

The triangle with vertices `(4, 2)`, `(3, 6)` and `(7, 5)` can be drawn inside a `4 xx 4` square with vertices `(3, 2)`, `(7, 2)`, `(7, 6)` and `(3, 6)`, dividing it into `4` triangles...

The two smaller triangles are right angled triangles with legs of lengths `1` and `4`, so their total area is `4`.

The larger isosceles right angled triangle has area `1/2 * 3 * 3 = 9/2`

So the total area of the given triangle is:

`4^2 - 4 - 9/2 = 15/2`

The pyramid has volume:

`1/3 * "base" * "height" = 1/3 * 15/2 * 8 = 20`

Answer 2

`20`

Explanation:

The area of a triangle with vertices `(x_1, y_1)`, `(x_2, y_2)`, `(x_3, y_3)` is given by the formula:

`"Area" = 1/2 abs(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)`

(see https://socratic.org/s/aK6h7PjW)

So putting:

`{ ((x_1, y_1) = (4, 2)), ((x_2, y_2) = (3, 6)), ((x_3, y_3) = (7, 5)) :}`

we find that the base of our pyramid has area:

`1/2 abs((4)(6)+(3)(5)+(7)(2)-(4)(5)-(3)(2)-(7)(6))`

`=1/2 abs(24+15+14-20-6-42)`

`=1/2 abs(-15)`

`=15/2`

Then the volume of the pyramid is:

`1/3 xx "base" xx "height" = 1/3 * 15/2 * 8 = 20`