Question

The base of a triangular pyramid is a triangle with corners at `(7 ,5 )`, `(6 ,9 )`, and `(3 ,4 )`. If the pyramid has a height of `4`, what is the pyramid's volume?

Answer 1

`color(crimson)("Volume of Pyramid " V_p = (1/3) * A_b * h = 11.36 " cubic units"`

Explanation:

`color(violet)("Volume of Pyramid " V_p = (1/3) * A_b * h`

`Area of base triangle " A_b = sqrt(s (s-a) (s-b) (s-c)), " using Heron's formula`

`A(7,5), B(6,9), C(3,4), h = 4`

`a = sqrt((6-3)^2 + (9-4)^2) = 5.83`

`b = sqrt((7-3)^2 + (5-4)^2) = 4.12`

`c = sqrt((6-7)^2 + (9-5)^2) = 4.12`

It's an isosceles triangle with sides b & c equal.

`"Semi-perimeter " s = (a + b + c) / 2`

`s = (5.83 + 4.12 + 4.12) / 2 ~~ 7.04`

`A_b = sqrt(7.04 * (7.04 - 5.83) * (7.04 - 4.12) * (7.04 - 4.12)) = 8.52`

`color(crimson)("Volume of Pyramid " V_p = (1/3) * A_b * h = (1/3) * 8.52 * 4 = 11.36 " cubic units"`