Question

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

Answer 1

24.47

Explanation:

• First find the length of each line using formula `sqrt ((y_2-y_1)^2 + (x_2-x_1)^2)` where 1 and 2 are x and y coordinates of the two points

Distance between (6,4) and (2,5) = `sqrt((5-4)^2+(2-6)^2)`
= `sqrt17`
Distance between (2,5) and (3,2) = `sqrt((2-5)^2 +(3-2)^2`
= `sqrt10`
Distance between (3,2) and (6,5) = `sqrt((5-2)^2 + (6-3)^2)`
= `3sqrt2`

• Then find base area
  Cos `theta` = `(b^2+c^2-a^2)/(2*b*c)`
  = `((sqrt 17)^2 +(sqrt 10)^2 - (3 sqrt 2)^2)/(2*sqrt10*sqrt17)`
  `theta` = `Cos^-1`(0.345134245)
  = `69.8^@`
  Area = `1/2 *sqrt17 *sqrt10 Sin69.8`
  = 6.11822

• Calculate volume using formula:
  volume = Base area * height

= `6.11822*4`
= `24.47`