Question

The base of a triangular pyramid is a triangle with corners at `(6 ,7 )`, `(3 ,1 )`, and `(4 ,2 )`. If the pyramid has a height of `8`, what is the pyramid's volume?

Answer 1

V=4

Explanation:

`V=1/6*a*h_a*H`
`H` is the hight into the 3-dimension.
`p_1=(6,7)`
`p_2=(3,1)`
`p_3=(4,2)`
`H=8`
To find the distanc `a` between `p_1` and `p_2` we use pythagoras:
`a^2+b^2=c^2`
Notice that the `c` from pythagoras equals our `a`.
`((x_1-x_2)^2+(y_1-y_2)^2)^(1/2)=a`
`((6-3)^2+(7-1)^2)^(1/2)=a`
`(9+36)^(1/2)=a`
`sqrt(45)=a`
`((x_1-x_3)^2+(y_1-y_3)^2)^(1/2)=b`
`sqrt(29)=b`
`((x_2-x_3)^2+(y_2-y_3)^2)^(1/2)=c`
`sqrt(2)=c`

`c^2=a^2+b^2-2abcos(gamma)`
`2=45+29-2sqrt(45*29)cos(gamma)`
`2-45-29=-2sqrt(1305)cos(gamma)`
`cos^-1(36/sqrt(1305))=gamma`
`4,76°~~gamma`
`sin(gamma)=h_a/b|*b`
`sin(gamma)*b=h_a`
`0.45~~h_a`

`V=1/6*sqrt(45)*0.45*8~~4`