Equations of Planes — Question 9

Question 9

A tetrahedron has base vertices $B=(0,0,0)$, $C=(4,0,0)$, $D=(1,3,0)$ and apex $A=(2,1,6)$. Find the base plane, the height, and the volume. Then find the plane parallel to the base that cuts the tetrahedron at half its height.

Question 9 – Solution

Strategy The base is horizontal. Volume is one third of base area times perpendicular height.

Base and height The base plane is $\boxed{z=0}$. Its triangular area is $\frac 12(4)(3)=6$, and the apex is 6 units from the plane.

Volume Thus $V=\frac 13(6)(6)=\boxed{12}$.

Half-height plane The plane parallel to $z=0$ halfway from the base to the apex is $\boxed{z=3}$.

Verification Distance from $z=3$ to both $z=0$ and $A$’s level $z=6$ is 3.