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Calculus IIChapter 6: 3-Dimensional SpaceSection 6-3

Equations of Planes — Question 9

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Question 9

A tetrahedron has base vertices B=(0,0,0)B=(0,0,0), C=(4,0,0)C=(4,0,0), D=(1,3,0)D=(1,3,0) and apex A=(2,1,6)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.

Original worksheet page 1: question and worked solution for 6-3-009
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Question 9 – Solution

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

See the diagram in the original worksheet below.

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

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

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

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

Original worksheet page 2: question and worked solution for 6-3-009

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

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