Altitude of a Tetrahedron Given Its Edges

a

FE`a$$3826308556724520344453704386324872670757

b

1

c

1

d

FE`d$$3826308556724520344453704386324872670757

e

1

f

1

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This Demonstration constructs an altitude of a tetrahedron

T=ABCD

given its edge lengths

a

,

b

,

c

,

d

,

e

,

f

,

g

. (In the figure, the edge length of

BC

is

e

.) Suppose the altitude is from vertex

D

to the opposite face

ABC

. First, construct the net of

T

with the triangle

ABC

in the center (unfold completely). Normals from the vertex

D

to the sides

AB

,

BC

,

CA

meet at a point

E

. This is the 3D orthogonal projection of vertex

D

. In 3D, the lines

DF

,

FE

and the altitude form a right triangle with

DF

as its hypotenuse. So we can construct the altitude as a leg of the triangle.