Question

Consider a regular tetrahedron whose face is an equilateral triangle of side 7.

Find the area of the horizontal cross section A at the level z=3.

A= ?

Find the volume of the tetrahedron.

Consider a regular tetrahedron whose face is an equilateral triangle of side 7.

Find the area of the horizontal cross section A at the level z=3.

A=

Find the volume of the tetrahedron.

Consider a regular tetrahedron whose face is an equilateral triangle of side 7.

Find the area of the horizontal cross section A at the level z=3.

A= ?

Find the volume of the tetrahedron.

V=?

My work:
height = 7sin(60)= 7/2sqrt(3)

Answer 1

A=The area of the horizontal cross section: A(y)=sqrt(3)/4*a^(2)

a/(h-y)=s/h --> a=(h-y)(s)/(h)

The height for a tetrahedron is:
h= sqrt(2/3)(s) --> sqrt(2/3)(7) = 5.7154

a=(h-y)(s)/(h) -->(5.7154-3)(7)/(5.7154) = 3.3257

A(y)=sqrt(3)/4*a^(2) --> sqrt(3)/4*3.3257^(2) = 4.789414

A=4.7894

Volume: V=(s^3)/(6sqrt(2)) --> (7^3)/((6sqrt(2))= 40.42297

V=40.4229