Answer:
[tex]\frac{4}{3}[/tex]
Step-by-step explanation:
The area of a regular octahedron is given by:
area = [tex]2\sqrt{3}\ *edge^2[/tex]. Let a is the length of the edge (diagonal).
area = [tex]2\sqrt{3}\ *a^2[/tex]
Given that the diagonal of the octahedron is equal to height (h) of the tetrahedron i.e.
a = h, where h is the height of the tetrahedron and a is the diagonal of the octahedron. Let the edge of the tetrrahedron be e. To find the edge of the tetrahedron, we use:
[tex]h=\sqrt{\frac{2}{3} } e\\but\ h=a\\a=\sqrt{\frac{2}{3} } e\\e=\sqrt{\frac{3}{2} }a[/tex]
The area of a tetrahedron is given by:
area = [tex]\sqrt{3}\ *edge^2[/tex] = [tex]\sqrt{3} *(\sqrt{\frac{3}{2} }a)^2=\frac{3}{2}\sqrt{3} *a^2[/tex]
The ratio of area of regular octahedron to area tetrahedron regular is given as:
Ratio = [tex]\frac{2\sqrt{3}\ *a^2}{\frac{3}{2} \sqrt{3}*a^2} =\frac{4}{3}[/tex]
