Question 1108536
Let us pretend that this sketch really looks like the cube {{{drawing(400,400,-6,4,-0.5,9.5,
line(0,0,-5,0),line(0,5,-5,5),
line(0,0,0,5),line(-5,5,-5,0),
line(0,0,2.4,3.2),line(0,5,2.4,8.2),
line(2.4,8.2,2.4,3.2),line(-5,5,-2.4,8.2),
line(2.4,8.2,-2.4,8.2),locate(-4.9,5,A),
locate(-0.3,5,B),locate(-0.3,0.4,C),
locate(-4.9,0.4,D),locate(-2.5,8.6,E),
locate(2.3,8.6,F),locate(2.5,3.4,G)
)}}}
The seven visible vertices are labeled A through G.
Vertex H is in back. In the attempted perspective view,
H is directly behind D, and directly below E.
{{{s=AB=BC=CD=DG=FG=AE}}} is the length of a side of the cube.
{{{AE=sqrt(s^2+s^2)=sqrt(2)s}}} is the length of the diagonal of a cube face.
A cross-section (cutting through A, F, G, and D, would look like this
{{{drawing(400,280,-0.85,9.15,-1,6,
green(triangle(0,0,8.66,0,8.66,5)),
green(rectangle(8.66,0,8.36,0.3)),
line(0,0,8.66,O),line(0,5,8.66,5),
line(0,0,0,5),line(8.66,0,8.66,5),
locate(-0.1,0,D),locate(8.56,0,G),
locate(-0.1,5.4,A),locate(8.56,5.4,F),
locate(4.1,0.5,sqrt(2)s),locate(8.4,2.7,s)
)}}}
{{{DF=sqrt((sqrt(2)s)^2+s^2)=sqrt(2s^2+s^3)=sqrt(3s^2)=sqrt(3)s}}} is the length of the diagonal of the cube,
the longest possible distance between two points of the cube.
For the cube to fit in the sphere, that has to be the diameter of the sphere.
So, {{{2r=sqrt(3)s}}} --> {{{highlight(r/s=sqrt(3)/2)}}} .
The ratio r:s is {{{highlight(sqrt(3)/2)}}} (about 0.866).