Question 1015275
<pre>
The other tutor's solution above is incorrect.
 
{{{drawing(400,275,-3,13,-3,8,
green(locate(-1.5,.5,x)),
line(-2,-1,1,-1),
line(1,-1,1,2),
line(1,2,-2,2),
line(-2,2,-2,-1),
locate(-2,-1,A), locate(1,-4,B), locate(-.9,1.4,C),
red(locate(6.6,-1,d)),
locate(-.6,4.1,E),green(locate(-.9,2.8,x)),
red(line(-.5,3.5,1,-1),locate(.3,1.6,d)),
line(1,-1,2.5,.5),
line(2.5,.5,2.5,3.5),
line(2.5,3.5,-.5,3.5),
line(-.5,3.5,-2,2),
line(1,2,2.5,3.5),

green(line(-.5,3.5,-.5,1), line(-.5,1,-2,-1),
line(-.5,1,1,-1)),


red(line(4,-1,9,-1)),
line(9,-1,9,4),
line(9,4,4,4),
line(4,4,4,-1),locate(1,-1,B),

line(9,-1,11.5,1.5),
line(11.5,1.5,11.5,6.5),
line(11.5,6.5,6.5,6.5),
line(6.5,6.5,4,4),
line(11.5,6.5,9,4),locate(-.5,-1,x)

)}}}

Let x be the length of every edge of the smaller cube.

&#916;ABC and &#916;BCE are right triangles. (They don't look it 
because they're drawn in perspective to make a 2D drawing 
look like 3D.)  But &#8736;ACB is the right angle in right 
triangle &#916;ABC, and &#8736;BCE is the right angle in right
triangle &#916;BCE.

BC is the hypotenuse of right triangle &#916;ABC, and it is
also the bottom leg of triangle ABC,

Using the Pythagorean theorem on right triangle &#916;ABC:

{{{AB^2+AC^2=BC^2}}}
{{{x^2+x^2=BC^2}}}
{{{2x^2=BC^2}}}
  
Using the Pythagorean theorem on right triangle &#916;ABC:

{{{BC^2+CE^2=BE^2}}}
{{{2x^2+x^2=d^2}}}
{{{3x^2=d^2}}}
{{{sqrt(3x^2)=d}}}
{{{x*sqrt(3)=d}}}
{{{x=d/sqrt(3)}}}

Using the volume of a cube formula for the smaller cube: 

{{{V = edge^3}}}
{{{V = x^3}}}
{{{V = (d/sqrt(3))^3}}}

Using the volume of a cube formula for the larger cube:
{{{V = edge^3}}}
{{{V = d^3}}}

The ratio of the volume of the larger cube to the volume
of the smaller cube:

{{{matrix(1,3,d^3,":",(d/sqrt(3))^3)}}}

or the fraction:

{{{d^3/(d/sqrt(3))^3)}}}{{{""=""}}}{{{d^3/((d^3/sqrt(3)^3)))}}}{{{""=""}}}{{{d^3}}}{{{""*""}}}{{{((sqrt(3)^3)/d^3))}}}{{{""=""}}}{{{sqrt(3)^3}}} to 1

So the volume of the larger cube is {{{sqrt(3)^3}}} or about 5.2 times
the volume of the smaller cube.
Edwin</pre>