Question 487782
<pre><font size = 4 color = "indigo"><b>
Detailed explanation:

Let's find the diagonal AC of the bottom square, drawn in green below:

{{{drawing(400,400,-1,10,-1,10,
rectangle(2,3,8.5,9.5), locate(2.1,6.25,5),
line(0,0,6.5,0), line(6.5,0,6.5,6.5),line(0,6.5,6.5,6.5),line(0,0,0,6.5),
locate(3.25,0,5), locate(1.2,1.8,5),
triangle(0,0,2,3,0,0), line(0,6.5,2,9.5),line(2,9.5,8.5,9.5),
line(6.5,6.5,8.5,9.5), line(8.5,3,8.5,9.5), line(6.5,0,8.5,3),
locate(2,10,B), locate(6.5,0,A), locate(0,0,D), locate(2.2,3.5,C), 
green(line(2,3,6.5,0))

 )}}}

Triangle ADC is a right triangle, with right angle ADC so we use the
Pythagorean theorem:

AC² = AD² + DC² 
AC² = 5² + 5²
AC² = 25 + 25
AC² = 50
       __    ____    __  _     _ 
 AC = <font face = "symbol">Ö</font>50 = <font face = "symbol">Ö</font>25×2 = <font face = "symbol">Ö</font>25×<font face = "symbol">Ö</font>2 = 5<font face = "symbol">Ö</font>2
                                                            _
That's the first answer.  So we label that green diagonal 5<font face = "symbol">Ö</font>2

{{{drawing(400,400,-1,10,-1,10,
rectangle(2,3,8.5,9.5),locate(2.1,6.25,5),

line(0,0,6.5,0), line(6.5,0,6.5,6.5),line(0,6.5,6.5,6.5),line(0,0,0,6.5),
locate(3.25,0,5), locate(1.2,1.8,5),
triangle(0,0,2,3,0,0), line(0,6.5,2,9.5),line(2,9.5,8.5,9.5),
line(6.5,6.5,8.5,9.5), line(8.5,3,8.5,9.5), line(6.5,0,8.5,3),
locate(2,10,B), locate(6.5,0,A), locate(0,0,D), locate(2.2,3.5,C), 
green(line(2,3,6.5,0)), locate(4,2.2,5sqrt(2))

 )}}}

Let's draw the cube's diagonal AB in green also

{{{drawing(400,400,-1,10,-1,10,
rectangle(2,3,8.5,9.5),locate(2.1,6.25,5),

line(0,0,6.5,0), line(6.5,0,6.5,6.5),line(0,6.5,6.5,6.5),line(0,0,0,6.5),
locate(3.25,0,5), locate(1.2,1.8,5),
triangle(0,0,2,3,0,0), line(0,6.5,2,9.5),line(2,9.5,8.5,9.5),
line(6.5,6.5,8.5,9.5), line(8.5,3,8.5,9.5), line(6.5,0,8.5,3),
locate(2,10,B), locate(6.5,0,A), locate(0,0,D), locate(2.2,3.5,C), 
green(line(2,3,6.5,0),line(2,9.5,6.5,0)), locate(4,2.2,5sqrt(2))

 )}}}


Triangle ABC is a right triangle with right angle ACB. So we use
the Pythagorean theorem to find AB

AB² = AC² + BC²
         _
AB² = (5<font face = "symbol">Ö</font>2)² + 5²
          _
AB² = 5²(<font face = "symbol">Ö</font>2)² + 25

AB² = 25(2) + 25

AB² = 50 + 25

AB² = 75
       __    ____    __  _     _
 AB = <font face = "symbol">Ö</font>75 = <font face = "symbol">Ö</font>25×3 = <font face = "symbol">Ö</font>25×<font face = "symbol">Ö</font>3 = 5<font face = "symbol">Ö</font>3

Edwin</pre>