Answer
a=d/$\sqrt 3$
surface area=$2d^{2}$
volume=$d^{3}$/3$\sqrt 3$
Work Step by Step
Applying Pythagorean theorem, edge diagonal,D= $\sqrt 2$ a.
Applying Pythagorean theorem again will yield body diagonal,d=$ \sqrt 3$a
[$d^{2}$=$a^{2}$+($\sqrt 2a$)^2]
Now we have a as function of d.
surface area=6$a^{2}$
volume=$a^{3}$
Substitute a in terms of d in the above two equations.
