Question

20 points!!!!!!!

Answer 1

See the attached figure to better understand the problem
let
L-----> length side of the cuboid
W----> width side of the cuboid
H----> height of the cuboid

we know that
One edge of the cuboid has length 2 cm----->  I'll assume it's L
so
L=2 cm
[volume of a cuboid]=L*W*H-----> 2*W*H
40=2*W*H------> 20=W*H-------> H=20/W------> equation 1

[surface area of a cuboid]=2*[L*W+L*H+W*H]----->2*[2*W+2*H+W*H]

100=2*[2*W+2*H+W*H]---> 50=2*W+2*H+W*H-----> equation 2
substitute 1 in 2
50=2*W+2*[20/W]+W*[20/W]----> 50=2w+(40/W)+20
multiply by W all expresion
50W=2W²+40+20W------> 2W²-30W+40=0

using a graph tool------> to resolve the second order equation
see the attached figure

the solutions are
13.52 cm x 1.48 cm
so the dimensions of the cuboid are
2 cm x 13.52 cm x 1.48 cm
or
2 cm x 1.48 cm x 13.52 cm

Find the length of a diagonal of the cuboid
diagonal=√[(W²+L²+H²)]------> √[(1.48²+2²+13.52²)]-----> 13.75 cm

the answer is
 the length of a diagonal of the cuboid is 13.75 cm