Answer
(a) Diagonal of any face = $x\sqrt 2$
(b) Diagonal that passes through th center of the cube = $x\sqrt 3$
Work Step by Step
(a) Given length of each edge is $'x'$
Each face of a cube is a square and a diagonal divides it into two 45°–45°–90° triangles. Therefore-
In 45°–45°–90° triangle
Hypotenuse = $\sqrt 2 \times $ Shorter Side
Therefore-
Diagonal of a face = $\sqrt 2 \times $ EDGE
Diagonal of a face = $\sqrt 2 \times x$ ($ x $ is the edge of cube )
Diagonal of a face = $x\sqrt 2$
(b) Diagonal that passes through the center of the cube, diagonal of face and respective edge of the cube make a right triangle together . Applying Pythagorean theorem-
$Center Diagonal^{2} $ = $Face Digonal^{2} $ + $Edge^{2} $
$Center Diagonal^{2} $ = $(x\sqrt 2)^{2} $ + $x^{2} $
[Substituting values of Face Diagonal and Edge]
$Center Diagonal^{2} $ = $2x^{2} + x^{2} $ = $3x^{2}$
Therefore-
Center Diagonal = $\sqrt (3x^{2})$
Or
Center Diagonal = $x\sqrt 3$
