(a) Six cubes, each with 12 cm edge, are joined end to end. Find the surface area of the resulting cuboid.

(b) The diagonal of a cube is $16\sqrt3$ cm. Find its surface area and volume.

(c) The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V; prove that : V² = xyz.

(a) Each cube has an edge length of 12 cm. When 6 cubes are joined end to end in a straight line, the resulting cuboid has:

Length = 6 x 12 = 72 cm

Breadth = 12 cm

Height = 12 cm

The surface area of a cuboid = 2(lb + bh + hl)

= 2(72 x 12 + 12 x 12 + 12 x 72)

= 2 x (864 + 144 + 864)

= 2 x 1872

= 3744 cm²

Hence, the surface area of the resulting cuboid = 3744 cm².

(b) Let the side length of the cube be a.

The formula for the diagonal of a cube is:

Diagonal = a $\sqrt{3}$

⇒ $16\sqrt3$ = a $\sqrt{3}$

⇒ a = 16 cm

Surface area of the cube = 6 x side²

= 6 x (16)²

= 6 x 256

= 1536 cm²

Volume of the cube = side³

= (16)³

= 4096 cm³

Hence, the surface area of the cube = 1536 cm² and its volume = 4096 cm³.

(c) Given that the areas of three adjacent faces of a cuboid are x, y, and z, we define:

⇒ x = lb, y = bh and z = lh

We also know that the volume of a cuboid = l x b x h

Squaring both sides:

⇒ V² = (lbh)²

⇒ V² = l² $\times$ b² $\times$ h²

⇒ V² = (lb) $\times$ (bh) $\times$ (hl)

⇒ V² = x $\times$ y $\times$ z

⇒ V² = xyz

Hence, proved V² = xyz.