# Proof of Surface area of a Cuboid

## Surface area of a Cuboid

• Cuboid is the three-dimensional object which has 6 rectangular faces having length, breadth and height. Also, cuboid has 8 vertices or corners, 12 edges and 4 diagonals as shown in following figure.
• If the cuboid has equal length, breadth and height then it is called as the cube.
• The surface area of the cuboid is nothing but the sum of surface area of all its 6 rectangular surfaces.

### Derivation for surface area of cuboid:

• Let us consider the cuboid with 6 rectangular surfaces having height h, length l and breadth b as shown in figure.
• But when we open this cuboid totally then the structure formed will be of the form as shown in figure.
• We can see that there are three pairs of rectangular surfaces. We denote them by 1, 2 and 3.

• So, pair number 1 having surface area = 2(h*l)
• Surface area of pair number 2 = 2(b*l)
• Surface area of pair number 3 = 2(b*h)

Thus, total surface area of cuboid = sum of surfaces area of pairs 1, 2 and 3

= 2(h*l) + 2(b*l) + 2(b*h)

= 2 (b*l + l*h + h*b)

Thus, the surface area of cuboid having length l, breadth b and height h is given by,

Surface area of cuboid = 2 (b*l + l*h + h*b)

Hence proved.

Updated: September 14, 2021 — 9:50 pm