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.

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