Cube and cuboid surface area

A carton, a water tank, a matchbox, a classroom — every box-shaped object has an "outside" made of flat faces, and knowing how much sheet metal, cardboard, glass, or paint covers that outside is a genuinely useful skill, not just a syllabus item. Surface area answers exactly this: add up the area of every face and you know how much material wraps the solid. The cube and cuboid are the two simplest 3-D solids, and both are built entirely from rectangles and squares — shapes whose area you already know. That makes their surface area formulas a short, logical step up from 2-D mensuration: unfold the solid flat into its six faces (a "net"), find each rectangle's area, and add. This article builds the total surface area (TSA) and lateral surface area (LSA) formulas from that net, extends them to open boxes and hollow containers, and applies them to real costing problems — painting a tank, covering a box in tin sheet — while flagging the traps ICSE examiners repeatedly test.

The core insight

The total surface area of a cuboid, $2(lb+bh+hl)$ , is simply the sum of its three pairs of matching rectangular faces, and it collapses to $6a^2$ for a cube and adjusts by removing or adding single-face areas for open or hollow solids.

The core idea

A cuboid has six faces in three matching pairs. Look at any box: the top matches the bottom ( $l \times b$ each), the front matches the back ( $l \times h$ each), and the two sides match each other ( $b \times h$ each). Adding all six and doubling each pair gives the total surface area:

\text{TSA of cuboid} = 2(lb + bh + hl)

where: $l$ is length, $b$ is breadth, and $h$ is height, all in the same unit, so the answer comes out in that unit squared.

A cube is a cuboid where every edge is equal, $l = b = h = a$ . Substituting into the cuboid formula, all three products become $a^2$ , giving a much shorter formula:

\text{TSA of cube} = 6a^2

where: $a$ is the edge length of the cube.

Lateral surface area (LSA) leaves out the top and bottom — only the four "walls" count. This matters whenever a problem is about painting the sides of a room, wrapping a pillar, or any container that is open at both top and bottom:

\text{LSA of cuboid} = 2h(l+b), \qquad \text{LSA of cube} = 4a^2

Open solids (no lid) simply subtract the one missing face from the TSA. A cuboid open at the top loses its $lb$ face; a cube open at the top loses one $a^2$ face:

\text{Cuboid, open at top} = 2(lb+bh+hl) - lb, \qquad \text{Cube, open at top} = 6a^2 - a^2 = 5a^2

Hollow boxes (a container with wall thickness, so inner and outer dimensions differ) do not have one single formula — treat the inner surface and the outer surface as two separate cuboids with their own $l, b, h$ , then add or subtract their areas depending on what the question asks (material used = outer TSA plus inner TSA plus the visible rim edges; wall cross-section = the difference between outer and inner face areas).

Figure

A cuboid net unfolded into six rectangles showing the three matching pairs top and bottom front and back left and right that together give the total surface area

Derivation of the cuboid total surface area

Derivation ATotal surface area of a cuboid from its net

The goal of this derivation is to justify, face by face from the unfolded net of a cuboid, why the formula groups the three distinct face-areas $lb$ , $bh$ , and $hl$ and multiplies the sum by exactly $2$ rather than by any other factor, so that the formula can be trusted and reconstructed instead of merely memorised.

Step 1. Identify the six faces of a cuboid with length $l$ , breadth $b$ , and height $h$ . Opposite faces of a cuboid are always congruent rectangles: the top and bottom are both $l \times b$ , the front and back are both $l \times h$ , and the left and right sides are both $b \times h$ . This gives three distinct face-shapes, each appearing exactly twice.

Step 2. Compute the area of one face from each pair. Top/bottom area $= lb$ . Front/back area $= lh$ (written $hl$ by convention). Left/right area $= bh$ . These are just length-times-breadth for each rectangle, using the two edges that meet at that face.

Step 3. Add all six faces, using two copies of each distinct area. Total $= lb + lb + lh + lh + bh + bh = 2(lb) + 2(lh) + 2(bh)$ . Factor out the common $2$ :

\text{TSA} = 2(lb + bh + hl)

Step 4. Specialise to a cube by setting $l = b = h = a$ . Every product $lb$ , $bh$ , $hl$ becomes $a \times a = a^2$ , so the three terms are identical: $\text{TSA} = 2(a^2 + a^2 + a^2) = 2(3a^2) = 6a^2$ , matching the familiar six-equal-faces picture of a cube directly (six square faces, each of area $a^2$ ).

Worked example

Problem. A closed cuboidal water tank measures $2.5$ m long, $2$ m wide, and $1.6$ m high. Find its total surface area, and the cost of painting it at ₹120 per square metre. Then find the tin sheet needed, in square metres, for an open cubical box of edge $40$ cm, and its cost at ₹250 per square metre.

Step 1. For the tank, identify $l = 2.5$ m, $b = 2$ m, $h = 1.6$ m, and apply the closed-cuboid TSA formula: $TSA = 2(lb+bh+hl) = 2[(2.5\times2)+(2\times1.6)+(1.6\times2.5)] = 2[5+3.2+4] = 2\times12.2 = 24.4\ \text{m}^2$ .

Step 2. Multiply by the rate to get the painting cost: $\text{Cost} = 24.4 \times 120 = ₹2928$ .

Step 3. For the open cubical box, convert the edge to metres first: $a = 40\ \text{cm} = 0.4\ \text{m}$ . Since it is open at the top, use the 5-face formula: $\text{Area} = 5a^2 = 5\times(0.4)^2 = 5\times0.16 = 0.8\ \text{m}^2$ .

Step 4. Multiply by the tin-sheet rate: $\text{Cost} = 0.8 \times 250 = ₹200$ .

The tank's total surface area is 24.4 m², costing ₹2928 to paint; the open box needs 0.8 m² of tin sheet, costing ₹200.

Common traps

Forgetting the factor of 2 in $2(lb+bh+hl)$ — this factor already accounts for the pair of each face, so it must appear exactly once, never dropped and never doubled again.
Applying the closed-solid formula to an open box — always check for words like "open," "no lid," or "without a top," and subtract the correct single face ( $lb$ for a cuboid, $a^2$ for a cube) before finishing.
Mixing units mid-calculation — convert every dimension to the same unit (as with the 40 cm to 0.4 m step above) before substituting into any formula, never after.
Confusing LSA with TSA — "painting the walls of a room" usually needs LSA (four walls, sometimes plus ceiling), while "material for a closed box" needs the full TSA of all six faces.
Treating a hollow box's material as just the outer TSA — a hollow container with wall thickness also has an inner surface and visible edges, so material problems for such boxes usually need the outer and inner surfaces considered together, not the outer alone.

Ask the tutor: Unsure whether your problem needs TSA or LSA, or want a walkthrough of a hollow-box wall-thickness question? Ask the AI tutor to work through it with your own numbers.

How is lateral surface area different from total surface area, in practice?

Total surface area covers all six faces of a cuboid (or all six faces of a cube), while lateral surface area covers only the four "standing" side faces and excludes the top and bottom. Use LSA whenever a problem is really about the walls of something — painting the four walls of a room, wrapping the side of a box — and use TSA whenever the object is fully closed or the question asks for the material to cover it completely, top and bottom included.

Why does an open box subtract just one face, and how do I know which one?

An open box is a closed box missing exactly one face — usually the top, since most "open boxes" or "tanks without a lid" are open upward. So you take the full TSA and subtract the area of that single missing face: $lb$ for a cuboid open at the top (since the top and bottom are both $l\times b$ ), or $a^2$ for a cube. Always re-read the problem to confirm which face is missing; if it says "open on all sides except the base," you are removing the top face, not a side.

For a hollow box with wall thickness, why can't I just use the outer dimensions?

Because a hollow box, like a wooden crate or a metal tank with real wall thickness, has two surfaces — an outer one (the outside dimensions) and an inner one (the outside dimensions minus twice the thickness on each side). If the question asks how much material was used to build the walls, you generally need both surfaces (or the exposed edge strips) accounted for, not just the outer TSA, since the material forms the visible inner lining too. Read carefully whether the question wants only the outer surface (for painting the outside) or the full material used (both surfaces).

Cube and cuboid surface area

The core idea

Derivation of the cuboid total surface area

Worked example

Common traps

Before this

Where this leads