Cube and cuboid volume

A water tanker, a matchbox, and a brick are all the same shape mathematically — a cuboid — and the everyday question "how much fits inside" is really the question "what is its volume." Volume measures the space a solid occupies: litres a tank can hold, bricks that fit in a container, or metal needed to cast a block. For the cube and cuboid, ICSE Class 9 mensuration builds every volume formula from one idea already familiar from area — that stacking a flat base upward through a height sweeps out a solid, so volume equals base area times height. This section stays inside that scope: computing volumes of solid cubes and cuboids, using the cross-section-times-height shortcut, and handling hollow boxes where outer and inner dimensions differ. Cylinders, cones, and spheres belong to Class X and are not covered here.

The core insight

A cuboid's volume is base area times height ( $V = l \times b \times h$ , and $V = a^3$ for a cube of edge $a$ ), and this same base-times-height idea extended to a hollow box gives the material volume as outer volume minus inner volume.

The core idea

Cuboid and cube. For a cuboid with length $l$ , breadth $b$ , and height $h$ , and for a cube with edge $a$ :

V_{\text{cuboid}} = l \times b \times h, \qquad V_{\text{cube}} = a^3

where: $l, b, h$ are the three edge lengths of the cuboid in the same unit, and $a$ is the single edge length of the cube (all edges of a cube are equal, so length, breadth, and height are all $a$ ).

Cross-section times height. A cuboid has the same rectangular cross-section ( $l \times b$ ) at every level along its height, so its volume can equally be written as

V = (\text{area of base}) \times h

This view matters whenever a problem describes a solid built by stacking a fixed cross-sectional shape upward, or asks how far a liquid level rises in a tank of known base area when a known volume is poured in — the rise in level is simply (volume poured) divided by (base area).

Hollow cuboid or cube. If a box has outer dimensions $L, B, H$ and a uniform wall thickness $t$ all round, each inner dimension is reduced by $2t$ (the wall sits on both sides of that dimension), giving inner dimensions $L-2t,\ B-2t,\ H-2t$ . The volume of material used to build the box — or equivalently the difference between its outer and inner capacities — is

V_{\text{material}} = LBH - (L-2t)(B-2t)(H-2t)

The capacity of a hollow box (what it can actually hold) always uses the inner dimensions alone, never the outer ones.

Cost per unit volume. Once a volume is known, applied problems often multiply it by a rate — cost of casting a metal block per cubic centimetre, or cost of filling a tank per litre. The volume calculation and the cost calculation are two separate steps: get the volume right first (in the unit the rate is quoted in), then multiply by the rate. Keeping these steps separate avoids errors where a unit conversion gets skipped halfway through a rate calculation.

Figure

A solid cuboid sliced into equal horizontal layers stacked from base to top height, showing that volume builds up as base area repeated height times

Derivation Amaterial volume of a hollow box

Show that when a closed box has outer dimensions $L, B, H$ and a uniform wall thickness $t$ measured on every face, the volume of material used to construct it equals the outer cuboid volume minus the inner cuboid volume, that is $V_{\text{material}} = LBH - (L-2t)(B-2t)(H-2t)$ , and this correctly reduces to zero material when $t = 0$ .

Step 1. Picture the box as two nested cuboids: an outer shell of dimensions $L \times B \times H$ , and the empty space it encloses, which is itself a smaller cuboid — the inner cavity. The material of the box (wood, metal, or cardboard) fills exactly the region between these two nested solids.

Step 2. Find the inner cuboid's dimensions. The wall thickness $t$ is built onto every face, so along the length direction there is a wall of thickness $t$ at one end and another wall of thickness $t$ at the opposite end, together removing $2t$ from the outer length $L$ . The same reasoning applies independently to the breadth and the height, since the wall runs uniformly all round. This gives inner length $L - 2t$ , inner breadth $B - 2t$ , and inner height $H - 2t$ .

Step 3. Apply the cuboid volume formula to each nested solid separately: the outer volume is $V_{\text{outer}} = LBH$ , and the inner (cavity) volume is $V_{\text{inner}} = (L-2t)(B-2t)(H-2t)$ , both by the base-times-height rule already established.

Step 4. The material occupies exactly the outer solid with the inner cavity removed, so by simple subtraction of the two volumes, $V_{\text{material}} = V_{\text{outer}} - V_{\text{inner}} = LBH - (L-2t)(B-2t)(H-2t)$ .

Step 5. Check the limiting case: if $t = 0$ (no wall thickness), then $L - 2t = L$ , $B - 2t = B$ , $H - 2t = H$ , so $V_{\text{inner}} = LBH = V_{\text{outer}}$ and $V_{\text{material}} = 0$ — correctly, a box with walls of zero thickness uses no material.

Worked example

Problem. A cuboidal tank is 3 m long, 2.5 m wide, and 2 m deep. Find how many litres of water it can hold when full, and then find how many centimetres the water level rises if a further 3000 litres are poured in.

Step 1 — Volume of the tank. Using $V = l \times b \times h$ ,

V = 3 \times 2.5 \times 2 = 15 \text{ m}^3

Step 2 — Convert to litres. Since $1 \text{ m}^3 = 1000$ litres,

15 \text{ m}^3 \times 1000 = 15000 \text{ litres}

Step 3 — Convert the extra water to cubic metres. Reversing the same conversion, $3000$ litres $= 3000 / 1000 = 3 \text{ m}^3$ .

Step 4 — Use base area times height to find the rise in level. The tank's base area is $l \times b = 3 \times 2.5 = 7.5 \text{ m}^2$ , and since $V = (\text{base area}) \times (\text{rise})$ , the rise is

\text{rise} = \frac{3 \text{ m}^3}{7.5 \text{ m}^2} = 0.4 \text{ m} = 40 \text{ cm}

The tank holds 15,000 litres when full, and pouring in a further 3000 litres raises the level by 40 cm.

Common traps

Subtracting the wall thickness only once instead of twice when finding inner dimensions — a wall on both sides of a dimension removes $2t$ , not $t$ .
Mixing units before multiplying — combining metres and centimetres in one volume calculation gives a badly wrong answer; convert everything to a single unit first.
Forgetting the litre conversion factor — $1 \text{ m}^3 = 1000$ litres, equivalently $1000 \text{ cm}^3 = 1$ litre; check which direction the problem needs.
Using the cross-section shortcut on a non-uniform solid — $V = (\text{base area}) \times h$ only holds when every horizontal slice has the same shape and size, which is guaranteed for a cuboid but should never be assumed for an irregular solid.

Ask the tutor: Want help setting up a hollow-box or liquid-level volume problem step by step, or checking a unit conversion? Ask the AI tutor.

Why do we subtract 2t and not just t when finding the inner dimensions of a hollow box?

Because the wall thickness is built onto both ends of that dimension, not just one. Think of the length: there is a wall of thickness $t$ on the left face and another wall of thickness $t$ on the right face, and both eat into the same outer length $L$ . That removes $t + t = 2t$ in total, leaving inner length $L - 2t$ . The same applies separately to breadth and height.

Is the volume of material used the same as the capacity of the box?

No, and mixing these up is a very common mistake. Capacity is what the box can hold, which depends only on the inner dimensions: $V_{\text{inner}} = (L-2t)(B-2t)(H-2t)$ . The volume of material used to build the box is the difference between the outer and inner volumes, $V_{\text{outer}} - V_{\text{inner}}$ , because that difference is exactly the solid region occupied by the walls themselves.

Why does volume equal base area times height instead of just multiplying all three dimensions in any order?

They give the same number for a cuboid, since $l \times b \times h$ can be grouped as $(l \times b) \times h$ regardless of which face you call the base. But thinking of it as base area times height is more useful, because it extends directly to any solid with a uniform cross-section swept through a height — you find the cross-sectional area once, then multiply by however far it is repeated, which is also how a rise in liquid level is calculated.

Cube and cuboid volume

The core idea

Worked example

Common traps

Before this

Where this leads