Cuboid: Volume, Surface Area, Diagonals and Fitting Things In

What is a cuboid, and how is it different from a cube?

A cuboid, also called a rectangular box or a rectangular prism, is a solid bounded by six rectangles that meet at right angles; its opposite faces are equal and parallel. Every corner is a right-angled corner, with three faces and three edges meeting there. The 12 edges come in three groups of four equal parallel edges, called the length l, the width w and the height h, and the whole solid is fixed by those three numbers.

The count obeys Euler’s formula, V-E+F=8-12+6=2, exactly as for the cube. The difference between the two is only in the edges: a cube is a cuboid whose three edges are equal, so every cube is a cuboid, and a cuboid is a cube only when l=w=h. In Indonesian the box is balok and the cube kubus, and schools write its edges as panjang p, lebar l and tinggi t. Place a corner at the origin and the opposite corner is (l,w,h): the eight corners are all the points whose coordinates are 0 or the full edge, which is why the cube and the box share their counts. A brick, a book, a shoe box, a room and a shipping container are all cuboids.

How do you find the volume, the surface area and the lateral area of a cuboid?

The volume of a cuboid is V=l× w× h, the surface area is S=2(lw+lh+wh), and the lateral area is 2h(l+w). The volume counts unit cubes: a floor of l× w of them, piled h layers high, so V is the base area times the height. For a box 5×3×2: V=30, and S=2(15+10+6)=62.

The surface formula comes from the net, the six rectangles laid flat. They come in three equal pairs, lw, lh and wh, so the surface is twice their sum; for 4×3×2 the pairs are 12, 8 and 6, and 2(12+8+6)=52. The lateral area leaves out the top and the bottom and keeps the four walls: they make one long rectangle as wide as the perimeter of the base, 2(l+w), and h high, so the area is 2h(l+w). For a box 5×3×2 that is 2·2·8=32, and the surface is the walls plus top and bottom: 32+2·15=62.

Everyday uses. To paint the walls of a room 6 m long, 4 m wide and 3 m high you need the lateral area 2·3·(6+4)=60 m² (minus doors and windows); with the ceiling and the floor the whole surface is 108 m². An aquarium 80 cm by 35 cm by 40 cm holds 80·35·40=112 000 cm³ =112 litres, since 1000 cm³ is one litre. Scaling: multiplying every edge by k multiplies the surface by k^2 and the volume by k^3; multiplying only one edge by k multiplies the volume by k. Slide the three edges below to see every measure at once.

What are the diagonals of a cuboid, and what angles do they make?

A cuboid has face diagonals of three lengths, √(l^2+w^2), √(l^2+h^2) and √(w^2+h^2), and four space diagonals, all of the length d=√(l^2+w^2+h^2). The face diagonal is the hypotenuse in the face; the space diagonal is the hypotenuse of a second right triangle on a face diagonal and an edge, so d^2=(l^2+w^2)+h^2 (this is the Pythagorean theorem in space). For the box 6×3×2 the three face diagonals are √(45)≈6.71, √(40)≈6.32 and √(13)≈3.61, and the space diagonal is √(36+9+4)=7 exactly.

The angles with the edges. The diagonal runs from (0,0,0) to (l,w,h), so its projection on the edge along l is l and \cosα=\frac ld; likewise \cosβ=\frac wd and \cos\gamma=\frac hd. For 6×3×2 the cosines are \frac67,\frac37,\frac27 and the angles 31.0^\circ, 64.6^\circ and 73.4^\circ. Their squares add up to (36+9+4)/(49)=1 for every box: \cos^2α+\cos^2β+\cos^2\gamma=1. In a cube all three angles are \arccos\frac1√3≈54.74^\circ.

Two space diagonals. The diagonals (l,w,h) and (-l,w,h) have dot product -l^2+w^2+h^2, so they cross at the acute angle with cosine \frac|-l^2+w^2+h^2|d^2; the other two kinds of pair give (|l^2-w^2+h^2|)/(d^2) and (|l^2+w^2-h^2|)/(d^2). For 6×3×2 these are (23)/(49),(31)/(49),(41)/(49), that is 62.0^\circ, 50.8^\circ and 33.2^\circ. In a cube every pair gives \frac13, the 70.53^\circ of the cube.

Which boxes have whole-number diagonals: Pythagorean quadruples, Euler bricks and the perfect cuboid?

A box with whole-number edges has a whole-number space diagonal when a^2+b^2+c^2=d^2, a Pythagorean quadruple, such as 2×3×6 with d=7 or 3×4×12 with d=13. Most boxes do not: the box 5×3×2 has d=√(38). The quadruples with d\le15 that cannot be shrunk to a smaller one are:

Multiplying a quadruple by a whole number gives another, such as 2×4×4 with d=6. There is a formula that makes them from any four whole numbers m,n,p,q: take a=m^2+n^2-p^2-q^2, b=2(mq+np), c=2(nq-mp) and d=m^2+n^2+p^2+q^2, and a^2+b^2+c^2=d^2. With (m,n,p,q)=(2,1,1,1) it gives (3,6,-2) and d=7, the box 2×3×6.

A harder wish is that the three face diagonals should be whole numbers. Such a box is an Euler brick, and the smallest is 44×117×240, whose face diagonals are √(44^2+117^2)=125, √(44^2+240^2)=244 and √(117^2+240^2)=267. Its space diagonal, though, is √(73 225)≈270.6, not whole. A perfect cuboid would have all three face diagonals and the space diagonal whole. Nobody has found one, nobody has proved that none exists, and computer searches through very large boxes have found none: a search through all edges up to 1500 finds eight Euler bricks and no perfect cuboid. Type edges in the widget to see which diagonals are whole.

Which cuboid has the largest volume for a given surface area?

For a given surface area S the cube holds the most: every cuboid satisfies V\le\left(\frac S6\right)^3/2, with equality only when l=w=h. The reason is the inequality of arithmetic and geometric means: for three positive numbers the geometric mean is at most the arithmetic mean. Apply it to the three face areas lw, lh and wh. Their product is (lw)(lh)(wh)=(lwh)^2=V^2 and their sum is \frac S2, so V^2/3=(lw)(lh)(wh)^(1/3)\le(lw+lh+wh)/(3)=\frac S6, and V\le\left(\frac S6\right)^3/2, with equality only if lw=lh=wh, which means l=w=h. For S=96 the bound is 16^3/2=64, reached by the cube of edge 4. Other boxes with the same surface:

The same fact the other way round: for a given volume the cube has the smallest surface, S\ge6V^2/3. For V=64 that is 96 for the cube, against 112 for 8×4×2, 136 for 16×2×2 and 160 for 8×8×1. And for a wire frame, where the twelve edges together are fixed, 4(l+w+h)=48 gives l+w+h=12 and V\le\left((12)/(3)\right)^3=64: again the cube. The same box problems with calculus, an open-top box cut from a sheet, are in applied optimization. Move the base of a box below and watch how far it falls short of the cube.

How do you know whether a rod, a plank or a pile of smaller boxes fits inside a box?

A thin straight rod fits inside a box exactly when its length is at most the space diagonal d=√(l^2+w^2+h^2), the longest straight segment the box contains; it just fits, corner to corner, when it is equal to d. A 1 m pole against a box 60×40×40 cm: d=√(3600+1600+1600)=√(6800)≈82.5 cm, so it does not fit, however it is tilted. In a box 80×50×40 cm, d=√(10 500)≈102.5 cm, so it does, leaning from a bottom corner towards the opposite top corner. A rod longer than every face diagonal must lie across the inside, never along a face.

A flat sheet is different: a rectangle fits in a box if it fits through some cross-section of it, and the largest is the diagonal rectangle through two opposite edges, l×√(w^2+h^2), of area l√(w^2+h^2); for the box 6×3×2 that is 6×√(13).

Smaller boxes. To know how many boxes a× b× c fit in a big box L× W× H when they are all laid the same way, count how many fit along each side: \left\lfloorL/a\right\rfloor·\left\lfloorW/b\right\rfloor·\left\lfloorH/c\right\rfloor, and try the six ways to turn the small box, taking the best. Bricks 3×2×1 in a box 10×6×4 fit 5·2·4=40 times, using each of the 240 units of volume (40·6=240); cartons 30×20×10 cm fill a space 120×80×60 cm with 4·4·6=96 cartons. The volume bound is always an upper limit: (V_big)/(V_small), and when the edges do not divide evenly, the leftover space is wasted. Turning some of the boxes the other way can sometimes fit more than any single arrangement, which is why packing is a hard problem.

How do you work with a cuboid in a program?

A program computes the measures of a box from three numbers, and finds whole-number diagonals by searching: loop over a\le b\le c, take the integer square root of a^2+b^2+c^2, and keep the boxes for which squaring it gives the sum back. Use whole numbers and an exact integer square root (isqrt in Python), not a floating-point sqrt rounded and compared, because a rounded square root of a large integer can be off by one. The programs below print the measures of 5×3×2, the quadruples with d\le9, whether 44×117×240 and 2×3×6 are Euler bricks, and whether the first has a whole space diagonal.

from math import isqrt, sqrt


def box(l, w, h):
    return {
        'volume': l * w * h,
        'surface': 2 * (l * w + l * h + w * h),
        'lateral': 2 * h * (l + w),
        'space diagonal': round(sqrt(l * l + w * w + h * h), 3),
    }


def is_square(n):
    return isqrt(n) ** 2 == n


def quadruples(max_d):
    """The boxes a <= b <= c whose space diagonal d <= max_d is a whole number: a² + b² + c² = d²."""
    found = []
    for a in range(1, max_d):
        for b in range(a, max_d):
            for c in range(b, max_d):
                s = a * a + b * b + c * c
                if s <= max_d ** 2 and is_square(s):
                    found.append((a, b, c, isqrt(s)))
    return sorted(found, key=lambda q: (q[3], q[:3]))


def euler_brick(a, b, c):
    """All three face diagonals are whole numbers."""
    return all(is_square(x * x + y * y) for x, y in ((a, b), (a, c), (b, c)))


print(box(5, 3, 2))                  # {'volume': 30, 'surface': 62, 'lateral': 32, 'space diagonal': 6.164}
print(quadruples(9))                 # [(1, 2, 2, 3), (2, 4, 4, 6), (2, 3, 6, 7), (1, 4, 8, 9), (3, 6, 6, 9), (4, 4, 7, 9)]
print(euler_brick(44, 117, 240), euler_brick(2, 3, 6))    # True False
print(is_square(44 ** 2 + 117 ** 2 + 240 ** 2))           # False: no perfect cuboid here

const box = (l, w, h) => ({ volume: l * w * h, surface: 2 * (l * w + l * h + w * h), lateral: 2 * h * (l + w), spaceDiagonal: +Math.sqrt(l * l + w * w + h * h).toFixed(3) })
const isSquare = (n) => Number.isInteger(Math.sqrt(n))

// the boxes a <= b <= c whose space diagonal d <= maxD is a whole number: a² + b² + c² = d²
function quadruples(maxD) {
  const found = []
  for (let a = 1; a < maxD; a++)
    for (let b = a; b < maxD; b++)
      for (let c = b; c < maxD; c++) {
        const s = a * a + b * b + c * c
        if (s <= maxD * maxD && isSquare(s)) found.push([a, b, c, Math.sqrt(s)])
      }
  return found.sort((p, q) => p[3] - q[3] || p[0] - q[0] || p[1] - q[1])
}

// all three face diagonals are whole numbers
const eulerBrick = (a, b, c) => [[a, b], [a, c], [b, c]].every(([x, y]) => isSquare(x * x + y * y))

box(5, 3, 2)                                   // { volume: 30, surface: 62, lateral: 32, spaceDiagonal: 6.164 }
quadruples(9)                                  // [[1,2,2,3], [2,4,4,6], [2,3,6,7], [1,4,8,9], [3,6,6,9], [4,4,7,9]]
;[eulerBrick(44, 117, 240), eulerBrick(2, 3, 6)]   // [true, false]
isSquare(44 ** 2 + 117 ** 2 + 240 ** 2)        // false

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

// 1 when n is a perfect square
int isSquare(int n) {
  int r = (int)(sqrt((double)n) + 0.5);
  return r * r == n ? 1 : 0;
}

// 1 when all three face diagonals are whole numbers
int eulerBrick(int a, int b, int c) {
  return isSquare(a * a + b * b) && isSquare(a * a + c * c) && isSquare(b * b + c * c) ? 1 : 0;
}

int main() {
  int l = 5;
  int w = 3;
  int h = 2;
  printf("volume %d; surface %d; lateral %d\n", l * w * h, 2 * (l * w + l * h + w * h), 2 * h * (l + w));
  printf("space diagonal %.3f\n", sqrt((double)(l * l + w * w + h * h)));

  // a <= b <= c with a * a + b * b + c * c = d * d, d up to 9
  for (int d = 2; d <= 9; d++) {
    for (int a = 1; a < d; a++) {
      for (int b = a; b < d; b++) {
        for (int c = b; c < d; c++) {
          if (a * a + b * b + c * c == d * d) printf("%d %d %d %d\n", a, b, c, d);
        }
      }
    }
  }
  printf("%d %d\n", eulerBrick(44, 117, 240), eulerBrick(2, 3, 6));
  printf("%d\n", isSquare(44 * 44 + 117 * 117 + 240 * 240));
  return 0;
}

/* output:
volume 30; surface 62; lateral 32
space diagonal 6.164
1 2 2 3
2 4 4 6
2 3 6 7
1 4 8 9
3 6 6 9
4 4 7 9
1 0
0
*/

Where does the study of boxes and their diagonals come from?

People have measured boxes for as long as there has been building, storing and trading, and the questions about their diagonals turned into number theory.

• About 300 BCE. Euclid’s Elements, Book XI, studies solids bounded by parallelograms and proves that two such solids of the same height are in the ratio of their bases (XI.32): the root of volume equals base times height.
• 1719. Paul Halcke finds the box 44×117×240 whose three face diagonals are whole numbers, the smallest of the bricks that bear Euler’s name; Euler later studied them and found families of others.
• The 18th and 19th centuries. The Pythagorean quadruples, the boxes with a whole space diagonal, are described and their formulas found, including the one that makes them from four whole numbers.
• The perfect cuboid. Whether a box can have all three face diagonals and the space diagonal whole at once is still open: not one is known, and searches by computer through very large boxes have found none.
• Today. The same formulas run inside every shipping calculator, 3D engine and storage layout: the bounding box of a shape in a game is a cuboid, and its diagonal is the quickest measure of its size.

What are the common mistakes with the cuboid?

The most common mistakes with the cuboid are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: the cuboid at a glance

• Parts: 6 faces in three equal pairs, 12 edges in three groups of four, 8 corners; a cube is the cuboid with l=w=h.
• Measures: V=lwh, S=2(lw+lh+wh), lateral area 2h(l+w), edges together 4(l+w+h); scaling by k gives k^2 and k^3.
• Diagonals: faces √(l^2+w^2), √(l^2+h^2), √(w^2+h^2); space d=√(l^2+w^2+h^2); \cosα=\frac ld and the three squared cosines add to 1.
• Whole diagonals: quadruples a^2+b^2+c^2=d^2 such as (2,3,6,7); Euler bricks such as 44×117×240; no perfect cuboid known.
• Extremes: for a given surface V\le\left(\frac S6\right)^3/2, for a given volume S\ge6V^2/3, equality only for the cube (AM–GM).
• Fitting: a rod fits if it is at most d; small boxes laid alike fit \lfloor L/a\rfloor\lfloor W/b\rfloor\lfloor H/c\rfloor, best of six turns.

Glossary: key terms

Cuboid

A solid with six rectangular faces meeting at right angles, in three pairs of equal faces; the box.

Length, width, height

The three edge lengths of a cuboid, written l, w and h, which fix the whole box.

Volume

The space inside the box: length times width times height, in cubic units.

Surface area

The total area of the six faces: twice the sum of l w, l h and w h.

Lateral area

The area of the four walls without the top and the bottom: 2 h times l plus w.

Space diagonal

The segment between opposite corners through the inside; its length is the square root of l squared plus w squared plus h squared.

Pythagorean quadruple

Four whole numbers a, b, c, d with a squared plus b squared plus c squared equal to d squared: a box with a whole space diagonal.

Euler brick

A box whose edges and three face diagonals are all whole numbers, such as 44 by 117 by 240.

Perfect cuboid

A box whose edges, face diagonals and space diagonal are all whole numbers; none is known.

AM–GM inequality

The geometric mean of positive numbers is at most their arithmetic mean, with equality only when they are all equal.

Frequently asked questions

References

1. Peter R. Cromwell (1997). Polyhedra. Cambridge University Press.
2. Thomas L. Heath (translator) (1956). The Thirteen Books of Euclid’s Elements (Book XI). Dover.
3. Richard K. Guy (2004). Unsolved Problems in Number Theory (3rd ed.), section on Euler bricks and the perfect cuboid. Springer.
4. J. Michael Steele (2004). The Cauchy–Schwarz Master Class. Cambridge University Press.
5. James Stewart, Lothar Redlin, Saleem Watson (2016). Precalculus: Mathematics for Calculus (chapter on geometry formulas). Cengage Learning.
6. Carl B. Boyer, Uta C. Merzbach (2011). A History of Mathematics (3rd ed.). Wiley.
7. Christer Ericson (2005). Real-Time Collision Detection. Morgan Kaufmann.

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