Trigonometry in 3D: Cuboids

Find lengths through a cuboid with Pythagoras twice, find the angle a diagonal makes with a face, and work back from an angle to a height or a volume.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Edexcel Further Pure Maths 4PM1, Cambridge O Level Maths 4024 · 49 real past-paper questions below

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A Cuboid Hides Right Triangles

A length that goes through a cuboid, corner to corner, is the long side of a right-angled triangle. So is every angle question.

You find it in two stages: first a diagonal across one face, then the line through the middle.

Formula

d^{2} = a^{2} + b^{2} + c^{2}

$a$ , $b$ and $c$ are the length, width and height. $d$ is the diagonal through the cuboid.

Key idea

The formula is just Pythagoras twice in one line. Draw the flat triangle each time, and put the right angle in first.

Key idea

Puzzle: A box is 12 cm by 4 cm by 3 cm. Will a 12.5 cm pencil fit inside? Answer: yes. Corner to corner through the box, $1 2^{2} + 4^{2} + 3^{2} = 169 = 13$ cm, so it fits on the long diagonal.

The Length Through the Cuboid

The 3-mark length question. Cuboid $A B C D E F G H$ has $E H = 4$ cm, $H G = 4$ cm and the edge $A E = 7$ cm. Find $A G$ .

Worked example

1
$E G^{2} = 4^{2} + 4^{2}$ — Triangle $E H G$ lies flat on the face $E F G H$ , with its right angle at $H$ . $E G$ is the longest side, so square the two short sides and add.
2
$E G^{2} = 16 + 16 = 32$ — Square each edge: 4 × 4 = 16. Then add the two squares.
3
$E G = 32$ — Take the square root, but leave it as a root. Rounding now would make the final answer slightly out.
4
$A G^{2} = 7^{2} + (32)^{2}$ — Now triangle $A E G$ . $A E$ stands straight up from the face, so its right angle is at $E$ . $A G$ is the longest side.
5
$A G^{2} = 49 + 32$ — Squaring a square root cancels it: $(32)^{2} = 32$ . Also 7 × 7 = 49.
6
$A G^{2} = 81$ — Add the two squares.
7
$A G = 9 cm$ — Take the square root to undo the squaring. A length cannot be negative, so only the positive root counts.

Common mistake

Adding the edges: 4 + 4 + 7 = 15. Lengths do not add like that. Square, add, then square root.

Key idea

Show both triangles. Most of the marks are for the working, so the face diagonal line still earns a mark even if the last line goes wrong.

The Angle With a Face

The 4-mark question: find the angle between $A G$ and the face $E F G H$ . Use the same triangle $A E G$ .

Worked example

1
$triangle A E G, right angle at E$ — $A G$ meets the face $E F G H$ at $G$ . Its shadow on the face is $E G$ , so the angle $θ$ we want is at $G$ , between $A G$ and $E G$ .
2
$sin θ = \frac{7}{9}$ — $A E$ is across from $θ$ , so it is the opposite. $A G$ is the longest side, the hypotenuse. Opposite and hypotenuse mean sin.
3
$θ = sin^{- 1} (\frac{7}{9})$ — We know the ratio and want the angle. Inverse sine, $sin^{- 1}$ , undoes sin.
4
$θ = 51. 1^{\circ}$ — Use the calculator in degree mode. Round angles to 1 decimal place.

Key idea

You can use tan instead: $tan θ = \frac{7}{32}$ gives the same 51.1°. Use whichever two sides you already have.

Common mistake

Using the edge $H G = 4$ as the adjacent side. The side on the face must run from G to the point below A, which is the diagonal.

Working Back From an Angle

The 5-mark question gives the angle and asks for a height or a volume. Find the face diagonal, then use tan to get the height.

A cuboid's base is 5 cm by 4 cm. The diagonal through it makes 35° with the base. Find the volume.

Worked example

1
$d^{2} = 5^{2} + 4^{2}$ — The diagonal of the base is the longest side of a right triangle with sides 5 and 4. Square each and add.
2
$d = 41$ — 25 + 16 = 41. Take the square root and leave it exact.
3
$tan 3 5^{\circ} = \frac{h}{41}$ — The long diagonal, its shadow $d$ on the base, and the height $h$ make a right triangle. $h$ is across from the 35° angle (opposite) and $d$ is next to it (adjacent). Opposite and adjacent mean tan.
4
$h = 41 \times tan 3 5^{\circ}$ — $h$ is divided by $41$ , so multiply both sides by $41$ .
5
$h = 4.4835 \dots$ — Type it into the calculator in degree mode. Keep the full number for the next line.
6
$V = 5 \times 4 \times 4.4835$ — The volume of a cuboid is length × width × height.
7
$V = 89.7 cm^{3}$ — Multiply, then round to 3 significant figures and add the unit.

Common mistake

Using the base edge 5 instead of the diagonal. The angle sits between the long diagonal and the base diagonal.

Finding an Edge From the Diagonal

Exams often give you the long diagonal and ask for an edge. You work backwards, and the sign changes.

Formula

c^{2} = d^{2} - a^{2} - b^{2}

$d$ is the diagonal through the cuboid. $a$ and $b$ are two edges you know. $c$ is the edge you want. The diagonal is the longest side, so it is the one you subtract from.

A cuboid $A B C D E F G H$ has $C G = 6$ cm and a diagonal $A G = 21$ cm. The base has $A B = 2 \times B C$ . Find $A B$ .

Worked example

1
$let B C = x, A B = 2 x$ — Call the short base edge $x$ . The question says $A B$ is twice as long.
2
$A C^{2} = (2 x)^{2} + x^{2}$ — Triangle $A B C$ is in the base, with its right angle at $B$ . $A C$ is the base diagonal, the longest side.
3
$A C^{2} = 4 x^{2} + x^{2} = 5 x^{2}$ — Squaring $2 x$ gives $4 x^{2}$ : the 2 is squared too. Then collect the like terms.
4
$2 1^{2} = 5 x^{2} + 6^{2}$ — Triangle $A C G$ has its right angle at $C$ , because $C G$ stands straight up. $A G^{2} = A C^{2} + C G^{2}$ .
5
$441 = 5 x^{2} + 36$ — 21 × 21 = 441 and 6 × 6 = 36.
6
$5 x^{2} = 405$ — Take 36 from both sides, to leave the $x^{2}$ term on its own.
7
$x^{2} = 81$ — Divide both sides by 5.
8
$x = 9$ — Take the square root. A length cannot be negative, so ignore −9.
9
$A B = 2 \times 9 = 18 cm$ — The question asked for $A B$ , which is twice $x$ . Check: $1 8^{2} + 9^{2} + 6^{2} = 441 = 2 1^{2}$ .

Common mistake

Adding the squares when you want an edge. The diagonal is the longest side, so its square is the biggest: subtract the others from it.

Key idea

If the question says “show that”, write the unrounded value first, for example 13.04…, then the rounded one. The unrounded line is what proves you worked it out.

Which Method? Read the Question

Find the flat right triangle first. Then pick the tool.

You see	You do
Length corner to corner	face diagonal, then Pythagoras again
Angle with a face or base	right triangle with the face diagonal
Diagonal given, find an edge	subtract the other squares from the diagonal squared
A prism with a sloping top	Angle Between a Line and a Plane
Angle given, find a height	face diagonal, then tan
The line starts mid edge	Angle Between a Line and a Plane
A pyramid or a cone	Trigonometry in 3D: Pyramids and Cones
A flat right triangle only	Using Trigonometry and Pythagoras Together

Key idea

Keep the face diagonal as a square root, such as $32$ . It stays exact for the next step.

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