Pythagoras Theorem in 3D
Pythagoras' theorem in 3D finds lengths inside solid objects by using two right-angled triangles in turn. To find the space diagonal of a rectangular prism, first find a diagonal of the base, then use it with the height. In NSW Year 10 Advanced maths this is applied to prisms, cubes, cylinders, cones and pyramids (NESA outcome MA5‑TRG‑P‑01).
You will find space diagonals of boxes and cubes, the height of a traffic cone, the radius of a jar from a chopstick that fits inside it, and the sloping edges, slant heights and heights of square pyramids. Practical problems include a cable from the middle of a ceiling to a floor corner, a crowbar inside a toolbox and a flagpole at the corner of a sports field. Harder questions form and solve a quadratic equation for the dimensions of a box. It is part of Year 10 Advanced Measurement.
Every question with a fully worked solution.
- Pythagoras Theorem in 3D - Video - Pythagorean Theorem 3D Watch
Theory
Pythagoras' theorem in 3D finds lengths inside solids by spotting right-angled triangles, often in two steps: across a face, then up through the solid. This Year 10 Advanced maths page covers content that comes up if you are heading for Standard or Advanced maths in senior years: space diagonals of boxes and cubes, square pyramids, cones and cylinders, and equations built from \(D^2 = a^2 + b^2 + c^2\).
A face diagonal joins opposite corners of one face of a solid. A space diagonal joins opposite corners of the whole solid and passes through its inside.
In 3D problems the right angle is often hidden: a vertical edge is at right angles to every line drawn on the floor through its foot. So the space diagonal, the base diagonal and a vertical edge form a right-angled triangle, even though it does not look right-angled in a drawing.
For a square pyramid with apex \(T\) above the centre \(O\) of the base, the perpendicular height \(TO\) is vertical, the slant height runs from \(T\) to the midpoint of a base edge, and a sloping edge runs from \(T\) to a corner.
For a cone, the height, a radius and the slant height make a right-angled triangle, with the slant height as hypotenuse.
Space diagonal of a rectangular prism with sides \(a\), \(b\) and \(c\):
Cube with edge \(a\):
Square pyramid with base edge \(2a\), height \(h\), slant height \(s\) and sloping edge \(e\):
How to solve a 3D Pythagoras problem
- Sketch the solid and mark the length you want.
- Find a right-angled triangle that contains it. Look for a vertical line meeting the floor, or a line from the apex to the middle of an edge.
- Find the missing side of that triangle first, often a diagonal across a face, in a second right-angled triangle.
- Apply Pythagoras twice. Keep squared values or exact surds between steps; round only at the end.
Common set-ups
- Box or room: floor diagonal, then up to the opposite top corner. From the centre of a ceiling, go half the length and half the width along the floor first.
- Point on an edge: drop it straight down to the floor, find the floor distance, then go up (e.g. to the midpoint of a top edge).
- Wedge or ramp: find the length of the sloping face first, then use it with the width.
- Square pyramid: use \(TOM\) for height and slant height, \(TMC\) for slant height and sloping edge, or \(TOC\) with half a diagonal for height and sloping edge.
- Cone and cylinder: use the radius for a cone; for a rod across a cylinder use the diameter.
- Working backwards: if the hypotenuse is known, subtract squares: \(h^2 = 2.1^2 - (1.6^2 + 0.9^2)\).
- Unknown dimensions: write each side in terms of \(x\), use \(D^2 = a^2 + b^2 + c^2\), expand and solve. Reject negative lengths (Example 4).
(i) the exact length of the base diagonal \(AC\)
(ii) the length of the space diagonal \(AG\).
Sketch the box and the two right-angled triangles: \(ABC\) on the base, then \(ACG\) standing up, with the right angle at \(C\).
(i)
| \(AC^2\) | \(=\) | \(12^2 + 4^2 = 144 + 16 = 160\) |
| \(AC\) | \(=\) | \(\sqrt{160} = \mathbf{4\sqrt{10}}\text{ cm}\) |
(ii) Use \(AC^2 = 160\) directly; there is no need to round \(AC\).
| \(AG^2\) | \(=\) | \(AC^2 + CG^2\) |
| \(\) | \(=\) | \(160 + 3^2 = 169\) |
| \(AG\) | \(=\) | \(\mathbf{13}\text{ cm}\) |
Check with the one-step formula: \(\sqrt{12^2 + 4^2 + 3^2} = \sqrt{169} = 13\).
(i) the slant height \(TM\)
(ii) the perpendicular height \(TO\), correct to \(1\) decimal place.
(i) In the face \(TBC\), \(TM\) meets \(BC\) at right angles and \(MC = 5\) cm, half a base edge. The sloping edge \(TC\) is the hypotenuse.
| \(TM^2\) | \(=\) | \(13^2 - 5^2 = 169 - 25 = 144\) |
| \(TM\) | \(=\) | \(\mathbf{12}\text{ cm}\) |
(ii) Triangle \(TOM\) stands upright with the right angle at \(O\). \(OM\) is also half a base edge, \(5\) cm, and now \(TM\) is the hypotenuse.
| \(TO^2\) | \(=\) | \(12^2 - 5^2 = 144 - 25 = 119\) |
| \(TO\) | \(=\) | \(\sqrt{119} = 10.908\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{10.9}\text{ cm}\) |
The height is shorter than the slant height, which is shorter than the sloping edge, as it must be.
The longest pencil runs from the bottom rim to the opposite point on the top rim. It is the hypotenuse of a vertical triangle whose other sides are a diameter of the base and the height.
| \(\text{diameter}\) | \(=\) | \(2 \times 4 = 8\text{ cm}\) |
| \(L^2\) | \(=\) | \(8^2 + 15^2 = 64 + 225 = 289\) |
| \(L\) | \(=\) | \(\mathbf{17}\text{ cm}\) |
Using the radius instead of the diameter gives \(\sqrt{241} \approx 15.5\) cm, which is too short.
For a box with sides \(a\), \(b\) and \(c\), the space diagonal satisfies \(D^2 = a^2 + b^2 + c^2\).
| \(x^2 + (x + 1)^2 + 2^2\) | \(=\) | \((x + 5)^2\) |
| \(x^2 + x^2 + 2x + 1 + 4\) | \(=\) | \(x^2 + 10x + 25\) |
| \(2x^2 + 2x + 5\) | \(=\) | \(x^2 + 10x + 25\) |
| \(x^2 - 8x - 20\) | \(=\) | \(0\) |
Factorise and solve:
| \((x - 10)(x + 2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(10 \text{ or } x = -2\) |
A length cannot be negative, so \(x = 10\). The box is \(\mathbf{10}\) cm by \(\mathbf{11}\) cm by \(\mathbf{2}\) cm with a space diagonal of \(\mathbf{15}\) cm. Check: \(100 + 121 + 4 = 225 = 15^2\ \checkmark\)
Common pitfalls
Frequently asked questions
How do you use Pythagoras' theorem in 3D?
Find a right-angled triangle inside the solid that contains the length you want. If one of its other sides is unknown, find it first with Pythagoras in another right-angled triangle, usually on a face, then apply Pythagoras again.
What is the formula for the space diagonal of a rectangular prism?
\(D = \sqrt{a^2 + b^2 + c^2}\), where \(a\), \(b\) and \(c\) are the length, width and height. For a cube with edge \(a\) it is \(a\sqrt{3}\).
How do you find the height of a square pyramid?
Use the right-angled triangle from the apex to the centre of the base to the midpoint of an edge: \(h^2 = s^2 - a^2\), where \(a\) is half the base edge. Or use half a diagonal with a sloping edge.
What is the difference between the slant height and the height of a pyramid?
The height is measured straight down from the apex to the centre of the base. The slant height is measured down the middle of a triangular face to the midpoint of a base edge, so it is longer.
How do you find the longest rod that fits in a box?
The longest straight rod lies along a space diagonal. Find it with \(\sqrt{\ell^2 + w^2 + h^2}\).