Pythagoras Theorem in 3D
Every question with a fully worked solution.
- Pythagoras Theorem in 3D - Video - Pythagoras Theorem in 3D Watch
Theory
To use Pythagoras' theorem in 3D, find a right-angled triangle inside the solid, often in two steps. This Year 9 maths page (the harder end of spatial problems) shows how to find the space diagonal of a box, the longest rod in a cylinder, and the height, slant height and sloping edge of cones and square pyramids, with answers as decimals or surds.
Pythagoras' theorem says that in a right-angled triangle \(c^{2} = a^{2} + b^{2}\), where \(c\) is the hypotenuse, the side opposite the right angle. In a 3D problem the triangle is hidden inside a solid. Your job is to find it and draw it flat.
The base diagonal of a rectangular prism runs across the base from corner to opposite corner. The space diagonal runs through the inside of the solid from a bottom corner to the opposite top corner. It is the longest straight line that fits inside.
The perpendicular height of a cone or pyramid goes straight down from the apex to the centre of the base. The slant height runs down the sloping surface: for a cone, from the apex to the edge of the base; for a pyramid, from the apex to the midpoint of a base edge. The sloping edge of a pyramid runs from the apex to a corner.
Every one still comes down to a right-angled triangle, drawn flat.
For a rectangular prism with length \(l\), width \(w\) and height \(h\):
For a cone with radius \(r\), height \(h\) and slant height \(s\):
| Solid | Right-angled triangle | Hypotenuse |
|---|---|---|
| Rectangular prism (base) | length, width, base diagonal | base diagonal |
| Rectangular prism (space) | base diagonal, height, space diagonal | space diagonal |
| Cylinder | diameter, height, longest diagonal | diagonal |
| Cone | radius, height, slant height | slant height |
| Square pyramid (face) | half the side, height, slant height | slant height |
| Square pyramid (edge) | half the base diagonal, height, sloping edge | sloping edge |
How to solve a 3D Pythagoras problem
- Sketch the solid and mark the length you want.
- Find a right-angled triangle that contains it. If that triangle has only one known side, find a second triangle first (usually across the base).
- Draw the triangle flat and label the hypotenuse, the side opposite the right angle.
- Apply Pythagoras: add the squares for the hypotenuse, subtract for a shorter side.
- Keep exact values (such as \(d^{2} = 41\) or a surd) between steps and round only at the end.
Useful facts
- Square pyramid: the centre of the base is half the side from each edge, and half the base diagonal from each corner.
- Cylinder or cone: use the radius from the centre and the diameter across the whole base.
- Will it fit? Compare the object's length with the space diagonal or the cylinder's diagonal.
- Square base of side \(x\): the space diagonal gives \(D^{2} = x^{2} + x^{2} + h^{2} = 2x^{2} + h^{2}\). Substitute and solve for \(x^{2}\).
- Wedge or ramp: the sloping face is a rectangle. Find its sloping length first (from the base length and the height), then use it with the width to get the diagonal of that face.
- Exact answers: if the total under the root is not a perfect square, leave the root sign in, such as \(\sqrt{41}\).
Find the base diagonal \(d\) first, then use it with the height. Keep \(d^{2}\) as it is; there is no need to find \(d\).
| \(d^{2}\) | \(=\) | \(10^{2} + 6^{2}\) |
| \(\) | \(=\) | \(136\) |
| \(D^{2}\) | \(=\) | \(d^{2} + 5^{2}\) |
| \(\) | \(=\) | \(136 + 25 = 161\) |
| \(D\) | \(=\) | \(\sqrt{161}\) |
| \(\) | \(=\) | \(12.6885\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{12.7}\text{ cm}\) |
Shortcut: \(D^{2} = l^{2} + w^{2} + h^{2}\).
(i) exactly, as a surd
(ii) correct to one decimal place.
The radius is half the diameter: \(r = 4\) cm. The height, the radius and the slant height form a right-angled triangle, with the slant height as the hypotenuse.
| \(s^{2}\) | \(=\) | \(4^{2} + 6^{2}\) |
| \(\) | \(=\) | \(16 + 36 = 52\) |
| \(\text{(i) } s\) | \(=\) | \(\mathbf{\sqrt{52}}\text{ cm}\) |
| \(\) | \(=\) | \(7.2111\ldots\) |
| \(\text{(ii) } s\) | \(\approx\) | \(\mathbf{7.2}\text{ cm}\) |
\(52\) is not a perfect square, so the exact answer keeps the root sign. \(7.2\) cm is longer than the height, as a slant height must be.
(i) the slant height \(TM\)
(ii) the length of the sloping edge \(TC\), correct to one decimal place.
(i) \(OM\) is half the side, \(4\) m. Triangle \(TOM\) has its right angle at \(O\).
| \(TM^{2}\) | \(=\) | \(3^{2} + 4^{2} = 25\) |
| \(TM\) | \(=\) | \(\mathbf{5}\text{ m}\) |
(ii) Triangle \(TMC\) has its right angle at \(M\), and \(MC\) is half the side, \(4\) m.
| \(TC^{2}\) | \(=\) | \(5^{2} + 4^{2} = 41\) |
| \(TC\) | \(=\) | \(\sqrt{41}\) |
| \(\) | \(=\) | \(6.4031\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{6.4}\text{ m}\) |
Check another way: half the base diagonal gives \(OC^{2} = 4^{2} + 4^{2} = 32\), and \(TC^{2} = 3^{2} + 32\) \(= 41\).
The rod is the hypotenuse. The height and the diameter are the two shorter sides, so subtract.
| \(d^{2} + 24^{2}\) | \(=\) | \(30^{2}\) |
| \(d^{2}\) | \(=\) | \(900 - 576\) |
| \(\) | \(=\) | \(324\) |
| \(d\) | \(=\) | \(\sqrt{324} = \mathbf{18}\text{ cm}\) |
The radius would be \(9\) cm. Any container narrower than \(18\) cm across is too small for the rod.
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 it only has one known side, first find another side in a second triangle, usually across the base. Then apply \(c^{2} = a^{2} + b^{2}\).
How do you find the space diagonal of a rectangular prism?
Square the length, width and height, add them, and take the square root: \(D = \sqrt{l^{2} + w^{2} + h^{2}}\). This is the same as finding the base diagonal first and then using it with the height.
What is the difference between slant height and perpendicular height?
The perpendicular height goes straight down from the apex to the centre of the base. The slant height runs down the sloping surface. The slant height is the hypotenuse of the triangle, so it is always longer.
How do you find the slant height of a cone?
Use the radius and the perpendicular height as the two shorter sides: \(s^{2} = r^{2} + h^{2}\).
What is the longest object that fits in a box?
The longest straight object lies along the space diagonal, from a bottom corner to the opposite top corner: \(\sqrt{l^{2} + w^{2} + h^{2}}\).