Pythagoras' theorem in three dimensions

A shipping crate is six metres long, four metres wide and three metres tall. A rod lying flat on the floor, from one corner to the opposite corner, fits inside. Tilt the rod up so its top end reaches the far top corner, and a longer rod fits. How long is the longest rod the crate can hold? The answer is hiding in a flat right-angled triangle inside the box, and spotting that triangle comes before any sum.

This is video eight of eleven in Pythagoras, Trigonometry and Coordinate Geometry. Every step here is flat Pythagoras, so if that needs a refresh, Pythagoras' theorem in two dimensions is the one to watch first.

Pythagoras in three dimensions is set at Higher, and at Extended on courses with Core and Extended tiers.

Pythagoras in three dimensions starts with a hunt, not a formula. Somewhere in the solid is a flat right-angled triangle with the length you want as one of its sides. The rod runs from a bottom corner of the crate to the top corner diagonally opposite. That line is called the space diagonal, because it cuts through the space inside the box rather than along a face. From the rod's top end, go straight down the upright edge of the crate to the floor. That edge is three metres, and it lands on the far floor corner. So, that gives a triangle: the floor diagonal along the bottom, the three-metre edge standing up, and the rod as the long side. Which corner is the right angle: the rod's lower end, its upper end, or the upright's foot? The foot of the upright. It rises straight up, so it makes ninety degrees with any line on the ground beneath it. Every angle faces its own side - look straight across. Stand in that square corner at the foot of the upright and look across the triangle: the side you see is the rod, the hypotenuse, the one you're after. The floor diagonal isn't given, so it needs a triangle of its own first. On the floor, the six-metre length and the four-metre width meet at a square corner, with the floor diagonal as the long side.

Two flat triangles, then: one lying on the floor, and one standing up through the middle of the box. The floor triangle comes first, and the floor diagonal is its longest side, so the squares are added. What is the floor diagonal squared, from the six and the four? Fifty-two. Six squared is thirty-six, four squared is sixteen, and thirty-six plus sixteen is fifty-two. The floor diagonal is root fifty-two metres. Leave it as a root and don't round it, because it's about to be squared again, and a rounded value squared drifts away from fifty-two. Now the upright triangle. The rod is its longest side, so add the squares again: root fifty-two squared, plus three squared. What does root fifty-two squared plus three squared come to? Sixty-one. Root fifty-two squared is fifty-two, straight back, and three squared is nine. The rod squared is sixty-one, so the rod is root sixty-one metres. Press the square root key, sixty-one, equals, and the display shows root sixty-one. The S to D key turns it into seven point eight one zero two four nine six seven six. To one decimal place, the longest rod the crate can hold is seven point eight metres. Right, look at what happened to the fifty-two. It came from the six and the four, and it went into the next step without ever being rooted. Now, what single line gives the rod straight from the six, the four and the three? The square root of six squared plus four squared plus three squared. That's root thirty-six plus sixteen plus nine, root sixty-one again. That, right there, is Pythagoras in three dimensions. For a cuboid with edges a, b and c, the space diagonal d equals the square root of a squared plus b squared plus c squared. It works because the floor diagonal squared is a squared plus b squared, and the upright triangle adds c squared on top. Two flat triangles fold into one line.

Now, a tent is a square-based pyramid: a square floor, with four sloping sides meeting at a top point called the apex. This one has a square base of side four metres. A pole three metres tall rises from the centre of the base straight up to the apex. A sloping edge runs from a base corner up to the apex, and its length is wanted. More than one triangle stands inside a tent, and they don't all help. Which triangle holds the pole and the sloping edge together? The one with the pole, half the base diagonal, and the sloping edge. Its square corner is at the foot of the pole. The triangle through the middle of a base edge holds the pole but not the sloping edge. Going that way adds an extra triangle before the edge appears. A student writes the sloping length squared as two squared plus three squared, with the two metres taken from the base. What is wrong with the two metres in that student's line? It's half the side, the distance from the centre to the middle of an edge. A corner is further out, at half the diagonal. Half the diagonal isn't given, so it has to be calculated. From the centre, a corner is two metres across and two metres along, with a square corner between those two moves. Two squared plus two squared is four plus four, which is eight. Half the diagonal is root eight metres. Root eight is about two point eight three, but it stays as root eight. The next step squares it, and root eight squared is exactly eight. Now the upright triangle: root eight along the floor, the three-metre pole standing up, and the sloping edge as the longest side. What is the sloping edge squared, using root eight and the three? Seventeen. Root eight squared is eight, three squared is nine, and eight plus nine is seventeen. The sloping edge is root seventeen. Press the square root key, seventeen, equals, then the S to D key, and the display reads four point one two three one zero five six two six. To one decimal place, each sloping edge is four point one metres. One examiner's report, on a pyramid question, says this. "Many candidates started by trying to find BE but took length BO as two point eight centimetres - these candidates scored a maximum of one mark." Those students used an in-between length - one they had not calculated. The fix is to calculate every in-between length, never read one off the diagram, even when a guess looks close. Further on, the same report adds this. "Only the very most able candidates used Pythagoras on the triangle with vertices E, O and the midpoint of BC." That triangle is the pole, the centre and the middle of a base edge, and it gives the height of a face. The habit, then, is to pick the triangle that holds the length you want next to lengths you know.

The crate's rod came out of two flat triangles folded into one line. Now some boxes and tents you haven't seen. What is Pythagoras in three dimensions, for the space diagonal of a cuboid? The square root of a squared plus b squared plus c squared, with a, b and c the length, width and height. Now try a box two by three by six centimetres. How long is its space diagonal? Exactly seven centimetres, since four, nine and thirty-six add to forty-nine. Now the same tent: how high is a face, from the middle of a base edge up to the apex? About three point six metres. That triangle is the pole and half a side, two metres. Two squared plus three squared is thirteen, so the face is root thirteen, the same sum the student wrongly used for the edge.

No need to see this twice once it's landed: a thumbs-up files it as finished, so it never comes back round in your revision. If it's still hazy, save it for later. Meanwhile, run a finger from corner to corner of any cardboard box, because seeing the space diagonal in your hands makes the drawing easier.

Next in the chapter: Finding lengths in three-D using Pythagoras and trigonometry.

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