From rectangle to parallelogram: area survives rearrangement

0:00Area is a number, and a number has to count something. So before anything else, take a square one unit on a side and agree that its area is one. That square is the unit, and every area after it is reported against it. A length can be laid against a ruler; an area has to be laid against a tile. Once the tile is chosen, finding an area becomes counting - and counting is something you can check.

0:31That one idea, pushed hard, produces a formula nobody had to guess. Start with the easiest shape there is. A rectangle five units by three: lay the unit squares in, row by row, and count them. Three rows of five, so fifteen. The strip is the step people skip. One by seven is seven units in a line, which is obvious; five by seven is five of those strips stacked, so thirty-five.

1:03For every rectangle from one by one up to eight by eight - sixty-four of them - laying the units out and adding gives exactly what multiplying the sides gives. Length times width is a shortcut for a count, and it is a shortcut for a rectangle. Now the two facts everything else leans on. First: pick a piece up, slide it, turn it, flip it, and its area comes with it unchanged.

1:33Second: cut a figure into pieces that do not overlap, and the pieces add up to the whole. Say them out loud: every argument from here is one of the two, or both. And they are not the same fact. Push the top of a six-by-four rectangle sideways so it leans: the area is still twenty-four, but the slanted sides are longer than the upright ones, so that move was not rigid.

2:02Stretch it instead and the area jumps to seventy-two. Area surviving a move is weaker than the shape surviving it, and that gap is where the next argument works. Here is a parallelogram: base six, top side pushed two to the right, sitting four above the base. Drop a perpendicular from the top-left corner down to the base. That cuts a right triangle off the left end. Carry it across - move it right by exactly the base - and set it down against the other end.

2:38It fits. What stands there now is a rectangle, six wide and four tall. One piece moved rigidly, two pieces that never overlapped were added up, and nothing was gained or lost. So the parallelogram has the rectangle's area, and that is a count: six fours, twenty-four. Base times height - and over two hundred different bases, heights and slants, it is right every time. That is the argument almost everyone is shown, and it has a hole in it.

3:14Lean the parallelogram much further over: base two, height three, top side pushed seven to the right. Drop the perpendicular from the top-left corner again. It lands five units past the end of a base only two long. There is no triangle inside the figure to cut off, and nowhere to carry it to. The construction has run out. And this is not a rare accident: over a spread of bases and slants, twenty-seven work and fifty-one do not.

3:47Anyone shown only the tidy picture has been handed a special case and told it was the argument. The repair is one move, and it is the lean from a moment ago. Slide the whole top side along its own line, to the left, by one unit. Neither the base nor the height moves, and a thin triangle drops off one end while an identical one appears at the other.

4:15The foot of the perpendicular has crept one unit closer. Five slides and it is on the base, with the area at six the whole way. You are usually told to repeat as needed, and left there. It stops: the overhang is one fixed length and each slide takes a fixed bite. Take it all at once and one slide does it; take a hundredth at a time and five hundred do.

4:42No choice of step goes on for ever. Now the error this material is famous for: height is not a side. Take a parallelogram with sides six and five and area twenty-four. Choose the six as base and the height is four - the perpendicular distance across to the opposite side, which is not an edge of the figure. Choose the five instead, and the same figure has height four and four fifths.

5:13The area is twenty-four either way, because it is the same figure. Pick a base and the height is decided for you; you do not get to choose it and you cannot read it off a side. Multiplying the two sides together gives thirty, which is a different number about a different shape. If that is not convincing, hinge it. Four strips of card, six, five, six, five, pinned at the corners.

5:42Stand it square and it is a rectangle with area thirty. Now push. The four sides never change - they are strips of card - but the figure leans, the height drops, and the area goes with it. Forty positions, one set of four side lengths, forty different areas. By the fortieth it is under a tenth of where it began, and it keeps going: name any area, however small, and a position is below it.

6:12Push all the way and it lies flat, no area at all. Four side lengths do not determine an area; for a rectangle they do, only because the corners are pinned square. One consequence is worth more than it looks. Two triangles on the same base, with their tips anywhere on one line parallel to it, have the same area. Not roughly - exactly, because the height is the gap between the two lines and sliding the tip cannot change it.

6:46Fix a parallelogram's top side as base and slide the apex along the bottom: thirteen positions, twelve every time. So asking for the ratio of two such triangles is asking for one to one - a shared height, not a calculation. The same idea wears a disguise. Put a point anywhere on a diagonal and join it to the corners either side. The two are equal at all eleven positions tried, because those corners sit the same distance from the diagonal.

7:24A trapezium's formula is not a new thing to remember either. Cut along a diagonal: two triangles, one on each parallel side, same height. Or turn a second copy half way round the midpoint of a slanted side and push them together. They close into a parallelogram of the same height on the two parallel sides end to end, so the trapezium is half of it. Both routes give half of a plus b, times the height, over four hundred and eighty trapeziums.

7:59A third route is the one usually drawn: split off a parallelogram and leave a triangle beside it. That one carries the same hole - the cut has to land on the base, and in a hundred and twenty of the four hundred and eighty it does not. Same failure, same repair: slide the top along its own line. Two shapes with the same area need not have the same perimeter, and the other way round.

8:29But what if both agree? For rectangles the answer is yes, and it is not geometry, it is arithmetic. Equal perimeters fix the two sides added; equal areas fix them multiplied. A pair of numbers with a known sum and product is pinned down. Over three hundred rectangles, not one pair agrees on both and differs in shape. For parallelograms the answer is the opposite, and it is the hinge again: same four sides, same perimeter, areas thirty and twenty-four.

9:06The right angles were doing all the work. One last use of the same two facts, and it is not about shapes. Draw a square of side five plus three and rule it into four cells: twenty-five, fifteen, fifteen and nine. They overlap nowhere and reach the edges, so they add to the whole square, sixty-four. That is the square of a sum, proved by adding areas. Three terms gives nine cells, which is where the six cross terms come from.

9:40The difference of two squares is the other property: cut the small square out, carry the leftover strip round, and a rectangle is left. Twenty-five take nine is sixteen, and the rectangle is eight by two. Nothing here was a new rule. Pick a unit, move pieces without stretching them, add pieces that do not overlap - the rest is bookkeeping.