The Fence Doesn't Know What It's Fencing Perimeter and Area

Akshi and Toshi run laps on two rectangular tracks drawn one inside the other. Akshi runs the outer track, 70 m by 40 m, and completes 5 rounds. Toshi runs the inner track, 60 m by 30 m, and completes 7 rounds. They cannot simply compare 5 and 7, because a round means a different distance on each track. To settle this fairly, the first thing needed is the distance travelled in exactly one round of each track, which is nothing but the perimeter, the total distance covered walking once all the way around a closed shape. For a shape made of straight sides, like a rectangle, perimeter is just the sum of the lengths of all the sides, so it can always be checked by literally adding up every side, before any formula is used at all.

For Akshi's rectangular track, the perimeter is AB + BC + CD + DA. Since opposite sides of a rectangle are always equal, this is the same as length + breadth + length + breadth, which is 2 x length + 2 x breadth, or 2 x (length + breadth). For a 70 m by 40 m track, that is 2 x (70 + 40) = 2 x 110 = 220 m for one round. A square is just a rectangle whose length and breadth happen to be equal, so its perimeter collapses even further, to side + side + side + side = 4 x side. A triangle has no such shortcut in general; its perimeter is simply the sum of its three sides, whatever they happen to be. The same idea extends to any regular polygon, a closed shape whose sides and angles are all equal to each other: since every side has the same length, the perimeter is always just the number of sides multiplied by the length of one side. Now Akshi and Toshi's puzzle can be finished. Akshi's track is 70 m by 40 m, so one round is 2 x (70+40) = 220 m, and 5 rounds cover 5 x 220 = 1100 m. Toshi's track is 60 m by 30 m, so one round is 2 x (60+30) = 180 m, and 7 rounds cover 7 x 180 = 1260 m. Toshi ran the longer distance, even though Toshi ran on the shorter track, simply because Toshi ran more rounds of it.

Akshi says the perimeter of a certain triangle shape, drawn on square-dotted paper, is 9 units. Toshi disagrees, saying it will be more than 9 units. Toshi is right, and the reason is worth stating clearly: some of that triangle's sides run straight along the grid lines, but at least one side runs diagonally, cutting across a grid square corner to corner, and a diagonal line is always longer than a straight grid-line of the same visual 'span', because it is the longest side of a right-angled corner, not one of the two shorter sides. So a shape with straight sides worth, say, 6 units and diagonal sides worth what looks like 3 more grid-steps will actually have a perimeter of more than 9 units once the diagonals are properly measured, not exactly 9. Rather than trying to force diagonal lengths into the same units as straight lines, this chapter simply keeps them labelled separately: a shape's perimeter can be written as, for example, 6 straight units + 3 diagonal units, in short 6s + 3d, without needing to convert diagonal units into straight ones at all. The same idea settles a farmer's fencing problem cleanly: a rectangular field 230 m by 160 m is fenced with 3 full rounds of rope. One round is 2 x (230+160) = 2 x 390 = 780 m, so 3 rounds need 3 x 780 = 2340 m of rope in total -- here every side is straight, so ordinary units work throughout.

Perimeter measures a boundary; area measures the region enclosed by that boundary, how much surface is inside. The standard way to measure area is to lay a shape over a grid of unit squares and count how many unit squares it covers, following four simple counting rules: a full square counts as 1 full unit of area; a square that is less than half covered gets ignored entirely (counted as 0); a square that is more than half covered gets counted as a whole 1 unit anyway; and a square that is covered exactly half gets counted as exactly 1/2 a unit. Applying this convention to four sample figures gives areas of 4, 9, 10 and 11 square units respectively, even though the figures are irregular and were never built purely out of whole squares. For rectangles and squares this grid-counting method simplifies into exact formulas, since a rectangle l units long and b units wide contains exactly l rows of b unit squares each, giving l x b unit squares total, i.e., Area of a rectangle = length x breadth, and Area of a square = side x side. A floor 5 m by 4 m has area 20 sq m; laying a 3 m square carpet on part of it covers 3 x 3 = 9 sq m, leaving 20 - 9 = 11 sq m of floor uncarpeted. A 12 m by 10 m plot of land has area 120 sq m; four square flower beds of side 4 m placed in its corners take up 4 x (4x4) = 64 sq m, leaving 120 - 64 = 56 sq m for everything else.

Squares are used as the standard unit for area, not circles, triangles, or any other shape, and there is a real reason for this choice, not just convention. Packing the same rectangle with circles instead of squares, two different honest attempts at packing it tightly give 42 circles in one arrangement and 44 in another, different counts for exactly the same region, because circles are curved and can only ever touch each other at single points, always leaving small curved gaps between them that cannot be filled by any whole number of same-sized circles. Squares have no such problem: a square's four right-angled corners fit perfectly against their neighbours' corners with zero gap and zero overlap, and every row of squares lines up exactly with the row below it, which is exactly why a rectangle's area works out to a clean length x breadth with nothing left over and nothing double-counted.

Draw any rectangle and cut it along one diagonal: the two triangles produced always overlap each other exactly, meaning they always have exactly equal areas, and since together they make up the whole rectangle, each one is exactly half of it. This works for every rectangle tried, of any dimensions, and for a square too, since a square is just a special rectangle. This gives a first, genuine fact about a triangle's area, even without any base-times-height formula yet: a right-angled triangle formed by a rectangle's diagonal always has area equal to exactly half of that rectangle's area. What about a triangle whose corner sits somewhere in the middle of one side of the enclosing rectangle, rather than exactly at a corner? Splitting such a triangle with a straight line from its top vertex straight down produces two smaller right-angled triangles, each one again exactly half of its own smaller enclosing rectangle by the same diagonal argument, and adding these two halves back together shows the original, less obviously-shaped triangle is still exactly half of the full enclosing rectangle. So regardless of exactly where a triangle's top vertex sits along the top side, its area always comes out to exactly half of the rectangle built around it. This chapter deliberately stops here, at 'exactly half of the enclosing rectangle', without yet writing this as a base x height formula -- that generalisation, valid for every triangle including ones where no enclosing rectangle is obvious at all, is built properly in Class 8.

Two facts, both demonstrated repeatedly in this chapter's puzzles, sit at the heart of why perimeter and area have to be studied as two genuinely separate ideas, not one standing in for the other. First: the same area can have very different perimeters. Take 9 unit squares and arrange them, edge-to-edge with no gaps or holes, into different connected shapes: a compact 3x3 square arrangement gives the smallest possible perimeter, 12 units, while a single straight 9x1 strip gives the largest possible, 20 units, and every whole number in between (such as 18) can also be reached by a suitably chosen shape, all while every single arrangement still covers exactly the same area of 9 square units. The tangram set of 7 puzzle pieces makes the same point differently: rearranging the identical 7 pieces from a perfect square into a long rectangle keeps the total area exactly the same, since it is the same pieces, but changes the total perimeter, since a different outer boundary is exposed. Second, and just as important: the same perimeter can enclose very different areas. Among all rectangles with whole-number sides and a fixed perimeter, a shape close to a square always encloses more area than a long, thin, stretched-out shape with the same perimeter. Put the two facts together and a single principle emerges, worth remembering long after the specific numbers are forgotten: for a fixed area, a shape closer to a square always has a smaller perimeter than a long, thin, spread-out shape covering that same area, and the reverse holds too, for a fixed perimeter, a shape closer to a square always encloses more area than a spread-out one.