Measuring Space: Perimeter and Area: Class 9 Ganita Manjari · Chapter 6

Why do the runners start apart?

Picture a 400 m race on sports day in which every runner keeps to her own lane for the whole lap. The outermost lane swings round a much wider curve than lane 1, so if everyone started on one line, the outside runner would have many more metres to cover. That is why the starting marks are spread out round the first bend, while all the athletes finish at one common line.

The offset between the starting marks of neighbouring lanes is called a stagger. How long must it be so that nobody gains or loses? Does it depend on the size of the track, or on which lanes we compare? Fig. 6.1 takes the track apart to find out.

The bends are pieces of circles, so we first need a way to measure the distance round a circle. The chapter then moves from lengths to areas: rectangles, parallelograms, triangles, four-sided figures and, finally, circles and their slices.

6.1 Perimeter of a Shape

The perimeter (परिमाप) of a flat shape is the length of its boundary. Lay a piece of string exactly along the edge of a leaf or a coin, then pull the string straight and measure it: that length is the perimeter. For a shape with straight edges you can skip the string and simply add up the side lengths.

For many shapes a short formula does the adding for us. A rectangle with length $a$ and breadth $b$ has two sides of each length, so its perimeter is $2(a+b)$. A regular polygon has $n$ equal sides; if each is $a$ long, its perimeter is $na$. With $n=3$ this gives $3a$ for an equilateral triangle, with $n=6$ it gives $6a$ for a regular hexagon, and with $n=4$ it gives $4a$ for a square.

The square belongs to both families. It is the regular polygon with $n=4$, and it is also the rectangle whose breadth equals its length. Put $b=a$ into $2(a+b)$ and you get $2(2a)=4a$ again, as you should. Getting a formula from a more general one by adding a condition such as $b=a$ is called taking a special case; you will meet the idea again in Brahmagupta’s formula for a cyclic 4-gon.

Now look at the curved end of the running track, or at a bangle. A circle has no sides to add up, so how do we find its perimeter? Here is a first clue. Draw regular hexagons of different sizes, as in Fig. 6.2. When the side doubles, every side doubles, so the perimeter doubles too. When the side triples, the perimeter triples. The perimeter divided by the side is always 6, however big or small the hexagon is.

The same holds for squares (the ratio is always $4:1$) and for equilateral triangles ($3:1$). Every circle is also just a scaled copy of every other circle. The perimeter of a circle is called its circumference (परिधि). So we may expect the circumference divided by the diameter to give one and the same number for all circles. The next section finds out what that number is.

6.2 Perimeter of a Circle — The C/D Ratio

Take a coin, a steel plate and a bicycle wheel. Divide each one’s circumference $C$ by its diameter $D$, and you get almost exactly the same number three times. People in many ancient cultures spotted this. We will call the number the C/D ratio, and the rest of this section is about pinning down its value.

There are two ways in. You can measure real circles carefully. Or you can reason with geometry and squeeze the number between two values you can calculate exactly. Try the first way yourself.

Measurement can never be perfectly exact. To pin the number down, mathematicians turned to pure reasoning. Today the C/D ratio is written $\pi$ (पाई), the Greek letter pi, which rhymes with sky. By definition,

The long hunt for π

The oldest reasoning is also the simplest. Draw a regular hexagon inside a circle of radius 1, with its corners on the circle. It splits into six equilateral triangles, so each side equals the radius, 1, and the perimeter is 6. The circle bulges outside every side, so it must be longer than 6. Since the circumference is $2\pi$, we get $2\pi >6$, that is, $\pi >3$. Mathematicians in Mesopotamia knew this around 1900 BCE and used $3\frac{1}{8}=3.125$.

Around 250 BCE, Archimedes of Syracuse sharpened this trap. He kept doubling the number of sides: 6, 12, 24, 48 and finally 96. Each doubling squeezes the two polygons closer to the circle. With 96 sides and patient hand calculation he reached the bounds $3\frac{10}{71}<\pi <3\frac{1}{7}$, that is, $\pi$ lies between about 3.1408 and 3.1429. Try his doubling yourself in Fig. 6.4.

Many others joined the hunt. The timeline lists some of the milestones. Notice how a better value needed either a cleverer method or many more sides.

Polygons can only ever give bounds. Around 1400 CE, Mādhava of Saṅgamagrāma in Kerala found something of a different kind: an exact formula. Add and subtract the reciprocals of the odd numbers, for ever, and the total is exactly a quarter of $\pi$:

Stop after any number of terms and you do not get $\pi$; keep going and the running total closes in on $\frac{\pi }{4}$ without end. A value reached in this step-by-step way is called a limit, and limits are the central idea of calculus, which you will study in later classes. On its own the series crawls, so Mādhava also worked out a correction to add at the point where you stop. With it he found $\pi$ correct to 11 decimal places. Later mathematicians such as Nīlakaṇṭha, Machin and Ramanujan found even faster routes. Compare both versions in the second tab of Fig. 6.4.

6.3 π Is Irrational

Could some clever fraction, perhaps with a huge numerator and denominator, be exactly equal to $\pi$? Decimals give a way to think about this. Divide 3 by 7 by long division: at every step the remainder is one of 0, 1, 2, …, 6. With only seven possible remainders, one of them must come back by the eighth step at the latest, and from then on the same digits come round again: $\frac{3}{7}=0.428571428571\dots$ The same reasoning works for every fraction, so its decimal either stops, like $\frac{3}{8}=0.375$, or settles into a repeating block, like $\frac{5}{11}=0.4545\dots$

Now look at $\pi =3.14159265358979323846\dots$ Computers have produced trillions of its digits, and they never settle into a repeating block. So no fraction can equal $\pi$. A number like this, which cannot be written as $\frac{a}{b}$ with integers $a$ and $b$, is called irrational (अपरिमेय). Fig. 6.5 puts the two kinds of decimal side by side.

Watching digits is not a proof, of course: a pattern might start after the trillionth digit. A real proof that no fraction equals $\pi$ was found by Johann Lambert in 1761, using tools far beyond this book. Long before that, Āryabhaṭa had called his value 3.1416 āsanna, “nearly”, as if he already sensed that no fraction would be exact; Zu Chongzhi seems to have felt the same. You have already met another irrational number, $\sqrt{2}$.

Now look down the last column of the table. $\frac{22}{7}$ misses $\pi$ by about a thousandth, $\frac{333}{106}$ by less than a ten-thousandth, and $\frac{355}{113}$ by only a few ten-millionths. Is there a closest fraction of all? No. Whatever fraction you pick is not equal to $\pi$, so a small gap separates the two. Cut the decimal of $\pi$ off after enough digits and you get a fraction, such as $\frac{3141592}{1000000}$, that lands inside that gap. So every fraction for $\pi$ is only an approximation. We write $\pi \approx \frac{22}{7}$ (“is close to”), and $\pi \mathrm{\ne }\frac{22}{7}$ is true as well.

For everyday work $\frac{22}{7}$ or $3.14$ is good enough, and in this chapter we use $\pi \approx \frac{22}{7}$ unless a question says otherwise. Engineers who need more accuracy use more digits.

6.4 Length of an Arc of a Circle

A piece of a circle between two of its points is called an arc (चाप). A swing, a windscreen wiper and the tip of a clock hand all travel along arcs. How long is an arc? Start with the easy pieces.

The key is the circle’s perfect roundness. Turn a circle about its centre through any angle and it lands exactly on itself. So if two arcs make the same angle at the centre, a turn carries one onto the other, and they have the same length. Equal angles at the centre give equal arcs.

Now share out the full turn. Split the 360° at the centre into 4 equal angles of 90°: the circle splits into 4 equal arcs, so each is $\frac{1}{4}$ of $2\pi r$, that is $\frac{1}{2}\pi r$. Split it into 2 angles of 180° and each half-circle is $\pi r$; into 6 angles of 60° and each arc is $\frac{1}{3}\pi r$.

In every case the arc gets the same share of the circumference as its angle gets of 360°. We say the arc subtends that angle at the centre. Fig. 6.6 shows the pattern, which holds for any angle $\theta$, not only for neat fractions of a turn.

A closer look at a running track

Now we can return to the runners. A standard track has two straight stretches joined by two semicircular bends, and the two bends of any lane together make one full circle, as in Fig. 6.1. So the length of one lap in a lane is twice a straight plus the circumference of that lane’s circle. Rules say the distance is measured along a line just a little way out from the inner edge of the lane, roughly where a runner’s feet fall.

Look at the last step with letters instead of numbers. If lane 1 bends have radius $r$ and the lanes are $w$ wide, the circles have circumferences $2\pi r$ and $2\pi (r+w)$. The difference is

The radius $r$ has cancelled! The extra distance depends only on the lane width and on how much turning the race contains, not on the size of the track. So the stagger between lanes 2 and 3 is the same as between lanes 1 and 2, and so on outwards. On a full-size track with lanes 1.22 m wide, one lap in lanes needs a stagger of $2\pi \times 1.22\approx 7.67\mathrm{m}$ per lane.11Real rules measure lane 1 at 30 cm from the kerb but the other lanes at 20 cm from their inner line, so the first stagger is slightly smaller, about 7.04 m. The stagger is no gift: it exactly cancels the longer bends.

6.5 Problems, Puzzles, and Paradoxes on Perimeter

Arc lengths lead to some surprising puzzles. Try each problem before opening its solution.

Fig. 6.8 keeps going: 8, 16, 32 semicircles. The wiggly road hugs the straight segment $PQ$ more and more tightly, so you might expect its length to approach 12. It does not. It stays $6\pi$, about 18.9, for every number of semicircles. Looking almost the same and having almost the same length are two different things.

6.6 Area of a Rectangle

Perimeter tells you how much fencing a field needs; area (क्षेत्रफल) tells you how much seed it needs. Area is the amount of flat surface inside a boundary, and we measure it by counting squares of side 1 unit. Each such square has an area of 1 square unit (1 unit²).

A rectangle 4 units long and 3 units wide holds 3 rows of 4 unit squares, so its area is $4\times 3=12$ square units. In the same way a rectangle with sides $a$ and $b$ has area $ab$, and a square of side $a$, the special case $b=a$, has area $a^2$. The rule still works when the sides are not whole numbers. A 2.5 by 1.5 rectangle holds 2 whole squares, 3 half squares and a quarter square: $2+1.5+0.25=3.75=2.5\times 1.5$.

6.7 Area of a Parallelogram

A parallelogram has two pairs of parallel sides. Pick one side as the base $b$. The height $h$ is the perpendicular distance between the base and the side opposite it. It is not the length of the slanting side.

Fig. 6.12 shows why the area is $bh$. Cut along a height from one top corner. A right-angled triangle falls away. Slide it across to the other end and it fits exactly into the gap, turning the parallelogram into a rectangle with sides $b$ and $h$. No paper was added or thrown away, so the areas are equal.

The cut-and-slide trick needs the height from a top corner to land on the base. For a steeply leaning, thin parallelogram it does not, so we need another argument. Enclose the parallelogram in a rectangle of height $h$ and width $b+x$, where $x$ is how far the top side overhangs. What lies outside the parallelogram is two right triangles with legs $x$ and $h$. So the area is $(b+x)h-2\times \frac{1}{2}xh=bh$. This argument works for every parallelogram, thin or not.

6.8 Area of a Triangle

You have used area of a triangle = ½ × base × height for years; here is why it holds. A right-angled triangle and a copy of it make a rectangle, so it is half that rectangle. For other triangles, drop the height from the top vertex. If it lands inside the base, it cuts the triangle into two right triangles with bases $b_1$ and $b_2$, where $b_1+b_2=b$, and the total is $\frac{1}{2}{b}_{1}h+\frac{1}{2}{b}_{2}h=\frac{1}{2}bh$.

That argument quietly assumed the height lands inside the base. In a triangle with an obtuse angle at one end of the base it does not: it lands outside, on the base line extended. Then we subtract instead of adding. In the figure below, triangle $PQR$ is the big right triangle $PSR$ with the small right triangle $PSQ$ removed.

There is also a slicker route that needs no cases at all. Make an identical copy of the triangle and give it half a turn about the midpoint of one side. The copy slots against the original to form a parallelogram, as Fig. 6.15 shows.

Here is a challenge. Draw any triangle and one straight cut from a corner to the side facing it. Where must the cut meet that side so that the two pieces have equal areas? Decide before you read the theorem below. A segment from a corner to the middle point of the opposite side is called a median (माध्यिका); every triangle has three.

The two pieces in Fig. 6.16 show how little the proof needs. One piece is narrow and upright, the other broad and leaning, and no turning or flipping makes one sit on the other: equal area does not require matching shapes. The proof used only equal bases and a shared height. So any triangles on one base whose third corners lie on one line parallel to that base have equal areas too. Fig. 6.17 lets you test this.

6.8.1 Heron’s formula

Suppose a farmer has measured the three sides of a triangular field with a tape, but nobody knows its height. The area can still be found from the sides alone, using a formula recorded by Heron of Alexandria, a Greek engineer and teacher of the first century CE. The first step is to halve the perimeter; the result is called the semi-perimeter $s$.

Why should the square root of a product of four numbers be an area? Before worrying about a proof, let us collect evidence: apply the formula to triangles whose areas we can find another way, and see whether the answers agree.