Measuring Space: Perimeter and Area

Why This Matters

Look at the start of a 4 × 100 m relay race. The runners are not standing on the same line. The runners in the outer lanes start a little ahead of the runners in the inner lanes. This head-start is called the stagger.

Why is this fair? Because the finish is the same for everyone, but the outer lanes curve around a bigger bend. A runner on the outside has to cover a longer curve. To make the race fair, the organisers give the outer runners a small head-start so that everyone runs exactly the same total distance.

But here is the real question. How do the organisers know exactly how much head-start to give? They cannot just guess. They need to measure the length of a curve. They need to know the distance around a circle.

That is what this chapter is about. We will learn to measure two things for any flat shape:

• Its perimeter — the distance all the way around its edge.
• Its area — the amount of flat space it covers.

We will go from simple squares and triangles, all the way to circles and curved tracks. And we will meet a beautiful, almost magical formula — Heron’s formula — that finds the area of any triangle if you just know its three sides.

The Big Idea

Two ideas run through this whole chapter, and they are different from each other.

The Big Idea: Perimeter is the length around a shape (a 1-dimensional distance, measured in cm or m). Area is the flat space inside a shape (a 2-dimensional amount, measured in cm² or m²). For circles, both depend on one special number, π (pi) ≈ 3.14. And for triangles, even when you do not know the height, you can still find the area from the three sides alone, using Heron’s formula.

Perimeter: the distance around a shape

The perimeter of any shape is the total length of its border. Here is an easy way to picture it. Imagine a tiny ant walking along the edge of the shape. It never turns back. It keeps walking until it returns to where it started. The total distance the ant walks is the perimeter.

Let’s see this for the two simplest shapes. Figure 6.1 shows a square and an equilateral triangle.

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So for a square with side a:

Perimeter of a square = 4a

For an equilateral triangle with side a:

Perimeter of an equilateral triangle = 3a

What about a rectangle? A rectangle has a length a and a width b. It has two sides of length a and two sides of length b. So the ant walks a + b + a + b.

Perimeter of a rectangle = 2(a + b)

Notice something neat. A square is just a rectangle where the length equals the width (a = b). Put a = b into the rectangle formula: 2(a + a) = 2(2a) = 4a. That is exactly the square formula. So the square formula is a special case of the rectangle formula. This idea — one formula being a special case of a more general one — comes up again and again in maths.

The perimeter of a circle, and the number π

A circle has no straight sides, so we cannot just add up side lengths. The distance around a circle has a special name: the circumference.

How do we find it? Mathematicians thousands of years ago noticed something wonderful. For any circle, if you divide the circumference (C) by the diameter (D), you always get the same number. It does not matter how big or small the circle is. Figure 6.2 shows this.

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This fixed ratio is given the special name π (we say “pie”, as in apple pie). So:

C ÷ D = π, which means C = π × D

Since the diameter is twice the radius (D = 2r), we can also write:

C = 2πr

How big is π? You can estimate it yourself at home. Wrap a thin thread tightly around a round tin 20 times. Unwind it, measure that length L, and divide by 20 times the diameter. You will get a number close to 3.14.

Why is the C/D ratio the same for every circle? Think about a square first. If you double the side of a square, the perimeter also doubles, so the ratio perimeter : side stays fixed at 4 : 1 for all squares. The same logic applies to a circle. If you make a circle twice as wide, every part of it — including the curve around — also doubles. So C and D grow together by the same factor, and their ratio never changes. That fixed ratio is π.

The exact value of π is 3.14159265… and the digits never end and never repeat. For most school problems we use the handy approximations π ≈ 22/7 or π ≈ 3.14.

A quick note: π is irrational

You learned in Grade 8 that some numbers, like √2, cannot be written as a neat fraction. They are called irrational numbers.

It turns out π is irrational too. Its digits march on forever with no pattern. This means there is no perfect fraction equal to π. The fraction 22/7 is close, but it is not exactly π. So we write π ≈ 22/7, with the ”≈” sign meaning “approximately equal to”, and we also write π ≠ 22/7. In the same way, √2 ≈ 1.414 but √2 ≠ 1.414.

Length of an arc of a circle

An arc is just a part of the circle’s boundary — a piece of the curve, not the whole loop. How long is a piece of a circle?

Start with the easy pieces. Figure 6.3 shows how symmetry gives us the length of a semicircle (half the circle) and a quarter circle.

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So:

Length of a semicircle = 2πr ÷ 2 = πr

Length of a quarter circle = 2πr ÷ 4 = πr/2

Now look closer. A semicircle is half a circle, and a half is the angle 180° out of the full 360°. So its length is 2πr × (180 ÷ 360). A quarter is 90° out of 360°, so its length is 2πr × (90 ÷ 360). Do you see the pattern? The arc length is just the full circumference times the fraction of the circle the arc covers.

In general, suppose an arc bends through an angle θ at the centre of the circle. (We say the arc “subtends” the angle θ at the centre — that is just the maths word for “makes” the angle.) Then the arc covers the fraction θ out of 360 of the whole circle. Figure 6.4 shows this.

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Length of arc = 2πr × (θ° ÷ 360°)

Let’s use this on a real timetable-style problem.

We saw the relay-race stagger at the start. Here is the idea worked out. A 400 m running track has two straight parts of 84.39 m each and two curved ends that together make one full circle. The inner edge has radius 36.5 m. Let’s check that one lap really is 400 m. The straights give 2 × 84.39 = 168.78 m. The two curved ends make a full circle. If the runner stays 0.3 m out from the edge, the circle’s radius is 36.8 m, so its circumference is 2 × 3.1416 × 36.8 ≈ 231.22 m. Adding up: 168.78 + 231.22 = 400 m. Perfect — one lap is 400 m.

Now you can see why staggers are needed. A runner in the second lane runs the same straights, but her curved ends have a bigger radius. So her circle is longer. The stagger gives her a head-start to cancel out exactly that extra curve.

Here is a full worked example of an arc-length calculation.

Area: the space inside a shape

Now we switch from the distance around a shape to the flat space inside it. That is the area.

To measure area, we need a unit. The unit is a small square that is 1 unit wide and 1 unit tall. Its area is 1 square unit (written 1 unit² or 1 sq. unit). The area of any shape is how many of these unit squares fit inside it.

Area of a rectangle and a square

You met this in Grade 8. A rectangle with sides a and b holds a rows of b squares, so it fits a × b unit squares.

Area of a rectangle = a × b

A square is a rectangle with a = b, so its area is a × a:

Area of a square = a²

Area of a parallelogram

You might guess the area of a parallelogram is base times the slanted side. That is wrong. The trick is to turn the parallelogram into a rectangle. Figure 6.5 shows how.

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So:

Area of a parallelogram = base × height = b × h

The height h is the perpendicular distance, the straight up-down gap. This is why the slanted side does not appear in the formula.

Why can’t we use the side lengths alone? Think about it. Take four sticks and pin them into a parallelogram. Now push the top sideways. The sides stay the same length, but the shape flattens — the height shrinks, so the area shrinks. The sides did not change, yet the area did. So side lengths alone cannot tell you the area of a parallelogram. You need the height.

Area of a triangle

You also met this in Grade 8. There is a lovely way to prove it. Take a triangle, make an exact copy, flip the copy, and fit the two together. They form a parallelogram. Figure 6.6 shows this.

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Area of a triangle = ½ × base × height = ½ bh

This proof is satisfying because it shows why the “½” is there — a triangle is literally half of a parallelogram.

A beautiful bonus: the median splits a triangle into two equal areas

Here is a surprising fact that drops right out of the ½bh formula. We can state it as a theorem and prove it.

Theorem: A median of a triangle divides it into two triangles of equal area.

Look at Figure 6.7. AD is a median, so D is the midpoint of BC.

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Proof. Triangle ABD has base BD. Triangle ACD has base DC. Since D is the midpoint of BC, BD = DC. Both triangles have their top corner at A, so both have the same height h (the perpendicular distance from A to the line BC). Using area = ½ × base × height:

• Area of ABD = ½ × BD × h
• Area of ACD = ½ × DC × h

Since BD = DC, the two areas are equal. Proved.

This should surprise you. The two triangles ABD and ACD usually look completely different in shape. Yet they cover exactly the same amount of space.

Heron’s formula: area of a triangle from its three sides

Here is a problem. Suppose you have a triangular field. You can measure its three sides easily by walking along them with a tape. But there is no obvious height to measure — no straight pole standing inside. So the ½bh formula seems stuck. How do you find the area?

A Greek mathematician named Heron (who taught in Alexandria, in ancient Egypt) found a wonderful answer. His formula gives the area from the three sides alone — no height needed.

Suppose a triangle has sides of length a, b and c. First find the semi-perimeter, called s. “Semi” means half, so s is half of the perimeter:

s = (a + b + c) ÷ 2

Then the area is given by Heron’s formula:

Area = √[ s(s − a)(s − b)(s − c) ]

Figure 6.8 lays out the two steps.

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It looks strange at first — a square root with four factors multiplied inside. But it works. Let’s test it on triangles whose area we already know.

Test on the 3-4-5 triangle. A triangle with sides 3, 4 and 5 is special. Because 3² + 4² = 9 + 16 = 25 = 5², the converse of the Pythagoras theorem tells us the angle between the sides 3 and 4 is a right angle. So we can use it as base 3 and height 4, giving area ½ × 3 × 4 = 6. Now let’s check that Heron’s formula agrees. Figure 6.9 shows both.

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Both ways give 6. Heron’s formula passes the test.

Why does the semi-perimeter appear? Many students find it odd that we halve the perimeter. Here is the gentle intuition. Notice the three pieces inside the formula: (s − a), (s − b), (s − c). Add them up: (s − a) + (s − b) + (s − c) = 3s − (a + b + c) = 3s − 2s = s. So the three “left-over” pieces add back up to s itself. The semi-perimeter is the natural “balance point” that makes all four factors fit together neatly. (The full proof uses the Pythagoras theorem and algebra, and you will see it in Grade 10.)

Here is a worked example on a triangle where you could not easily find a height.

Two more triangle-area formulas (just so you know)

There are even more ways to find a triangle’s area, and they connect to circles. Every triangle has a circumcircle — the one circle that passes through all three corners — and an incircle — the one circle that fits snugly inside, touching all three sides. If R is the circumcircle’s radius and r is the incircle’s radius, then:

Area = (a × b × c) ÷ (4R)

Area = r × (a + b + c) ÷ 2 = r × s

You do not need to use these in Class 9, but it is good to know that the same area can be found in several beautiful ways.

Can we find the area of a quadrilateral from its four sides?

A natural next question: Heron’s formula gives a triangle’s area from three sides — so can four sides give a quadrilateral’s (“4-gon’s”) area? The answer is No, and it is the same reason as the parallelogram.

Take four rods of fixed lengths, say 3, 3, 3, 3, and pin them at the corners. You can squash and stretch the shape into a tall thin rhombus (small area) or a flat square (large area). The four sides stay the same, but the area changes a lot. So four side lengths alone do not fix the area. You need extra information — one angle, or one diagonal, or a special property.

One such special property is being a cyclic quadrilateral — one whose four corners all lie on a single circle. For a cyclic 4-gon, the ancient Indian mathematician Brahmagupta (628 CE) found a formula that does work from the four sides. If the sides are a, b, c, d and the semi-perimeter is s = (a + b + c + d) ÷ 2, then:

Area = √[ (s − a)(s − b)(s − c)(s − d) ]

Look how similar this is to Heron’s formula! That is no accident. A triangle is like a 4-gon whose fourth side has shrunk to zero (d = 0). Put d = 0 into Brahmagupta’s formula and the (s − d) factor becomes (s − 0) = s, turning it straight back into Heron’s formula. So Heron’s formula is a special case of Brahmagupta’s. The bigger formula contains the smaller one.

Area of a circle

The last big formula. How much flat space does a circle cover? The answer is famous:

Area of a circle = πr²

But why? Here is a beautiful argument, first found by the Indian mathematician Nīlakaṇṭha around 1500 CE. Slice the circle into many thin wedges, like cutting a pizza. Then lay the wedges out in a row, pointing alternately up and down. Figure 6.10 shows the trick.

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The rectangle has base πr (half the circumference) and height r (the radius). So:

Area = base × height = πr × r = πr²

The thinner you cut the slices, the more perfectly the wobbly edge straightens into a true rectangle. That is the whole idea.

Area of a sector

A sector is a “pizza slice” of a circle — the region between two radii and the arc between them. Finding its area uses the very same idea as arc length. A sector with central angle θ covers the fraction θ out of 360 of the whole circle. Figure 6.11 shows a sector.

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Area of a sector = πr² × (θ° ÷ 360°)

Check it on the easy cases. A semicircle is θ = 180°, giving πr² × (180/360) = ½πr², half the circle. A quarter circle is θ = 90°, giving πr² × (90/360) = ¼πr², a quarter of the circle. Both make perfect sense.

Quick reminder of the two circle formulas, side by side, so you do not mix them up:

Common Mistakes

Watch out for these traps. They catch many students.

Quick Check

Test yourself before the practice problems.

Practice Problems

Try each one yourself before opening the solution.

Easy

Medium

Challenge

Summary

After this chapter, you can now explain:

• What perimeter means (the length around a shape) and find it for a square (4a), a rectangle (2(a + b)) and an equilateral triangle (3a).
• Why every circle has the same C/D ratio, the number π ≈ 22/7, so the circumference is C = 2πr, and why π is irrational.
• How to find the length of an arc: 2πr × (θ ÷ 360), and use it to understand why relay-race tracks need staggers.
• How to find the area of a rectangle (a × b), a parallelogram (base × height) and a triangle (½ × base × height), and prove the triangle formula by fitting two copies into a parallelogram.
• Why a median splits a triangle into two equal-area pieces.
• How to use Heron’s formula, Area = √[s(s − a)(s − b)(s − c)] with s = (a + b + c) ÷ 2, to find a triangle’s area from its three sides alone.
• Why you cannot find a quadrilateral’s area from its four sides alone, and how Brahmagupta’s formula handles the special cyclic case.
• How to find the area of a circle (πr²) and a sector (πr² × (θ ÷ 360)), and the picture-proof that slices a circle into a rectangle.

What’s Next

So far, every answer in this chapter has been certain — a perimeter, an area, an exact number. But a lot of real life is uncertain. Will it rain tomorrow? Will the coin land heads? Next we learn to measure uncertainty itself with numbers. Get ready for Chapter 7 — The Mathematics of Maybe: Introduction to Probability.