Class 9 Maths Chapter 6 – Measuring Space: Perimeter and Area

Learning Objectives

After studying this chapter you should be able to:

• Explain the idea of perimeter as the total length of the boundary of a shape, and write the perimeters of a square \((P = 4a)\), a rectangle \((P = 2(a+b))\) and an equilateral triangle \((P = 3a)\).
• Describe the constant ratio of the circumference \(C\) to the diameter \(D\) of a circle, and hence use \(C = \pi d = 2\pi r\).
• Tell the story of \(\pi\) — from ancient estimates to Mādhava’s exact infinite series — and explain why \(\pi\) is irrational with \(\pi \approx 22/7\) but \(\pi \neq 22/7\).
• Find the length of an arc of a circle using \(l = 2\pi r \times \dfrac{\theta^\circ}{360^\circ}\).
• Compute areas of a square, rectangle, parallelogram, triangle, circle and sector, knowing when the height is perpendicular to the chosen base.
• Prove that a median divides a triangle into two triangles of equal area, and use this result.
• Apply Heron’s formula, the circumradius formula \(\dfrac{abc}{4R}\) and the inradius formula \(\dfrac{r(a+b+c)}{2}\) for the area of a triangle.
• Apply Brahmagupta’s formula for the area of a cyclic quadrilateral, and relate it to Heron’s formula as a special case.
• Follow Baudhāyana’s Śhulbasūtra construction that squares a rectangle, and see why it works.

1. The 4 × 100 m Relay: Why Do Outer Lanes Start Ahead?

Look at the photograph of athletes assembled at the start of a 4×100 m relay race. The tracks are laid out on the field, and every runner is ready to sprint — but the athletes are not standing in one straight line. Those in the outer lanes seem to be starting ahead of those in the inner lanes, while the finish line is the same for all of them. What could be the reason for this?

Each runner must cover exactly the same total distance. Over the two straight sections of the track this is automatic, because a straight stretch has the same length in every lane. But the track also has two curved (semicircular) bends, and a runner in an outer lane travels around a bend of larger radius. A larger circle has a larger circumference, so the outer-lane runner covers more distance around the bends. To compensate, the outer lanes are given a small head start.

Do you think the stagger gives anyone (outer lanes or inner lanes) an unfair advantage? It should not — every racer, in every lane, runs the same total distance. On what basis can the organisers work out the length of the stagger between lanes? To answer this we must know how to find the length around a circle. That is exactly the mathematics of this chapter!

2. Perimeter of a Shape

Let us recall the perimeter formulas you already know, and notice a beautiful pattern: the ratio of the perimeter to the side stays fixed for a family of similar shapes.

The formula for the perimeter of a square is a special case of the formula for the perimeter of a rectangle with \(a = b\): indeed \(P = 2(a+a) = 4a\). You will see this idea of a “special case of a general formula” repeated throughout mathematics — and it appears again at the end of this chapter when Heron’s formula emerges as a special case of Brahmagupta’s formula.

If we double the side of a square, its perimeter also doubles; for every square the ratio of perimeter to side is \(4:1\). For every equilateral triangle the ratio of perimeter to side is \(3:1\). This fixed ratio is a first glimpse of the same idea we will meet for circles: every circle has the same ratio of circumference to diameter, and that constant ratio is called \(\pi\).

Perimeter is always measured in units of length (cm, m, km). Compare this with area, which is measured in square units (cm², m²) — we will stress this contrast later in the chapter.

3. Perimeter of a Circle — The C/D Ratio

Here is a circle with radius \(r\) units. What is its perimeter? If we simply call it the circumference, we have only given it a name — we still need a formula to compute it. In Question 1 we saw that a square’s perimeter is always \(4\) times its side; the same trick works for circles: find the ratio of the circumference \(C\) to the diameter \(D\).

4. The Story of \(\pi\)

We say “pi” as in “pie”, not as in “pizza”. This one constant, \(\pi\), is a bridge between many areas of mathematics — it connects the straight-edged world of polygons with the infinite curves of nature. Long before anyone called it \(\pi\), many civilisations had studied the C/D ratio. Here is the adventurous journey described in Ganita Manjari.

Where did the name \(\pi\) come from? The perimeter of a circle is the boundary length around it, and the Greek word for perimeter is perimetros, which begins with the letter \(\pi\). In 1706 the Welsh mathematician William Jones used the Greek symbol \(\pi\) to denote the C/D ratio; the symbol was made popular by the Swiss-German mathematician Leonhard Euler (say “oiler”). We still use it today.

5. \(\pi\) Is Irrational

The digits of \(\pi\) go on forever with no visible pattern. Fractions give decimal expansions with a rhythmic pattern, e.g., \(\frac{1}{3} = 0.33333\dots\), \(\frac{1}{11} = 0.09090909\dots\), \(\frac{1}{7} = 0.142857\ 142857\dots\). But for \(\pi\) there is no such pattern. It turns out that \(\pi\) cannot be written as a ratio of two integers — that is, \(\pi\) is an irrational number.

You already know from Grade 8 that \(\sqrt{2}\) is irrational. From their writings it seems that Āryabhaṭa and Zu Chongzhi already regarded \(\pi\) as irrational. Centuries later (1761), the mathematician Lambert showed that \(\pi\) is indeed irrational — but proving it requires more advanced mathematics, which you will learn in later grades. This Grade 9 chapter simply establishes the fact that \(\pi\) is irrational, without giving the full advanced proof.

6. Length of an Arc of a Circle

The whole circumference is \(2\pi r\). What part of it is one semicircle or one quarter-circle? Mathematicians use the symmetry of the circle to answer — and then generalise to any arc subtending a given angle at the centre.

7. A Closer Look at a 400 m Athletics Track

The figure below is a schematic diagram of a 400 m athletics track, exactly as described in Ganita Manjari. It has two straight sections of length 84.39 m each, and two curved portions that are semicircles with a common centre; the inner semicircle on each side has radius 36.5 m. The width of each lane is 1.22 m.

Now imagine an athlete running one complete circuit of the track. Where exactly does she run? A runner cannot hug the lane border itself, so let us assume she runs at a distance of 0.3 m from the inner border of her lane.

8. Problems, Puzzles and Paradoxes on Perimeter

Perimeter behaves in wonderfully surprising ways. Here are two classic problems from the chapter — reason through them before you read the solutions!

Why does this happen? The length of a semicircle depends only on its diameter (length \(=\frac{\pi d}{2}\)). If the small semicircles gobble up the whole diameter \(PQ\) in pieces, the total of their diameters is exactly \(PQ\) — so the total of their arcs equals the arc of the single semicircle, no matter how the pieces are arranged.

9. Area of a Rectangle

From perimeter we move to area: the amount of space occupied by a two-dimensional region in a plane. Measurement is always relative to a unit. For length the unit is, say, a centimetre; for area the unit is a \(1 \times 1\) square, whose area is taken to be \(1\) sq. unit (written \(1\) unit²).

Why is multiplication the right operation? If a rectangle is exactly laid out with unit squares, a rectangle of sides \(a\) and \(b\) fits \(a\) unit squares along one side and \(b\) unit squares along the other, giving \(a \times b\) unit squares in all. The square is again the special case with \(b = a\).

10. Area of a Parallelogram

You know from Grade 8 that the area of a parallelogram is base × height. The key subtlety this year is to understand what “height” means.

Why the formula works (cut-and-shift): Given a parallelogram \(ABCD\), cut off a right triangle at one end and shift it to the other end. The pieces fit together to make a rectangle with the same base \(b\) and the same height \(h\) (see the figure — \(A'B'C'D'\) is a copy of \(ABCD\)). The two shapes are different, but their areas are the same: \(bh\).

The “thin” parallelogram: what if the parallelogram is so thin that the foot of the perpendicular from a vertex does not lie on the side itself? The cut-and-shift construction then seems to fail. Here is the fix from the source: take a point \(D'\) on \(DA\) close to \(D\), and a point \(A'\) on \(DA\) produced, with \(A'A = D'D\). Then \(A'BCD'\) is a parallelogram with the same area as \(ABCD\) (because \(\triangle CDD' \cong \triangle BAA'\)). Repeat this step as many times as needed!

11. Area of a Triangle

Why it works: Enclose the triangle in a rectangle with base \(b\) and height \(h\). The rectangle is split into two smaller rectangles, and the triangle is exactly half of each, so it is half of the whole rectangle: \(\frac{1}{2} bh\).

The obtuse-angle gap: if the triangle has an obtuse angle, the perpendicular may fall outside the triangle. Then use the extension of the base; the enclosed rectangle still works, at the cost of a small adjustment — you are invited to work out the details.

A nicer derivation: two congruent copies of any triangle can be fitted together (matching equal angles) to make a parallelogram — in fact a rhombus when the triangle is isosceles. Since the parallelogram has base \(b\), height \(h\) and area \(bh\), the single triangle has area \(\frac{1}{2} bh\).

12. Median and Equal-Area Triangles

From the area formula we reach a simple but beautiful conclusion about the medians of a triangle.

Why: In \(\triangle ABC\), let \(AD\) be a median, so \(BD = DC\). Draw the altitude from \(A\) to \(BC\); call its length \(h\). Both \(\triangle ABD\) and \(\triangle ACD\) have height \(h\), and their bases are equal (\(BD = DC\)). So each area is \(\frac{1}{2} \times\) (base) \(\times h\), with equal bases — hence equal areas, each \(\frac{ah}{2}\) sq. units.

Can we cut \(\triangle ABD\) into pieces and rearrange them to exactly cover \(\triangle ACD\)? Surprisingly, yes — the chapter invites you to find the (surprisingly small) number of pieces required. Equal-area polygons are always dissectable into one another in this way; try this with a square and a non-square rectangle of equal area, or two differently shaped triangles of equal area, and formulate your own conjecture.

13. Heron’s Formula for the Area of a Triangle

So far, to find a triangle’s area we need a base and a height. But a triangle is completely determined by its three sides alone — so it must be possible to compute the area from the three sides with no height at all. This is exactly what Heron’s formula does. It is named after the Greek mathematician-scientist Heron of Alexandria (c. 62 CE), who worked on mechanics, pneumatics, geometry and geodesy.

14. Circumradius and Inradius of a Triangle

Every triangle fits inside a circle (the circumcircle, through all three vertices) and contains one (the incircle, touching all three sides). The corresponding radii are the circumradius \(R\) and inradius \(r\). Both are elegantly linked to the area of the triangle.

15. Brahmagupta’s Formula for Cyclic Quadrilaterals

Can we find the area of a quadrilateral from its four sides alone? Usually no: for example, a rhombus can have four equal sides yet different areas, depending on its angles. The area of a rectangle is a product of its sides, so two different rectangles with equal perimeters have equal areas; but they are congruent too. For general quadrilaterals the four sides are not sufficient. However, there is a glorious exception: a quadrilateral whose four vertices lie on a circle.

Around 628 CE, the great Indian mathematician Brahmagupta gave the area of a cyclic quadrilateral purely in terms of its sides.

16. From Brahmagupta to Heron

Brahmagupta’s formula for the area of polygons is extremely influential: it eventually remarked that “the area is the root of the product of the deviations of the sides”. One of its most celebrated consequences is a deduction of Heron’s formula.

The trick: regard a triangle as a degenerate cyclic quadrilateral with one side of length \(d = 0\). Then points \(A\), \(B\), \(C\) (and the degenerate \(D\)) still lie on a circle, and Brahmagupta’s formula reads

\(\sqrt{(s-a)(s-b)(s-c)(s-0)} = \sqrt{s(s-a)(s-b)(s-c)}\),

with \(s = \frac{a+b+c+0}{2} = \frac{a+b+c}{2}\). That is exactly Heron’s formula!

17. Squaring a Rectangle: A Showcase of Constructive Geometry

Here is a lovely problem from ancient Indian geometry: can we draw a square with exactly the same area as a given rectangle? The Indo-Arabic tradition answers with an explicit construction based on a circle arc. The earliest rules of this kind appear in the Śhulba-sūtras (c. 800 BCE — the “rules of the cord”, used to lay out Vedic fire-altars). The problem’s construction is given below as in the Gujarati geometry commentary (Arabic-speaking tradition influenced by Āryabhaṭa and Brahmagupta).

18. Area of a Circle

Ancient peoples needed the area of a circle long before anyone had a formula for it — for altars, storage vessels, land borders and more. Their approximations were all built from the same idea: measure the diameter, then compare with whatever square the circle “falls between”.

• Babylonians (c. 1900 BCE) used the circumference: \(A \approx \dfrac{C^2}{12}\), i.e. taking \(\pi\) to be 3.
• Egyptians used \(\dfrac{C^2}{4} \times \left(\dfrac{8}{9}\right)^2\), that is, \(A \approx \left(\dfrac{8d}{9}\right)^2 = \dfrac{64}{81}d^2\).
• Śhulba-sūtras (c. 800 BCE) gave an equivalent square construction based on about \(\dfrac{7}{8}\) of the diameter.
• Archimedes (c. 250 BCE) discovered a route that is exact in principle: cut the circle into a large number of thin sectors and lay them out as a near-parallelogram. He found the area is \(A = \dfrac{1}{2}Cr\).
• Nīlakaṇṭha (Kerala, c. 1500) gave a beautiful proof by the same “slicing” idea, which you can reproduce below.

19. Area of a Sector and of a Segment

By the same “fraction of the circle” reasoning used for arc length, a sector that subtends \(\theta^\circ\) at the centre fills the fraction \(\theta/360\) of the whole circle.

20. Perimeter vs Area: Units Matter

Perimeter and area often get confused in the exam hall. Keep this table in mind:

21. Key Definitions at a Glance

22. Formula Bank

23. More Solved Examples

24. Applications

Perimeter and area are everywhere. Here are nine typical real-world uses — work each one with the formulas above:

• A1 · Fencing a garden: a rectangular garden 30 m by 20 m needs fencing all around — find perimeter \(= 100\) m.
• A2 · Circular park: a park of diameter 14 m needs a boundary fence — circumference \(= 44\) m; at ₹120 per metre the fence costs ₹5280.
• A3 · Tiling: a 6 m × 4 m room is tiled with 50 cm square tiles. Area \(= 24\) m² \(= 240{,}000\) cm²; each tile \(= 2500\) cm² ⇒ 96 tiles.
• A4 · Painting: painting a parallelogram-shaped wall needs its area \(\text{base} \times \text{height}\).
• A5 · Triangular field: a field of sides 13, 14, 15 m has semiperimeter \(s = 21\) and area \(\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84\) m² (Heron).
• A6 · Pizza slice: a slice (sector) of central angle \(60^\circ\) from a 14 cm-radius pizza has area \(\approx 102.67\) cm².
• A7 · Jogging track: a circular track of radius 35 m needs its circumference \(= 220\) m per lap; 5 laps ⇒ 1100 m.
• A8 · Farm land: a parallelogram farm with base 120 m and height 80 m has area 9600 m².
• A9 · Semicircular garden: a semicircular garden of radius 7 m needs a fence along its boundary: arc 22 m + diameter 14 m = 36 m, plus grass area \(\frac{1}{2}\pi r^2 = 77\) m².

25. Competency-Based Questions

These questions check your understanding, not just your recall of formulas.

26. Case-Based Questions

27. Common Mistakes to Avoid

28. Exam Focus

• Read the units. If the radius is given in cm and the answer expected in m (or vice versa), convert before substituting.
• State formulas first. Write \(A = \frac{1}{2}bh\) (or Heron) before plugging in — it scores method marks even if the arithmetic hiccups.
• Heron: compute \(s\) first and write the three factors explicitly. A slip in \((s-a)\) is the most common way to lose marks.
• Fractions vs decimals. With \(\pi \approx 22/7\), cancelling is usually cleaner; with \(\pi = 3.14\), decimals are natural. Keep the answer form consistent with what is asked.
• Perimeter of a sector = arc + 2 radii — it comes up again and again as a trap.
• Brahmagupta needs a cyclic quadrilateral; Heron needs a triangle. State which you are using and why it applies.
• Check the plausibility of the answer. An area can never be a tiny number if the sides are measured in metres; a perimeter can never be in square units.
• Track problems: straights + two semicircles = straights + one circle. That single trick converts a scary word problem into a one-line computation.

29. Quick Revision Cards

30. Self-Practice Problems (SP 1–32)

Attempt all of these on paper before checking Section 31. They are arranged from foundational to challenging.

1. Find the perimeter and area of a square of side 15 cm.
2. Find the perimeter and area of a rectangle 12 cm by 7 cm.
3. The circumference of a circle is 88 cm. Find its diameter.
4. The area of a circle is 616 cm². Find its radius.
5. Find the circumference of a circle of radius 21 m.
6. Find the length of an arc that subtends \(90^\circ\) at the centre of a circle of radius 10.5 cm.
7. Find the perimeter of a semicircular region of radius 7 cm (include the diameter).
8. Four semicircles of diameters 1, 2, 3 and 4 cm are drawn alternately above and below a line. Find the total length of the four arcs.
9. A “stadium” shape is made from a rectangle 20 cm by 14 cm with two semicircular ends on the 14 cm sides. Find its perimeter.
10. A parallelogram has base 18 cm and height 8 cm. Find its area.
11. A triangle has base 20 cm and height 9 cm. Find its area.
12. Find the area of a triangle whose sides are 5, 6 and 7 cm.
13. Find the area of a triangle whose sides are 10, 17 and 21 m.
14. Find the area of an equilateral triangle of side 10 cm.
15. An isosceles triangle has equal sides 13 cm and base 10 cm. Find its area.
16. A right triangle has area 54 cm² and one leg 12 cm. Find the other leg, the hypotenuse and the perimeter.
17. The sides of a triangle are in the ratio 2 : 3 : 4 and the perimeter is 45 cm. Find its area.
18. A trapezium has parallel sides 26 cm and 10 cm, and equal legs 17 cm. Find its area.
19. A cyclic quadrilateral has sides 4, 5, 6 and 7 cm. Find its area.
20. A cyclic quadrilateral has sides 3, 4, 5 and 6 cm. Find its area.
21. A right triangle has sides 6, 8, 10. Find its circumradius.
22. A 7-24-25 triangle has area 84. Find its inradius.
23. Find the area of a sector of a circle of radius 21 m with central angle \(60^\circ\).
24. Find the area of a quadrant of a circle of radius 10.5 cm.
25. Find the perimeter of the sector of Question 23.
26. A chord subtends \(60^\circ\) at the centre of a circle of radius 14 cm. Find the minor and major segment areas. (Use \(\pi = 22/7\), \(\sqrt{3} \approx 1.73\).)
27. The minute hand of a clock is 8 cm long. Find the area it sweeps in 15 minutes.
28. A wheel of diameter 70 cm makes how many revolutions to cover 11 km?
29. A ring-shaped path surrounds a circular park of radius 35 m; the path is 5 m wide. Find the path area.
30. Four semicircles of radius 7 cm are drawn outward on the sides of a square of side 14 cm. Find the total area of the four semicircles. At least what fraction of the square’s area do they cover?
31. A square and a circle have the same perimeter \(P = 44\). Which has larger area, and by how much?
32. Two rectangles have the same perimeter and the same area. Are they necessarily congruent? Justify. (Hint: if sides are \(x\) and \(y\), both \(x+y\) and \(xy\) are fixed.)

31. Answers to Self-Practice

1. \(P = 60\) cm; \(A = 225\) cm².
2. \(P = 38\) cm; \(A = 84\) cm².
3. \(d = 88 \times \frac{7}{22} = 28\) cm.
4. \(r = \sqrt{616 \times \frac{7}{22}} = \sqrt{196} = 14\) cm.
5. \(C = 2 \times \frac{22}{7} \times 21 = 132\) m.
6. \(l = 2 \times \frac{22}{7} \times 10.5 \times \frac{1}{4} = 16.5\) cm.
7. Arc 22 cm + diameter 14 cm ⇒ 36 cm.
8. \(\frac{\pi}{2}(1+2+3+4) = 5\pi \approx 15.7\) cm.
9. \(2 \times 20 + \pi \times 14 = 40 + 44 = 84\) cm.
10. \(A = 18 \times 8 = 144\) cm².
11. \(A = \frac{1}{2} \times 20 \times 9 = 90\) cm².
12. \(s = 9\); \(A = \sqrt{9 \times 4 \times 3 \times 2} = 6\sqrt{6} \approx 14.70\) cm².
13. \(s = 24\); \(A = \sqrt{24 \times 14 \times 7 \times 3} = \sqrt{7056} = 84\) m².
14. \(A = \frac{10^2\sqrt{3}}{4} = 25\sqrt{3} \approx 43.3\) cm².
15. Height \(= \sqrt{13^2 - 5^2} = 12\); \(A = \frac{1}{2} \times 10 \times 12 = 60\) cm².
16. Other leg \(= 9\) cm; hypotenuse \(= 15\) cm; \(P = 12 + 9 + 15 = 36\) cm.
17. Sides 10, 15, 20; \(s = 22.5\); \(A = \sqrt{22.5 \times 12.5 \times 7.5 \times 2.5} = \frac{75\sqrt{15}}{4} \approx 72.62\) cm².
18. Overhang \(8\) cm; height \(= \sqrt{17^2 - 8^2} = 15\); \(A = \frac{26+10}{2} \times 15 = 270\) cm².
19. \(s = 11\); \(A = \sqrt{7 \times 6 \times 5 \times 4} = 2\sqrt{210} \approx 28.98\) cm².
20. \(s = 9\); \(A = \sqrt{6 \times 5 \times 4 \times 3} = 6\sqrt{10} \approx 18.97\) cm².
21. \(R = \frac{abc}{4A} = \frac{480}{96} = 5\) units.
22. \(r = \frac{A}{s} = \frac{84}{28} = 3\) units.
23. \(\frac{22}{7} \times 441 \times \frac{1}{6} = 231\) m².
24. \(\frac{1}{4} \times \frac{22}{7} \times (10.5)^2 = 86.625\) cm².
25. Arc \(= 22\) m; perimeter \(= 2 \times 21 + 22 = 64\) m.
26. Sector \(= 616 \times \frac{1}{6} \approx 102.67\); triangle \(= \frac{1.73 \times 196}{4} = 84.77\); minor \(\approx 17.9\), major \(\approx 598.1\) cm².
27. \(15\) min \(= 90^\circ\); \(A = \frac{1}{4} \times \frac{22}{7} \times 64 = \frac{352}{7} \approx 50.29\) cm².
28. \(C = \pi \times 70 = 220\) cm; \(11\) km \(= 1{,}100{,}000\) cm; \(1{,}100{,}000 \div 220 = 5000\) revolutions.
29. \(A = \pi(40^2 - 35^2) = \frac{22}{7} \times 375 = \frac{8250}{7} \approx 1178.57\) m².
30. Four semicircles \(= \) two circles of radius 7: \(2\pi r^2 = 2 \times 154 = 308\) cm². Square area \(= 196\) cm², so the semicircles cover about \(308/196 \approx 1.57\) times the square area.
31. Square: \(A = 11^2 = 121\). Circle: \(r = \frac{44}{2\pi} = 7\), \(A = 154\). The circle is larger by \(154 - 121 = 33\) — as found in Section 18.
32. Yes — they are congruent. If the sides are \(x, y\), the fixed values of \(x + y = \frac{P}{2}\) and \(xy = A\) force \(x, y\) to be the two (unique) roots of \(t^2 - \frac{P}{2}t + A = 0\), so the unordered pair \(\{x, y\}\) is the same in both rectangles.

32. Chapter Checklist

Before moving to the practice paper, revisit each item and tick it off.

• I can explain why outer lanes start ahead in a relay race (stagger).
• I know the perimeter formulas: square, rectangle, equilateral triangle, circle.
• I know that \(C/D\) is constant and equals \(\pi\), and that \(\pi\) is irrational with \(\pi \approx 22/7\) but \(\pi \neq 22/7\).
• I can retell at least three stories from the history of \(\pi\) (Mesopotamia, Archimedes, Zu Chongzhi, Āryabhaṭa, Mādhava, …).
• I can find an arc length and the perimeter of a sector, semicircle and quadrant.
• I can solve the 400 m track problem, including the stagger computation.
• I can solve the two-circle and semicircle perimeter puzzles (Section 8).
• I can find the area of a rectangle, parallelogram (with perpendicular height) and triangle.
• I know that a median creates two equal-area (not congruent) triangles.
• I can apply Heron’s formula and its equilateral/isosceles special cases.
• I can use \(A = abc/4R\) and \(A = rs\) (circumradius and inradius).
• I can apply Brahmagupta’s formula to a cyclic quadrilateral and explain why a rhombus is not always cyclic.
• I can explain how Heron’s formula follows from Brahmagupta’s with \(d = 0\).
• I can describe the Śhulba construction that squares a rectangle (\(\text{side} = \sqrt{ab}\)).
• I know \(A = \pi r^2\) and can justify it by Nīlakaṇṭha’s slicing argument.
• I can find sector and segment areas, and I never confuse the two.
• I keep perimeter in length units and area in square units.
• I attempted all 32 self-practice problems and checked my answers.

Well done! You have now measured space in its two dimensions: the one-dimensional length of every boundary (perimeter and circumference) and the two-dimensional extent of every region (area). The ideas in this chapter — the constant \(\pi\), Heron’s and Brahmagupta’s formulas, and the ancient constructions — will recur throughout geometry. Now test yourself with the practice paper, quiz and test below.