Measuring Space: Perimeter and Area

Find the perimeter of polygons, the circumference of a circle, the length of an arc and the perimeter of a sector
Explain what π is, trace how its value was found from Mesopotamia to Mādhava, and why π is irrational
Work out the stagger on a 400 m track and solve perimeter puzzles made of arcs
Find areas of rectangles, parallelograms and triangles, and use the median theorem
Use Heron’s formula, the circumcircle and incircle formulas, and Brahmagupta’s formula for a cyclic 4-gon
Square a rectangle by Baudhāyana’s construction, and find the area of a circle, a sector and a segment

perimetercircumferencestaggerπ (pi)irrationalarcsectorsegmentmediansemi-perimetercircumcircleincirclecyclic 4-gonspecial casegeneralisationŚhulbasūtra

The chapter opens at a 4 × 100 m relay race. The athletes are not on one starting line: those in the outer lanes start ahead of those in the inner lanes, yet the finish line is the same for everyone. To see why this is fair, we need to know how to find the length around a circle.

Learn by heartDefinition 1

The stagger is the distance between the starting points of adjacent lanes on a running track.

1 Perimeter of a Shape

Learn by heartDefinition 2

The perimeter of a shape is the total length around its border. Picture a tiny insect walking around the border, never turning back, until it returns to its starting point: the distance it travels is the perimeter.

Shape	Perimeter	Ratio perimeter : side
Square, side $a$	$4a$	4 : 1 for every square
Equilateral triangle, side $a$	$3a$	3 : 1 for every equilateral triangle
Rectangle, length $a$ , width $b$	$2(a+b)$	depends on the shape of the rectangle
Circle, diameter $D$	circumference $C$	$C : D$ is the same for every circle

Double the side of a square (2 units → 4 units) and the perimeter also doubles (8 units → 16 units). As the square grows or shrinks, the ratio stays fixed at 4 : 1.
The formula for the perimeter of a square is a special case of the rectangle formula: put $b = a$ in $2(a+b)$ to get $4a$ .
The perimeter of a circle is usually called its circumference.

2 Perimeter of a Circle: the C/D Ratio

In ancient days people realised that the ratio of the circumference to the diameter of a circle does not change when the size of the circle changes. The book calls it the C/D ratio. Today this constant is written π (said “pie”, as in pie, not as in pizza).

Home measurement, step 1Take a cotton reel with thin thread and measure its diameter

D

as accurately as you can

↓

Step 2Unwrap the thread, then wrap it tightly around the reel 20 times

↓

Step 3Unwrap again and measure its length

L

; the C/D ratio is

\dfrac{L}{20D}

(the thread must be very thin)

The answer comes out between 3 and 4, and with care between 3.1 and 3.2. The ratio can also be estimated by pure geometry, with no measurement at all, as the next section shows.

2.1 Why π is more than 3 and less than 2√3

Figure 1: The circle lies between an inscribed hexagon (inside) and a circumscribed hexagon (outside), so its circumference lies between their perimeters. — **Figure 1**: The circle lies between an inscribed hexagon (inside) and a circumscribed hexagon (outside), so its circumference lies between their perimeters.

Inscribed hexagon = 6 equilateral triangles of side

r

, so perimeter

= 6r = 3 \times (2r)

each triangle has two radii as sides and a 60° angle at O

Circle is a little longer:

C \gt 3D

, so

\pi \gt 3

this is the Mesopotamian argument

Circumscribed hexagon: side

s

with

\frac{\sqrt3}{2}s = r

, so

s = \frac{2r}{\sqrt3}

and perimeter

= \frac{12r}{\sqrt3} = 4\sqrt3\,r

Baudhāyana-Pythagoras theorem on half of one triangle

Circle is shorter:

C \lt 4\sqrt3\,r = 2\sqrt3 \times D

, so

\pi \lt 2\sqrt3 \approx 3.46

Archimedes’ first bounds:

3 \lt \pi \lt 2\sqrt3

2.2 C/D’s adventurous journey: from ancient approximations to Mādhava’s exact formula

Mathematicians in every region have been fascinated by π. It is a bridge between many areas of mathematics, linking the straight-edged world of polygons with the curves of nature. Early civilisations used “good-enough” values for building and trade; later, finding π more and more exactly became a benchmark of how advanced a mathematical theory was.

When	Who / where	What they found
c. 1900 BCE	Mesopotamia	Circle slightly longer than its inscribed hexagon, so π > 3; took $\pi = 3 + \frac18 = 3.125$
by 250 BCE	Archimedes of Syracuse	“Trapped” π between inscribed and circumscribed polygons: hexagons gave 3 and $2\sqrt3 \approx 3.46$ ; with 96-sided polygons, $3\frac{10}{71} \lt \pi \lt 3\frac17$
about 150 CE	Ptolemy of Alexandria	Refined Archimedes’ work for his astronomical tables: $\frac{377}{120} \approx 3.14167$
263 CE	Liu Hui (China)	“Circle-cutting method”, the groundwork for Zu Chongzhi
480 CE	Zu Chongzhi (China)	Polygons up to 24,576 sides; Yuelü (Approximate Ratio) $\frac{22}{7} \approx 3.1428$ and Miü (Close Ratio) $\frac{355}{113} \approx 3.1415929$
499 CE	Āryabhaṭa	$\frac{62832}{20000} = 3.1416$ , called asanna (“approaching”, “approximate”)
628 CE	Brahmagupta	$\sqrt{10} \approx 3.1622$ , less accurate but chosen for elegance and ease in equations
(after them)	Mādhava of Sangamagrāma	First exact formula for π, an infinite series; π to 11 decimal places, 3.14159265358
c. 1500, 1706, 1914, 1988	Nīlakaṇṭha, Machin, Ramanujan, the Chudnovsky brothers	Faster and faster series; π is now known to neels (hundreds of trillions) of digits
1706	William Jones (Welsh)	First used the Greek letter π for the C/D ratio
1761	Lambert	Proved that π is irrational

Zu Chongzhi’s 355/113 stayed the most accurate value of π in the world for over 800 years. No fraction with a denominator less than 15,000 is as close to π.
Āryabhaṭa’s word asanna hinted that the ratio cannot be written exactly as one simple fraction.
Brahmagupta’s √10 became the main approximation in the Arab world and medieval Europe for centuries.
Until Mādhava, no value was as accurate as 355/113, or as simple as 22/7, 3.1416 and √10.
Jones chose π because it is the first letter of the Greek word perimetros (perimeter). The Swiss German mathematician Leonhard Euler (said “oiler”) made the symbol popular.

Mādhava’s exact formula for π

\frac{\pi}{4} = 1 – \frac13 + \frac15 – \frac17 + \cdots

Mādhava saw that π is not a number to be approximated by fractions but a limit to be reached. Moving from cutting circles to adding up numbers, this infinite series gave birth to calculus.

3 π Is Irrational

Learn by heartDefinition 3

A rational number is a number that can be written as a fraction $\frac{a}{b}$ , where $a$ and $b$ are integers and $b \ne 0$ . Examples: $1\frac23$ , $\frac{-7}{11}$ , $1.4$ .

Learn by heartDefinition 4

A number that cannot be written as a ratio of two integers is called irrational. π and $\sqrt2$ are irrational.

3.1 Fractions versus π

Fractions (rational)

Decimal digits repeat in a rhythm
$\frac13 = 0.33333\ldots$
$\frac{1}{11} = 0.09090909\ldots$
$\frac17 = 0.142857\,142857\ldots$

π (irrational)

Digits go on forever
No visible pattern at all
Āryabhaṭa and Zu Chongzhi seem to have regarded π as irrational
Lambert proved it in 1761 (needs advanced mathematics)

Because π is irrational there is no “best fraction” for it: for any fraction close to π, we can find another that is closer.
$\frac{22}{7}$ is good enough for most practical uses, and $\frac{355}{113}$ is much better.

Common Mistake

π and $\frac{22}{7}$ are “close but not equal”. Write $\pi \approx \frac{22}{7}$ and $\pi \ne \frac{22}{7}$ , never $\pi = \frac{22}{7}$ . In the same way, $\sqrt2 \approx 1.414$ and $\sqrt2 \ne 1.414$ .

Fun fact

Count the letters in each word of “How I wish I could recollect pi”: 3, 1, 4, 1, 5, 9, 2, so π ≈ 3.141592… Because π ≈ 3.14, 14 March (written “3-14” in North America) is Pi Day; because π ≈ 22/7, 22 July (written “22-7” in India) is Pi Approximation Day.

4 Length of an Arc of a Circle

Formula

Circumference

C = \pi d = 2\pi r

d

= diameter,

r

= radius (

d = 2r

)

4.1 Semicircle, quarter circle and any arc

Semicircle

Reflect the circle in a diameter AB (or rotate it through 180°, a half-turn): the two halves swap places, so they are equal. Length $= 2\pi r \div 2 = \pi r = 2\pi r \times \frac{180°}{360°}$ .

Quarter circle

Rotate the circle through 90°: each quarter covers the next, so all four are equal. Length $= 2\pi r \div 4 = \frac{\pi r}{2} = 2\pi r \times \frac{90°}{360°}$ .

Figure 2: Arc AB subtends an angle θ° at the centre O. — **Figure 2**: Arc AB subtends an angle θ° at the centre O.

Formula

Length of arc AB

= 2\pi r \times \dfrac{\theta°}{360°}

\theta

= the angle the arc makes (“subtends”) at the centre O, in degrees

Given

Radius $r = 6.3$ m, angle at the centre $= 120°$

Find

Arc length (use $\pi \approx \frac{22}{7}$ )

Solution

Arc $= 2 \times \frac{22}{7} \times 6.3 \times \frac{120}{360} = 2 \times 22 \times 0.9 \times \frac13 = \;$ $13.2$ m

Perimeter of a sector

A sector’s boundary is the curved arc plus the two radii. Perimeter $= 2\pi r \times \frac{\theta}{360} + 2r$ . Forgetting the $2r$ is the most common slip.

4.2 A closer look at a 400 m athletics track

Figure 3: A 400 m track: two straights of 84.39 m, and two semicircular bends with centres A and B; the innermost bend has radius 36.5 m and each lane is 1.22 m wide (lanes drawn wider for clarity). — **Figure 3**: A 400 m track: two straights of 84.39 m, and two semicircular bends with centres A and B; the innermost bend has radius 36.5 m and each lane is 1.22 m wide (lanes drawn wider for clarity).

Given

Straights 84.39 m each; inner radius 36.5 m; runner keeps 0.3 m from the inner border; $\pi \approx 3.1416$

Find

Distance she runs in one full lap of lane 1

Solution

Two straights $= 2 \times 84.39 = 168.78$ m. Her bends have radius $36.5 + 0.3 = 36.8$ m, and the two semicircles together make one full circle: $2 \times 3.1416 \times 36.8 = 231.22$ m.

Total $= 168.78 + 231.22 = \;$ $400$ m

On the straights, runners in every lane run the same distance.
On the bends, the runner in lane 2 runs further, because her semicircles have a larger radius. The stagger exists to make up exactly this difference, so it gives no one an unfair advantage.

Think and Reflect, worked out

Lane 1 vs lane 2: if each runner keeps the same distance from her own lane’s inner border, the radii differ by one lane width, 1.22 m. Over the two bends (one full circle) the extra distance is $2\pi(36.8 + 1.22) – 2\pi(36.8) = 2\pi \times 1.22 \approx 7.67$ m, so the lane-2 runner starts about 7.67 m ahead for one full lap.

Lane 3 vs lane 2: the radius grows by the same 1.22 m again, so the same stagger is needed between every pair of adjacent lanes.

The 200 m school track: each semicircular bend adds $\pi \times 1.22 \approx 3.83$ m per lane, whatever the radius of the bend. So the stagger depends on how many bends are run in lanes, not on how big the track is; a smaller track does not by itself mean a smaller stagger.

4.3 Problems, puzzles and paradoxes on perimeter

Figure 4: Two equal circles, each passing through the other's centre. The outer red arcs form the shape's perimeter; the dashed arcs inside are ignored. ∠CAD = 120°. — **Figure 4**: Two equal circles, each passing through the other’s centre. The outer red arcs form the shape’s perimeter; the dashed arcs inside are ignored. ∠CAD = 120°.

Example 1 (from the book)

Q. Two circles of radius $r$ each pass through the other’s centre. Find the perimeter of the shape they form (ignore the parts inside).

Step 1: $AB = AC = BC = r$ , so △ABC is equilateral and ∠CAB = 60°. Similarly ∠BAD = 60°, so ∠CAD = 120° (and ∠CBD = 120°).
Step 2: each dashed arc is $\frac{120}{360} = \frac13$ of its circle, so each red arc is $\frac23$ of its circle.
Answer: perimeter $= 2 \times \frac23 \times 2\pi r = \;$ $\frac83\pi r$

Example 2 (from the book)

Q. Points P and Q are joined by one big semicircle $a$ , and also by three smaller semicircles $b$ , $c$ , $d$ placed along PQ. Which path is longer?

Let the radii be $a', b', c', d'$ . Path 1 $= \frac12(2\pi)a' = \pi a'$ ; path 2 $= \pi(b' + c' + d')$ . Since $PQ = 2a'$ and also $PQ = 2b' + 2c' + 2d'$ , we get $a' = b' + c' + d'$ .

Answer: the two paths have equal length, however the smaller semicircles are sized.

5 Area of a Rectangle and a Parallelogram

Learn by heartDefinition 5

Area is the amount of space occupied by a two-dimensional region in a plane. It is measured against a unit: a $1 \times 1$ square, whose area is 1 unit² (also written 1 sq. unit).

Shape	Area	Why
Rectangle, sides $a$ , $b$	$ab$ sq. units	$a$ rows of $b$ unit squares (Grade 8)
Square, side $a$	$a^2$ sq. units	special case of the rectangle, $b = a$
Parallelogram, base $b$ , height $h$	$bh$	cut and move into a rectangle with the same base and height
Triangle, base $b$ , height $h$	$\frac12 bh$	half of a parallelogram (Section 6)
Trapezium, parallel sides $a$ , $b$ , height $h$	$\frac12(a+b)h$	two copies make a parallelogram of base $a+b$ (end-of-chapter exercises)
Rhombus or kite, diagonals $d_1$ , $d_2$	$\frac12 d_1 d_2$	two triangles on one diagonal (end-of-chapter exercises)

Parallelogram: in the book’s Fig. 6.17, a copy A′B′C′D′ of parallelogram ABCD is turned into rectangle EB′C′F with the same base $b$ and the same height $h$ . The shapes differ, but the area is the same: base × height $= bh$ .
The “thin” parallelogram gap: if the parallelogram is so slanted that the foot of the perpendicular from C to AD does not land on side AD, the cut does not work directly. Fix: choose D′ on DA close to D and A′ on DA extended with A′A = D′D. Then A′BCD′ is a parallelogram, and since △CDD′ ≅ △BAA′ it has the same area as ABCD. Repeat this step as many times as needed.
Can the sides alone give a parallelogram’s area? No. Keep the side lengths fixed and change the angle between adjacent sides: the height changes, so the area changes. (The rectangle, with a 90° angle, has the largest area.)

6 Area of a Triangle

6.1 Three ways to the formula

Enclose it in a rectangle

A right-angled triangle is half of a $b \times h$ rectangle. For any other triangle, the height splits the base into $b_1 + b_2 = b$ , giving two halves of rectangles: $\frac12 b_1 h + \frac12 b_2 h = \frac12 bh$ .

Obtuse-angled triangle

If the angle at F is obtuse, the height from E falls outside, at M on GF produced (FM = $x$ ). Then △EFG = △EMG − △EMF $= \frac12(b+x)h – \frac12 xh = \frac12 bh$ .

Two congruent copies

Two congruent copies of a triangle fit together into a parallelogram (∠B′C′A′ = ∠BCA are alternate angles, so the sides are parallel). The triangle is half: $\frac12 bh$ .

Figure 5: Two congruent triangles fit together to make a parallelogram, so a triangle's area is half of base × height. — **Figure 5**: Two congruent triangles fit together to make a parallelogram, so a triangle’s area is half of base × height.

Formula

Area of a triangle

= \dfrac12 \times \text{base} \times \text{height} = \dfrac12 bh

6.2 A property of the median

Learn by heartDefinition 6

A median of a triangle is the line segment joining a vertex to the midpoint of the opposite side.

Figure 6: AD is a median of △ABC, so BD = DC = a. △ABD and △ACD share the height h. — **Figure 6**: AD is a median of △ABC, so BD = DC = a. △ABD and △ACD share the height h.

Theorem

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

Proof: △ABD and △ACD have equal bases (BD = DC = $a$ ) and the same height $h$ from A, so each has area $\frac{ah}{2}$ sq. units. This is surprising, because the two triangles are, in general, different shapes (not congruent).

Think and Reflect: △ABD can be cut with straight cuts into pieces that rearrange to cover △ACD exactly. The book says this is possible but leaves the least number of pieces for you to find.
Think and Reflect: given any two polygons of equal area, can one always be cut into pieces that rearrange into the other? Try a square and a non-square rectangle, two different triangles, a triangle and a square, then form your own conjecture.
Rectangles with perimeter 40 units: $a + b = 20$ , so there are infinitely many (sides need not be integers). The largest area is the square $10 \times 10 = 100$ sq. units. There is no smallest: a $19.9 \times 0.1$ rectangle has area only 1.99, and the area can be made as small as we like without ever reaching 0.

7 Heron’s Formula

Heron was a Greek mathematician and inventor who taught at the Museum in Alexandria, a city in ancient Egypt on the banks of the Nile. His formula finds a triangle’s area from its three sides alone, with no height needed.

Learn by heartDefinition 7

The semi-perimeter $s$ of a triangle is half its perimeter: $s = \frac12(a + b + c)$ .

Heron’s formula

Area of △ABC

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

, where

s = \dfrac{a+b+c}{2}

a = BC

b = CA

c = AB

Book example	Heron’s formula gives	Check with ½ × base × height
Example 3: equilateral, side $a$	$s = \frac32 a$ ; area $= \sqrt{\frac32 a \cdot \frac12 a \cdot \frac12 a \cdot \frac12 a} = \frac{\sqrt3}{4}a^2$	$h^2 = a^2 – \frac{a^2}{4} = \frac{3a^2}{4}$ , so $h = \frac{\sqrt3}{2}a$ ; area $= \frac12 \cdot a \cdot \frac{\sqrt3}{2}a = \frac{\sqrt3}{4}a^2$
Example 4: isosceles, equal sides $a$ , base $2b$	$s = a + b$ ; area $= \sqrt{(a+b)(b)(b)(a-b)} = b\sqrt{a^2 – b^2}$	$h = \sqrt{a^2 – b^2}$ ; area $= \frac12 \cdot 2b \cdot \sqrt{a^2-b^2} = b\sqrt{a^2-b^2}$
Example 5: sides 3, 4, 5	$s = 6$ ; area $= \sqrt{6 \times 3 \times 2 \times 1} = 6$	$3^2 + 4^2 = 5^2$ , so right-angled at B (converse of the Baudhāyana-Pythagoras theorem); area $= \frac12(3 \times 4) = 6$

Note

The symbol ∴ means and is read as “therefore”. Heron’s formula has many proofs; one uses the Baudhāyana-Pythagoras theorem and repeated use of $a^2 – b^2 = (a-b)(a+b)$ . You will study proofs in Grade 10.

Given

A triangle with sides 13 cm, 14 cm, 15 cm

Find

Its area by Heron’s formula

Solution

$s = \frac{13+14+15}{2} = 21$ . Area $= \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = \;$ $84$ cm²

7.1 Two more formulas: the circumcircle and the incircle

Learn by heartDefinition 8

The circumcircle of △ABC is the one circle that passes through all three vertices of the triangle. Its radius is $R$ .

Learn by heartDefinition 9

The incircle of △ABC is the one circle that fits tightly inside the triangle, touching all three sides. Its radius is $r$ .

Figure 7: The circumcircle passes through A, B and C (radius R); the incircle touches all three sides (radius r). — **Figure 7**: The circumcircle passes through A, B and C (radius R); the incircle touches all three sides (radius r).

Formulas

Area of △ABC

= \dfrac{abc}{4R}

and Area of △ABC

= \dfrac{r(a+b+c)}{2}

Both have a beautiful symmetry. The second is

r \times s

; its proof needs a result from Grade 10.

Check on the 3, 4, 5 triangle

It is right-angled, so its circumcircle has the hypotenuse as diameter: $R = 2.5$ . Then $\frac{abc}{4R} = \frac{3 \times 4 \times 5}{10} = 6$ . Its inradius is $r = 1$ , and $\frac{r(a+b+c)}{2} = \frac{1 \times 12}{2} = 6$ . Both agree with Heron’s answer, 6 sq. units.

8 Brahmagupta’s Formula for the Area of a Cyclic 4-gon

Heron finds a triangle’s area from its three sides. Can we find a 4-gon’s area from its four sides? No.

4-gon with sides 3, 3, 3, 3 (a rhombus)	Area
Square shape	9
Slightly pushed over	8.01
Pushed over further	5.41

The book drew these in GeoGebra and measured them with its “Area” tool (you can also join four rods at their ends). Same sides, different areas. So we need one more piece of information: one angle, or the length of one diagonal, or the angle at which the diagonals cut each other, or a key geometric property of the figure, such as being cyclic.

Learn by heartDefinition 10

A cyclic 4-gon (cyclic quadrilateral) is a 4-gon whose four vertices all lie on one circle.

Brahmagupta’s formula (628 CE)

Area of a cyclic 4-gon

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

, where

s = \dfrac{a+b+c+d}{2}

Discovered by Brahmagupta (598-668 CE) in 628 CE. Valid only for a cyclic 4-gon. Notice how much it looks like Heron’s formula.

8.1 Checking the formula on special 4-gons

Example 6: rectangle

All rectangles are cyclic. Sides $a, b, a, b$ : $s = a + b$ , so area $= \sqrt{(b)(a)(b)(a)} = \sqrt{ab \cdot ab} = ab$ . Correct.

Example 7: isosceles trapezium

All isosceles trapezia are cyclic. Parallel sides $2a$ , $2b$ , equal legs $c$ : $s = a + b + c$ , and the formula simplifies to $(a+b)\sqrt{c^2 – (b-a)^2}$ . The height is $h = \sqrt{c^2 – (b-a)^2}$ , so this is $(a+b)h$ , exactly half the sum of the parallel sides × the distance between them.

Given

A cyclic 4-gon with sides 5, 5, 12, 12 units

Find

Its area

Solution

$s = \frac{5+5+12+12}{2} = 17$ . Area $= \sqrt{(17-5)(17-5)(17-12)(17-12)} = \sqrt{12 \times 12 \times 5 \times 5} = 12 \times 5 = \;$ $60$ sq. units

8.2 Special cases and generalisation in mathematics

Learn by heartDefinition 11

A special case is what a general result becomes when we apply some extra condition. The general result is then called a generalisation of the special case. Generalisation is an extremely important part of mathematics.

General result	Extra condition	Special case
Rectangle: $A = ab$ , perimeter $2(a+b)$	$b = a$	Square: $A = a^2$ , perimeter $4a$
Right-angled △ with ∠C = 90°: $a^2 + b^2 = c^2$	$a = b$ (isosceles)	$c^2 = 2a^2$ , so $c = a\sqrt2$
$(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$	$c = 0$	$(a+b)^2 = a^2+b^2+2ab$
Brahmagupta: $\sqrt{(s-a)(s-b)(s-c)(s-d)}$	$d = 0$	Heron: $\sqrt{s(s-a)(s-b)(s-c)}$

8.3 Brahmagupta’s formula generalises Heron’s formula

Step 1A triangle with sides

a, b, c

is a 4-gon ABCD whose fourth side has length

d = 0

: AB =

a

, BC =

b

, CD =

c

, and vertices A and D coincide

↓

Step 2Every triangle is cyclic (a circle can be drawn through any three points not on one line), so this 4-gon is cyclic and Brahmagupta’s formula applies

↓

Step 3

s = \frac12(a+b+c+0) = \frac12(a+b+c)

, the triangle’s own semi-perimeter, and

s – d = s

↓

ResultArea

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

: this is Heron’s formula

Did you know?

If the vertices were planets, A and D coinciding would be two planets colliding to form one large planet. The book notes that this actually happened in the early history of the solar system.

9 Squaring a Rectangle

Learn by heartDefinition 12

In ancient times, to square a shape meant to construct a square equal in area to that shape. The shape could be a rectangle, a triangle or a circle.

The ancient Indian mathematician Baudhāyana squared a rectangle in his Śhulbasūtra (800 BCE). Given a rectangle with sides $a$ and $b$ ( $a \gt b$ ), we must construct a square of area $ab$ . Here is a slightly simplified form of his construction.

GivenRectangle ABCD with AD =

a

, AB =

b

a \gt b

↓

Step 1Locate E on AD so that AE = AB

↓

Step 2Locate the midpoint F of ED

↓

Step 3Draw square AFGH with side AF, vertex H on AB produced

↓

Step 4With centre H draw arc AG; let it cut side BC at K

↓

Step 5Draw a line through K parallel to AH; let it cut GH at P

↓

ResultDraw square HPQS with side HP: it has the same area as rectangle ABCD

Figure 8: Baudhāyana's construction drawn for a = 9, b = 4. The shaded square HPQS has side 6, so its area 36 equals the rectangle's area 9 × 4. — **Figure 8**: Baudhāyana’s construction drawn for a = 9, b = 4. The shaded square HPQS has side 6, so its area 36 equals the rectangle’s area 9 × 4.

Why it works: the construction is a geometrical translation of the identity $\left(\dfrac{a+b}{2}\right)^2 – \left(\dfrac{a-b}{2}\right)^2 = ab$ .

AF = \dfrac{AE + AD}{2} = \dfrac{a+b}{2}

F is the midpoint of ED and AE =

b

HG = \dfrac{a+b}{2} = HK

side of square AFGH; HK is a radius of the arc

BH = AH – AB = AF – AB = \dfrac{a+b}{2} – b = \dfrac{a-b}{2}

AH = AF (square)

HP^2 = HK^2 – BH^2 = \dfrac{a^2+2ab+b^2}{4} – \dfrac{a^2-2ab+b^2}{4} = ab

Baudhāyana-Pythagoras in right △HKP, where KP = BH

Area of square HPQS

= HP^2 = ab

= area of rectangle ABCD

Think and Reflect: squaring a triangle

First turn the triangle into a rectangle of the same area (base $b$ and height $\frac{h}{2}$ , since $\frac12 bh = b \times \frac{h}{2}$ ), then square that rectangle with Baudhāyana’s construction.

10 Area of a Circle

A circle is easy to draw with a rope and a peg, and people have used circular shapes in settlements, buildings and tools for thousands of years. Early on they needed its area: how much grain a cylindrical tower (circular cross-section) can hold, how much land a circular garden or building takes up, for planning cities and for taxes.

10.1 Why area must be proportional to the square of the circumference

Shape	Perimeter P	Area A	P² : A
Square, side $a$	$4a$	$a^2$	$16a^2 : a^2 = 16 : 1$
Equilateral triangle, side $a$	$3a$	$\frac{\sqrt3}{4}a^2$	$9a^2 : \frac{\sqrt3}{4}a^2 = 36 : \sqrt3$

For any shape, the ratio P² : A depends only on the shape, not on its scale. So for a circle too, C² : A must be a fixed constant. The question was: which constant?

Well before 1500 BCE: BabyloniansBy actual measurement, C² : A ≈ 12 : 1, so

A \approx \dfrac{C^2}{12}

Around 1500 BCE: ancient EgyptiansA much more accurate formula:

A \approx \left(\dfrac{8d}{9}\right)^2 = \dfrac{64}{81}(4r^2) = \dfrac{256}{81}r^2

800 BCE: Baudhāyana ŚhulbasūtraThe same formula appears, through a geometric method of constructing a square with (approximately) the same area as a circle

Ancient GreeksKnew that

\dfrac{A}{r^2}

is a constant, but not its value; two civilisations had approximated it as

\dfrac{256}{81}

c. 250 BCE: ArchimedesShowed the constant is exactly π, the same constant as in the circumference:

A = \pi r^2

c. 1500: Nīlakaṇṭha SomayājīGave the slice-and-rearrange explanation in his commentary on the Āryabhaṭīya

Check the Babylonian value

With $C = 2\pi r$ and $A = \pi r^2$ , $C^2 : A = 4\pi^2 r^2 : \pi r^2 = 4\pi$ . Taking this as 12 is the same as taking π ≈ 3.

10.2 Archimedes: the circle as a triangle

“The area of a circle is equal to the area of a right angle triangle with sides equal to the radius and the circumference of the circle.”

Archimedes, as quoted in the book

Formula

Area of a circle

= \dfrac12 \times \text{circumference} \times \text{radius} = \dfrac12 \times 2\pi r \times r = \pi r^2

His proof uses a property of every regular polygon: its area $= \frac12 \times$ perimeter $\times$ radius of the circle that fits tightly inside it. (This extends the triangle formula area $= r \times s$ with the incircle.)
He then did a thought experiment: let the number of sides grow larger and larger. The polygon approaches the circle, its perimeter approaches $2\pi r$ , and the area becomes $\frac12 \times 2\pi r \times r = \pi r^2$ .

10.3 Nīlakaṇṭha’s visual explanation

Figure 9: A circle sliced into sectors. As the slices get thinner, laying them side by side approaches a parallelogram, giving Area = base × height = πr × r = πr². — **Figure 9**: A circle sliced into sectors. As the slices get thinner, laying them side by side approaches a parallelogram, giving Area = base × height = πr × r = πr².

Step 1Cut the circle into many thin equal slices and lay them side by side, alternately point-up and point-down

↓

Step 2As the slices get smaller, the curved edges get closer and closer to straight lines: the figure approaches a parallelogram

↓

Step 3Base = half the circumference =

\pi r

(half the slices’ arcs lie along the bottom); height = radius

r

↓

ResultArea of the circle = area of the parallelogram = base × height =

\pi r \times r = \pi r^2

11 Area of a Sector and a Segment

Learn by heartDefinition 13

A sector of a circle is the region bounded by an arc and the two radii containing the endpoints of the arc.

Learn by heartDefinition 14

A segment of a circle is the region bounded by an arc of the circle and the chord joining the endpoints of the arc.

Figure 10: A sector (left) is cut off by two radii; a segment (right) is cut off by a chord. — **Figure 10**: A sector (left) is cut off by two radii; a segment (right) is cut off by a chord.

Region	Fraction of the circle	Area	Symmetry used
Semicircular disc	$\frac{180}{360} = \frac12$	$\frac12\pi r^2$	reflection symmetry
Quarter circular disc (quadrant)	$\frac{90}{360} = \frac14$	$\frac14\pi r^2$	quarter-turn symmetry
Sector of angle $\theta°$	$\frac{\theta}{360}$	$\pi r^2 \times \frac{\theta}{360}$	rotational symmetry

Formula

Area of a sector

= \pi r^2 \times \dfrac{\theta°}{360°}

Same idea as arc length: the sector is the fraction

\frac{\theta}{360}

of the whole disc

A chord cuts the circle into a minor part (the smaller angle $\theta$ ) and a major part (angle $360° – \theta$ ); this is true for both sectors and segments.
Minor segment = sector − the triangle formed by the two radii and the chord. Major segment = whole circle − minor segment.

Given

Radius $= 14$ cm, sector angle $= 75°$

Find

Area of the sector (use $\pi \approx \frac{22}{7}$ )

Solution

Area $= \frac{22}{7} \times 14^2 \times \frac{75}{360} = \frac{22}{7} \times 196 \times \frac{5}{24} = \;$ $128.33$ cm² (approx.)

Given

Radius $10$ cm; a chord subtends $60°$ at the centre ( $\pi \approx 3.14$ , $\sqrt3 \approx 1.73$ )

Find

Area of the minor segment

Solution

Sector $= 3.14 \times 100 \times \frac{60}{360} = 52.33$ cm² (approx.). The two radii and the chord form an equilateral triangle (two sides 10 and a 60° angle between them), area $= \frac{\sqrt3}{4} \times 10^2 = 43.25$ cm².

Minor segment $= 52.33 – 43.25 = \;$ $9.08$ cm² (approx.)

Exam Tip: scaling

Double every length of a shape and its area becomes $2^2 = 4$ times; triple every length and it becomes $3^2 = 9$ times. That is why circumferences in the ratio 3 : 2 give areas in the ratio 9 : 4.

Figure 11: An area model: a square of side a + b splits into a², ab, ab and b², so (a + b)² = a² + 2ab + b². — **Figure 11**: An area model: a square of side a + b splits into a², ab, ab and b², so (a + b)² = a² + 2ab + b².

Check yourself before the exam

Can I explain why π is more than 3 using an inscribed hexagon, and give Archimedes’ bounds?
Can I match each name (Zu Chongzhi, Āryabhaṭa, Brahmagupta, Mādhava, Jones, Lambert) to what they did with π?
Can I find arc length, the perimeter of a sector and the stagger between two lanes?
Can I prove that a median divides a triangle into two equal areas?
Can I use Heron’s formula and Brahmagupta’s formula, and show Heron’s is the special case $d = 0$ ?
Can I write the steps of Baudhāyana’s construction and prove $HP^2 = ab$ ?
Can I find the area of a circle, a sector, and a minor or major segment?

Quick Revision: read this the night before the exam

π = C/D, the same for all circles; $\pi \approx \frac{22}{7}$ or $3.14$ ; π is irrational (Lambert, 1761)
Circumference $C = 2\pi r$ ; arc length $l = 2\pi r \times \frac{\theta°}{360°}$
Estimates of π: Archimedes $3\frac{10}{71} \lt \pi \lt 3\frac17$ ; Zu Chongzhi $\frac{355}{113}$ ; Āryabhaṭa 3.1416; Mādhava $\pi = 4\left(1 – \frac13 + \frac15 – \frac17 + \cdots\right)$
Stagger makes up for the longer bends of outer lanes
Parallelogram = $bh$ ; triangle = $\frac12 bh$ ; a median splits a triangle into two equal areas
Heron: $\sqrt{s(s-a)(s-b)(s-c)}$ , $s = \frac12(a+b+c)$ ; also $\frac{abc}{4R}$ and $\frac{r(a+b+c)}{2}$
Brahmagupta (cyclic 4-gon): $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ , $s = \frac12(a+b+c+d)$ ; $d = 0$ gives Heron
Baudhāyana squared a rectangle using $\left(\frac{a+b}{2}\right)^2 – \left(\frac{a-b}{2}\right)^2 = ab$
Area of a circle $A = \pi r^2$ ; area of a sector $= \pi r^2 \times \frac{\theta°}{360°}$

Practice Questions

1 markFind the circumference of a circle of radius 21 cm (use $\pi \approx \frac{22}{7}$ ).
2 marksFind the length of an arc of radius 10.5 cm subtending 60° at the centre.
3 marksFind the area of a triangle with sides 8 cm, 15 cm, 17 cm using Heron’s formula, and verify using base × height.
2 marksFind the area of a cyclic 4-gon with sides 9, 9, 16, 16 units.
2 marksFind the area of a sector of radius 21 cm with a central angle of 60°.
2 marksThe diameter of a wheel is 63 cm. How far does it travel in 500 revolutions?
1 markAD is a median of △ABC, whose area is 48 sq. cm. Find the area of △ABD.
2 marksFind the perimeter of a sector of radius 7 cm with a central angle of 90°.
3 marksA park is a rhombus with all sides 13 m and one diagonal 24 m. Find its area using Heron’s formula (split it into two triangles).
2 marksThe ratio of the circumferences of two circles is 3 : 2. Find the ratio of their areas.

Multiple Choice

1 markπ is:
(a) exactly 22/7(b) a rational number(c) an irrational number(d) an integer
1 markThe area of a sector with θ = 180° is:
(a) πr²(b) πr²/2(c) πr²/4(d) 2πr²
1 markBrahmagupta’s formula applies to:
(a) any 4-gon(b) any parallelogram(c) only a cyclic 4-gon(d) only a square

Assertion (A): A median of a triangle always divides it into two triangles of equal area.
Reason (R): The two triangles formed have equal bases and share the same height from the opposite vertex.

Case Study

A circular flower bed has radius 14 m. A gardener plants flowers in a sector of the bed with a central angle of 90°, and builds a path along the sector’s curved and straight edges.

1 markWhat is the arc length of the sector (use $\pi \approx \frac{22}{7}$ )?
1 markWhat is the perimeter of the sector (including the two straight edges)?
2 marksWhat area is planted with flowers?

Answer Key

2 \times \frac{22}{7} \times 21 = 132

2 \times \frac{22}{7} \times 10.5 \times \frac{60}{360} = 11

s = 20

; area

= \sqrt{20 \times 12 \times 5 \times 3} = \sqrt{3600} = 60

sq. cm; also right-angled (

8^2 + 15^2 = 17^2

\frac12(8)(15) = 60

sq. cm

s = 25

; area

= \sqrt{16 \times 16 \times 9 \times 9} = 16 \times 9 = 144

sq. units

\frac{22}{7} \times 21^2 \times \frac{60}{360} = 231

sq. cm

Q6Circumference

= \frac{22}{7} \times 63 = 198

cm; distance

= 198 \times 500 = 99000

= 990

Q724 sq. cm (half of 48)

Q8Arc

= 2 \times \frac{22}{7} \times 7 \times \frac{90}{360} = 11

cm; perimeter

= 11 + 7 + 7 = 25

Q9The 24 m diagonal makes two triangles with sides 13, 13, 24:

s = 25

, area

= \sqrt{25 \times 12 \times 12 \times 1} = 60

; rhombus

= 2 \times 60 = 120

sq. m

Q10Radii are in the ratio 3 : 2, so areas are in the ratio

3^2 : 2^2 = 9 : 4

MCQ 1-3(c) · (b) · (c)

A/R(a) Both A and R are true, and R is the correct explanation of A

Case Study22 m · 50 m · 154 sq. m