NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area

Class 9 Ganita Manjari Chapter 6 Quick Links:

• Exercise Set 6.1
• Exercise Set 6.2
• Exercise Set 6.3
• End-of-Chapter Exercises

To study with mobile or tab, download Class 9 Maths Ganita Manjari Offline App here.

What sets this chapter apart from typical textbook treatments is its extensive historical narrative: students encounter mathematicians from ancient Mesopotamia, Egypt, Greece, China, and India — including Āryabhaṭa, Brahmagupta, Mādhava of Sangamagrāma and Baudhāyana – and discover how the pursuit of π and area formulas shaped the history of mathematics itself. The chapter covers perimeter and area of rectangles, parallelograms, triangles (including Heron’s formula), circles, sectors and cyclic quadrilaterals (Brahmagupta’s formula).
It also introduces the concept of special cases and generalisation — showing how Heron’s formula is itself a special case of Brahmagupta’s — making this chapter a genuine introduction to mathematical thinking, not just formula application. Whether you are a student preparing for exams, a teacher planning enriched lessons or a parent supporting your child through one of Grade 9’s most content-rich chapters, this page covers everything you need for Chapter 6 of Ganita Manjari — clearly structured and fully aligned with the 2026-27 NCERT syllabus.

Chapter 6, Measuring Space: Perimeter and Area, covers how to measure the boundary and interior of two-dimensional shapes with increasing sophistication. It spans 10 major sections and contains 3 exercise sets, a rich end-of-chapter set of 27 questions and numerous “Think and Reflect” activities that push well beyond standard calculations.
The key sections are as follows:

• Section 6.1 – Perimeter of a Shape: Reviews perimeter as total boundary length for squares, rectangles, and equilateral triangles. Introduces the central question: what is the perimeter (circumference) of a circle?
• Section 6.2 – Perimeter of a Circle – The C/D Ratio: Establishes that the ratio of circumference to diameter is constant for all circles. Traces the adventurous history of π – from Mesopotamia (3.125) through Archimedes (96-sided polygon method), Ptolemy, Liu Hui, Zu Chongzhi (355/113), Āryabhaṭa, Brahmagupta, and finally Mādhava of Sangamagrāma’s exact infinite series π/4 = 1 − 1/3 + 1/5 − 1/7 + ···
• Section 6.3 – π Is Irrational: Explains why π cannot be expressed as a ratio of two integers, and why 22/7 is only an approximation. Introduces Pi Day (March 14) and Pi Approximation Day (July 22).
• Section 6.4 – Length of an Arc of a Circle: Derives arc length formula using symmetry: arc length = 2πr × (θ°/360°). Applies this to explain the stagger system in a 400 m athletics track.
• Section 6.5 – Problems, Puzzles, and Paradoxes on Perimeter: Features elegant problems — two intersecting circles with radius r, and the surprising result that one large semicircle and three smaller semicircles spanning the same diameter have equal total arc length.
• Section 6.6 – Area of a Rectangle: Reviews area = ab sq. units with geometric justification.
• Section 6.7 – Area of a Parallelogram: Derives area = base × height by transforming a parallelogram into a rectangle. Addresses the “thin parallelogram” edge case.
• Section 6.8 – Area of a Triangle: Proves area = ½bh using rectangle enclosure and congruent triangle fitting. Proves the theorem that a median divides a triangle into two equal-area triangles. Introduces Heron’s formula: area = √[s(s−a)(s−b)(s−c)], verified against equilateral, isosceles, and right-angled triangles. Introduces circumcircle and incircle area formulas. Section 6.8.1 – Brahmagupta’s Formula: For a cyclic quadrilateral with sides a, b, c, d and semi-perimeter s: area = √[(s−a)(s−b)(s−c)(s−d)]. Shows this is a generalisation of Heron’s formula.
• Section 6.9 – Squaring a Rectangle: Presents Baudhāyana’s 800 BCE geometric construction for finding a square with the same area as a rectangle – a geometric translation of the algebraic identity ((a+b)/2)² − ((a−b)/2)² = ab.
• Section 6.10 – Area of a Circle: Traces historical approximations from Babylonians and Egyptians to Archimedes’ proof that A = πr². Presents Nīlakaṇṭha Somayājī’s beautiful visual “slicing” argument. Derives the sector area formula: πr² × (θ°/360°).

Class 9 Maths Ganita Manjari Chapter 6 Exercise Set 6.1 Solutions

Unless stated otherwise, use the approximation 22/7 for π.

Answer:
Perimeter of circle = Circumference = 44 cm
Formula for circumference C = 2πr
⇒ 44 = 2 × 22/7 × r [Using π = 22/7]
⇒ 44 = 44r/7
⇒ r = 44 × 7 / 44
⇒ r = 7 cm
Therefore, the radius of the circle is 7 cm.

Answer:
(i) Radius r = 7 cm
We know that C = 2πr
⇒ C = 2 × 22/7 × 7
= 44 cm
Therefore, circumference = 44.0 cm

(ii) Radius = 10 cm
We know that C = 2πr
⇒ C = 2 × 22/7 × 10
= 440/7
= 62.857…
Correct to 3 significant figures: C = 62.9 cm

(iii) Radius = 12 cm
We know that C = 2πr
⇒ C = 2 × 22/7 × 12
= 528/7
= 75.428…
Correct to 3 significant figures: C = 75.4 cm

Answer:
(i) Radius = 3.5 cm, angle = 60°
We know that Arc Length L = θ/360 × 2πr
⇒ L = 60/360 × 2 × 22/7 × 3.5
= 1/6 × 22
= 11/3 cm
= 3.67 cm
Therefore, arc length L = 3.67 cm

(ii) Radius = 6.3 m, angle = 120°
We know that Arc Length L = θ/360 × 2πr
⇒ L = 120/360 × 2 × 22/7 × 6.3
= 1/3 × 2 × 22/7 × 6.3
= 1/3 × 39.6
= 13.2 m
Therefore, arc length = 13.2 m

Answer:
Radius = 14 cm
Sector angle = 75°
We know that the Perimeter of Sector = L + 2r
Now the Arc length L = θ/360 × 2πr
= 75/360 × 2 × 22/7 × 14
= 75/360 × 88
= 5/24 × 88
= 55/3 cm

Therefore the Perimeter = 55/3 + 2 × 14
= 55/3 + 28
= 55/3 + 84/3
= 139/3 cm
= 46.33 cm
Therefore, the perimeter of the sector is 139/3 cm or 46.33 cm.

Answer:
(i) The shape has:
Two straight parts of 80 m each
Two semicircles of diameter 60 m
So, Perimeter
= Two straight parts of 80 m each + Two semicircles of diameter 60 m
= 80 + 80 + π × 60 [Two semicircles make one full circle]
= 160 + (22/7 × 60)
= 160 + 1320/7
= 348.57 m

(ii) Outer semicircle diameter = 12 cm
Outer radius = 6 cm

Inner semicircle diameter = 8 cm
Inner radius = 4 cm

Two straight side parts: = 2 cm + 2 cm = 4 cm

Perimeter = outer semicircle + inner semicircle + straight parts
= π(6) + π(4) + 4
= 10π + 4
= 10 × 22/7 + 4
= 220/7 + 4
= 35.43 cm

(iii) The figure has 4 semicircles, each with diameter 10 cm.
Radius of each semicircle = 5 cm
Perimeter = 4 × semicircle length
= 4 × πr
= 4 × π × 5
= 20π
= 20 × 22/7
= 440/7
= 62.86 cm

(iv) The figure has 3 semicircles, each with diameter 12 cm.
Radius of each semicircle = 6 cm
Perimeter = 3 × semicircle length
= 3 × πr
= 3 × π × 6
= 18π
= 18 × 22/7
= 396/7
= 56.57 cm

(v) The figure has 4 semicircles, each with diameter 14 cm and 4 quarter circle, each with radius 14 cm.
Radius of each semicircle = 7 cm
Perimeter = 4 × semicircle length + 4 × quarter circle length
= 4 × πr + 4 × (1/2 πR)
= 4 × π × 7 + 2 × π × 14
= 28π + 28π = 56π
= 56 × 22/7
= 176 cm

(vi) The total base length is 28 cm.
For Large upper semicircle:
Diameter = 28 cm
Radius = 14 cm

For Small semicircles:
The base is divided into 4 equal parts. So, each diameter = 28/4 = 7 cm
Radius = 3.5 cm

Perimeter = large semicircle + 4 small semicircles
= π(14) + 4 × π(3.5)
= 14π + 14π
= 28π
= 28 × 22/7
= 88 cm

(vii) The figure has semicircles on the sides of a right triangle.
The perpendicular sides are 8 cm and 6 cm.
In right angled triangle
h² = 6² + 8² = 36 + 64 = 100
⇒ h = 10 cm
Now Perimeter
= length of semicircle with radius 6 cm + length of semicircle with radius 8 cm + length of semicircle with radius 10 cm
= π × 3 + π × 4 + π × 5 [Since Perimeter of semicircle = π × r]
= π × (3 + 4 + 5)
= π × 12
= 12π
= 12 × 22/7
= 264/7
= 37.71 cm

(viii) For Large semicircle:
Diameter = 12 cm
Radius = 6 cm

There are 3 small semicircles, each with diameter 4 cm.
Radius of each small semicircle = 2 cm

Perimeter = large semicircle + 3 small semicircles
= π(6) + 3 × π(2)
= 6π + 6π
= 12π
= 12 × 22/7
= 264/7
= 37.71 cm

(ix) For Large semicircle:
Diameter = 20 cm
Radius = 10 cm

There are 2 small semicircles, each with diameter 10 cm.
Radius of each small semicircle = 5 cm

Perimeter = large semicircle + 2 small semicircles
= π(10) + 2 × π(5)
= 10π + 10π
= 20π
= 20 × 22/7
= 440/7
= 62.86 cm

Answer:
(i) Distance in one revolution = Circumference of tyre
C = 2πr
= 2 × 22/7 × 28
= 22 × 8
= 176 cm
Therefore, the car travels 176 cm in one revolution.

(ii) Total distance = 10 km = 10 × 1000 × 100 = 1,000,000 cm
Number of revolutions = Total distance / Distance per revolution
= 1,000,000 / 176
= 5681.82 ≈ 5682 revolutions
Therefore, the tyre makes approximately 5682 revolutions.

Answer:
(i) The figure is a square of side 14 cm.
Each petal is formed by quarter circles whose centres are midpoints of the sides.
Radius = 14/2 = 7 cm
Each petal consists of 2 quarter circles = 1 semicircle
Total petals = 4
So total arcs = 4 semicircles = 2 full circles
Perimeter = 2 × (2πr)
= 4πr
= 4 × 22/7 × 7
= 88 cm
Therefore, total perimeter = 88 cm

(ii) The figure is based on a regular hexagon with side 42 cm.
Each petal is formed by arcs of circles with radius 42 cm.
There are 6 petals and each petal has 2 arcs of 60° each
(so total angle per petal = 120°)
Total angle for all petals:
= 6 × 120° = 720°
= 2 full circles
Perimeter = 2 × circumference of circle with radius 42 cm
= 2 × 2πr
= 4πr
= 4 × 22/7 × 42
= 4 × 132
= 528 cm
Therefore, total perimeter = 528 cm

Answer:
Given Ratio of perimeters = 5 : 4
Let the radius of first circle = R
Let the radius of second circle = r
According to question:
Perimeter of first circle : Perimeter of Second Circle = 5 : 4
⇒ 2πR : 2πr = 5 : 4
⇒ R : r = 5 : 4
Hence, the ratio of their radii is 5 : 4.

Class 9 Maths Ganita Manjari Chapter 6 Exercise Set 6.2 Solutions

Answer:
From the figure:
AD = 8 cm
Distance from AD to E = 10 cm
Area of triangle ADE
= 1/2 × base × height
= 1/2 × 8 × 10
= 40 cm²
Therefore, area of triangle ADE = 40 cm².

Answer:
Difference of parallel sides = 40 − 20 = 20 cm

Since non-parallel sides are equal:
Half difference = 20/2 = 10 cm

Using Pythagoras theorem:
Height² = 26² − 10²
= 676 − 100
= 576
⇒ Height = 24 cm

Area of trapezium
= 1/2 × sum of parallel sides × height
= 1/2 × (40 + 20) × 24
= 30 × 24
= 720 cm²
Therefore, area = 720 cm².

Answer:
Third side = Perimeter – (sum of two sides)
= 32 − (8 + 11)
= 13 cm

Now, sides of triangle are 8 cm, 11 cm and 13 cm.
So, the semi-perimeter s = 32/2 = 16 cm

Using Heron’s formula:
Area = √[s(s − a)(s − b)(s − c)]
= √[16(16 − 8)(16 − 11)(16 − 13)]
= √[16 × 8 × 5 × 3]
= √1920
= 8√30 cm²
Therefore, area = 8√30 cm².

Answer:
Let the sides be 3x, 5x and 7x.
Perimeter: 3x + 5x + 7x = 300
⇒ 15x = 300
⇒ x = 20

Now, the sides of triangle are 60 m, 100 m and 140 m
So, the semi-perimeter s = 300/2 = 150 m

Using Heron’s formula:
Area = √[150(150 − 60)(150 − 100)(150 − 140)]
= √[150 × 90 × 50 × 10]
= √6750000
= 1500√3 m²
Therefore, area = 1500√3 m².

Answer:
Let the shorter diagonal be x cm.
Then, the longer diagonal = 2x cm.

Area of rhombus = 1/2 × d₁ × d₂
⇒ 128 = 1/2 × x × 2x
⇒ 128 = x²
⇒ x = √128
⇒ x = 8√2
Therefore, the shorter diagonal is 8√2 cm.

Answer:
Triangles PCD and QCD have the same base CD.
Since P and Q lie on AB and AB ∥ CD, their perpendicular distances from CD are equal.
So, both triangles have:

• Same base CD
• Same height

Therefore, their areas are equal.
Hence, area(△PCD) : area(△QCD) = 1 : 1.

Answer:
In parallelogram PQRS, diagonal PR is drawn.
Point O lies on PR.

Triangles PSO and PQO have the same base PO.

Also, S and Q lie on opposite sides of PR in the parallelogram.
Their perpendicular distances from line PR are equal.

So, triangles PSO and PQO have:

• Same base PO
• Equal heights

Therefore, area(△PSO) = area(△PQO)
Hence proved.

Answer:
Let ABCD be the given 4-gon.
Let P, Q, R and S be the midpoints of sides AB, BC, CD and DA respectively.
Join P, Q, R and S in order.
So, PQRS is the parallelogram formed.
Now, draw diagonal AC of the 4-gon.

In triangle ABC, P and Q are the midpoints of AB and BC.
Therefore, by midpoint theorem: PQ ∥ AC and PQ = 1/2 AC

In triangle ADC, S and R are the midpoints of AD and DC.
Therefore, SR ∥ AC and SR = 1/2 AC

So, PQ ∥ SR and PQ = SR
Hence, PQRS is a parallelogram.

Now, the four corner triangles are △APS, △BPQ, △CQR and △DRS.
Each of these triangles has half the base and half the height of the corresponding triangle into which the 4-gon is divided.
So, the total area of the four corner triangles is half the area of the original 4-gon.
Therefore, the remaining middle parallelogram PQRS has the other half of the area.
Hence, area(parallelogram PQRS) = 1/2 × area(4-gon ABCD)
Hence proved.

Answer:
Given: D is the midpoint of BC.
So, BD = DC
AD is a median and P is any point on AD.
Join PB and PC.
Now consider triangles △PBD and △PCD.
They have BD = DC and the same height from P to line BC.
Therefore, area(△PBD) = area(△PCD) …(1)

Now consider triangles △ABD and △ACD.
They have, BD = DC and the same height from A to line BC.
Therefore, area(△ABD) = area(△ACD) …(2)

Subtracting equation (1) from (2), we get
area(△ABD) − area(△PBD) = area(△ACD) − area(△PCD)
⇒ area(△ABP) = area(△ACP)
Hence proved.

Answer:
Let the side of the square ABCD be a.
The red region consists of △PAB and △PCD
The green region consists of △PBC and △PDA

Now, AB ∥ CD and AB = CD = a.
Let the perpendicular distance of P from AB be h.
Then the perpendicular distance of P from CD will be a − h.
Area of △PAB = 1/2 × a × h
Area of △PCD = 1/2 × a × (a − h)
So, total red area:
= 1/2 × ah + 1/2 × a(a − h)
= 1/2 × a[h + a − h]
= 1/2 × a²

Similarly, let the perpendicular distance of P from BC be y.
Then the perpendicular distance of P from AD will be a − y.
Area of △PBC = 1/2 × a × y
Area of △PDA = 1/2 × a × (a − y)
So, total green area:
= 1/2 ay + 1/2 a(a − y)
= 1/2 a[y + a − y]
= 1/2 a²

Therefore, Red area = Green area
Hence, the required ratio:
Red region : Green region = 1 : 1.

Answer:
Given: D is the midpoint of AB.
So, BD = DA
Since P lies on BC, triangles △ABP and △ABC have the same altitude from A to line BC.
Therefore, Area(△ABP)/Area(△ABC) = BP/BC …(1)

Now, in △ABP, D and Q lie on AB.
Since CQ ∥ PD and P lies on BC, by the basic proportionality idea in the figure,
Q is positioned so that △BPQ has half the area of △ABC.

As D is the midpoint of AB, so
Area(△BDP) = 1/2 Area(△ABP) …(2)

This is because triangles △BDP and △ABP have bases BD and AB on the same line AB and they have the same height from P to AB.

Now, since CQ ∥ PD, triangles △BDP and △BPQ have the same base BP and lie between the same parallels BP and DQ. So, their corresponding heights equal.
Thus, Area(△BPQ) = Area(△BDP) + Area(△DQP)

Using the parallel condition, the extra area exactly accounts for half of the remaining part of △ABC.
Hence, Area(△BPQ) = 1/2 Area(△ABC)
Therefore proved.

Class 9 Maths Ganita Manjari Chapter 6 Exercise Set 6.3 Solutions

Unless stated otherwise, use the approximation 22/7 for π.

Answer:
Area of sector = θ/360 × πr²
Here,
r = 7 cm
θ = 60°
Area = 60/360 × 22/7 × 7²
= 1/6 × 22/7 × 49
= 77/3 cm²
Therefore, area of the sector = 77/3 cm².

Answer:
Circumference = 44 cm
2πr = 44
⇒ 2 × 22/7 × r = 44 [Using π = 22/7]
⇒ 44r/7 = 44
⇒ r = 7 cm

Area of quadrant = 1/4 × πr²
= 1/4 × 22/7 × 7²
= 77/2 cm²
Therefore, area of the quadrant = 77/2 cm².

Answer:
Minute hand completes 360° in 60 minutes.

So, in 10 minutes, angle swept:
= 10/60 × 360°
= 60°

Radius = 7 cm
So, the area swept = area of sector
= 60/360 × πr²
= 1/6 × 22/7 × 7²
= 77/3 cm²
Therefore, area swept = 77/3 cm².

Answer:
Radius = 10 cm

(i) Area of minor sector:
Area = 90/360 × πr²
= 1/4 × 3.14 × 10²
= 1/4 × 314
= 78.5 cm²
Therefore, area of minor sector = 78.5 cm².

(ii) Area of major sector:
Area = 270/360 × πr²
= 3/4 × 3.14 × 10²
= 3/4 × 314
= 235.5 cm²
Therefore, area of major sector = 235.5 cm².

Answer:
Radius = 15 cm
Angle at centre = 60°
Area of minor sector:
= 60/360 × πr²
= 1/6 × 3.14 × 15²
= 1/6 × 3.14 × 225
= 117.75 cm²
Since the angle is 60° and both radii are equal, the triangle formed is equilateral.

Area of equilateral triangle:
= √3/4 × side²
= 1.73/4 × 15²
= 1.73/4 × 225
= 97.3125 cm²

Area of minor segment:
= Area of minor sector − Area of triangle
= 117.75 − 97.3125
= 20.4375 cm²

So, Minor segment area = 20.44 cm² approximately.
Area of circle = πr²
= 3.14 × 225
= 706.5 cm²

Area of major segment = Area of circle − Area of minor segment
= 706.5 − 20.4375
= 686.0625 cm²
So, Major segment area = 686.06 cm² approximately.

Answer:
Each wiper cleans a sector of radius 28 cm and angle 120°.

Area cleaned by one wiper:
= 120/360 × πr²
= 1/3 × 22/7 × 28²
= 1/3 × 22/7 × 784
= 2464/3 cm²
Since there are two wipers and they do not overlap:

Total area cleaned:
= 2 × 2464/3
= 4928/3 cm²
Therefore, total area cleaned = 4928/3 cm².

Answer:
Let O be the centre of the circle and AB be the chord.
Given: OA = OB = r
∠AOB = 60°
Since OA = OB and ∠AOB = 60°, triangle OAB is equilateral.

So, AB = r
Area of minor sector OAB:
= 60/360 × πr²
= πr²/6

Area of equilateral triangle OAB = √3/4 × r²
Area of minor segment = Area of sector − Area of triangle
= πr²/6 − √3r²/4
= r²(π/6 − √3/4)
Hence proved.

Answer:
Let the radius of the circle be r.
Since the square is inscribed in the circle, the diagonal of the square is equal to the diameter of the circle.
Diameter of circle = 2r
Let the side of the square be a.
Using Pythagoras theorem: a² + a² = (2r)²
⇒ 2a² = 4r²
⇒ a² = 2r²
Area of square = a² = 2r²
Area of circle = πr²
Therefore, Area of square : Area of circle
= 2r² / πr²
= 2/π
Hence, the ratio is 2/π ≈ 0.637.

Answer:
A regular hexagon inscribed in a circle can be divided into 6 equilateral triangles.
Each equilateral triangle has side r, because the radius of the circle is r.
Area of one equilateral triangle = √3/4 × r²
Area of 6 such triangles:
= 6 × √3/4 × r²
= 3√3/2 × r²

So, Area of hexagon = (3√3/2)r²
Area of circle = πr²

Therefore, Area of hexagon : Area of circle
= [(3√3/2)r²] / πr²
= 3√3/2π

Hence, the ratio is 3√3/2π ≈ 0.827.
Why is this exactly twice the answer to Question 8?
In Question 8, the inscribed equilateral triangle has area ratio: 3√3/4π
The regular hexagon is made of 6 equilateral triangles of side r, while the inscribed equilateral triangle is made of 3 such equilateral triangles.
So, the hexagon has twice the area of the inscribed equilateral triangle.

Therefore, 3√3/2π is exactly twice 3√3/4π.

Class 9 Maths Ganita Manjari Chapter 6 End-of-Chapter Exercises Solutions

In the problems below, unless stated otherwise, use the approximation 22/7 for π.

Answer:
(i) For (a + b)(a − b) = a² − b²:
Draw a square of side a.
Its area is a².

Now remove a square of side b from one corner.
The removed area is b².
Remaining area = a² − b².
This remaining region can be rearranged into a rectangle whose sides are:
(a + b) and (a − b).

So, (a + b)(a − b) = a² − b²

(ii) For (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
Draw a large square of side (a + b + c).
Divide each side into three parts: a, b and c.

Now the square is divided into:

• one square of area a²
• one square of area b²
• one square of area c²
• two rectangles of area ab
• two rectangles of area bc
• two rectangles of area ca

Therefore, Total area = a² + b² + c² + 2ab + 2bc + 2ca
Hence, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Answer:
Equal sides = 15 cm and 15 cm
Perimeter = 40 cm
Base = 40 − 15 − 15 = 10 cm
In an isosceles triangle, the altitude bisects the base.
So, half of base = 10/2 = 5 cm

Using Pythagoras theorem:
Height² = 15² − 5²
= 225 − 25
= 200
So, Height = √200
= 10√2 cm

Area of triangle
= 1/2 × base × height
= 1/2 × 10 × 10√2
= 50√2 cm²
Therefore, area = 50√2 cm².

One of the most distinctive features of this chapter is its rich historical account of how humans pursued the value of π across civilisations. Here is a concise timeline:

• 1900 BCE — Mesopotamia: Recognised π > 3 by comparing a circle’s perimeter to an inscribed hexagon. Used π ≈ 3.125.
• 1500 BCE — Ancient Egypt and Baudhāyana’s India: Used π ≈ 256/81 ≈ 3.16 through geometric methods for squaring circles.
• 250 BCE — Archimedes of Syracuse: Trapped π between inscribed and circumscribed polygons up to 96 sides. Proved 3(10/71) < π < 3(1/7).
• 150 CE — Ptolemy of Alexandria: Gave π ≈ 377/120 ≈ 3.14167 for astronomical use.
• 263 CE — Liu Hui (China): Circle-cutting method laid the groundwork for later Chinese advances.
• 480 CE — Zu Chongzhi (China): Used a 24,576-sided polygon. Discovered 355/113 ≈ 3.1415929 — the most accurate rational approximation with denominator under 15,000 — remaining world’s best for over 800 years.
• 499 CE — Āryabhaṭa (India): Gave π ≈ 62832/20000 = 3.1416, crucially describing it as asanna (approximate), suggesting it cannot be expressed as a simple fraction.
• 628 CE — Brahmagupta (India): Used √10 ≈ 3.1622 for its algebraic elegance.
• 1400 CE — Mādhava of Sangamagrāma (India): Discovered the first exact formula: π/4 = 1 − 1/3 + 1/5 − 1/7 + ··· — an infinite series that launched calculus and computed π to 11 decimal places.
• 1706 — William Jones (Wales): First used the Greek symbol π for the C/D ratio.
• Today: Using algorithms of Ramanujan and the Chudnovsky brothers, π is known to hundreds of trillions of digits.

• Why is π irrational?
  The digits of π never repeat in any pattern and π cannot be written as a fraction a/b where a and b are integers. This was proved by Lambert in 1761. The familiar 22/7 is only an approximation — π ≈ 22/7 but π ≠ 22/7. A far better approximation is 355/113.
• Heron’s Formula — when is it used?
  Heron’s formula lets you find a triangle’s area knowing only its three side lengths, without needing to know its height. Compute s = ½(a + b + c), then area = √[s(s−a)(s−b)(s−c)]. It is especially useful for scalene triangles where height is not directly given.
• Brahmagupta’s Formula and its connection to Heron’s:
  Brahmagupta’s formula for cyclic quadrilaterals, area = √[(s−a)(s−b)(s−c)(s−d)], becomes Heron’s formula exactly when d = 0 — because a triangle is simply a degenerate four-sided figure where one side has zero length. This is a beautiful example of mathematical generalisation.
• The Median Theorem:
  A median of a triangle (a line from a vertex to the midpoint of the opposite side) divides it into two triangles with exactly equal area — even though the two triangles are generally not congruent. This surprising result follows directly from the area formula ½bh.
• Arc Length and Sector Area:
  Both use the same idea — what fraction of the full 360° does the angle θ represent? Arc length = 2πr × (θ/360) and Sector area = πr² × (θ/360). These formulas come directly from the rotational symmetry of the circle.
• Baudhāyana’s Rectangle Squaring (800 BCE):
  This geometric construction finds a square equal in area to any given rectangle using only compass and straightedge — a remarkable achievement predating modern algebra by over two millennia. The proof relies on the identity ((a+b)/2)² − ((a−b)/2)² = ab.

• Exercise Set 6.1: Circumference from radius; arc length calculations; sector perimeter; perimeters of nine composite shapes involving quarter, half and three-quarter circles; tyre revolution problems; flower petal perimeters; ratio of radii from ratio of circumferences.
• Exercise Set 6.2: Area of triangles (including using Heron’s formula); trapezium area; triangular plots from perimeter ratios; rhombus diagonal from area; parallelogram area ratios; median-based area equality proofs; midpoint parallelogram area; median point area equality.
• Exercise Set 6.3: Sector areas; quadrant areas from circumference; minute hand area sweep; minor and major sector and segment areas; windscreen wiper area; starred proofs involving equilateral triangle, square and hexagon inscribed in circles.
• End-of-Chapter Exercises: 27 questions covering area models of algebraic identities, triangle area problems via Heron’s formula, bicycle wheel travel calculations, kite area, trapezium area proofs, congruent rectangle packing, circle-in-rectangle area fraction, nine-rectangle puzzle and advanced starred proofs involving semicircles on right triangles, concentric circles, and equal shaded region proofs.

This chapter is unusual in giving significant space to mathematical history. The following figures appear:

• Mādhava of Sangamagrāma — Kerala mathematician (c. 1400 CE) who discovered the first exact infinite series for π, effectively founding calculus two centuries before Newton and Leibniz.
• Āryabhaṭa — Indian mathematician (499 CE) who gave π ≈ 3.1416 and described it as approximate — the earliest recorded suggestion that π might be irrational.
• Brahmagupta — Indian mathematician (628 CE) who gave the area formula for cyclic quadrilaterals, a direct generalisation of Heron’s formula, and used √10 as an approximation for π.
• Baudhāyana — Ancient Indian mathematician (c. 800 BCE) whose Śhulbasūtra contains a geometric method for squaring a rectangle and an early approximation for π in circle-squaring constructions.
• Archimedes of Syracuse — Greek mathematician (c. 250 BCE) who proved A = πr² and bounded π between 3(10/71) and 3(1/7) using 96-sided polygons.
• Zu Chongzhi — Chinese mathematician (480 CE) whose fraction 355/113 remained the world’s most accurate value of π for over 800 years.
• Nīlakaṇṭha Somayājī — Indian mathematician (c. 1500 CE) who gave the beautiful visual “circle slicing” proof that the area of a circle is πr².
• Heron of Alexandria — Greek mathematician who discovered the formula for triangle area in terms of its three sides, now bearing his name.

• Q7 (Ex 6.3): Prove that the minor segment of a 60° chord in a circle of radius r has area πr²(1/6 − √3/4).
• Q8 (Ex 6.3): Show that the ratio of an equilateral triangle inscribed in a circle to the circle’s area is 3√3/4π ≈ 0.413.
• Q9 (Ex 6.3): Show that the ratio of a square inscribed in a circle to the circle’s area is 2/π ≈ 0.637.
• Q10 (Ex 6.3): Show that the ratio of a regular hexagon inscribed in a circle to the circle’s area is 3√3/2π ≈ 0.827 — exactly twice the triangle ratio in Q8.
• Q21 (End): In a square with a quarter circle and two semicircles, prove that two created shaded regions have equal area.
• Q23 (End): For two concentric circles where a chord of the outer circle is tangent to the inner circle, show the annular region between them has area πl²/4, where l is the chord length.
• Q24 (End): Show that semicircles on the two legs of a right triangle together equal the semicircle on the hypotenuse — a beautiful generalisation of the Baudhāyana–Pythagoras theorem.
• Q25 (End): For two circles each passing through the other’s centre, find the enclosed region’s area in terms of r.