✦ Educational Innovator & Creator

I craft engaging mathematical journeys and creative learning spaces, students love.

A dedicated and disciplined Mathematics educator at Sainik School Nalanda, recognized as a Gemini Certified Educator and Google AI Certified Educator. Dedicated to transforming abstract math into joyful learning through interactive digital tools, NEP‑aligned resources, and creative publications.

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✦ My Process

From idea to impact.

A structured, student‑centered process I follow to turn a classroom gap into a working digital resource.

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1. Explore

Identifying student learning gaps and where a concept needs more clarity.

✎

2. Formulate

Crafting digital TLMs and NEP‑aligned worksheets around that gap.

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3. Execute

Implementing interactive, NEP 2020‑aligned methodologies in the classroom.

✦

4. Inspire

Achieving academic rigor and clarity that students genuinely enjoy.

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Have an innovative mathematical
project or idea in mind? Let's bring it to life!

✉ Send Me a Message

Quick Questions

What kind of worksheets do you design?+
I design interactive, NEP-aligned digital worksheets tailored for conceptual clarity, self-paced practice, and immediate feedback.
Can you build TLMs for my chapter?+
Yes! I specialize in creating digital Teaching-Learning Models (TLMs) and virtual visual tools.
How do you integrate technology into math?+
I leverage ICT tools, interactive Live Worksheets, dynamic geometric models, and activity-based learning aligned with NEP 2020.
Can educators reach out to discuss ideas?+
Absolutely! I am always happy to connect, share insights, and discuss innovative math pedagogy with fellow educators.
Ganita Manjari • Class 9 • Ch-6 Measuring Space ⬅Chapter Hub
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GANITA MANJARI • CLASS 9 • CHAPTER 6
01

Measuring Space:
Perimeter and Area

From athletics tracks to ancient Indian formulas

Designed with Dedication & Love For Mathematics

Akash Srivastva, TGT (Mathematics)

Ganita Manjari • Class 9 • Chapter 6 Akash Srivastva | www.akashmaths.online
INTRODUCTION • ATHLETICS TRACK STAGGER
02

Why are the starting lines staggered?

Outer lanes have larger radius on the curves. Without stagger, outer-lane athletes would run a longer distance.
Stagger = forward offset that makes the total distance equal for every lane.

Think & Reflect

If a school has a 200 m track instead of 400 m, does it need a smaller stagger for the same 4×100 m race?

Book p.118 Akash Srivastva | www.akashmaths.online
6.1 PERIMETER OF A SHAPE
03

Perimeter = Distance around the boundary

Square

Perimeter = 4a

Ratio 4 : 1

Equilateral Triangle

Perimeter = 3a

Ratio 3 : 1

Rectangle

Perimeter = 2(a+b)

Think & Reflect

What is the connection between perimeter of a circle and the stagger calculation on a track?

Book p.119 Akash Srivastva | www.akashmaths.online
6.2 CIRCUMFERENCE & THE RATIO π
04

C ÷ D is always constant

For every circle: C / D = π
C = 2πr = πD
D C ≈ 3.14 × D Circumference unrolled
Book p.120–121 Akash Srivastva | www.akashmaths.online
6.2 HISTORICAL JOURNEY OF π
05

Ancient Approximations

Mesopotamia (~1900 BCE)
π ≈ 3.125
Archimedes (~250 BCE)
3.1408 < π < 3.1428
Zu Chongzhi (480 CE)
355/113 ≈ 3.1415929
Mādhava (~1400 CE)
11 correct decimal places
Book p.121–123 Akash Srivastva | www.akashmaths.online
6.2 MĀDHAVA’S INFINITE SERIES
06

From Geometry to Infinite Series

π/4 = 1 − 1/3 + 1/5 − 1/7 + 1/9 − …

Mādhava of Saṅgamagrāma (Kerala School, ~1400 CE) calculated π correct to 11 decimal places using this series.

Book p.123 Akash Srivastva | www.akashmaths.online
6.3 CALCULATING TRACK STAGGERS
07

Stagger depends only on lane width

Extra distance on a half-turn = πw
(w = lane width)
For w = 1.22 m → Stagger ≈ 3.83 m

Inner radius does not affect the stagger.

r r+w Extra length = πw
Book p.124 Akash Srivastva | www.akashmaths.online
6.4 ARC LENGTH
08

Length of an Arc

Arc Length = 2πr × (θ / 360°)
90° → πr/2
180° → πr
θ Arc length
Book p.125 Akash Srivastva | www.akashmaths.online
6.5 PERIMETER PARADOX
09

Surprising Equality of Paths

A large semicircle and several smaller semicircles drawn on the same diameter have the same total length.

Because the sum of the smaller diameters equals the large diameter, their semicircular paths add up to the same value.

Book p.126 Akash Srivastva | www.akashmaths.online
6.6–6.7 AREA OF RECTANGLE & PARALLELOGRAM
10

Area Formulas

Rectangle

Area = length × breadth

Parallelogram

Area = base × height

A parallelogram can be cut and rearranged into a rectangle of the same area.

Book p.127–128 Akash Srivastva | www.akashmaths.online
6.8 AREA OF A TRIANGLE
11

½ × base × height

Area = ½ × b × h

Any triangle occupies exactly half the area of the rectangle that has the same base and height.

Book p.129 Akash Srivastva | www.akashmaths.online
6.8.1 HERON’S FORMULA
12

Area using only three sides

Area = √[s(s−a)(s−b)(s−c)]

where s = (a+b+c)/2

No height is required. Discovered by Heron of Alexandria.

Book p.130 Akash Srivastva | www.akashmaths.online
6.8.2 BRAHMAGUPTA’S FORMULA
13

Area of a Cyclic Quadrilateral

Area = √[(s−a)(s−b)(s−c)(s−d)]

where s = (a+b+c+d)/2

When one side becomes zero, Brahmagupta’s formula reduces to Heron’s formula.
Book p.131 Akash Srivastva | www.akashmaths.online
6.9 BAUDHĀYANA’S SQUARING
14

Squaring a Rectangle (Śulbasūtra)

Baudhāyana (~800 BCE) gave a geometric construction to make a square equal in area to a given rectangle.

Side of the square = √(ab)

Using the identity: ab = [(a+b)/2]² − [(a−b)/2]²

Book p.132–133 Akash Srivastva | www.akashmaths.online
6.9 SQUARING A TRIANGLE
15

Two-stage process

1. Convert the triangle into a rectangle of equal area (base × half-height).
2. Apply Baudhāyana’s method to square that rectangle.

Think & Reflect

Why is converting a triangle into a rectangle first a helpful step?

Book p.133 Akash Srivastva | www.akashmaths.online
6.10 AREA OF A CIRCLE
16

A = πr²

Archimedes: Area of circle = Area of a right triangle with base = circumference (2πr) and height = radius (r)
½ × (2πr) × r = πr²
πr² ½ × 2πr × r
Book p.134 Akash Srivastva | www.akashmaths.online
6.10 NĪLAKAṆṬHA’S VISUAL PROOF
17

Pie-Slice → Rectangle

Thin sectors of the circle are rearranged alternately to form a near-rectangle.

Base ≈ πr    Height = r

Area = πr × r = πr²

≈ πr r Sectors rearranged
Book p.135 Akash Srivastva | www.akashmaths.online
KEY TAKEAWAYS & THANK YOU
18

Thank You

From tracks to temples — measuring space connects mathematics with the real world.

“Geometry is the foundation of all painting.”

— Albrecht Dürer

Designed with Dedication & Love For Mathematics

Akash Srivastva, TGT (Mathematics)

Ganita Manjari • Class 9 Akash Srivastva | www.akashmaths.online
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