MEASURING SPACE:
PERIMETER & TRACK STAGGERS
In a $4 \times 100\text{ m}$ relay race, athletes in outer lanes start ahead of inner lanes (staggered positions), yet all run the exact same distance to a common finish line[cite: 1]! Outer lanes possess a larger curvature radius requiring geometric offset calibration[cite: 1].
Perimeter Fundamentals & Fixed Ratios
Perimeter is the total length around the outer boundary of any 2D closed geometric shape[cite: 1]. Regular polygons establish fixed invariant ratios with their side lengths[cite: 1].
Square (Side $a$)
$\text{Perimeter} = 4a$[cite: 1]
Fixed Ratio Perimeter : Side = $4 : 1$[cite: 1]
Equilateral Triangle ($a$)
$\text{Perimeter} = 3a$[cite: 1]
Fixed Ratio Perimeter : Side = $3 : 1$[cite: 1]
Rectangle ($a, b$)
$\text{Perimeter} = 2(a + b)$[cite: 1]
Square is a special case where $a = b$[cite: 1]
Perimeter of a Circle: The $C/D$ Ratio & $\pi$
Constant Ratio $\pi$:
Regardless of size, all circles maintain an identical constant ratio of Circumference ($C$) to Diameter ($D$):
Formula: $C = \pi D = 2\pi r$ (where $r$ is radius)[cite: 1]. As a circle scales, $C$ and $D$ change proportionally keeping $\pi$ invariant[cite: 1].
Historical Journey of $\pi$ & Ancient Approximations
Mesopotamia (c. 1900 BCE)
Compared circle to inscribed hexagon ($6r = 3D$), setting $\pi \approx 3 + \frac{1}{8} = 3.125$[cite: 1].
Archimedes (c. 250 BCE)
Trapped $\pi$ using 96-sided inscribed/circumscribed polygons: $3\frac{10}{71} < \pi < 3\frac{1}{7}$ ($3.1408 < \pi < 3.1428$)[cite: 1].
Zu Chongzhi (480 CE)
Used 24,576-sided polygon to calculate $\frac{355}{113} \approx 3.1415929$ (accurate for 800 years!)[cite: 1].
Mādhava’s Infinite Series for $\pi$ (Kerala School)
The Dawn of Calculus (c. 1400 CE):
Mādhava of Saṅgamagrāma shifted focus from geometric polygon cutting to infinite numerical series analysis[cite: 1].
Calculated exact values of $\pi$ correct to 11 decimal places ($3.14159265358$)[cite: 1].
Calculating Athletics Track Staggers
Stagger Formulae:
• Full Circle Turn: $S = 2\pi w$[cite: 1]
• Half-Turn Curve: $S = \pi w$ (where $w$ is lane width)[cite: 1]
Standard Calculation ($w = 1.22\text{ m}$):
Stagger per lane on half-turn curve = $\pi \times 1.22 \approx 3.83\text{ m}$[cite: 1].
Key Insight: Stagger depends solely on lane width $w$, independent of inner track radius[cite: 1]!
Arc Length Formula & Track Curves
An arc represents a fraction of circle circumference subtended by central angle $\theta$[cite: 1]:
• Semicircle ($\theta = 180^\circ$): $l = \pi r$[cite: 1] | Quarter Circle ($\theta = 90^\circ$): $l = \frac{\pi r}{2}$[cite: 1]
Area of Rectangles & Parallelograms
Rectangle Area
Measured relative to $1 \times 1$ square unit tiles[cite: 1].
$\text{Area} = a \times b$
Parallelogram Area
Transformed into an equal-area rectangle by cutting/shifting right triangles[cite: 1].
$\text{Area} = \text{base} \times \text{height} = b \cdot h$[cite: 1]
Triangle Area & Heron’s Formula
Standard Area
Enclosing triangle in $b \times h$ bounding rectangle yields half area[cite: 1]:
$\text{Area} = \frac{1}{2}bh$[cite: 1]
Heron's Formula
Calculates area using three side lengths $a, b, c$ without altitude $h$[cite: 1]:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$[cite: 1]
where semi-perimeter $s = \frac{a+b+c}{2}$[cite: 1].
Brahmagupta’s Formula for Cyclic Quadrilaterals
Formulated in 628 CE by Brahmagupta for cyclic 4-gons inscribed in a circle with sides $a, b, c, d$[cite: 1]:
Grand Generalisation: Setting fourth side $d = 0$ transforms Brahmagupta's formula directly into Heron's Formula[cite: 1]!
Baudhāyana’s Śulbasūtra: Squaring a Rectangle
Geometric Construction:
Constructing a square equal in area to an $a \times b$ rectangle ($a > b$)[cite: 1].
Side of constructed square equals $\sqrt{ab}$[cite: 1].
Algebraic Identity:
Demonstrates ancient Vedic geometry linking areas with right triangles[cite: 1].
Area of a Circle: $\pi r^2$ & Nīlakaṇṭha's Proof
Archimedes' Theorem:
Circle area equals a right triangle with base equal to circumference ($2\pi r$) and height equal to radius ($r$)[cite: 1]:
$\text{Area} = \frac{1}{2} \times 2\pi r \times r = \pi r^2$[cite: 1]
Nīlakaṇṭha Somayājī (c. 1500 CE):
Visual pie-slice dissection rearranging circle sectors into an interlocking rectangular strip of base $\pi r$ and height $r$[cite: 1].
Chapter Summary: Formulas at a Glance
THANK YOU!
"Space and shapes connect straight-edged polygons with the infinite curves of nature."
Visit www.akashmaths.online →