NCERT Class 9 Maths Chapter 6 full notes on perimeter, area of triangles, parallelograms, circles, and composite figures.
At the start of a 4 × 100 m relay race, runners in outer lanes are placed ahead of runners in inner lanes. Why? Because the outer semicircular curves have a larger radius! The distance between starting points is called the Stagger.
Perimeter is the total boundary distance around a 2D shape:
• Square: \( 4a \)
• Equilateral Triangle: \( 3a \)
• Rectangle: \( 2(a + b) \)
For every circle, large or small, the ratio of its Circumference (\( C \)) to its Diameter (\( D \)) is a constant value called \( π \) (Pi):
By rotational symmetry, an arc subtending an angle \( θ^° \) at the center has length proportional to \( θ^°360^° \):
Problem: On a standard 400 m track, the lane width is \( w = 1.22 m \). What is the stagger distance needed for a runner in Lane 2 relative to Lane 1 over a full 400 m lap?
Solution:
Both runners complete two straight stretches of equal length.
The difference arises on the two semicircular bends (which make 1 full circle).
Let Lane 1 radius be \( r \), then Lane 2 radius is \( r + w \).
\[ Stagger = 2π(r + w) - 2π r = 2π w \]
Using \( w = 1.22 m \):
\[ Stagger = 2 × 3.1416 × 1.22 = \mathbf{7.66 metres} \]
The runner in Lane 2 must start 7.66 m ahead of Lane 1!
Proof:
Let large semicircle \(a\) have radius \(a'\). Its arc length is \(L_a = \pi a'\).
Let small semicircles \(b, c, d\) have radii \(b', c', d'\). Their combined arc length is \(L_{b+c+d} = \pi b' + \pi c' + \pi d' = \pi(b' + c' + d')\).
Since total diameter \( PQ = 2a' = 2b' + 2c' + 2d' \implies a' = b' + c' + d' \).
Therefore, \(L_a = \pi a' = \pi(b' + c' + d') = L_{b+c+d}\). The paths are EQUAL!
\( Area = Base × Height = bh \). (Transformed into an equal-area rectangle).
\( Area = 12 × Base × Height = 12 bh \). (Two congruent triangles fit together to make a parallelogram!).
When height is unknown, Heron's Formula calculates triangle area directly from side lengths \( a, b, c \):
In 628 CE, Brahmagupta discovered a profound formula for the area of a cyclic quadrilateral with sides \( a, b, c, d \):
Problem: Find the area of a triangle with side lengths 7 cm, 24 cm, and 25 cm.
Solution:
Semi-perimeter \( s = 7 + 24 + 252 = 562 = 28 cm \).
\[ Area = √28(28 - 7)(28 - 24)(28 - 25) = √28 × 21 × 4 × 3 \]
\[ = √(7 × 4) × (7 × 3) × 4 × 3 = √7^2 × 4^2 × 3^2 = 7 × 4 × 3 = \mathbf{84 sq. cm} \]
In 800 BCE, Indian mathematician Baudhāyana gave a geometric construction to build a square whose area equals a given rectangle of sides \( a \) and \( b \):
By constructing a right triangle with hypotenuse \( a+b2 \) and base \( a-b2 \), its altitude is exactly \( √ab \), creating a square of area \( ab \)!
In 1500 CE, Nīlakaṇṭha Somayājī (Kerala School) presented a visual proof: slice a circle into infinite tiny sectors and interlock them. They form a parallelogram with base \( = half circumference = π r \) and height \( = r \):
A sector subtending an angle \( θ^° \) at the center has area proportional to \( θ^°360^° \):
Problem: A clock's minute hand is 7 cm long. Find the area swept by it in 10 minutes.
Solution:
In 60 minutes, minute hand completes \( 360^° \).
In 10 minutes, angle swept \( θ = 1060 × 360^° = 60^° \).
\[ Area = π r^2 × 60^°360^° = 227 × 7^2 × 16 = 22 × 76 = 1546 = \mathbf{25.67 sq. cm} \]
Additional worked examples and questions from the previous edition