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 \( \pi \) (Pi):
By rotational symmetry, an arc subtending an angle \( \theta^\circ \) at the center has length proportional to \( \frac{\theta^\circ}{360^\circ} \):
Problem: On a standard 400 m track, the lane width is \( w = 1.22 \text{ m} \). 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 \( r + w \).
\[ \text{Stagger} = 2\pi(r + w) - 2\pi r = 2\pi w \]
Using \( w = 1.22 \text{ m} \):
\[ \text{Stagger} = 2 \times 3.1416 \times 1.22 = \mathbf{7.66 \text{ metres}} \]
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!
\( \text{Area} = \text{Base} \times \text{Height} = bh \). (Transformed into an equal-area rectangle).
\( \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} bh \). (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 \( s = \frac{7 + 24 + 25}{2} = \frac{56}{2} = 28 \text{ cm} \).
\[ \text{Area} = \sqrt{28(28 - 7)(28 - 24)(28 - 25)} = \sqrt{28 \times 21 \times 4 \times 3} \]
\[ = \sqrt{(7 \times 4) \times (7 \times 3) \times 4 \times 3} = \sqrt{7^2 \times 4^2 \times 3^2} = 7 \times 4 \times 3 = \mathbf{84 \text{ sq. cm}} \]
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 \( \frac{a+b}{2} \) and base \( \frac{a-b}{2} \), its altitude is exactly \( \sqrt{ab} \), 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 \( = \text{half circumference} = \pi r \) and height \( = r \):
A sector subtending an angle \( \theta^\circ \) at the center has area proportional to \( \frac{\theta^\circ}{360^\circ} \):
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 \( 360^\circ \).
In 10 minutes, angle swept \( \theta = \frac{10}{60} \times 360^\circ = 60^\circ \).
\[ \text{Area} = \pi r^2 \times \frac{60^\circ}{360^\circ} = \frac{22}{7} \times 7^2 \times \frac{1}{6} = \frac{22 \times 7}{6} = \frac{154}{6} = \mathbf{25.67 \text{ sq. cm}} \]