Complete guide to area formulas for all shapes, Heron's formula for triangles, π and the irrationality of π, area of circles and sectors, arc length, and Brahmagupta's formula for cyclic quadrilaterals — all newly expanded for Class 9. Chapter 8 · 8 Marks.
📌 Chapter 8 · 8 Marks
🆕 EXPANDED
All Basic Area Formulas
Heron's Formula
Area of Circle & Sectors ★
Brahmagupta's Formula ★
Arc Length
50 Practice Qs
Section 8.1
Area & Perimeter Formulas — All Shapes
These are the standard formulas tested at every mark level. Know all of them cold — especially the triangle and quadrilateral variants.
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Rectangle
Area = l × b Perimeter = 2(l+b) Diagonal = √(l²+b²)
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Square
Area = a² Perimeter = 4a Diagonal = a√2
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Parallelogram
Area = b × h Perimeter = 2(a+b) h = perpendicular height
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Triangle
Area = ½ × b × h Equilateral: (√3/4)a² Right: ½ × leg₁ × leg₂
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Rhombus
Area = ½ × d₁ × d₂ (d₁, d₂ = diagonals) Perimeter = 4a
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Trapezium
Area = ½(a+b)×h a, b = parallel sides h = height between them
Area formulas for all common 2D shapes — h always means perpendicular height, not slant side
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Height ≠ Side: In parallelogram Area = b×h, h is the perpendicular height, NOT the slant side. Same for trapezium. Drawing a perpendicular from vertex to base gives h.
Section 8.2
Heron's Formula
When you know all three sides of a triangle but not the height, Heron's formula gives the area directly — no need to find the perpendicular height.
📐 Heron's Formula — Area of a Triangle
s = (a+b+c)/2 Area = √[s(s−a)(s−b)(s−c)]
where s = semi-perimeter = half the perimeter a, b, c = three sides of the triangle Named after Heron of Alexandria (c. 10–75 CE)
Example 1
Find the area of a triangle with sides 13cm, 14cm, 15cm using Heron's formula. [3M]
1
Semi-perimeter: s = (13+14+15)/2 = 42/2 = 21 cm
2
s−a = 21−13 = 8, s−b = 21−14 = 7, s−c = 21−15 = 6
3
Area = √[21×8×7×6] = √[7056] = 84 cm²
✓ Area = 84 cm²
Example 2
A triangular park has sides 120m, 80m, 50m. Find the area and cost of fencing at ₹20/m. [3M]
1
s = (120+80+50)/2 = 125
2
s−a=5, s−b=45, s−c=75 Area = √[125×5×45×75] = √[2109375] ≈ 1452 m²
Heron's checks: (1) Verify triangle inequality before computing: a+b>c, b+c>a, a+c>b. (2) Always simplify the product under √ by factoring. (3) Common perfect squares to recognise: 7056=84², 1764=42², 8100=90².
Section 8.3
Circles, π, and Irrationality
The 2026-27 syllabus includes the irrationality of π — connecting this chapter with Chapter 3 (Number Systems). π is the ratio of circumference to diameter and is irrational (proven in 1761 by Lambert).
📐 Circle Formulas
Circumference = 2πr Area = πr²
r = radius = d/2 (d = diameter) | π ≈ 3.14159… (irrational, non-terminating non-repeating) Use π = 22/7 for approximation unless told to use 3.14 or leave in terms of π
📌 Why is π Irrational?
π = 3.14159265358979… is non-terminating and non-repeating — it cannot be expressed as p/q where p, q are integers and q≠0.
Proved by Johann Lambert (1761) using continued fractions. The proof is beyond Class 9 but the fact is tested: π is irrational. Therefore 2πr and πr² are irrational for rational r (unless r=0).
Common mistake: 22/7 ≠ π. 22/7 is a rational approximation of π. π itself is irrational.
Example 3
Find area and circumference of circle with radius 7cm. [2M]
1
Circumference = 2πr = 2×(22/7)×7 = 44 cm
2
Area = πr² = (22/7)×49 = 154 cm²
Example 4
A circular field has circumference 440m. Find its area. [3M]
1
2πr=440 → r = 440×7/(2×22) = 70 m
2
Area = π×70² = (22/7)×4900 = 15400 m²
Section 8.4 — NEW from Class 10
Sectors and Arc Length ★
A sector is a "pizza slice" of a circle. The arc length and sector area are proportional to the central angle θ.
θ = central angle (in degrees) | r = radius Perimeter of sector = 2r + arc length | Minor sector: θ < 180°; Major sector: θ > 180°
A sector is a fractional "slice" of a circle — area and arc length both scale linearly with the angle θ
Example 5
A sector has radius 14cm and angle 90°. Find (a) arc length (b) area (c) perimeter. [3M]
a
Arc = (90/360)×2πr = (1/4)×2×(22/7)×14 = (1/4)×88 = 22 cm
b
Area = (90/360)×πr² = (1/4)×(22/7)×196 = (1/4)×616 = 154 cm²
c
Perimeter = 2r + arc = 2×14 + 22 = 28+22 = 50 cm
✓ Arc = 22cm, Area = 154cm², Perimeter = 50cm
Example 6
Find the area of a ring (annulus) between two concentric circles of radii 10cm and 6cm. [2M]
1
Area of ring = π(R²−r²) = π(10²−6²) = π(100−36) = 64π cm²
2
= 64×(22/7) ≈ 201.1 cm²
✓ Area = 64π ≈ 201.1 cm²
Section 8.5 — NEW from Class 10 (India's Contribution)
Brahmagupta's Formula ★
Indian mathematician Brahmagupta (598–668 CE) derived a beautiful generalisation of Heron's formula for any cyclic quadrilateral (a quadrilateral inscribed in a circle).
👨🔬 Who was Brahmagupta?
Brahmagupta (598–668 CE) was an Indian mathematician and astronomer from Bhinmal (Rajasthan). His work Brāhmasphuṭasiddhānta (628 CE) was the first book to treat zero as a number. He also gave rules for arithmetic with negative numbers and this formula for cyclic quadrilaterals. His contributions were 1000+ years ahead of European mathematics.
📐 Brahmagupta's Formula — Area of a Cyclic Quadrilateral
s = (a+b+c+d)/2 Area = √[(s−a)(s−b)(s−c)(s−d)]
where s = semi-perimeter, a, b, c, d = four sides of the cyclic quadrilateral A cyclic quadrilateral is one whose all four vertices lie on a circle When d=0, this reduces to Heron's formula for triangles!
Example 7
Find area of cyclic quadrilateral with sides 3cm, 4cm, 6cm, 5cm. [3M]
1
s = (3+4+6+5)/2 = 18/2 = 9
2
s−a=6, s−b=5, s−c=3, s−d=4
3
Area = √(6×5×3×4) = √360 = 6√10 ≈ 18.97 cm²
🔗 Connection to Heron's Formula
If d → 0 (cyclic quadrilateral degenerates to triangle): s = (a+b+c)/2 Brahmagupta: √[(s−a)(s−b)(s−c)(s−0)] = √[s(s−a)(s−b)(s−c)] = Heron's formula!
Heron is a special case of Brahmagupta.
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Important: Brahmagupta's formula only works for CYCLIC quadrilaterals (all vertices on a circle). For non-cyclic quadrilaterals, a general formula doesn't exist without diagonal length. CBSE problems will always specify "cyclic" or inscribed in a circle.
NCERT Style · CBSE Pattern
More Worked Examples
Example 8
A square and equilateral triangle have equal perimeters. If the equilateral triangle has side 12cm, find the ratio of their areas. [3M]