1. Chapter at a Glance
A farmer wants to fence a triangular field and also know how much seed to buy. Fencing needs the perimeter — the length around the boundary. Seed depends on the area — the surface inside. Sometimes we know only the three sides of the field, not its height. How do we find the area then? Mathematicians over thousands of years — Heron of Alexandria, Āryabhaṭa, Brahmagupta, Archimedes — found clever answers. This chapter derives the key formulas for triangles, quadrilaterals and circles, and shows how they connect.
- Perimeter and area of triangles and quadrilaterals (recap and derivation).
- Heron's formula — area from three sides.
- Area of a quadrilateral by splitting into triangles; Brahmagupta's formula (enrichment).
- Circumference and area of a circle; why A = πr².
- Arc length, sector area and ring (annulus) area.
- How area changes when lengths are scaled.
2. Perimeter and Area — Recap
| Shape | Perimeter | Area |
|---|---|---|
| Rectangle l × b | 2(l + b) | lb |
| Square side a | 4a | a² |
| Triangle | a + b + c | ½ × base × height |
| Parallelogram | 2(a + b) | base × height |
| Trapezium | sum of sides | ½(a + b)h |
| Rhombus | 4a | ½ d₁d₂ |
Special triangles:
- Equilateral (side a): height = (√3/2)a, area = (√3/4)a².
- Right triangle with legs p and q: area = ½pq.
3. Heron's Formula
When only the three sides a, b, c are known, use the semi-perimeter s = (a + b + c)/2.
Area = √[s(s − a)(s − b)(s − c)]
Heron of Alexandria recorded it around 60 CE. In India, Brahmagupta (628 CE) gave a more general version for cyclic quadrilaterals.
Find the area of a triangle with sides 13 cm, 14 cm, 15 cm.
Solution: s = 21. Area = √(21 × 8 × 7 × 6) = √7056 = 84 cm².
Use Heron's formula for the 3-4-5 right triangle.
Solution: s = 6. √(6 × 3 × 2 × 1) = √36 = 6, and ½ × 3 × 4 = 6 ✓.
An isosceles triangle has equal sides 10 cm and base 12 cm. Find its area.
Solution: s = 16. √(16 × 6 × 6 × 4) = √2304 = 48 cm². (Height method: √(100 − 36) = 8; ½ × 12 × 8 = 48 ✓.)
Why it works (outline): drop the height h onto side c, splitting it into x and c − x. Then a² − x² = h² = b² − (c − x)². Solve for x, substitute to get h in terms of a, b, c, and simplify ½ch — the algebra factorises neatly into s(s − a)(s − b)(s − c) using the identity p² − q² = (p − q)(p + q).
4. Area of a Quadrilateral
Split it along a diagonal into two triangles and add their areas (Heron's formula for each, if needed).
Quadrilateral ABCD has AB = 9, BC = 40, CD = 28, DA = 15 (in m) and ∠ABC = 90°. Find its area.
Solution: AC = √(81 + 1600) = 41. △ABC = ½ × 9 × 40 = 180. △ACD: sides 41, 28, 15; s = 42; area = √(42 × 1 × 14 × 27) = √15876 = 126. Total = 306 m².
For a cyclic quadrilateral with sides a, b, c, d and s = (a + b + c + d)/2: Area = √[(s − a)(s − b)(s − c)(s − d)]. Setting d = 0 gives Heron's formula.
5. The Circle
For every circle, circumference ÷ diameter is the same number, called π (≈ 3.14159…). So:
- Circumference C = 2πr = πd
- Area A = πr²
Why A = πr²: cut a circle into many thin equal sectors and arrange them alternately point-up and point-down. They form a shape close to a rectangle of length half the circumference (πr) and height r. Area ≈ πr × r = πr². With more sectors the shape becomes an exact rectangle.
Āryabhaṭa (499 CE) stated that half the circumference multiplied by half the diameter gives the area of a circle — exactly (πr)(r).
A circular park has circumference 176 m. Find its area (π = 22/7).
Solution: 2πr = 176 → r = 28 m. Area = (22/7) × 784 = 2464 m².
6. Arcs, Sectors and Rings
A sector with central angle θ is the fraction θ/360 of the whole circle.
| Quantity | Formula |
|---|---|
| Arc length | (θ/360) × 2πr |
| Sector area | (θ/360) × πr² = ½ × arc × r |
| Ring (annulus) area | π(R² − r²) = π(R + r)(R − r) |
| Semicircle perimeter | πr + 2r |
Find the area of a sector of radius 21 cm and angle 60°, and its arc length (π = 22/7).
Solution: Area = (1/6) × (22/7) × 441 = 231 cm². Arc = (1/6) × 2 × (22/7) × 21 = 22 cm.
A circular running track has inner radius 63 m and is 7 m wide. Find its area (π = 22/7).
Solution: R = 70. Area = (22/7)(70 + 63)(70 − 63) = 22 × 133 = 2926 m².
7. Scaling and Generalisation
If every length of a figure is multiplied by k:
- Perimeter is multiplied by k.
- Area is multiplied by k².
Doubling the radius of a pizza doubles its crust length but gives four times as much pizza.
Among all rectangles of the same perimeter, the square has the largest area; among all shapes of the same perimeter, the circle encloses the most area.
8. Solved Problems (Board Pattern)
A triangular field has sides 50 m, 120 m and 130 m. Find the cost of fencing at ₹20 per m and the area.
Solution: Perimeter 300 m → ₹6,000. 50² + 120² = 130², so it is right-angled; area = ½ × 50 × 120 = 3000 m².
The sides of a triangle are in the ratio 25 : 14 : 12 and its perimeter is 510 m. Find its area.
Solution: 51k = 510 → k = 10; sides 250, 140, 120. s = 255. Area = √(255 × 5 × 115 × 135) = √19794375 ≈ 4449.1 m².
A wheel of diameter 70 cm makes how many revolutions to cover 1.1 km?
Solution: Circumference = (22/7) × 70 = 220 cm. 110000 ÷ 220 = 500.
9. Master Formula Table
| Result | Formula |
|---|---|
| Semi-perimeter | s = (a + b + c)/2 |
| Heron's formula | √[s(s − a)(s − b)(s − c)] |
| Equilateral triangle | (√3/4)a² |
| Brahmagupta (cyclic) | √[(s − a)(s − b)(s − c)(s − d)] |
| Circumference | 2πr |
| Circle area | πr² |
| Arc length | (θ/360) × 2πr |
| Sector area | (θ/360) × πr² |
| Ring area | π(R² − r²) |
| Scaling by k | Perimeter × k, area × k² |
10. Master Summary Table
| Figure | Key method | Exam focus |
|---|---|---|
| Triangle (3 sides known) | Heron's formula | Fields, ratio of sides |
| Quadrilateral | Split by a diagonal | Combined areas |
| Circle | π = C/d; sector rearrangement | Wheels, tracks, parks |
| Sector / arc | Fraction θ/360 | Clock hands, pizza slices |
| Scaling | Area grows as k² | Compare sizes |
Chart — area enclosed by a 40 cm boundary:
11. Mind Map
- ½bh
- Heron's formula
- Split by diagonal
- Brahmagupta
- 2πr, πr²
- Sector, ring
- Scaling k²
- Max area: circle
12. Tips and Tricks
- Before using Heron's formula, check whether the triangle is right-angled — ½ × leg × leg is quicker.
- Factorise inside the square root: 21 × 8 × 7 × 6 = (7 × 3)(2³)(7)(2 × 3) = 2⁴ × 3² × 7² → √ = 84.
- Choose π = 22/7 when the radius is a multiple of 7.
- Using the perimeter instead of the semi-perimeter in Heron's formula.
- Mixing up diameter and radius in circle formulas.
- Ring area is π(R² − r²), not π(R − r)².
13. Important One-liners
- Heron's formula needs only the three sides.
- Area of an equilateral triangle = (√3/4)a².
- π is the ratio of circumference to diameter for every circle.
- Āryabhaṭa: area of a circle = half circumference × radius.
- Scaling lengths by k scales area by k².
- For a fixed perimeter, the circle encloses the greatest area.
14. Exam-pattern Questions with Detailed Answers
Find the area of an equilateral triangle of side 8 cm.
Show answer
(√3/4) × 64 = 16√3 ≈ 27.7 cm².
Answer: 16√3 cm²
The semi-perimeter of a triangle with sides 7, 8, 9 cm is:
- 24
- 12
- 8
- 16
Show answer
(7 + 8 + 9)/2 = 12.
Answer: (B)
Find the circumference of a circle of radius 14 cm (π = 22/7).
Show answer
2 × 22/7 × 14 = 88 cm.
Answer: 88 cm
Find the area of a triangle with sides 5 cm, 12 cm and 13 cm using Heron's formula.
Show answer
s = 15; √(15 × 10 × 3 × 2) = √900 = 30 cm².
Answer: 30 cm²
The area of a circle is 154 cm². Find its perimeter (π = 22/7).
Show answer
πr² = 154 → r² = 49 → r = 7. Perimeter = 44 cm.
Answer: 44 cm
The minute hand of a clock is 14 cm long. What area does it sweep in 15 minutes?
Show answer
15 minutes = 90°. Area = ¼ × (22/7) × 196 = 154 cm².
Answer: 154 cm²
An isosceles triangle has perimeter 30 cm and each equal side 12 cm. Find its area.
Show answer
Base = 6. s = 15. √(15 × 3 × 3 × 9) = √1215 = 9√15 ≈ 34.9 cm².
Answer: 9√15 cm²
If the radius of a circle is increased by 50%, by what percent does its area increase?
Show answer
k = 1.5; area × 2.25 → increase of 125%.
Answer: 125%
A park in the shape of quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. Find its area.
Show answer
BD = √(144 + 25) = 13. △BCD = ½ × 12 × 5 = 30. △ABD: sides 9, 8, 13; s = 15; √(15 × 6 × 7 × 2) = √1260 ≈ 35.5. Total ≈ 65.5 m².
Answer: ≈ 65.5 m²
Show using Heron's formula that the area of an equilateral triangle of side a is (√3/4)a².
Show answer
s = 3a/2; s − a = a/2 (three times). Area = √[(3a/2)(a/2)³] = √(3a⁴/16) = (√3/4)a². ∎
Answer: Derived as shown
A square is inscribed in a circle of radius 10 cm. Find the area between the circle and the square (π = 3.14).
Show answer
Square's diagonal = 20 → area = 20²/2 = 200. Circle = 314. Difference = 114 cm².
Answer: 114 cm²
A 4 m wide road runs around the outside of a circular garden of radius 31.5 m. Find the cost of paving the road at ₹50 per m² (π = 22/7).
Show answer
R = 35.5, r = 31.5. Area = (22/7)(67)(4) = 5896/7 ≈ 842.29 m². Cost ≈ ₹42,114.
Answer: ≈ 842.3 m²; ≈ ₹42,114
15. Quick Revision Summary
- Heron: √[s(s − a)(s − b)(s − c)], s = semi-perimeter.
- Quadrilaterals: split into triangles; cyclic ones — Brahmagupta.
- Circle: C = 2πr, A = πr² (sectors → rectangle).
- Sector = θ/360 of the circle; ring = π(R² − r²).
- Scale lengths by k → area by k².