1. Formulas
| Figure | Area | Perimeter |
|---|---|---|
| Rectangle | l × b | 2(l + b); diagonal √(l² + b²) |
| Square | a² = d²/2 | 4a; diagonal a√2 |
| Triangle | ½ × base × height; Heron: √(s(s − a)(s − b)(s − c)) | a + b + c |
| Equilateral triangle | (√3/4)a² | 3a; height (√3/2)a |
| Parallelogram | base × height | 2(a + b) |
| Rhombus | ½ d₁d₂ | 4a; side = ½√(d₁² + d₂²) |
| Trapezium | ½ (sum of parallel sides) × h | |
| Circle | πr² | 2πr |
| Semicircle | πr²/2 | πr + 2r |
| Sector (θ) | θ/360 × πr² | Arc = θ/360 × 2πr |
| Ring | π(R² − r²) | |
| Regular hexagon | (3√3/2)a² | 6a |
2. Results to remember
- A wire bent into different shapes keeps its length; the circle gives the largest area.
- If sides change by x% and y%, area changes by x + y + xy/100 %.
- Path of width w outside a rectangle: area = 2w(l + b) + 4w².
- Wheel: distance in one revolution = circumference.
3. Solved examples
(ASO 2015) Sides 8, 11, perimeter 32 → c = 13, s = 16 → area √(16 × 8 × 5 × 3) = 8√30.
(ASO 2012) Circle of radius 28 bent into a square → side 176/4 = 44 cm.
(ASO 2021) Circle and semicircle of radius 14: 88 : 72 = 11 : 9.
A wheel of radius 35 cm makes 1000 revolutions → 220 × 1000 cm = 2.2 km.
Diagonals 16 and 12 → rhombus side 10, area 96.
Solved examples by question type
Each type that comes in the ASO paper, worked step by step. Try it yourself first, then read the solution.
Basic level
Example 1. The area of a rectangle 20 cm long and 12 cm wide is:
(a) 32 cm² (b) 250 cm² (c) 64 cm² (d) 240 cm²
Answer: (d) 240 cm²
Solution: Area = l × b = 20 × 12 = 240 cm².
Example 2. The area of a circle of radius 7 cm is (π = 22/7):
(a) 616 cm² (b) 154 cm² (c) 77 cm² (d) 44 cm²
Answer: (b) 154 cm²
Solution: Area = πr² = 22/7 × 7² = 154 cm².
Example 3. The circumference of a circle of radius 14 cm is (π = 22/7):
(a) 44 cm (b) 616 cm (c) 95 cm (d) 88 cm
Answer: (d) 88 cm
Solution: C = 2πr = 2 × 22/7 × 14 = 88 cm.
Example 4. A rectangle has perimeter 118 m and length 50 m. Its area is:
(a) 450 m² (b) 2950 m² (c) 500 m² (d) 3481 m²
Answer: (a) 450 m²
Solution: Breadth = 59 − 50 = 9. Area = 50 × 9 = 450 m².
Example 5. The diagonal of a square is 23√2 cm. Its area is:
(a) 1058 cm² (b) 529 cm² (c) 530 cm² (d) 92 cm²
Answer: (b) 529 cm²
Solution: Side = 23 cm. Area = 529 cm².
Example 6. The diagonal of a square is 23√2 cm. Its area is:
(a) 530 cm² (b) 529 cm² (c) 92 cm² (d) 1058 cm²
Answer: (b) 529 cm²
Solution: Side = 23 cm. Area = 529 cm².
ASO exam level
Example 7. Two sides of a triangle are 10 cm and 17 cm and its perimeter is 48 cm. Its area is:
(a) 84 cm² (b) 90 cm² (c) 168 cm² (d) 85 cm²
Answer: (a) 84 cm²
Solution: Third side = 21. s = 24. Area = √[s(s−a)(s−b)(s−c)] = √7,056 = 84 cm².
Example 8. Two sides of a triangle are 13 cm and 14 cm and its perimeter is 42 cm. Its area is:
(a) 90 cm² (b) 84 cm² (c) 91 cm² (d) 168 cm²
Answer: (b) 84 cm²
Solution: Third side = 15. s = 21. Area = √[s(s−a)(s−b)(s−c)] = √7,056 = 84 cm².
Example 9. The area of a regular hexagon of side 4 cm is:
(a) 24√3 cm² (b) 48√3 cm² (c) 4√3 cm² (d) 24 cm²
Answer: (a) 24√3 cm²
Solution: Area = (3√3/2)a² = 3√3/2 × 16 = 24√3 cm².
Example 10. The area of a regular hexagon of side 4 cm is:
(a) 24 cm² (b) 4√3 cm² (c) 48√3 cm² (d) 24√3 cm²
Answer: (d) 24√3 cm²
Solution: Area = (3√3/2)a² = 3√3/2 × 16 = 24√3 cm².
Example 11. A wire in the shape of a circle of radius 28 cm is bent into a square. The side of the square is (π = 22/7):
(a) 44 cm (b) 48 cm (c) 176 cm (d) 56 cm
Answer: (a) 44 cm
Solution: Perimeter stays the same: 2πr = 176 cm. Side = 176/4 = 44 cm.
Example 12. A wire is bent first into a square and then into a circle. Which statement is true?
(a) It depends on the length of the wire (b) The square encloses more area (c) The circle encloses more area (d) Both enclose equal area
Answer: (c) The circle encloses more area
Solution: For a fixed perimeter, the circle encloses the largest possible area.
Example 13. A wire is bent first into a square and then into a circle. Which statement is true?
(a) Both enclose equal area (b) It depends on the length of the wire (c) The circle encloses more area (d) The square encloses more area
Answer: (c) The circle encloses more area
Solution: For a fixed perimeter, the circle encloses the largest possible area.
Example 14. A garden is 25 m by 15 m. A path 1 m wide runs around it outside. The area of the path is:
(a) 84 m² (b) 80 m² (c) 100 m² (d) 299 m²
Answer: (a) 84 m²
Solution: Outer = 27 × 17 = 459. Path = 459 − 375 = 84 m².
Example 15. Each side of a square is increased by 50%. The area increases by:
(a) 125% (b) 129% (c) 100% (d) 50%
Answer: (a) 125%
Solution: 50 + 50 + 50×50/100 = 125%.
Example 16. The circumference of a circle is 44 cm. Its area is (π = 22/7):
(a) 77 cm² (b) 154 cm² (c) 198 cm² (d) 308 cm²
Answer: (b) 154 cm²
Solution: 2πr = 44 → r = 7. Area = 22/7 × 7² = 154 cm².
Example 17. The circumference of a circle is 132 cm. Its area is (π = 22/7):
(a) 2,772 cm² (b) 693 cm² (c) 1,518 cm² (d) 1,386 cm²
Answer: (d) 1,386 cm²
Solution: 2πr = 132 → r = 21. Area = 22/7 × 21² = 1,386 cm².
Example 18. A trapezium has parallel sides 19 cm and 11 cm and height 6 cm. Its area is:
(a) 180 cm² (b) 96 cm² (c) 90 cm² (d) 104.5 cm²
Answer: (c) 90 cm²
Solution: Area = ½ (19 + 11) × 6 = 90 cm².
Example 19. A trapezium has parallel sides 16 cm and 8 cm and height 8 cm. Its area is:
(a) 64 cm² (b) 192 cm² (c) 96 cm² (d) 104 cm²
Answer: (c) 96 cm²
Solution: Area = ½ (16 + 8) × 8 = 96 cm².
Example 20. The diagonals of a rhombus are 28 cm and 96 cm. Its perimeter is:
(a) 204 cm (b) 248 cm (c) 200 cm (d) 124 cm
Answer: (c) 200 cm
Solution: Half diagonals 14 and 48 → side = √(196 + 2304) = 50. Perimeter = 200 cm.
Example 21. The diagonals of a rhombus are 16 cm and 30 cm. Its perimeter is:
(a) 68 cm (b) 46 cm (c) 72 cm (d) 92 cm
Answer: (a) 68 cm
Solution: Half diagonals 8 and 15 → side = √(64 + 225) = 17. Perimeter = 68 cm.
Advanced level
Example 22. The area of the largest triangle that can be inscribed in a semicircle of radius 6 cm is:
(a) 56.57 cm² (b) 36 cm² (c) 18 cm² (d) 72 cm²
Answer: (b) 36 cm²
Solution: Base = diameter 12, height = radius 6. Area = ½ × 12 × 6 = 36 cm² (r²).
Example 23. The area of the largest triangle that can be inscribed in a semicircle of radius 10 cm is:
(a) 200 cm² (b) 157.14 cm² (c) 100 cm² (d) 50 cm²
Answer: (c) 100 cm²
Solution: Base = diameter 20, height = radius 10. Area = ½ × 20 × 10 = 100 cm² (r²).
Example 24. The diagonals of a rhombus are 16 cm and 12 cm. Its perimeter is:
(a) 40 cm (b) 48 cm (c) 56 cm (d) 96 cm
Answer: (a) 40 cm
Solution: Half-diagonals 8 and 6 → side = √(64 + 36) = 10. Perimeter = 40 cm.
Example 25. The diagonals of a rhombus are 24 cm and 10 cm. Its perimeter is:
(a) 60 cm (b) 52 cm (c) 120 cm (d) 68 cm
Answer: (b) 52 cm
Solution: Half-diagonals 12 and 5 → side = √(144 + 25) = 13. Perimeter = 52 cm.
Example 26. Two concentric circles have radii 28 cm and 14 cm. The area of the ring between them is (π = 22/7):
(a) 88 cm² (b) 616 cm² (c) 1,870 cm² (d) 1,848 cm²
Answer: (d) 1,848 cm²
Solution: π(R² − r²) = 22/7 × (784 − 196) = 1,848 cm².
Example 27. Two concentric circles have radii 49 cm and 21 cm. The area of the ring between them is (π = 22/7):
(a) 6,160 cm² (b) 2,464 cm² (c) 176 cm² (d) 6,182 cm²
Answer: (a) 6,160 cm²
Solution: π(R² − r²) = 22/7 × (2401 − 441) = 6,160 cm².
Topic-wise practice for Mensuration: plane figures (Quant): Triangle, quadrilaterals, circle, sector. Useful for OPSC ASO, OSSC, OSSSC and other Odisha exams.
Question types covered
- Area and perimeter of triangle, rectangle, square
- Parallelogram, rhombus, trapezium
- Circle: area, circumference
- Sector, arc and segment
- Wheel revolutions
- Path around a field, cost of fencing or flooring