Mensuration
1. Perimeter and Area of Basic Shapes
Square
- Perimeter = 4 × side = 4s
- Area = side × side = s²
Example: Side = 8 cm. P = 32 cm. A = 64 cm².
Rectangle
- Perimeter = 2(length + breadth) = 2(l + b)
- Area = length × breadth = l × b
Example: l = 12 cm, b = 5 cm. P = 2(12 + 5) = 34 cm. A = 12 × 5 = 60 cm².
Triangle
- Perimeter = Sum of all three sides = a + b + c
- Area = ½ × base × height = ½ × b × h
Example: Base = 10 cm, height = 6 cm. A = ½ × 10 × 6 = 30 cm².
Parallelogram
- Perimeter = 2 × (sum of adjacent sides)
- Area = base × height = b × h
Important: Area = base × height (NOT base × slant side).
Circle
- Circumference = 2πr = πd
- Area = πr²
Use: π ≈ 22/7 or 3.14.
2. Area Between Rectangles (Paths)
Path Around a Rectangle (Outside)
If a path of width w is built around a rectangle of length L and breadth B:
- Outer length = L + 2w
- Outer breadth = B + 2w
- Area of path = Outer area - Inner area = (L + 2w)(B + 2w) - L × B
Path Inside a Rectangle (Border)
If a path of width w is built inside a rectangle:
- Inner length = L - 2w
- Inner breadth = B - 2w
- Area of path = Outer area - Inner area = L × B - (L - 2w)(B - 2w)
Cross Paths (Two Paths in the Middle)
If two paths of width w cross at right angles in the middle of a rectangle:
- Area of path along length = L × w
- Area of path along breadth = B × w
- Area of overlap (square) = w × w = w²
- Total area of paths = L × w + B × w - w²
Worked Example (ICSE 2024, 4 marks)
'A rectangular garden 30 m long and 20 m wide has a path of width 2 m around it on the outside. Find the area of the path.'
Solution: Inner area = 30 × 20 = 600 m². Outer length = 30 + 2(2) = 34 m. Outer breadth = 20 + 2(2) = 24 m. Outer area = 34 × 24 = 816 m². Area of path = 816 - 600 = 216 m².
3. Area of Composite Figures
To find the area of an IRREGULAR shape:
- Divide the shape into STANDARD shapes (rectangles, squares, triangles, circles).
- Find the area of each part.
- Add or subtract as needed.
Example (ICSE 2023, 4 marks)
'Find the area of the shaded region in a square of side 14 cm with a circle of radius 7 cm inscribed in it.'
Solution: Area of square = 14² = 196 cm². Area of circle = π × 7² = (22/7) × 49 = 154 cm². Shaded area = 196 - 154 = 42 cm².
4. Perimeter and Area: Units
| Quantity | Common Units |
|---|---|
| Length | m, cm, mm, km |
| Perimeter | m, cm, mm, km (same as length) |
| Area | m², cm², mm², km² |
Conversions
- 1 m = 100 cm. 1 m² = 10,000 cm².
- 1 km = 1000 m. 1 km² = 1,000,000 m².
- 1 cm = 10 mm. 1 cm² = 100 mm².
- 1 hectare = 10,000 m².
- 1 acre = 4,047 m² (approx).
5. ICSE Exam Focus
Common Mistakes
- Using diameter instead of radius in circle formulas (r = d/2).
- Forgetting unit² for area (saying 'cm' instead of 'cm²').
- Area of parallelogram = base × slant side (WRONG — use perpendicular height).
- Path area: subtracting wrong way (outer - inner, NOT inner - outer).
| Topic | Marks | Frequency |
|---|---|---|
| Area and perimeter of square/rectangle | 2 marks | Very High |
| Area of triangle and parallelogram | 2-3 marks | High |
| Circle (circumference and area) | 3 marks | Very High |
| Path problems (area between rectangles) | 3-4 marks | High |
| Composite figures | 3-4 marks | Medium |
Formula Reference Table
| Shape | Perimeter | Area |
|---|---|---|
| Square | 4s | s² |
| Rectangle | 2(l + b) | l × b |
| Triangle | a + b + c | ½ × b × h |
| Parallelogram | 2(a + b) | b × h |
| Circle | 2πr | πr² |
Self-Test (5 Questions)
Q1. Find area of a circle with radius 7 cm. (2 marks)
- A) 144 cm²
- B) 154 cm²
- C) 164 cm²
- D) 174 cm²
Q2. 'A rectangle has length 15 m and breadth 10 m. Find its perimeter.' (1 mark)
Q3. 'A rectangular park 50 m by 30 m has a 2 m wide path inside along its boundary. Find the area of the path.' (3 marks)
Q4. 'Find area of a triangle with base 12 cm and height 8 cm.' (2 marks)
Q5. 'Two paths of width 2 m cross at right angles through the centre of a rectangular field 40 m × 30 m. Find the total area of the paths.' (3 marks)
Answers
A1. B) 154 cm². (π × 49 = 22/7 × 49 = 154 cm².) A2. 50 m. (2(15 + 10) = 2 × 25 = 50 m.) A3. Inner length = 50 - 4 = 46 m. Inner breadth = 30 - 4 = 26 m. Inner area = 46 × 26 = 1196 m². Outer area = 1500 m². Path = 1500 - 1196 = 304 m². A4. 48 cm². (½ × 12 × 8 = 48 cm².) A5. 136 m². (Path along length: 40 × 2 = 80 m². Along breadth: 30 × 2 = 60 m². Overlap: 2 × 2 = 4 m². Total = 80 + 60 - 4 = 136 m².)
