Perimeter and Area - Class 7 Mathematics (CBSE)
Based on the 2025-26 NCERT syllabus for Class 7 Mathematics. This chapter builds on earlier perimeter and area concepts to include triangles, parallelograms, circles, and composite shapes including paths.
1. Why this chapter matters
Perimeter and area calculations are used in gardening, flooring, painting, construction, and land measurement. In CBSE exams, this chapter contributes 10-12 marks with a mix of formula-based and application questions, making it one of the highest-weightage chapters.
2. Perimeter and area of squares and rectangles (recap)
- Perimeter of rectangle = 2 x (length + breadth)
- Area of rectangle = length x breadth
- Perimeter of square = 4 x side
- Area of square = side x side
3. Area of a triangle
Area of triangle = (1/2) x base x height.
The height (altitude) is the perpendicular distance from the base to the opposite vertex.
Example: Base = 8 cm, Height = 5 cm. Area = (1/2) x 8 x 5 = 20 sq cm.
For right triangles
Base and height are the two legs (sides forming the right angle). Area = (1/2) x product of legs.
4. Area of a parallelogram
Area of parallelogram = base x height.
The height is the perpendicular distance between the two parallel sides.
Example: Base = 10 cm, Height = 6 cm. Area = 10 x 6 = 60 sq cm.
Parallelogram vs rectangle
A rectangle is a special type of parallelogram. The formula is the same: base x height.
| Shape | Perimeter formula | Area formula |
|---|---|---|
| Rectangle | 2(l + b) | l x b |
| Square | 4a | a x a |
| Triangle | Sum of all sides | (1/2) x b x h |
| Parallelogram | Sum of all sides | b x h |
5. Circumference of a circle
Circumference = 2 x pi x radius (C = 2 pi r) Circumference = pi x diameter (C = pi x d)
Value of pi (pi)
pi is the ratio of circumference to diameter. pi = 22/7 or 3.14 approximately.
6. Area of a circle
Area of circle = pi x radius x radius (A = pi r-squared)
Examples
Radius = 7 cm. Area = (22/7) x 7 x 7 = 154 sq cm. Radius = 7 cm. Circumference = 2 x (22/7) x 7 = 44 cm.
7. Area between two rectangles (paths)
When a path is constructed around or inside a rectangular area, the area of the path is the difference between the area of the outer rectangle and the inner rectangle.
Path outside a rectangle (around)
If a garden of length l and breadth b has a path of width w around it:
- Outer length = l + 2w
- Outer breadth = b + 2w
- Area of path = Outer area - Inner area = (l + 2w)(b + 2w) - l x b
Path inside a rectangle (within)
If a carpet of length l and breadth b has a border of width w:
- Inner length = l - 2w
- Inner breadth = b - 2w
- Area of border = l x b - (l - 2w)(b - 2w)
8. Conversion of units
| Unit | Equivalent |
|---|---|
| 1 m | 100 cm |
| 1 m x m | 10000 cm x cm |
| 1 km x km | 1,000,000 m x m |
| 1 hectare | 10000 m x m |
Always ensure units are consistent before substituting into formulas.
9. Worked examples
Example 1: Find the area of a triangle with base 12 cm and height 8 cm.
Area = (1/2) x 12 x 8 = 48 sq cm.
Example 2: The base of a parallelogram is 15 cm and area is 120 sq cm. Find the height.
Area = base x height. 120 = 15 x height. Height = 120/15 = 8 cm.
Example 3: Find the circumference and area of a circle of radius 14 cm.
Circumference = 2 x (22/7) x 14 = 2 x 22 x 2 = 88 cm. Area = (22/7) x 14 x 14 = 22 x 2 x 14 = 616 sq cm.
Example 4: A rectangular garden 20 m by 15 m has a 2 m wide path around it. Find the area of the path.
Outer length = 20 + 2 x 2 = 24 m. Outer breadth = 15 + 2 x 2 = 19 m. Outer area = 24 x 19 = 456 sq m. Inner area = 20 x 15 = 300 sq m. Path area = 456 - 300 = 156 sq m.
10. Common mistakes and how to fix them
| Mistake | Fix |
|---|---|
| Confusing circumference and area formula | Circumference = 2 pi r; Area = pi r-squared |
| Using diameter instead of radius in area | Area uses radius. If diameter d is given, radius = d/2 |
| Forgetting height is perpendicular in triangle | Height is NOT the slant side -- it is the altitude |
| Not converting units before calculation | Convert all measurements to the same unit |
| Adding 2w only once for outer path | Path around adds 2w to both length and breadth |
11. CBSE exam focus
| Question type | Marks | Frequency |
|---|---|---|
| Triangle/parallelogram area | 2-3 marks | 1 question |
| Circle circumference/area | 3 marks | 1 question |
| Path area (rectangular) | 3 marks | 1 question |
| Mixed application problems | 4 marks | 1 question |
| Conversion of units | 2 marks | 1 question |
12. Self-test
- Find the area of a triangle with base 16 cm and height 9 cm.
- A parallelogram has area 180 sq cm and base 12 cm. Find its height.
- Find the radius of a circle with circumference 88 cm. (Use pi = 22/7)
- Find the area of a circle with diameter 28 cm.
- A rectangular park 30 m by 24 m has a 3 m wide path around it inside the park. Find the area of the path.
- How many times will a wheel of radius 35 cm rotate to cover 22 m?
13. Answer key
- Area = (1/2) x 16 x 9 = 72 sq cm.
- Height = 180/12 = 15 cm.
- 2 x (22/7) x r = 88. r = (88 x 7)/(2 x 22) = 616/44 = 14 cm.
- Radius = 14 cm. Area = (22/7) x 14 x 14 = 616 sq cm.
- Inner length = 30 - 6 = 24 m. Inner breadth = 24 - 6 = 18 m. Inner area = 432 sq m. Outer area = 720 sq m. Path area = 720 - 432 = 288 sq m.
- Circumference = 2 x (22/7) x 35 = 220 cm = 2.2 m. Number of rotations = 22/2.2 = 10.
14. Quick revision
- Triangle area = (1/2) x base x height.
- Parallelogram area = base x height.
- Circle circumference = 2 pi r. Circle area = pi r-squared.
- Path area = outer area - inner area.
- Outer path adds 2w to length and breadth.
- Inner path subtracts 2w from length and breadth.
- Always use same units in calculations.
