Chapter 6
Perimeter and Area
Welcome to the comprehensive study guide for Chapter 6: Perimeter and Area. Learn key geometric principles, standard formulas, step-by-step solved solutions, visual figures, real-life practical applications, and practice exercises tailored for Class 6 Mathematics.
Introduction
Perimeter and Area are fundamental concepts in geometry that help us measure boundary lengths and surface regions of flat, closed figures.
Whether we are fencing a rectangular garden, tiling a room floor, calculating carpet dimensions, or designing architectural layouts, understanding perimeters and areas is essential in everyday life.
Figures constructed with the exact same number of unit squares have identical areas, but can possess completely different perimeters based on how the squares are connected!
Fencing & Boundaries
Determining total wire lengths for fencing parks, yards, and agricultural plots.
Flooring & Tiling
Calculating total floor coverage area to determine required tiles or carpeting.
Architecture & Design
Planning house layout dimensions, land plot comparisons, and interior spacing.
Framing & Ribbons
Measuring ribbon borders for photo frames, cards, and decorative boards.
Learning Objectives
After completing this chapter, you will be able to:
- Understand the precise definitions of perimeter and surface area.
- Apply standard perimeter formulas for rectangles, squares, and regular polygons.
- Calculate areas using square grids and unit square estimations.
- Solve multi-step real-world problems involving tiling, wire bending, and fencing.
- Compare geometric figures with fixed areas and variable perimeters.
- Analyze diagonal vs. straight line segment lengths on square grids.
1. Perimeter
The perimeter of any closed plane figure is the total distance covered along its boundary when going around it once. For any polygon, it is the sum of the lengths of all its sides.
P = 2 × (5 + 3) = 16 cm
P = 4 × 4 = 16 cm
P = 20 + 14 + 21 = 55 cm
Core Perimeter Formulas
| Shape | Formula | Property / Description |
|---|---|---|
| Rectangle | 2 × (length + breadth) | Twice the sum of length and width. |
| Square | 4 × side length | Quadruple the side length (all 4 sides equal). |
| Triangle | a + b + c | Sum of lengths of all 3 sides. |
| Equilateral Triangle | 3 × side length | Three times the length of one side. |
| Regular Polygon (n sides) | n × side length | Total sides multiplied by length of one side. |
2. Area
The area of a closed plane figure is the total measure of the region enclosed by its boundary, expressed in square units (e.g., cm², m²).
12 Unit Squares
Uncarpeted = 20 - 9 = 11 sq m
Core Area Formulas
| Shape | Area Formula |
|---|---|
| Square | side × side |
| Rectangle | length × width |
| Triangle | ½ × Area of enclosing rectangle |
Measuring Area Using Grids
- Full Square: Counted as 1 sq unit.
- More than Half Square: Counted as 1 sq unit.
- Exactly Half Square: Counted as ½ sq unit.
- Less than Half Square: Ignore completely.
Solved Solutions & Textbook Exercises
Section 6.1 Solutions
Q1. Find missing terms:
(a) Perimeter of rectangle = 14 cm, breadth = 2 cm ⇒ Length = (14/2) - 2 = 5 cm
(b) Perimeter of square = 20 cm ⇒ Side = 20/4 = 5 cm
(c) Perimeter of rectangle = 12 m, length = 3 m ⇒ Breadth = (12/2) - 3 = 3 m
Q2. Wire Bending: A wire forms a rectangle of sides 5 cm and 3 cm. Straightened and bent into a square, side length = ?
Perimeter = 2 × (5 + 3) = 16 cm ⇒ Square Side = 16 / 4 = 4 cm
Q3. Missing Side of Triangle: Perimeter = 55 cm, two sides = 20 cm and 14 cm.
Third Side = 55 - (20 + 14) = 55 - 34 = 21 cm
Q4. Cost of Fencing: Park 150 m long and 120 m wide at ₹40/m.
Perimeter = 2 × (150 + 120) = 540 m ⇒ Total Cost = 540 × ₹40 = ₹21,600
Section 6.2 Solutions
Q1. Garden Width: Area = 300 sq m, Length = 25 m ⇒ Width = 300 / 25 = 12 m
Q2. Cost of Tiling Plot: Plot 500 m × 200 m at ₹8 per 100 sq m.
Area = 500 × 200 = 100,000 sq m ⇒ Cost = (100,000 / 100) × 8 = ₹8,000
Q3. Coconut Grove Capacity: Grove 100 m × 50 m. Each tree requires 25 sq m.
Total Area = 5,000 sq m ⇒ Max Trees = 5,000 / 25 = 200 trees
Section 6.3 Advanced Applications & Paper Folding Puzzle
Carpet Problem: Floor is 5 m × 4 m (Area = 20 sq m). Square carpet is 3 m × 3 m (Area = 9 sq m).
Uncarpeted Area = 20 - 9 = 11 sq m
Paper Folding MCQ: A square paper is folded in half and cut into two equal rectangles.
Perimeter Ratio = 6x / 4x = 1.5
Conclusion: The sum of perimeters of both rectangles is 1.5 times the perimeter of the original square (Ratio 6x / 4x = 1.5).
Key Takeaways
- Perimeter measures the boundary; area measures surface space enclosed.
- Perimeter is expressed in linear units (cm, m), area in square units (cm², m²).
- Square shapes pack surface areas efficiently without leaving gaps.
- Regular polygons have perimeters equal to (number of sides × side length).
- Shapes with equal areas can have different perimeters.
Multiple Choice Questions
Click any option to reveal instant feedback!
1. The perimeter of a square of side 6 cm is:
2. The area of a rectangle measuring 8 cm long and 5 cm wide is:
3. If the perimeter of an equilateral triangle is 27 cm, its side length is:
4. A square of side 'x' is cut into two identical rectangles. The combined perimeter of both rectangles is:
Chapter Quiz
Test your understanding of Chapter 6.
Download Notes
Download printable revision notes for offline study.