Mensuration — Area and Perimeter of Plane Figures

Overview

Mensuration is the branch of mathematics that deals with measurement of geometric figures — their lengths, areas and volumes. For JKTET Paper I, the focus is on **plane figures** (2D shapes), specifically calculating their **perimeter** (boundary length) and **area** (surface covered). This topic forms a reliable scoring area in the mathematics section, with 2–4 questions typically appearing.

Students must master the standard formulas for common shapes and develop the ability to apply them in word problems involving real-life contexts — fencing a field, tiling a floor, finding the cost of painting a wall. The pedagogy component also expects you to understand how children learn measurement concepts through concrete experiences before moving to abstract formulas.

The key to success is not just memorising formulas but understanding what perimeter and area actually represent, recognising which formula applies to which shape, and handling unit conversions confidently.

Key Concepts

• **Perimeter** is the total length of the boundary of a closed figure. It is measured in linear units (cm, m, km).

• **Area** is the amount of surface enclosed within a closed figure. It is measured in square units (cm², m², km²).

• **Unit consistency** is essential — all measurements must be in the same unit before applying any formula.

• **Composite figures** are shapes made by combining two or more basic shapes. Find the area by adding or subtracting the areas of component shapes.

• **Relationship between perimeter and area** — two figures can have the same perimeter but different areas, and vice versa. A square has the maximum area for a given perimeter among rectangles.

• **For circles**, the perimeter is called **circumference**. The constant π (pi) ≈ 22/7 or 3.14 connects the diameter to the circumference.

• **Semi-circle perimeter** includes the curved part plus the diameter (the straight edge).

Formulas / Key Facts

### Rectangle

• Perimeter = 2 × (length + breadth) = 2(l + b)
• Area = length × breadth = l × b
• Diagonal = √(l² + b²)

### Square

• Perimeter = 4 × side = 4a
• Area = side × side = a²
• Diagonal = a × √2

### Triangle (General)

• Perimeter = sum of all three sides = a + b + c
• Area = ½ × base × height = ½ × b × h

### Right-angled Triangle

• Area = ½ × base × perpendicular
• Hypotenuse² = base² + perpendicular² (Pythagoras theorem)

### Equilateral Triangle

• Perimeter = 3 × side = 3a
• Area = (√3/4) × a²

### Parallelogram

• Perimeter = 2 × (sum of adjacent sides) = 2(a + b)
• Area = base × height = b × h

### Rhombus

• Perimeter = 4 × side = 4a
• Area = ½ × d₁ × d₂ (where d₁ and d₂ are diagonals)

### Trapezium

• Area = ½ × (sum of parallel sides) × height = ½ × (a + b) × h

### Circle

• Circumference = 2πr = πd (where r = radius, d = diameter)
• Area = πr²

### Semi-circle

• Perimeter = πr + 2r = r(π + 2)
• Area = ½ × πr²

### Unit Conversions

• 1 m = 100 cm → 1 m² = 10,000 cm²
• 1 km = 1000 m → 1 km² = 10,00,000 m²
• 1 hectare = 10,000 m²

Worked Examples

**Example 1: Rectangle Problem**

A rectangular garden is 25 m long and 18 m wide. Find the cost of fencing it at ₹45 per metre and the cost of planting grass at ₹12 per m².

*Solution:*

• Perimeter = 2(l + b) = 2(25 + 18) = 2 × 43 = 86 m
• Cost of fencing = 86 × 45 = ₹3,870
• Area = l × b = 25 × 18 = 450 m²
• Cost of planting grass = 450 × 12 = ₹5,400

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**Example 2: Composite Figure**

A rectangular sheet of paper measuring 28 cm by 14 cm has a circle of radius 7 cm cut out from it. Find the area of the remaining paper. (Use π = 22/7)

*Solution:*

• Area of rectangle = 28 × 14 = 392 cm²
• Area of circle = πr² = (22/7) × 7 × 7 = 22 × 7 = 154 cm²
• Remaining area = 392 − 154 = 238 cm²

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**Example 3: Finding a Missing Dimension**

The area of a triangle is 84 cm² and its base is 14 cm. Find the height.

*Solution:*

• Area = ½ × base × height
• 84 = ½ × 14 × h
• 84 = 7 × h
• h = 84 ÷ 7 = 12 cm

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**Example 4: Path Around a Rectangle**

A rectangular park 60 m by 40 m has a path 2 m wide running around it on the outside. Find the area of the path.

*Solution:*

• Outer dimensions (including path) = (60 + 4) by (40 + 4) = 64 m by 44 m
• Outer area = 64 × 44 = 2,816 m²
• Inner area (park) = 60 × 40 = 2,400 m²
• Area of path = 2,816 − 2,400 = 416 m²

Common Mistakes

• **Confusing perimeter and area formulas** → Remember: perimeter is a length (single unit like cm), area is a surface (square unit like cm²). Perimeter uses addition; area uses multiplication.

• **Forgetting to convert units** → A field 2 km by 500 m must be converted to the same unit (2000 m by 500 m) before calculation.

• **Using diameter instead of radius in circle formulas** → The formula πr² requires radius. If diameter is given, divide by 2 first.

• **Ignoring the straight edge in semi-circle perimeter** → Semi-circle perimeter = curved part (πr) + diameter (2r), not just πr.

• **Wrong formula for trapezium** → Students often forget the ½ factor. Area = ½ × (sum of parallel sides) × height, not just (a + b) × h.

• **Assuming all four-sided figures use l × b** → This works only for rectangles. Parallelograms need base × height; rhombus needs ½ × d₁ × d₂.

Quick Reference

• Rectangle: P = 2(l + b), A = l × b
• Square: P = 4a, A = a²
• Triangle: A = ½ × b × h
• Circle: C = 2πr, A = πr²
• Trapezium: A = ½ × (a + b) × h
• Always check units match before calculating
• Composite figures = add or subtract component areas