Mensuration — Area and Perimeter of Plane Figures

Overview

Mensuration is the branch of mathematics dealing with the measurement of geometric figures — their lengths, areas and volumes. For CG TET Paper I, the focus is strictly on **plane figures** (two-dimensional shapes), specifically calculating their **perimeter** (the total boundary length) and **area** (the space enclosed within the boundary).

This topic carries direct weightage in the mathematics section and also appears in pedagogy questions where you must explain how to teach these concepts to primary students. Questions typically involve rectangles, squares, triangles and circles — the shapes children encounter in Classes 3–5. Mastery requires memorising key formulas and knowing when to apply each one in word problems involving real-life contexts like fencing a field, tiling a floor or painting a wall.

The key skill tested is not just formula recall but the ability to identify which formula fits a given situation, convert units correctly and solve multi-step problems involving combined figures.

Key Concepts

• **Perimeter** is the total length of the boundary of a closed figure. Think of it as the length of wire needed to go around the shape exactly once.

• **Area** is the measure of the surface enclosed by a figure. Think of it as the number of unit squares needed to cover the shape completely.

• **Units matter**: Perimeter is measured in linear units (cm, m, km) while area is measured in square units (cm², m², km²). Converting between units is a common exam trap.

• **Rectangle vs Square**: A square is a special rectangle where all four sides are equal. This means square formulas are simplified versions of rectangle formulas.

• **Triangle types**: The basic formula (½ × base × height) works for all triangles, but the height must be perpendicular to the chosen base.

• **Circle terminology**: Radius (r) is the distance from centre to boundary; diameter (d) = 2r; π (pi) ≈ 22/7 or 3.14 for calculations.

• **Composite figures**: Many exam problems involve shapes made by combining or removing standard figures — solve by adding or subtracting individual areas.

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

• Perimeter = sum of all three sides = a + b + c
• Area = ½ × base × height = ½ × b × h
• Area of equilateral triangle = (√3/4) × a² (where a is the side)

### Circle

• Circumference (perimeter) = 2πr = πd
• Area = πr²

### Parallelogram

• Perimeter = 2(a + b) where a and b are adjacent sides
• 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

### Unit Conversions

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

Worked Examples

### Example 1: Rectangle — Fencing Problem **Problem**: A rectangular garden is 25 m long and 15 m wide. Find the cost of fencing it at ₹8 per metre.

**Solution**:

• Step 1: Find perimeter = 2(l + b) = 2(25 + 15) = 2 × 40 = 80 m
• Step 2: Cost = perimeter × rate = 80 × 8 = ₹640

**Answer**: ₹640

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### Example 2: Circle — Area Calculation **Problem**: Find the area of a circular playground with radius 14 m. (Use π = 22/7)

**Solution**:

• Area = πr²
• Area = (22/7) × 14 × 14
• Area = (22/7) × 196
• Area = 22 × 28 = 616 m²

**Answer**: 616 m²

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### Example 3: Composite Figure **Problem**: A square field of side 50 m has a circular pond of radius 7 m in the centre. Find the area of the field excluding the pond.

**Solution**:

• Step 1: Area of square = a² = 50 × 50 = 2500 m²
• Step 2: Area of circle = πr² = (22/7) × 7 × 7 = 154 m²
• Step 3: Remaining area = 2500 − 154 = 2346 m²

**Answer**: 2346 m²

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### Example 4: Triangle — Finding Height **Problem**: A triangle has area 48 cm² and base 12 cm. Find its height.

**Solution**:

• Area = ½ × base × height
• 48 = ½ × 12 × h
• 48 = 6h
• h = 48/6 = 8 cm

**Answer**: 8 cm

Common Mistakes

| Wrong Thinking | Correct Fix | |----------------|-------------| | Using perimeter formula when the question asks for area (e.g., "tiles needed" means area, not perimeter) | Read carefully — "fencing/boundary" = perimeter; "covering/painting/tiling" = area | | Forgetting to square the radius when finding circle area; writing πr instead of πr² | Always remember: circumference has r once, area has r squared | | Mixing up units — adding cm and m directly | Convert all measurements to the same unit before calculating | | Using diameter value directly in πr² formula | If given diameter, first find radius = diameter ÷ 2 | | For triangles, using slant side as height | Height must be perpendicular to the base — look for the right angle | | In composite figures, adding areas when you should subtract (or vice versa) | Visualise the figure — is the smaller shape inside (subtract) or attached outside (add)? |

Quick Reference

• **Rectangle**: P = 2(l+b), A = l×b
• **Square**: P = 4a, A = a²
• **Triangle**: A = ½ × base × height
• **Circle**: C = 2πr, A = πr²
• **Perimeter** → boundary/fencing; **Area** → covering/painting
• **π = 22/7** unless told otherwise; **1 hectare = 10,000 m²**