Mensuration: Area and Perimeter of Simple Plane Figures

Overview

Mensuration is the branch of mathematics dealing with measurement of geometric figures—their lengths, areas, and volumes. For OTET Paper I, the focus is strictly on **plane figures** (2D shapes), specifically calculating their perimeter (boundary length) and area (surface covered).

This topic appears consistently in OTET mathematics sections, typically carrying 2–4 questions. Questions range from direct formula application to word problems involving fields, gardens, paths, and picture frames. Mastery requires memorizing formulas and understanding when to apply each—particularly in composite figure problems where shapes combine or one is cut from another.

Students must be comfortable with unit conversions (cm to m, m² to cm²) and able to visualize real-world applications like fencing a plot (perimeter) or tiling a floor (area).

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Key Concepts

• **Perimeter** is the total length of the boundary of a closed figure. Think of it as the length of wire needed to form the shape's outline.

• **Area** is the amount of surface enclosed within 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); area is measured in square units (cm², m², km²). Converting 1 m = 100 cm means 1 m² = 10,000 cm².

• **Composite figures** are formed by combining or removing simple shapes. Calculate by adding or subtracting individual areas/perimeters appropriately.

• **Path problems** involve a path around or inside a rectangle/square. The path area equals outer area minus inner area.

• For **circles**, π (pi) is taken as 22/7 or 3.14 unless specified otherwise in the question.

• **Semi-circle perimeter** includes both the curved part (half circumference) and the diameter.

• In word problems, identify the shape first, extract dimensions, then apply the correct formula.

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Formulas / Key Facts

### Rectangle (length = l, breadth = b)

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

### Square (side = a)

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

### Triangle (sides a, b, c; base = b, height = h)

• Perimeter = a + b + c
• Area = (1/2) × base × height
• **Heron's formula**: Area = √[s(s−a)(s−b)(s−c)], where s = (a+b+c)/2

### Equilateral Triangle (side = a)

• Perimeter = 3a
• Area = (√3/4) × a²

### Right Triangle (legs = a, b; hypotenuse = c)

• Area = (1/2) × a × b
• Hypotenuse: c = √(a² + b²)

### Circle (radius = r, diameter = d = 2r)

• Circumference = 2πr = πd
• Area = πr²

### Semi-circle (radius = r)

• Perimeter = πr + 2r = r(π + 2)
• Area = (1/2)πr²

### Parallelogram (base = b, height = h)

• Perimeter = 2(sum of adjacent sides)
• Area = b × h

### Rhombus (diagonals d₁ and d₂)

• Perimeter = 4 × side
• Area = (1/2) × d₁ × d₂

### Trapezium (parallel sides a, b; height = h)

• Area = (1/2) × (a + b) × h

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Worked Examples

### Example 1: Basic Rectangle Problem **Question**: A rectangular field is 120 m long and 80 m wide. Find (a) the cost of fencing at ₹15 per metre, and (b) the cost of ploughing at ₹5 per m².

**Solution**:

• Perimeter = 2(120 + 80) = 2 × 200 = 400 m
• Cost of fencing = 400 × 15 = ₹6,000

• Area = 120 × 80 = 9,600 m²
• Cost of ploughing = 9,600 × 5 = ₹48,000

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### Example 2: Path Around a Rectangle **Question**: A rectangular garden is 50 m by 40 m. A path 5 m wide runs outside around it. Find the area of the path.

**Solution**:

• Outer rectangle dimensions: (50 + 5 + 5) by (40 + 5 + 5) = 60 m by 50 m
• Outer area = 60 × 50 = 3,000 m²
• Inner area (garden) = 50 × 40 = 2,000 m²
• Path area = 3,000 − 2,000 = **1,000 m²**

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### Example 3: Circle Problem **Question**: The radius of a circular park is 21 m. Find (a) the length of the boundary, and (b) the area. (Use π = 22/7)

**Solution**:

• Circumference = 2 × (22/7) × 21 = 2 × 22 × 3 = **132 m**
• Area = (22/7) × 21 × 21 = 22 × 63 = **1,386 m²**

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### Example 4: Using Heron's Formula **Question**: Find the area of a triangle with sides 13 cm, 14 cm, and 15 cm.

**Solution**:

• s = (13 + 14 + 15)/2 = 42/2 = 21 cm
• Area = √[21 × (21−13) × (21−14) × (21−15)]
• Area = √[21 × 8 × 7 × 6] = √7056 = **84 cm²**

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Common Mistakes

• **Confusing perimeter and area formulas** → Remember: perimeter adds lengths (linear), area multiplies dimensions (square units). Perimeter of rectangle is 2(l+b), not l×b.

• **Forgetting to double the path width** → When a path surrounds a shape, it adds to both sides. A 5 m path adds 10 m to both length and breadth of a rectangle.

• **Wrong units in answers** → If dimensions are in metres, area must be in m², not just "metres." Always write the unit.

• **Using wrong π value** → Check if the question specifies 22/7 or 3.14. Using 3.14 when 22/7 is expected gives different numerical answers.

• **Semi-circle perimeter error** → Students forget to add the diameter. Semi-circle perimeter = curved part (πr) + straight part (2r).

• **Applying triangle area formula without height** → Area = (1/2) × base × height requires the perpendicular height, not a slant side. Use Heron's formula when only three sides are given.

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Quick Reference

• Rectangle: Perimeter = 2(l+b); Area = l×b

• Square: Perimeter = 4a; Area = a²

• Triangle: Area = (1/2) × base × height; Heron's when only sides given

• Circle: Circumference = 2πr; Area = πr²

• Path area = Outer area − Inner area

• 1 m² = 10,000 cm²; 1 hectare = 10,000 m²