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 JTET Paper I, the focus is strictly on plane figures (2D shapes), specifically calculating their perimeter (boundary length) and area (surface enclosed).
Mastery of mensuration requires two things: memorizing the standard formulas and knowing when to apply which formula based on the shape described. Questions often present real-life contexts—fencing a field, tiling a floor, finding the cost of painting a wall—so recognizing the underlying geometric shape from word problems is essential.
Students must be comfortable with squares, rectangles, triangles, circles, and basic composite figures (shapes made by combining or removing simpler shapes). Speed and accuracy in calculation, especially with fractions and decimals, will determine success here.
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 measure of the surface enclosed within a closed figure. It is measured in square units (cm², m², km²).
- Units matter: Always check if the question gives dimensions in different units; convert them to the same unit before calculating.
- Composite figures are shapes formed by combining two or more basic shapes. Find the area by adding or subtracting the areas of component shapes.
- Path and border problems involve finding the area of a pathway around or inside a rectangle. This is calculated as: Area of outer rectangle − Area of inner rectangle.
- Cost-based problems multiply area or perimeter by the rate per unit to get total cost (e.g., cost of fencing = perimeter × rate per metre).
- Diagonal of a rectangle can be found using Pythagoras theorem: d = √(l² + b²), where l is length and b is breadth.
- For circles, remember that π ≈ 22/7 or 3.14 unless the question specifies otherwise.
Formulas / Key Facts
Rectangle
- Perimeter = 2(l + b)
- Area = l × b
- Diagonal = √(l² + b²)
Square
- Perimeter = 4 × side = 4a
- Area = side² = a²
- Diagonal = a√2
Triangle (General)
- Perimeter = a + b + c (sum of all sides)
- Area = ½ × base × height
Triangle (Using Heron's Formula)
- Semi-perimeter s = (a + b + c)/2
- Area = √[s(s−a)(s−b)(s−c)]
Right-angled Triangle
- Area = ½ × base × perpendicular
- Hypotenuse = √(base² + perpendicular²)
Equilateral Triangle
- Perimeter = 3a
- Area = (√3/4) × a²
- Height = (√3/2) × a
Circle
- Circumference = 2πr = πd
- Area = πr²
Semicircle
- Perimeter = πr + 2r = r(π + 2)
- Area = πr²/2
Quadrant (Quarter Circle)
- Perimeter = (πr/2) + 2r
- Area = πr²/4
Parallelogram
- Perimeter = 2(a + b)
- Area = base × height
Rhombus
- Perimeter = 4 × side
- Area = ½ × d₁ × d₂ (where d₁ and d₂ are diagonals)
Trapezium
- Area = ½ × (sum of parallel sides) × height = ½ × (a + b) × h
Worked Examples
Example 1: Rectangle – Fencing Cost
Problem: A rectangular park is 80 m long and 60 m wide. Find the cost of fencing it at ₹25 per metre.
Solution:
- Perimeter = 2(l + b) = 2(80 + 60) = 2 × 140 = 280 m
- Cost = Perimeter × Rate = 280 × 25 = ₹7000
Answer: ₹7000
Example 2: Circle – Area Calculation
Problem: The radius of a circular garden is 14 m. Find its area. (Use π = 22/7)
Solution:
- Area = πr² = (22/7) × 14 × 14
- Area = (22/7) × 196 = 22 × 28 = 616 m²
Answer: 616 m²
Example 3: Composite Figure – Path Around Rectangle
Problem: A rectangular lawn 50 m by 40 m has a path 2 m wide running outside it. Find the area of the path.
Solution:
- Outer dimensions: Length = 50 + 2 + 2 = 54 m; Breadth = 40 + 2 + 2 = 44 m
- Outer area = 54 × 44 = 2376 m²
- Inner area (lawn) = 50 × 40 = 2000 m²
- Area of path = 2376 − 2000 = 376 m²
Answer: 376 m²
Example 4: Triangle – Heron's Formula
Problem: Find the area of a triangle with sides 13 cm, 14 cm, and 15 cm.
Solution:
- Semi-perimeter s = (13 + 14 + 15)/2 = 42/2 = 21 cm
- Area = √[s(s−a)(s−b)(s−c)] = √[21 × 8 × 7 × 6]
- Area = √[21 × 8 × 7 × 6] = √7056 = 84 cm²
Answer: 84 cm²
Common Mistakes
- Confusing perimeter and area: Perimeter is the boundary (linear), area is the surface (square). A question asking for "fencing" needs perimeter; "tiling" or "painting" needs area.
- Forgetting to square the radius for circle area: Students often calculate 2πr (circumference) when area (πr²) is asked. Read the question carefully.
- Not converting units: If length is in metres and breadth in centimetres, convert both to the same unit first. Otherwise, the answer will be wrong by a factor of 10 or 100.
- Wrong application of Heron's formula: Students forget to calculate semi-perimeter first or make arithmetic errors inside the square root. Write out each step.
- Path problems—adding width only once: For a path outside a rectangle, width is added on both sides. So outer length = inner length + 2 × path width, not just + path width.
Quick Reference
- Rectangle: P = 2(l+b), A = l×b
- Square: P = 4a, A = a²
- Circle: C = 2πr, A = πr²
- Triangle: A = ½ × base × height; Heron's formula for three sides
- Trapezium: A = ½ × (sum of parallel sides) × height
- Path area = Outer area − Inner area
- Always convert to same units before calculating