Mensuration
KTET Mathematics
Overview
Mensuration is the branch of mathematics that deals with the measurement of geometric figures — their lengths, areas, and volumes. For KTET, this topic carries significant weight across all categories, appearing in both the mathematics content section and pedagogy-based application questions.
You must be comfortable with two-dimensional figures (rectangles, triangles, circles, parallelograms, trapeziums) and three-dimensional solids (cubes, cuboids, cylinders, cones, spheres). Questions typically test direct formula application, unit conversions, and word problems involving real-life contexts like finding the cost of fencing a field or the volume of a water tank.
The key to mastering mensuration is not rote memorization but understanding the logic behind formulas. A rectangle's area is length × breadth because you're counting unit squares. A cylinder's volume is base area × height because you're stacking circular discs. This conceptual clarity will help you tackle unfamiliar variations confidently.
Key Concepts
- Perimeter is the total length of the boundary of a 2D figure. It is measured in linear units (cm, m, km).
- Area is the amount of surface enclosed within a 2D figure. It is measured in square units (cm², m², km²).
- Surface Area of a 3D solid is the total area of all its outer faces. Curved Surface Area (CSA) excludes the base(s); Total Surface Area (TSA) includes everything.
- Volume is the space occupied by a 3D solid, measured in cubic units (cm³, m³, litres where 1 litre = 1000 cm³).
- For composite figures, break them into standard shapes, calculate separately, then add or subtract as needed.
- Unit conversion is critical: 1 m = 100 cm, 1 km = 1000 m, 1 m² = 10000 cm², 1 m³ = 1000000 cm³ = 1000 litres.
- In word problems, identify what is being asked (fencing = perimeter, painting = area, filling = volume) before selecting the formula.
Formulas / Key Facts
Two-Dimensional Figures
| Figure | Perimeter | Area |
|---|---|---|
| Rectangle | 2(l + b) | l × b |
| Square | 4a | a² |
| Triangle | a + b + c | ½ × base × height |
| Right Triangle | a + b + c | ½ × base × perpendicular |
| Equilateral Triangle | 3a | (√3/4) × a² |
| Circle | 2πr (circumference) | πr² |
| Semicircle | πr + 2r | ½πr² |
| Parallelogram | 2(a + b) | base × height |
| Rhombus | 4a | ½ × d₁ × d₂ |
| Trapezium | sum of all sides | ½ × (a + b) × h |
Three-Dimensional Solids
| Solid | CSA | TSA | Volume |
|---|---|---|---|
| Cube | 4a² | 6a² | a³ |
| Cuboid | 2h(l + b) | 2(lb + bh + hl) | l × b × h |
| Cylinder | 2πrh | 2πr(r + h) | πr²h |
| Cone | πrl (l = slant height) | πr(r + l) | ⅓πr²h |
| Sphere | 4πr² | 4πr² | (4/3)πr³ |
| Hemisphere | 2πr² | 3πr² | (2/3)πr³ |
Key relationships:
- Slant height of cone: l = √(r² + h²)
- Diagonal of cuboid: d = √(l² + b² + h²)
- Diagonal of cube: d = a√3
Worked Examples
Example 1: Area and Perimeter (Rectangle)
Problem: A rectangular field is 120 m long and 80 m wide. Find the cost of fencing it at Rs 25 per metre and the cost of levelling it at Rs 5 per square metre.
Solution:
- Perimeter = 2(l + b) = 2(120 + 80) = 2 × 200 = 400 m
- Cost of fencing = 400 × 25 = Rs 10,000
- Area = l × b = 120 × 80 = 9600 m²
- Cost of levelling = 9600 × 5 = Rs 48,000
Example 2: Volume of Cylinder
Problem: A cylindrical water tank has radius 3.5 m and height 4 m. Find its capacity in litres. (Take π = 22/7)
Solution:
- Volume = πr²h = (22/7) × 3.5 × 3.5 × 4
- = (22/7) × 12.25 × 4 = (22/7) × 49 = 22 × 7 = 154 m³
- Wait, let me recalculate: (22/7) × (3.5)² × 4 = (22/7) × (49/4) × 4 = 22 × 7 = 154 m³
- Capacity = 154 × 1000 = 1,54,000 litres
Example 3: Composite Figure
Problem: A rectangular sheet of paper 44 cm × 20 cm is rolled along its length to form a cylinder. Find the volume of the cylinder.
Solution:
- When rolled along length, the length becomes the circumference of the base.
- Circumference = 2πr = 44, so r = 44 × (7/22) × (1/2) = 7 cm
- Height of cylinder = 20 cm (the breadth of sheet)
- Volume = πr²h = (22/7) × 7 × 7 × 20 = 22 × 7 × 20 = 3080 cm³
Common Mistakes
- Confusing perimeter with area: Students add all sides when area is required, or multiply dimensions when perimeter is asked. Fix: Always identify whether the question involves boundary (perimeter) or surface (area).
- Forgetting to square or cube units: Writing "Area = 25 cm" instead of "25 cm²". Fix: Area is always in square units, volume in cubic units — make this a reflex.
- Using diameter instead of radius: Formulas use radius, but problems often give diameter. Fix: Always halve the diameter before substituting.
- Mixing up CSA and TSA: Using curved surface area when total surface area is needed (e.g., for painting a closed box). Fix: Read whether the solid is open or closed, and whether bases are included.
- Incorrect unit conversion: Forgetting that 1 m² = 10000 cm² (not 100 cm²). Fix: Convert linear units first, then square or cube them. 1 m = 100 cm, so 1 m² = 100 × 100 = 10000 cm².
Quick Reference
- Rectangle area = l × b; Perimeter = 2(l + b)
- Circle area = πr²; Circumference = 2πr
- Cylinder volume = πr²h; TSA = 2πr(r + h)
- Cone volume = ⅓πr²h; CSA = πrl
- Sphere volume = (4/3)πr³; Surface area = 4πr²
- 1 m³ = 1000 litres; 1 litre = 1000 cm³