Mensuration — Study Notes for HSSC CET
Overview
Mensuration is the branch of mathematics dealing with measurement of geometric figures — their lengths, areas, and volumes.
The topic divides naturally into two parts: 2D figures (plane figures like triangles, circles, rectangles) where you calculate area and perimeter, and 3D figures (solids like cubes, cylinders, spheres) where you calculate surface area and volume. Exam questions often combine multiple concepts — for instance, finding how many small cubes can be cut from a larger cube, or calculating the cost of painting a room.
Mastery requires memorizing key formulas and developing the ability to visualize shapes. Most questions are direct formula applications, but some involve unit conversions or combining shapes. Speed and accuracy in calculations are crucial.
Key Concepts
- Perimeter is the total length of the boundary of a 2D figure. Think of it as the length of fence needed to enclose a plot.
- Area measures the surface covered by a 2D figure. Think of it as the amount of paint needed to cover a floor.
- Surface Area of a 3D solid is the total area of all its outer faces. Curved Surface Area (CSA) excludes the flat ends (bases), while Total Surface Area (TSA) includes everything.
- Volume measures the space occupied by a 3D solid. Think of it as the amount of water a container can hold.
- Unit consistency is critical: if dimensions are in metres, area is in m² and volume is in m³. Convert all dimensions to the same unit before calculating.
- For combined figures, break the shape into standard parts, calculate separately, then add or subtract as needed.
- When one solid is melted and recast into another shape, volume remains constant — this principle drives many exam questions.
Formulas / Key Facts
2D Figures (Plane Figures)
| Figure | Area | Perimeter |
|---|---|---|
| Rectangle | l × b | 2(l + b) |
| Square | a² | 4a |
| Triangle | ½ × base × height | Sum of all sides |
| Right Triangle | ½ × base × perpendicular | a + b + hypotenuse |
| Equilateral Triangle | (√3/4) × a² | 3a |
| Circle | πr² | 2πr (circumference) |
| Semicircle | ½πr² | πr + 2r |
| Parallelogram | base × height | 2(a + b) |
| Rhombus | ½ × d₁ × d₂ | 4a |
| Trapezium | ½ × (sum of parallel sides) × height | Sum of all sides |
Heron's Formula for triangle with sides a, b, c:
- s = (a + b + c)/2
- Area = √[s(s−a)(s−b)(s−c)]
3D Figures (Solids)
| Solid | Volume | CSA | TSA |
|---|---|---|---|
| Cube (side a) | a³ | 4a² | 6a² |
| Cuboid (l, b, h) | l × b × h | 2h(l + b) | 2(lb + bh + hl) |
| Cylinder (r, h) | πr²h | 2πrh | 2πr(r + h) |
| Cone (r, h, slant l) | ⅓πr²h | πrl | πr(r + l) |
| Sphere (radius r) | (4/3)πr³ | — | 4πr² |
| Hemisphere (r) | (2/3)πr³ | 2πr² | 3πr² |
Slant height of cone: l = √(r² + h²)
Diagonal of cuboid: √(l² + b² + h²)
Diagonal of cube: a√3
Worked Examples
Example 1: Area of a Path (Ring)
Problem: A circular park has radius 21 m. A path 3.5 m wide runs around it. Find the area of the path. (Use π = 22/7)
Solution:
- Inner radius (r) = 21 m
- Outer radius (R) = 21 + 3.5 = 24.5 m
- Area of path = πR² − πr² = π(R² − r²)
- = (22/7) × (24.5² − 21²)
- = (22/7) × (600.25 − 441)
- = (22/7) × 159.25
- = 22 × 22.75 = 500.5 m²
Example 2: Volume of Melted and Recast Solid
Problem: A metallic sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder.
Solution:
- Volume of sphere = (4/3)πr³ = (4/3)π(6)³ = (4/3)π × 216 = 288π cm³
- Volume of cylinder = πR²h = π(4)²h = 16πh
- Since volume is conserved: 288π = 16πh
- h = 288/16 = 18 cm
Example 3: Cost of Painting
Problem: A room is 8 m long, 6 m wide, and 4 m high. If the cost of painting the four walls is Rs 15 per m², find the total cost. (Ignore doors and windows)
Solution:
- Area of four walls = 2h(l + b) = 2 × 4 × (8 + 6) = 8 × 14 = 112 m²
- Total cost = 112 × 15 = Rs 1680
Common Mistakes
- Confusing CSA with TSA: Students use total surface area when only curved surface is asked (e.g., painting the lateral surface of a cylinder). → Always check whether the question asks for CSA or TSA.
- Forgetting unit conversion: Mixing cm and m in the same problem leads to answers off by powers of 10 or 100. → Convert all measurements to the same unit first.
- Using diameter instead of radius: Many questions give diameter, but all formulas use radius. → Divide diameter by 2 immediately when you read the problem.
- Wrong formula for cone/sphere volume: Forgetting the ⅓ in cone volume or 4/3 in sphere volume. → Associate: cone has ⅓ (it's "one-third" of a cylinder), sphere has 4/3.
- Ignoring slant height in cone TSA: Using height instead of slant height in πrl. → Calculate l = √(r² + h²) first if not given directly.
Quick Reference
- Rectangle area = l × b; Perimeter = 2(l + b)
- Circle area = πr²; Circumference = 2πr
- Volume of cube = a³; TSA = 6a²
- Volume of cylinder = πr²h; CSA = 2πrh
- Volume of sphere = (4/3)πr³; Surface area = 4πr²
- When solid is recast: Old volume = New volume