3.4 Mensuration (2D & 3D)

Why Mensuration Matters

RRB asks 2-4 mensuration questions testing area, perimeter, surface area and volume of standard shapes. Memorising the formula sheet is non-negotiable — most questions are direct formula applications, and the only real traps are unit mismatches and the "open vs closed" or "curved vs total" surface area distinctions.

2D Shapes — Area and Perimeter

3D Shapes — Volume and Surface Area

Worked Examples

Worked example 1 (rectangular path)

A rectangular field is 40 m × 30 m. A path 2 m wide runs outside along its boundary. Find the area of the path.

• Outer rectangle = (40 + 4) × (30 + 4) = 44 × 34 = 1496 m²
• Inner rectangle = 40 × 30 = 1200 m²
• Path area = 1496 − 1200 = 296 m²

(Add 2 m on both ends of each dimension to get the outer size, so 40 + 2 + 2 = 44, 30 + 2 + 2 = 34.)

Worked example 2 (closed cylinder)

Find the volume and total surface area of a closed cylinder of radius 7 cm and height 20 cm (use π = 22/7).

• Volume = π r² h = (22/7) × 49 × 20 = 22 × 7 × 20 = 3080 cm³
• TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 20) = 44 × 27 = 1188 cm²

Worked example 3 (cone — find height from slant)

A cone has radius 6 cm and slant height 10 cm. Find its volume (height first).

• h = √(l² − r²) = √(100 − 36) = √64 = 8 cm
• Volume = (1/3) π r² h = (1/3) × (22/7) × 36 × 8 = (22/7) × 96 ≈ 301.71 cm³

Worked example 4 (sphere inscribed in a cylinder)

A sphere fits exactly inside a cylinder so that it touches both ends. The ratio of the sphere's volume to the cylinder's volume is:

• Sphere radius = r; cylinder has radius r and height 2r
• V_sphere = (4/3) π r³
• V_cylinder = π r² × 2r = 2 π r³
• Ratio = (4/3) / 2 = 4/6 = 2/3 (Archimedes' result)

Diagram — Volume Relationships

graph TD
  CY["Cylinder<br/>V = π r² h"] -->|"Cone same base & height"| CO["Cone<br/>V = 1/3 π r² h"]
  CO -->|"Volume = 1/3 of cylinder"| CY
  SP["Sphere<br/>V = 4/3 π r³"] -->|"Inscribed in cylinder h = 2r"| CY
  SP -->|"Volume = 2/3 of cylinder"| CY

Worked Examples — Extended Scenarios

Worked example 5 (cost of painting a hemisphere)

Find the cost of painting the curved outer surface only of a solid hemisphere of radius 7 cm at 5 paise per cm² (use π = 22/7).

• A solid hemisphere has a curved surface (2πr²) plus a flat circular face (πr²) on the base, so the total surface is 3πr². Which of the two the question wants is decided entirely by its wording.
• Curved SA = 2 × (22/7) × 49 = 2 × 22 × 7 = 308 cm²
• Cost = 308 × 5 paise = 1540 paise = ₹15.40

Had the question said "the whole surface is painted", you would use 3πr² = 462 cm² and the cost would be ₹23.10. Read the wording before choosing between 2πr² and 3πr² — RRB puts both values in the options.

Worked example 6 (cone — curved surface area from height and radius)

A cone has radius 5 cm and height 12 cm. Find its curved (lateral) surface area (use π = 22/7).

• Slant height l = √(r² + h²) = √(25 + 144) = √169 = 13 cm
• Curved SA = πrl = (22/7) × 5 × 13 = (22/7) × 65 = 1430/7 ≈ 204.29 cm²

Worked example 7 (volume of a frustum-free composite)

A solid is made of a cylinder of radius 7 cm and height 10 cm, topped by a hemisphere of the same radius. Find its total volume (π = 22/7).

• Cylinder volume = πr²h = (22/7) × 49 × 10 = 22 × 7 × 10 = 1540 cm³
• Hemisphere volume = (2/3) πr³ = (2/3) × (22/7) × 343 = (2/3) × 22 × 49 = (2/3) × 1078 = 2156/3 ≈ 718.67 cm³
• Total volume ≈ 1540 + 718.67 = 2258.67 cm³

Quick Reference — Open vs Closed Surface Areas

Common Exam Traps

• "Open cylinder/cuboid" — exclude the top face from surface area.
• "Cost of painting" uses surface area; "cost of filling" uses volume — match the quantity to the question.
• Unit consistency: convert cm ↔ m so that area comes out in consistent square units; 1 m³ = 10⁶ cm³, 1 litre = 1000 cm³.
• A sphere inscribed in a cylinder has its diameter equal to the cylinder's height; sphere volume is 2/3 of the cylinder's volume.
• "Lateral/curved surface area" excludes the base(s); "total surface area" includes every face.
• For a cone, the slant height l, height h and radius r form a right triangle: l² = r² + h². Don't substitute h when l is given.
• A hemisphere's total surface area includes the flat circular base (πr²) on top of the curved part (2πr²) — total is 3πr², not 2πr².
• For a "path around a rectangular field", the outer rectangle's dimensions grow by TWICE the path width on each side (width × 2), because the path runs on both ends of every dimension.
• Cost rates given in "paise per cm²" must be divided by 100 to convert to rupees before stating the final answer.
• When a solid is a composite (cylinder + hemisphere, cone on cylinder), add the volumes of the parts, but for surface area subtract the contact area where the parts meet.