Lesson 8: Mensuration — Areas and Volumes

Mensuration is where Class X mathematics meets the physical world: paint for a wall, canvas for a tent, water in a tank, grain in a silo. NCERT devotes two chapters' worth of weight to it — areas related to circles, and surface areas and volumes of solids *and their combinations*. The formulas are few; the marks are lost on choosing the wrong one (curved vs total surface) and on unit conversions.

The formula sheet (with π = 22/7 unless stated)

For circles: area πr²; sector area (θ/360°) × πr²; segment = sector − triangle.

Worked example 1 (basic). A cylinder has radius 7 cm and height 10 cm. Find its curved surface area and volume. CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm². Volume = πr²h = (22/7) × 49 × 10 = 1540 cm³. Radius 7 with π = 22/7 is a deliberate gift — the 7s cancel. Recognise it and save the arithmetic.

Worked example 2 (intermediate). A cone has radius 7 cm and height 24 cm. Find its slant height and curved surface area. l = √(49 + 576) = √625 = 25 cm. CSA = πrl = (22/7) × 7 × 25 = 550 cm².

Worked example 3 (NCERT exemplar style — combination of solids). A solid consists of a cylinder of radius 6 cm and height 8 cm surmounted by a cone of the same radius and height 4 cm. Find its total volume. Cylinder: π × 36 × 8 = 288π. Cone: (1/3)π × 36 × 4 = 48π. Total = 336π = 336 × 22/7 = 1056 cm³. Strategy for all combinations: sketch, split into named solids, compute separately, add (volumes) — but for *surface* area, add only the surfaces actually exposed: the joining faces are internal and must be excluded.

Competency application: how much water does the tank hold?

A cylindrical overhead tank has diameter 1.4 m and height 2 m. Capacity = πr²h with r = 0.7 m: (22/7) × 0.49 × 2 = 3.08 m³. Water is billed in litres, and 1 m³ = 1000 L, so the tank holds 3080 litres. If a family uses 385 litres a day, one full tank lasts 3080/385 = 8 days. Every step here is a board-style demand: radius from diameter, formula, arithmetic, unit conversion, interpretation.

Hemisphere note. A hemispherical bowl of radius 7 cm has *curved* surface 2πr² = 308 cm², but if it is a solid hemisphere (like a paperweight), total surface = 3πr² = 462 cm² because the flat circular base counts. Read the object, not just the word "hemisphere".

Common mistakes

• Using diameter as radius. The single most common mensuration error. Halve first, write r = … explicitly.
• Counting hidden surfaces in combinations. Where a cone sits on a cylinder, the shared circular face is inside the solid — exclude it from surface area (but volumes simply add).
• Mixing units. A radius in cm and a height in m gives nonsense. Convert everything before substituting.
• Forgetting that 1 m³ = 1000 litres (not 100). Capacity answers are commonly expected in litres.
• Sector/segment confusion. Segment = sector − triangle; if you report the sector area for a segment question, the triangle is your missing mark.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. The total surface area of a solid hemisphere of radius 7 cm is: (a) 308 cm² (b) 462 cm² (c) 616 cm² (d) 154 cm². Answer: (b) 462 cm². 3πr² = 3 × (22/7) × 49 = 462. (308 is the curved surface alone — the trap.)

Q2. The volume of a cone is 77 cm³ and its height is 6 cm. Its base radius is: (a) 7 cm (b) 3.5 cm (c) 4.9 cm (d) 14 cm. Answer: (b) 3.5 cm. (1/3)(22/7)r²(6) = 77 gives (44/7)r² = 77, so r² = 77 × 7/44 = 12.25 and r = 3.5.

Section B — Short answer (3 marks) Q3. A cylindrical vessel of radius 4 cm and height 9 cm is full of water. Find the volume of water, and state how many full hemispherical bowls of radius 3 cm it can fill. (Use π = 22/7.) Marking scheme: 1 mark for cylinder volume = (22/7) × 16 × 9 = 3168/7 cm³; 1 mark for bowl volume = (2/3)(22/7)(27) = 396/7 cm³; 1 mark for the quotient (3168/7)/(396/7) = 8 bowls exactly.

Section C — Case-based long answer (5 marks) Q4. A tent is a cylinder of radius 3 m and height 2.5 m, surmounted by a cone of height 4 m (same radius). Canvas is used for all outer surfaces except the floor. (i) Find the slant height of the cone. (1 mark) (ii) Find the curved surface area of the cylindrical part. (2 marks) (iii) Find the total canvas required. (2 marks) Marking scheme: (i) l = √(3² + 4²) = 5 m (1 mark). (ii) CSA of cylinder = 2πrh = 2 × (22/7) × 3 × 2.5 = 330/7 ≈ 47.14 m² (1 mark for formula, 1 for value). (iii) Cone CSA = πrl = (22/7) × 3 × 5 = 330/7 m² (1 mark); total = 660/7 ≈ 94.3 m² (1 mark).

Going deeper: sectors, segments and arc length

A sector is a slice of the circle of angle θ: its area is (θ/360°) × πr² and its arc length is (θ/360°) × 2πr. For r = 6 cm and θ = 60°: area = (1/6) × (22/7) × 36 = 132/7 ≈ 18.86 cm²; arc = (1/6) × 2 × (22/7) × 6 = 44/7 ≈ 6.29 cm. The corresponding segment (sector minus the triangle formed by the two radii and the chord) needs the triangle's area: for θ = 60° the triangle is equilateral with side 6, area (√3/4) × 36 = 9√3 ≈ 15.59, so the minor segment ≈ 18.86 − 15.59 ≈ 3.27 cm². Clock problems are sector problems in uniform: the minute hand (length = radius) sweeps 6° per minute, so "area swept in 15 minutes" is a 90° sector.

Going deeper: the frustum — a cone with its top cut off

A bucket is a frustum: radii R (top) and r (bottom), height h. Its volume is (1/3)πh(R² + Rr + r²) and its curved surface is π(R + r)l with slant l = √(h² + (R − r)²). Example: R = 10 cm, r = 6 cm, h = 8 cm. Volume = (1/3)π × 8 × (100 + 60 + 36) = (1/3)π × 8 × 196 = 1568π/3 ≈ 1641 cm³ (with π = 3.14). Notice the formula is a "difference of two cones" compressed: if you ever forget it, complete the cone, subtract the missing tip, and re-derive — the same completion trick as Worked example 3's combinations.

Going deeper: melting and recasting — volume is conserved

When a solid is melted and recast, volume is unchanged but surface area is not. A sphere of radius 6 cm is melted into a cone of base radius 6 cm. Sphere volume = (4/3)π × 216 = 288π. Cone: (1/3)π × 36 × h = 12πh. Equate: 12πh = 288π, so h = 24 cm. Examination variants: one big sphere into many small spheres (divide volumes — the count is a cube ratio: radius halved means 8 balls), a cuboid into wire (equate to a long thin cylinder), water flowing through a pipe (volume per minute = cross-section × speed). In every case, one equation — old volume = new volume — does all the work.

One more discipline: units, the silent mark-loser

Volume conversions are *cubed*: 1 m = 100 cm, so 1 m³ = 100³ = 1,000,000 cm³ — not 100. Practical anchor: a tank 2 m × 1.5 m × 1 m holds 3 m³ = 3000 litres (1 m³ = 1000 L). Surface areas are *squared*: 1 m² = 10,000 cm². Board papers mix units deliberately (a pipe's radius in cm feeding a tank measured in metres); the defence is to convert everything to one unit *before* substituting, and to write the unit at each line. More mensuration marks are lost to units than to formulas — by a wide margin.

Final drills

Drill 1 — fencing a circular park. A circular park has radius 35 m. Its boundary length is 2πr = 2 × (22/7) × 35 = 220 m; at ₹24 per metre, fencing costs 220 × 24 = ₹5,280. Circumference questions are area questions' siblings — the error to avoid is reaching for πr² when the problem is about a *boundary* (perimeter, fencing, rope) rather than a *surface* (turf, paint, carpet).

Drill 2 — air in a dome. A hemispherical dome has radius 10.5 m. Volume = (2/3)πr³ = (2/3) × (22/7) × 10.5³. Since 10.5³ = 1157.625, the volume is (2 × 22 × 1157.625)/(3 × 7) = 50935.5/21 = 2425.5 m³. A conical tent instead: radius 3.5 m, slant height 5 m needs canvas πrl = (22/7) × 3.5 × 5 = 55 m² — two formulas, two objects, both decided by asking "surface or space?"