Chapter 19 — Mensuration Optional

The optional exercise does not introduce any new solid or any new formula. Every problem is solved with the formulas for the cuboid, cube, cylinder, cone, sphere and hemisphere, together with two principles: the volume of a composite object is the sum of the volumes of its parts (or the difference, for a cavity), and the surface area of a composite object counts only the surfaces exposed to the outside. What the exercise adds is length of reasoning. A problem may have three parts instead of two, may require a dimension to be found before the main calculation can begin, may involve a density or a cost, or may ask the question in reverse, giving the result and asking for a dimension.

The plan for every problem. (1) Draw the object and label every dimension that is given. (2) Name the parts and write down the dimension each part shares with its neighbour; find any missing dimension by subtraction or by Pythagoras. (3) Decide whether volume or surface area is asked, or both. (4) For volume, list the parts and add or subtract. For surface area, list the exposed surfaces only. (5) Write each formula, substitute, and compute, keeping π and fractions as long as possible. (6) Apply any final step: multiply by density for mass, by rate for cost, divide by rate for time. (7) State the answer with the unit and check that its size is sensible.

Reverse problems. Several problems give the volume or the surface area and ask for a radius or height. These lead to a simple equation: a linear one when the unknown is a height, a quadratic or a cube root when the unknown is a radius. For a sphere, r³ appears; for a cone with given height, r² appears. Solve exactly where possible: r³ = 729 gives r = 9, r² = 12.25 gives r = 3.5.

Units and density. Mass = volume × density, with density in g per cm³ or kg per m³. Since 1 m³ = 1,000,000 cm³, a density of 8 g/cm³ is 8000 kg/m³. Cost = area × rate, or volume × rate, with attention to whether the rate is per m², per 100 cm² or per litre. Time = volume ÷ rate of flow, with the flow in the same volume unit as the tank.

Checking. Before writing the final answer, ask whether it could be right. A toy a few centimetres across has a volume of hundreds of cm³, not tens of thousands. An iron pole two metres tall weighs several hundred kilograms, not a few grams. A ball that raises the water in a jar by a few centimetres has a radius of a few centimetres. If the answer fails the check, look first for a unit slip and then for a diameter used as a radius.

The remaining topics take the problem types of the exercise one at a time, state the principle, and work the standard cases with full arithmetic.

A golf ball is a sphere whose surface is covered with small hemispherical dimples. Each dimple removes a small circular patch of the spherical surface and replaces it with the curved inner surface of a hemisphere. The problem asks for the total surface area of the ball including the dimples.

Principle. For a sphere of radius R with n hemispherical dimples each of radius r: surface area = 4πR² − n × πr² + n × 2πr² = 4πR² + nπr². Each dimple adds a net πr², because the hemisphere's curved surface 2πr² is twice the flat circle πr² it replaces. The same formula holds for hemispherical bumps instead of dimples, since a bump also replaces a circle by a curved hemispherical surface.

Worked example 1. A golf ball has diameter 4.1 cm and its surface has 150 dimples, each a hemisphere of radius 2 mm. Find the total surface area exposed (π = 3.14). R = 2.05 cm, r = 0.2 cm. 4πR² = 4 × 3.14 × 4.2025 = 52.78 cm². nπr² = 150 × 3.14 × 0.04 = 18.84 cm². Total = 52.78 + 18.84 = 71.62 cm². The dimples increase the surface area by about 36%.

Worked example 2. A decorative ball of radius 7 cm has 20 hemispherical bumps of radius 1 cm. Surface area = 4 × (22/7) × 49 + 20 × (22/7) × 1 = 616 + 62.86 = 678.86 cm².

Worked example 3: a cuboid with hemispherical pits. A wooden block 20 cm × 15 cm × 5 cm has 6 hemispherical pits of radius 1 cm on its top face. Surface area = 2(300 + 75 + 100) + 6 × π × 1² = 950 + 18.86 = 968.86 cm². The volume of wood is 1500 − 6 × (2/3) × (22/7) × 1 = 1500 − 12.57 = 1487.43 cm³.

Volume of the golf ball. Each dimple removes a hemisphere of volume (2/3)πr³, so the volume of the ball is (4/3)πR³ − n(2/3)πr³ = (4/3)(3.14)(8.615) − 150 × (2/3)(3.14)(0.008) = 36.07 − 2.51 = 33.56 cm³.

An approximation to note. Strictly, the circular patch removed from a curved spherical surface is slightly larger than a flat circle πr², but for dimples that are tiny compared with the ball the flat-circle value is used, and the examination expects it.

The lesson is general: whenever a small hemisphere is added to or scooped from a surface, the surface area rises by πr² and the volume changes by (2/3)πr³, in the direction that the figure shows.

When a solid is completely immersed in a liquid, the liquid level rises by an amount that depends on the volume of the solid and the cross-section of the container. The volume of liquid displaced equals the volume of the solid. If the container is a cylinder of radius R and the level rises by h, then the displaced volume is πR²h.

Worked example 1. A cylindrical vessel of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped in and the water level rises by 6.75 cm. Find the radius of the ball. Volume displaced = π × 12² × 6.75 = 972π cm³. So (4/3)πr³ = 972π, r³ = 729, r = 9 cm. (The initial depth of 20 cm is not needed, except to confirm that the ball, of diameter 18 cm, is fully submerged in 26.75 cm of water.)

Worked example 2. A cylindrical bucket of radius 15 cm is filled with water to some depth; a solid cone of radius 5 cm and height 12 cm is immersed completely. Rise = volume of cone ÷ base area of bucket = (⅓ × π × 25 × 12)/(π × 225) = 100/225 = 4/9 cm ≈ 0.44 cm.

Worked example 3: many small solids. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing water. How many marbles raise the level by 5.6 cm? Displaced volume = (22/7) × 3.5² × 5.6 = 215.6 cm³. Volume of one marble = (4/3) × (22/7) × 0.7³ = 1.437 cm³. Number = 215.6/1.437 = 150.

Worked example 4: a cone-shaped vessel. A conical vessel of radius 5 cm and height 24 cm is full of water; a solid sphere of radius 3 cm is dropped in and water overflows. The overflow equals the sphere's volume, (4/3) × 3.14 × 27 = 113.04 cm³, as long as the sphere is fully submerged, which it is, since the vessel holds ⅓ × 3.14 × 25 × 24 = 628 cm³.

Worked example 5: a fall in level. A cylindrical tank of radius 7 m contains water; a solid hemispherical rock of radius 2 m lying on the bottom is removed. The level falls by (2/3)π(8)/(π × 49) = 16/147 m ≈ 0.109 m ≈ 10.9 cm.

Worked example 6: a hollow object. A hollow iron pipe (outer radius 4 cm, inner radius 3 cm, length 21 cm), standing upright, is placed in a wide cylindrical tank of radius 20 cm; the water fills the hollow too, so only the iron displaces water: (22/7)(16 − 9) × 21 = 462 cm³, and the rise is 462/((22/7) × 400) = 0.3675 cm.

In every displacement problem the equation is volume of solid = (cross-section of container) × (change in level), and the unknown may be on either side.

A solid toy, a pencil sharpened at one end with an eraser at the other, or a decorative pillar may be a cylinder with a cone attached at one end and a hemisphere at the other. All three parts share the same radius; the total length is the sum of the cone's height, the cylinder's height and the hemisphere's radius.

Worked example 1. A solid toy consists of a cylinder of radius 4.2 cm and height 12 cm, a cone of the same radius and height 7 cm at one end, and a hemisphere of the same radius at the other. Find its volume (π = 22/7).

r² = 17.64. Cylinder: πr²h = π × 17.64 × 12 = 211.68π. Cone: ⅓πr²h = ⅓ × π × 17.64 × 7 = 41.16π. Hemisphere: (2/3)πr³ = (2/3) × π × 74.088 = 49.392π. Total = (211.68 + 41.16 + 49.392)π = 302.232 × (22/7) = 949.87 cm³.

Its surface area. Exposed: curved surface of the cone, curved surface of the cylinder, curved surface of the hemisphere. l = √(4.2² + 7²) = √(17.64 + 49) = √66.64 ≈ 8.16 cm. Cone: πrl = π × 4.2 × 8.16 = 34.29π. Cylinder: 2πrh = 2π × 4.2 × 12 = 100.8π. Hemisphere: 2πr² = 35.28π. Total = 170.37π = 170.37 × (22/7) = 535.4 cm².

Worked example 2: a pencil. A pencil is a cylinder of radius 0.35 cm and length 14 cm, sharpened at one end into a cone of height 1.5 cm. The volume of the pencil = π(0.1225)(14 − 1.5) + ⅓π(0.1225)(1.5) = 1.53125π + 0.06125π = 1.5925π = 5.005 cm³. (The cone occupies part of the total length, so the cylinder is 12.5 cm long.)

Worked example 3: a solid with a hemisphere at both ends. A solid is a cylinder of radius 3.5 cm and length 10 cm with a hemisphere at each end. Volume = π × 12.25 × 10 + (4/3)π × 42.875 = 122.5π + 57.17π = 179.67π = 564.67 cm³. Surface = 2π × 3.5 × 10 + 4π × 12.25 = 70π + 49π = 119π = 374 cm².

Worked example 4: a pillar with a dome. A pillar model is a cylinder of radius 1 m and height 3 m topped by a hemisphere of radius 1 m. Volume = π × 1 × 3 + (2/3)π × 1 = (3 + 0.667)π = 3.667 × 3.14 = 11.51 m³. Exposed surface (the base stands on the ground) = 2π × 1 × 3 + 2π × 1 = 8π = 25.12 m². A cone cannot be placed on top of the hemisphere, because a hemisphere has no flat top to receive it; the student should notice when a described combination is geometrically impossible.

Method. A three-part problem is three two-line calculations followed by one addition. Take out the common factor πr² or π at the start; the arithmetic then involves only the heights and radii, and the multiplication by 22/7 or 3.14 happens once at the end.

The mass of a metal object is its volume multiplied by the density of the metal. Iron has density about 8 g/cm³, copper 8.9 g/cm³, aluminium 2.7 g/cm³, and the problem always gives the value to be used. Volume must be in cm³ when the density is in g/cm³.

Worked example 1: an iron pole. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm surmounted by another cylinder of height 60 cm and radius 8 cm. Find its mass, given that 1 cm³ of iron has mass 8 g (π = 3.14). Lower cylinder: π × 12² × 220 = 31,680π cm³. Upper cylinder: π × 8² × 60 = 3,840π cm³. Total = 35,520π = 35,520 × 3.14 = 111,532.8 cm³. Mass = 111,532.8 × 8 = 892,262.4 g ≈ 892.26 kg.

Worked example 2: a hollow pipe. An iron pipe has external diameter 8 cm, thickness 1 cm and length 21 cm. Volume of iron = π(4² − 3²) × 21 = (22/7) × 7 × 21 = 462 cm³. At 7.5 g/cm³ the mass is 3465 g = 3.465 kg.

Worked example 3: a spherical shell. A hollow copper sphere has internal diameter 8 cm and external diameter 10 cm. Volume of copper = (4/3)π(5³ − 4³) = (4/3)π × 61 = 255.5 cm³ (π = 3.14). Mass at 8.9 g/cm³ = 2274 g ≈ 2.27 kg. If the shell is melted and recast into a solid cylinder of radius 2 cm, its height is 255.5/(3.14 × 4) = 20.33 cm.

Worked example 4: a hemispherical bowl. A hemispherical bowl of steel has inner radius 5 cm and thickness 0.25 cm. Volume of steel = (2/3)π(5.25³ − 5³) = (2/3) × 3.14 × (144.70 − 125) = 41.25 cm³. Mass at 7.8 g/cm³ = 321.7 g.

Worked example 5: a metal cone. A solid metal cone of radius 7 cm and height 24 cm has mass 8.6 kg. Find the density. Volume = ⅓ × (22/7) × 49 × 24 = 1232 cm³. Density = 8600/1232 = 6.98 g/cm³.

Worked example 6: units in m³. A concrete pillar is a cylinder of radius 0.3 m and height 4 m; concrete has density 2400 kg/m³. Volume = 3.14 × 0.09 × 4 = 1.1304 m³; mass = 1.1304 × 2400 = 2713 kg. If the density had been given as 2.4 g/cm³ the student would convert 1.1304 m³ to 1,130,400 cm³ and obtain the same 2,713,000 g.

Write the volume with its unit first, then the multiplication by density, then convert grams to kilograms if the number is large. The examiner gives a mark for the final conversion.

A frequent problem gives a solid block and asks for the largest cone, cylinder, sphere or cube that can be carved from it, and often the volume of material wasted. The key is to see which dimension of the block limits the carved solid.

Largest cone from a cube of side a. The base of the cone is the largest circle in a face, radius a/2, and its height is the side a. Volume = ⅓π(a/2)²a = πa³/12. For a = 14 cm: ⅓ × (22/7) × 49 × 14 = 718.67 cm³. Wasted = 2744 − 718.67 = 2025.33 cm³, about 74%.

Largest cylinder from a cube of side a. Radius a/2, height a: volume = πa³/4. For a = 7 cm: (22/7) × 12.25 × 7 = 269.5 cm³, and the wasted wood is 343 − 269.5 = 73.5 cm².

Largest sphere from a cube of side a. Diameter a, volume = πa³/6. The ratio of the cube's volume to the sphere's is 6 : π ≈ 1.91 : 1. For a = 6 cm: (4/3) × 3.14 × 27 = 113.04 cm³ out of 216 cm³.

Largest cube from a sphere of radius r. The cube's diagonal a√3 equals the diameter 2r, so a = 2r/√3. For r = 3√3 cm, a = 6 cm and the cube's volume is 216 cm³; the sphere's volume is (4/3)π × 81√3 ≈ 587.7 cm³.

Largest cone from a cylinder of radius r and height h: the cone with the same base and height, volume ⅓πr²h, one third of the cylinder; two thirds is wasted. From a cylinder of radius 6 cm and height 8 cm the cone has volume 301.44 cm³ and 602.88 cm³ is removed.

Largest sphere from a cylinder of radius r and height h ≥ 2r: radius r, volume (4/3)πr³. If h < 2r the height limits it and the radius is h/2.

Largest hemisphere from a cube of side a: radius a/2, volume (2/3)π(a/2)³ = πa³/12, the same as the largest cone.

Worked example. A cubical block of side 7 cm is surmounted by a hemisphere; the greatest diameter the hemisphere can have is 7 cm, and the surface area of the solid is 6 × 49 + π(3.5)² = 294 + 38.5 = 332.5 cm². This was worked in the main chapter and reappears here with the phrase 'greatest diameter', which simply means the diameter equals the side.

The general idea, that a solid inscribed in another is limited by the smallest relevant dimension, also gives quick ratio answers: the largest sphere in a cube fills π/6 ≈ 52% of it; the largest cone in a cylinder fills a third; the largest cube in a sphere fills 2/(π√3) ≈ 37% of it.

Many one- and two-mark questions ask how the volume or surface area changes when a dimension changes, or ask for the ratio of two quantities without any actual dimensions. The rules follow directly from the formulas.

Scaling all dimensions. If every length of a solid is multiplied by k, every area is multiplied by k² and the volume by k³. Two spheres with radii in the ratio 2 : 3 have surface areas in the ratio 4 : 9 and volumes in the ratio 8 : 27. Conversely, if the surface areas of two spheres are in the ratio 4 : 9, their radii are in the ratio 2 : 3 and their volumes 8 : 27.

Scaling one dimension. If only the height of a cylinder or cone is doubled, the volume doubles. If only the radius is doubled, the volume becomes four times, because r appears squared. If the radius of a cylinder is doubled and the height halved, the volume becomes 4 × ½ = 2 times the original, while the curved surface area 2πrh stays the same.

Same base and height. A cone, a hemisphere and a cylinder with the same radius r and height r have volumes ⅓πr³, (2/3)πr³ and πr³, in the ratio 1 : 2 : 3. A cone and a cylinder of the same base and height are always 1 : 3.

Worked example 1. Two cones have the same height and radii in the ratio 2 : 3. Ratio of volumes = 4 : 9. If instead they have the same radius and heights in the ratio 2 : 3, the ratio is 2 : 3.

Worked example 2. The volumes of two spheres are in the ratio 64 : 27. Then r₁ : r₂ = 4 : 3 and their surface areas are in the ratio 16 : 9.

Worked example 3. A cylinder and a cone have equal radii and equal volumes; the height of the cone is 12 cm. Then πr²H = ⅓πr² × 12, so H = 4 cm: the cylinder is one third as tall.

Worked example 4. A sphere and a cube have equal surface areas. Then 4πr² = 6a², so r²/a² = 3/(2π) and the ratio of volumes (4/3)πr³ : a³ = (4/3)π(3/(2π))^(3/2) : 1 = √(6/π) : 1 ≈ 1.38 : 1. The sphere encloses more volume, which is a general property of the sphere.

Worked example 5. A cone of radius r and height h is cut by a plane through the midpoint of its axis parallel to the base. The small cone on top has radius r/2 and height h/2, so its volume is one eighth of the whole; the lower part, a frustum, has seven eighths. The ratio of the two parts is 1 : 7. This is the scaling rule with k = ½.

Worked example 6. If the radius of a sphere is increased by 10%, its surface area increases by (1.1)² − 1 = 21% and its volume by (1.1)³ − 1 = 33.1%.

Ratio questions are quick to answer once the student sees which power of the scale factor applies: 1 for a length, 2 for an area, 3 for a volume.

A cost problem is a surface-area or volume problem with one extra line: multiply by the rate. The only difficulty is to identify exactly which surface is to be paid for and to match the units of the rate.

Worked example 1: painting a dome. The inner surface of a hemispherical dome of diameter 14 m is to be painted at ₹5 per m² (π = 22/7). Area = 2πr² = 2 × (22/7) × 49 = 308 m². Cost = 308 × 5 = ₹1540. Only the curved surface is painted; the dome has no flat floor of its own.

Worked example 2: tin-plating a bowl. The inside of a hemispherical bowl of inner radius 10.5 cm is to be tin-plated at ₹16 per 100 cm². Area = 2 × (22/7) × 110.25 = 693 cm². Cost = 693 × 16/100 = ₹110.88.

Worked example 3: sheet metal for a closed tank. A closed cylindrical tank of radius 1.4 m and height 3 m is made of sheet metal costing ₹40 per m². Area = TSA = 2πr(r + h) = 2 × (22/7) × 1.4 × 4.4 = 38.72 m². Cost = ₹1548.80. If the tank were open at the top, only one circular end would be counted: 2πrh + πr² = 26.4 + 6.16 = 32.56 m², cost ₹1302.40.

Worked example 4: canvas for a conical tent with wastage. A conical tent of radius 7 m and height 24 m needs canvas for its curved surface: πrl = (22/7) × 7 × 25 = 550 m². If 10% extra is allowed for stitching and wastage, canvas bought = 550 × 1.1 = 605 m²; at ₹80 per m² the cost is ₹48,400.

Worked example 5: painting a composite solid. A wooden toy (cone of radius 3.5 cm and slant height 12.5 cm on a hemisphere of radius 3.5 cm) is to be painted at ₹0.50 per cm². Surface = 214.5 cm² (from the main chapter). Cost = ₹107.25.

Worked example 6: cost by volume. Milk is sold at ₹40 per litre from a cylindrical container of radius 14 cm and height 50 cm. Volume = (22/7) × 196 × 50 = 30,800 cm³ = 30.8 litres. Value = ₹1232.

Worked example 7: cost of digging. Digging a well of diameter 3 m to a depth of 14 m costs ₹40 per m³. Volume = (22/7) × 2.25 × 14 = 99 m³. Cost = ₹3960. Plastering its inner curved surface at ₹25 per m²: area = 2 × (22/7) × 1.5 × 14 = 132 m², cost ₹3300.

Always state which surfaces are included and why: 'the base is on the ground and not painted', 'the tank is open, so one end only', 'the inner surface only is plated'. That sentence earns the reasoning mark and prevents the commonest error.

A pipe delivers a fixed volume of liquid per unit time. The time to fill or empty a container is its volume divided by the rate. When the rate is given as a speed of flow through a pipe of known cross-section, the volume per unit time is (cross-section area) × (speed). When two pipes act together, their rates add if both fill, and subtract if one fills while the other empties.

Worked example 1: emptying a tank. A cylindrical tank of radius 7 m and depth 3 m, full of water, is emptied by a pipe that removes 7 litres per second. Volume = (22/7) × 49 × 3 = 462 m³ = 462,000 litres. Time = 462,000/7 = 66,000 s = 1100 minutes = 18 hours 20 minutes.

Worked example 2: a hemispherical tank. A hemispherical tank of radius 1.75 m is full and is emptied at 7 litres per second. Volume = (2/3) × (22/7) × 1.75³ = (2/3) × (22/7) × 5.359 = 11.23 m³ = 11,229 litres. Time = 11,229/7 ≈ 1604 s ≈ 26.7 minutes.

Worked example 3: flow through a pipe into a tank. Water flows through a pipe of diameter 14 cm at 15 km/h into a rectangular tank 50 m × 44 m. In how long does the level rise by 21 cm? Volume needed = 50 × 44 × 0.21 = 462 m³. Rate = (22/7) × 0.07² × 15,000 = (22/7) × 0.0049 × 15,000 = 231 m³ per hour. Time = 462/231 = 2 hours.

Worked example 4: two pipes. A cylindrical tank of volume 6000 litres has an inlet supplying 25 litres per minute and an outlet draining 10 litres per minute. With both open, the net rate is 15 litres per minute and the tank fills in 6000/15 = 400 minutes, that is 6 hours 40 minutes.

Worked example 5: a conical funnel emptying into a cylinder. A conical vessel of radius 12 cm and height 32 cm is full of water and is emptied into a cylindrical vessel of radius 8 cm. The water in the cylinder stands at height h where π × 64 × h = ⅓ × π × 144 × 32, so h = 1536/64 = 24 cm.

Worked example 6: rain filling a tank. Rain of 3.5 cm falls on a flat terrace 20 m × 15 m and all of it drains into a cylindrical tank of radius 1 m. The depth of water in the tank = (20 × 15 × 0.035)/((22/7) × 1) = 10.5/3.1429 = 3.34 m.

Worked example 7: speed of flow required. A cylindrical tank of radius 5 m and depth 2 m must be filled in 100 minutes through a pipe of radius 10 cm. Volume = (22/7) × 25 × 2 = 157.14 m³; rate needed = 157.14/100 = 1.5714 m³ per minute; cross-section = (22/7) × 0.01 = 0.031429 m²; speed = 1.5714/0.031429 = 50 m per minute = 3 km/h.

Convert litres and cubic metres with 1 m³ = 1000 litres, and hours and minutes with care; then the arithmetic is a single division.

A wire is a very long, very thin cylinder. Two kinds of problem occur: a solid is melted or drawn into a wire of given thickness and its length is required, or a wire of given thickness is wound around a cylinder and its length, volume or mass is required. Both use the cylinder formula V = πr²l with r tiny and l large, and both need careful unit conversion between millimetres, centimetres and metres.

Worked example 1: a sphere drawn into a wire. A copper sphere of radius 3 cm is drawn into a wire of radius 0.1 cm. Volume = (4/3)π × 27 = 36π cm³. Wire: π × 0.01 × L = 36π, so L = 3600 cm = 36 m.

Worked example 2: a rod drawn into a wire. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m. Find its thickness. Volume = π × 0.25 × 8 = 2π cm³. Wire: πr² × 1800 = 2π, r² = 1/900, r = 1/30 cm, so the thickness (diameter) is 1/15 cm ≈ 0.067 cm, or about 0.67 mm.

Worked example 3: a cylinder into a wire. A cylinder of radius 2 cm and height 45 cm is melted and drawn into a wire of diameter 4 mm. Volume = π × 4 × 45 = 180π cm³. Wire radius 0.2 cm: π × 0.04 × L = 180π, L = 4500 cm = 45 m.

Worked example 4: a wire wound on a cylinder. A copper wire of diameter 3 mm is wound around a cylinder of length 12 cm and diameter 10 cm so as to cover its whole curved surface. Find the length and mass of the wire, given copper has density 8.88 g/cm³. Each turn occupies 3 mm = 0.3 cm of the length, so the number of turns = 12/0.3 = 40. Each turn has length equal to the circumference, π × 10 = 31.4 cm. Total length = 40 × 31.4 = 1256 cm = 12.56 m. Volume of wire = π × (0.15)² × 1256 = 3.14 × 0.0225 × 1256 = 88.74 cm³. Mass = 88.74 × 8.88 = 788 g.

Worked example 5: a wire bent into a shape. A wire of length 44 cm is bent into a circle; its radius is 44/(2 × 22/7) = 7 cm and the area enclosed is 154 cm². Bent instead into a square, the side is 11 cm and the area is 121 cm². The same length of wire encloses more area as a circle, which connects with the sphere enclosing the most volume for a given surface.

Worked example 6: many wires. How many wires of diameter 1 mm and length 1 m can be drawn from a cube of copper of side 10 cm? Cube volume = 1000 cm³. One wire: π × (0.05)² × 100 = 0.7854 cm³. Number = 1000/0.7854 ≈ 1273.

Keep every length in centimetres while computing, and convert the answer to metres at the end; a wire's radius is often given as a diameter in millimetres, so halve it and divide by 10.

Problem 1. A golf ball has diameter 4.1 cm and 150 hemispherical dimples of radius 2 mm. Total exposed surface area (π = 3.14): 4π(2.05)² + 150π(0.2)² = 52.78 + 18.84 = 71.62 cm².

Problem 2. A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped in and the level rises by 6.75 cm. Radius of the ball: (4/3)πr³ = π × 144 × 6.75 = 972π, r³ = 729, r = 9 cm.

Problem 3. A solid toy is a cylinder of radius 4.2 cm and height 12 cm with a cone of height 7 cm at one end and a hemisphere at the other. Volume = π(17.64)(12) + ⅓π(17.64)(7) + (2/3)π(74.088) = (211.68 + 41.16 + 49.392)π = 302.232 × 22/7 = 949.87 cm³.

Problem 4. A hemispherical bowl of internal radius 15 cm is full of liquid, which is poured into cylindrical bottles of diameter 5 cm and height 6 cm. Number of bottles: bowl = (2/3)π × 3375 = 2250π cm³; bottle = π × 6.25 × 6 = 37.5π cm³; 2250/37.5 = 60 bottles.

Problem 5. A solid iron pole is a cylinder of height 220 cm and radius 12 cm with a cylinder of height 60 cm and radius 8 cm on top; iron has mass 8 g per cm³. Mass = (31,680 + 3,840) × 3.14 × 8 = 35,520 × 25.12 = 892,262 g ≈ 892.26 kg.

Problem 6. A container open at the top is a cylinder of radius 7 cm and height 10 cm standing on a hemispherical base of the same radius. Its capacity = (22/7)(49)(10) + (2/3)(22/7)(343) = 1540 + 718.67 = 2258.67 cm³ ≈ 2.26 litres. Its outer surface = 2π(7)(10) + 2π(49) = 440 + 308 = 748 cm².

Problem 7. A solid sphere of radius 6 cm is melted and recast into a single hollow cylinder of outer radius 5 cm and inner radius 3 cm; find its height. (4/3)π × 216 = π(25 − 9)h, 288 = 16h, h = 18 cm.

Problem 8. A cubical block of side 7 cm has a hemisphere of the greatest possible diameter placed on top. Surface area = 294 + 38.5 = 332.5 cm².

Problem 9. A conical vessel of radius 12 cm and height 32 cm is full of water; it is emptied into a cylinder of radius 8 cm. Height of water in the cylinder = (⅓ × 144 × 32)/64 = 1536/64 = 24 cm.

Problem 10. The rain of 2 cm on a roof 22 m × 20 m flows into a cylindrical tank of radius 1 m; depth in tank = 8.8/((22/7) × 1) = 2.8 m.

Problem 11. Two cubes each of volume 27 cm³ are joined end to end; surface area of the cuboid = 2(18 + 9 + 18) = 90 cm², compared with 108 cm² for the two cubes separately.

Problem 12. A hemispherical depression is cut from one face of a cubical block of side 7 cm such that the diameter of the hemisphere equals the edge; surface area of the remaining solid = 294 + 38.5 = 332.5 cm², the same as adding a hemisphere on top.

Strategy. Read the problem twice: once to picture the object, once to note the numbers and the question. Draw the object and label the parts. For each part write its radius and height on the figure; where a total length is given, subtract to find the part lengths; where a slant height is needed, compute it with Pythagoras at once. Decide what is asked: volume, surface, mass, cost, time or a dimension. Write the plan in words before the formulas: 'volume = cylinder + cone + hemisphere', 'surface = curved cylinder + curved cone + curved hemisphere'. Then compute, taking out common factors and multiplying by π only at the end.

The errors that cost marks. Using the diameter as the radius, especially for the golf ball (4.1 cm is a diameter) and the pipe (20 cm is a diameter). Adding hidden faces to a surface area. Using (4/3)πr³ for a hemisphere. Forgetting the ⅓ in a cone's volume or the slant height in its surface. Mixing millimetres, centimetres and metres, particularly in wire and pipe problems. Treating 1 m³ as 100 cm³ or 1000 cm³ instead of 1,000,000 cm³. Giving a mass in grams when it is nearly a tonne. Stopping at the volume when the question asks for the cost or the time. Omitting the unit.

Checking. Estimate with π ≈ 3 and round numbers; the estimate should agree with the answer to within about 10%. A composite volume must exceed its largest part; a remaining volume must be less than the whole; a wire drawn from a small solid is metres long but under a millimetre thick; a level rise in a wide vessel is small.

Presentation. The examiner awards marks for the figure, the formula, the substitution, the arithmetic and the answer with unit; a problem worth four marks usually gives one mark to each of the first four and the last mark to the final answer. Writing the formula in symbols before substituting is therefore never wasted.

What lies ahead. The frustum of a cone, a cone with its top cut off parallel to the base, is the shape of a bucket, a lampshade and a glass; its volume is ⅓πh(R² + Rr + r²) and its curved surface is π(R + r)l with l = √(h² + (R − r)²). It is treated in the Intermediate course and in other boards' Class 10 books, and the student who has mastered this exercise can derive it as the difference of two similar cones. Integration later proves the formulas accepted here; physics uses them for density, buoyancy and pressure; and every engineering estimate, from the concrete in a pillar to the paint on a tank, is a mensuration problem of exactly this kind.