Chapter 12 — Surface Areas and Volumes

Everything in this chapter is built on the formulas for the basic solids, so the first job is to have them exact and to know what each letter means.

For a cone, the slant height l is related to the radius r and the height h by l² = r² + h², because the axis, the radius of the base and the slant edge form a right triangle. Always compute l before using the curved surface area of a cone.

The distinction between curved surface area (CSA, also called lateral surface area) and total surface area (TSA) matters throughout: TSA includes the flat bases, CSA does not. A hemisphere's TSA of 3πr² is its curved 2πr² plus the flat circular face πr².

Units. Surface areas are in square units, volumes in cubic units. Capacity is often asked in litres: 1 litre = 1000 cm³, and 1 m³ = 1000 litres. Convert all lengths to one unit before substituting.

Worked example. A cone has radius 7 cm and height 24 cm. Its slant height is √(49 + 576) = 25 cm, CSA = (22/7) × 7 × 25 = 550 cm², TSA = 550 + 154 = 704 cm², and volume = ⅓ × (22/7) × 49 × 24 = 1232 cm³.

Worked example. A sphere of radius 2.1 cm has surface area 4 × (22/7) × 4.41 = 55.44 cm² and volume (4/3) × (22/7) × 9.261 = 38.81 cm³.

Worked example. A cylindrical tank of radius 1.75 m and height 2 m holds (22/7) × 1.75² × 2 = 19.25 m³ = 19250 litres.

A useful comparison: a cylinder, a cone and a hemisphere of the same radius r and height r have volumes in the ratio πr³ : ⅓πr³ : ⅔πr³ = 3 : 1 : 2.

Look at a few common objects. A capsule is a cylinder with a hemisphere at each end. A circus tent is a cylinder with a cone on top. A toy may be a cone standing on a hemisphere. A gulab jamun is a cylinder with hemispherical ends. A pencil sharpened at one end is a cylinder plus a cone. A rocket model is a cone on a cylinder. A wooden pen stand is a cuboid with conical or cylindrical depressions. A test tube is a cylinder closed by a hemisphere. In each case the object is a combination of basic solids.

The method for a combined solid is:

• Identify each basic solid in the combination.
• Find the dimensions of each part from the information given. Usually the parts share a common radius, and the total height is the sum of the heights of the parts; the height of a hemisphere is its radius.
• For volume, add the volumes of the parts (or subtract, if a part is scooped out).
• For surface area, add only the surfaces that are actually exposed. Where two parts are joined, the surfaces in contact are hidden and must not be counted.

Consider a toy made of a cone of radius 3.5 cm and height 12 cm mounted on a hemisphere of the same radius. The parts are a cone and a hemisphere with common radius 3.5 cm. Total height of the toy = 12 + 3.5 = 15.5 cm. The surfaces exposed are the curved surface of the cone and the curved surface of the hemisphere; the circular base of the cone and the flat face of the hemisphere are glued together and hidden.

Consider a capsule of total length 14 mm and diameter 5 mm. The parts are two hemispheres of radius 2.5 mm and a cylinder of radius 2.5 mm whose height is 14 − 2 × 2.5 = 9 mm. All exposed surface is curved: the cylinder's CSA plus the two hemispheres' CSA, which together make a full sphere's surface.

Consider a cuboidal block from which a hemisphere is scooped out of the top face. The volume is the cuboid minus the hemisphere; the surface area is the cuboid's TSA minus the circular area removed plus the curved surface of the hollow.

The examination figure will show the solid, or the text will describe it; draw it and mark the dimensions of each part before writing any formula. Misidentifying a hidden surface is the most common error in this chapter.

The total surface area of a combined solid is the sum of the curved or flat surfaces that are visible from outside. Let us work through the standard cases.

Cone on a hemisphere (a toy). A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius; the total height of the toy is 15.5 cm. Height of the cone = 15.5 − 3.5 = 12 cm, slant height l = √(3.5² + 12²) = √(12.25 + 144) = 12.5 cm. TSA = CSA of hemisphere + CSA of cone = 2πr² + πrl = πr(2r + l) = (22/7) × 3.5 × (7 + 12.5) = 11 × 19.5 = 214.5 cm².

Cube with a hemisphere on top. A cubical block of side 7 cm is surmounted by a hemisphere; the greatest diameter the hemisphere can have is 7 cm. Surface area = TSA of cube − area of the circle covered + CSA of hemisphere = 6 × 49 − π(3.5)² + 2π(3.5)² = 294 + π(3.5)² = 294 + 38.5 = 332.5 cm². Note that the hemisphere adds its curved surface but hides a circle of the cube's top; the net addition is πr².

Cube with a hemispherical depression. If the hemisphere is scooped out of a cube of side l with diameter l, the surface area is 6l² − π(l/2)² + 2π(l/2)² = 6l² + πl²/4 = (l²/4)(24 + π), the same net addition as before.

Capsule. Length 14 mm, diameter 5 mm: r = 2.5, cylinder height 9. Surface area = 2πrh + 2 × 2πr² = 2πr(h + 2r) = 2 × (22/7) × 2.5 × (9 + 5) = 220 mm².

Tent. A tent is a cylinder of diameter 4 m and height 2.1 m surmounted by a cone of slant height 2.8 m. Canvas area = 2πrh + πrl = πr(2h + l) = (22/7) × 2 × (4.2 + 2.8) = (22/7) × 14 = 44 m². At ₹500 per m² the cost is ₹22000. The ground is not covered, so no base is included.

Cone cut from a cylinder. From a solid cylinder of height 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and diameter is hollowed out. Remaining TSA = CSA of cylinder + area of one base + CSA of cone = 2πrh + πr² + πrl, with r = 0.7, h = 2.4, l = √(0.49 + 5.76) = 2.5: = (22/7) × 0.7 × (4.8 + 0.7 + 2.5) = 2.2 × 8 = 17.6 cm², nearest 18 cm².

Two cones on a cylinder (a decorative block). If cones are attached to both ends of a cylinder, the exposed area is the cylinder's CSA plus both cones' CSA; no base is exposed at all.

Factor out πr where possible; it shortens the arithmetic and reduces errors.

Volume is simpler than surface area: the volume of a combined solid is the sum of the volumes of its parts, because the parts occupy separate regions of space. When a part is hollowed out, its volume is subtracted.

Toy: cone on a hemisphere. A toy is a cone of radius 3.5 cm and height 12 cm on a hemisphere of the same radius (total height 15.5 cm). Volume = ⅓πr²h + ⅔πr³ = ⅓πr²(h + 2r) = ⅓ × (22/7) × 12.25 × (12 + 7) = ⅓ × 38.5 × 19 = 243.83 cm³.

Rocket: cone on a cylinder. A model rocket has a cylinder of radius 3 cm and height 12 cm with a cone of the same radius and height 4 cm on top. Volume = π × 9 × 12 + ⅓π × 9 × 4 = 108π + 12π = 120π ≈ 377.14 cm³.

Gulab jamun. A gulab jamun is a cylinder with two hemispherical ends, of total length 5 cm and diameter 2.8 cm; it contains sugar syrup up to 30% of its volume. Find the syrup in 45 gulab jamuns. r = 1.4, cylinder height = 5 − 2.8 = 2.2. Volume of one = πr²h + (4/3)πr³ = πr²(h + 4r/3) = (22/7) × 1.96 × (2.2 + 1.867) = 6.16 × 4.067 = 25.05 cm³. Syrup = 30% × 25.05 × 45 = 338.2 cm³, approximately 338 cm³.

Pen stand with conical depressions. A cuboid 15 cm × 10 cm × 3.5 cm has four conical depressions of radius 0.5 cm and depth 1.4 cm. Volume of wood = 15 × 10 × 3.5 − 4 × ⅓ × (22/7) × 0.25 × 1.4 = 525 − 1.47 = 523.53 cm³.

Solid iron pole. A pole consists of a cylinder of height 220 cm and base diameter 24 cm surmounted by a cylinder of height 60 cm and radius 8 cm. Volume = π(144 × 220 + 64 × 60) = π(31680 + 3840) = 35520π = 111532.8 cm³ (π = 3.14). At 8 g per cm³, mass = 892.26 kg.

Cone with hemisphere scooped out. A solid consists of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm, placed upright in a cylinder full of water so that it touches the bottom. The cylinder has radius 60 cm and height 180 cm. Water left = cylinder − (cone + hemisphere) = π × 3600 × 180 − (⅓π × 3600 × 120 + ⅔π × 216000) = 648000π − (144000π + 144000π) = 360000π ≈ 1131428.6 cm³ ≈ 1.131 m³.

Spherical glass vessel with a cylindrical neck. Neck: length 8 cm, diameter 2 cm; sphere: diameter 8.5 cm. Volume = π × 1 × 8 + (4/3)π × (4.25)³ = 8π + 102.35π = 110.35π = 346.51 cm³ (π = 3.14), so a child's measurement of 345 cm³ is nearly right.

The pattern is always: volume of each part, add or subtract, convert units if capacity is asked.

When a solid is melted and recast, or clay is remoulded, or a metal block is drawn into wire, its shape changes but its volume stays the same. Problems of this kind set the volume of the original solid equal to the volume of the new solid and solve for the unknown dimension or the number of pieces. (In practice a small loss occurs, and questions sometimes mention it as a percentage; otherwise assume none.)

Sphere to cylinder. A metallic sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find the height. (4/3)π(4.2)³ = π(6)²h, so h = (4/3) × 74.088/36 = 2.744 cm.

Spheres to sphere. Metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to form a single solid sphere. (4/3)πR³ = (4/3)π(216 + 512 + 1000) = (4/3)π × 1728, so R³ = 1728 and R = 12 cm.

Cone to spheres. A cone of height 24 cm and radius 6 cm is melted and made into spheres of radius 2 cm. How many spheres are formed? Volume of cone = ⅓π × 36 × 24 = 288π; volume of one sphere = (4/3)π × 8 = (32/3)π; number = 288 ÷ (32/3) = 27.

Cylinder to cones. A cylinder of radius 6 cm and height 15 cm is melted to make cones of radius 3 cm and height 12 cm. Each cone has volume ⅓π × 9 × 12 = 36π and the cylinder 540π, giving 15 cones; if the cones were of radius 3 and height 12 with hemispherical tops (ice cream cones) each has 36π + 18π = 54π and the number is 10.

Well to embankment. A well of diameter 3 m is dug 14 m deep and the earth is spread evenly to form a circular embankment of width 4 m around it. Find the height of the embankment. Earth dug = π × 1.5² × 14 = 31.5π m³. The embankment is a ring of inner radius 1.5 m and outer radius 5.5 m: area = π(30.25 − 2.25) = 28π. Height = 31.5π/28π = 1.125 m.

Well to platform. A well of diameter 3 m and depth 14 m; the earth forms a platform 22 m × 14 m. With π = 22/7, the volume is 31.5 × 22/7 = 99 m³, so the height = 99/(22 × 14) = 99/308 ≈ 0.32 m.

Wire drawn from a rod. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m. Find the thickness. π(0.5)² × 8 = πr² × 1800, so r² = 2/1800 = 1/900, r = 1/30 cm, diameter = 1/15 cm ≈ 0.067 cm.

Sphere to wire (a long thin cylinder). A sphere of diameter 6 cm is drawn into a wire of diameter 2 mm. Length L: (4/3)π × 27 = π × (0.1)² × L, so L = 36/0.01 = 3600 cm = 36 m.

State the principle in one line, volume before = volume after, then write both volumes, cancel π, and solve. Keep the unit of length uniform, especially when mm, cm and m are mixed.

Water flowing through a pipe of circular cross-section at a known speed delivers a cylinder of water each second: the cylinder's radius is the pipe's radius and its length is the distance the water moves in one second. So

Volume delivered per unit time = (area of cross-section) × (speed) = πr² × v.

Multiplying by the time gives the total volume delivered, which is then equated to the volume of the tank or of the layer of water in a field.

Filling a tank. Water flows at 3 km/h through a pipe of internal diameter 2 cm into a cylindrical tank of base radius 40 cm. In how much time will the water rise by 3.15 m? Speed = 3000 m/h = 300000 cm/h; pipe radius 1 cm; volume per hour = π × 1 × 300000 = 300000π cm³. Required volume = π × 1600 × 315 = 504000π cm³. Time = 504000/300000 = 1.68 h = 1 h 40.8 min ≈ 1 hour 41 minutes.

Irrigating a field. A canal is 6 m wide and 1.5 m deep and water flows at 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed? In 30 minutes the water advances 5 km = 5000 m; volume = 6 × 1.5 × 5000 = 45000 m³. Area = 45000/0.08 = 562500 m² = 56.25 hectares.

Filling a cuboidal tank from a river-fed pipe. A cylindrical pipe of diameter 14 cm delivers water at 15 km/h into a cuboidal pond 50 m × 44 m. Time to raise the level by 21 cm? Volume needed = 50 × 44 × 0.21 = 462 m³. Pipe: r = 0.07 m, speed 15000 m/h, volume per hour = (22/7) × 0.0049 × 15000 = 231 m³. Time = 2 hours.

Emptying a hemispherical tank. A hemispherical tank of internal diameter 3 m, radius 1.5 m, holds (2/3) × (22/7) × 3.375 = 7.071 m³ = 7071 litres. If it is emptied by a pipe at 3.5 litres per second, the time is 7071/3.5 ≈ 2020 s ≈ 33.7 minutes. If instead the tank is only half full, the time halves to about 16.8 minutes; read the data carefully and compute afresh.

Rainfall collected from a roof. Rain falling on a flat roof 22 m × 20 m is drained into a cylindrical vessel of diameter 2 m and height 3.5 m. If the vessel is just full, the rainfall is volume ÷ roof area = [(22/7) × 1 × 3.5]/(22 × 20) = 11/440 m = 2.5 cm.

Conversions used constantly: 1 km/h = 1000 m/h; 1 hour = 3600 s; 1 m³ = 1000 L; 1 hectare = 10000 m². The cross-section of a pipe is πr² with r the internal radius.

Take a right circular cone and cut it by a plane parallel to its base. The part between the base and the cutting plane, the shape of a bucket, a drinking glass, a lampshade or a flower pot, is called a frustum of the cone. It has two circular ends of different radii, the larger R and the smaller r, a height h (the perpendicular distance between the ends) and a slant height l along the sloping side.

Volume by subtraction. Let the whole cone have height H and radius R, and the small cone removed have height H − h and radius r. Since the two cones are similar, r/R = (H − h)/H, so H = Rh/(R − r) and H − h = rh/(R − r). Volume of the frustum = ⅓πR²H − ⅓πr²(H − h) = ⅓π[R² × Rh/(R − r) − r² × rh/(R − r)] = ⅓πh(R³ − r³)/(R − r) = ⅓πh(R² + Rr + r²), using R³ − r³ = (R − r)(R² + Rr + r²).

Volume of a frustum = ⅓πh(R² + Rr + r²).

Check: when r = R it becomes πR²h, a cylinder; when r = 0 it becomes ⅓πR²h, a cone.

The slant height, needed for surface area, comes from the right triangle formed by the height h and the difference of radii R − r: l = √[h² + (R − r)²].

Worked example. A drinking glass is in the shape of a frustum of height 14 cm with diameters 4 cm and 2 cm at its ends. Capacity = ⅓ × (22/7) × 14 × (4 + 2 + 1) = (44/3) × 7 = 102.67 cm³.

Worked example. A bucket has height 24 cm and radii 15 cm and 5 cm. Volume = ⅓ × (22/7) × 24 × (225 + 75 + 25) = (22/7) × 8 × 325 = 57200/7 = 8171.43 cm³ ≈ 8.17 litres.

Worked example. A container in the shape of a frustum of height 16 cm with radii 20 cm and 8 cm is full of milk; find the cost at ₹20 per litre. Volume = ⅓ × 3.14 × 16 × (400 + 160 + 64) = ⅓ × 3.14 × 16 × 624 = 10449.92 cm³ ≈ 10.45 L; cost ≈ ₹209.

Worked example. A cone of height 20 cm and vertical angle 60° is cut midway by a plane parallel to the base; the frustum's ends have radii 10 tan 30° = 10/√3 and 20 tan 30° = 20/√3. Its volume = ⅓π × 10 × (400/3 + 200/3 + 100/3) = ⅓π × 10 × 700/3 = 7000π/9 cm³. If drawn into a wire of diameter 1/16 cm, the length is (7000π/9) ÷ [π(1/32)²] = (7000/9) × 1024 = 796444 cm ≈ 7964 m.

Metallic frustum problems (recasting, wire-drawing) combine this formula with the equal-volume principle of the earlier topic.

The curved surface of a frustum is the curved surface of the whole cone minus that of the small cone removed. With the notation of the previous topic, the whole cone has slant height L = Rl/(R − r) and the small cone has slant height L − l = rl/(R − r), by similarity. So

CSA of frustum = πRL − πr(L − l) = π[R²l/(R − r) − r²l/(R − r)] = πl(R² − r²)/(R − r) = πl(R + r).

Curved surface area of a frustum = π(R + r)l, where l = √[h² + (R − r)²].

Total surface area of a frustum = π(R + r)l + πR² + πr², adding both circular ends. For an open container such as a bucket, only the bottom is added: CSA + πr² (or πR² if the larger end is the base).

Check: with r = R the CSA becomes 2πRl = 2πRh, a cylinder; with r = 0 it becomes πRl, a cone.

Worked example. The slant height of a frustum is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the CSA. 2πR = 18 and 2πr = 6, so π(R + r) = 12 and CSA = 12 × 4 = 48 cm².

Worked example. A fez cap is a frustum with radii 4 cm and 10 cm and slant height 15 cm, closed at the top (the smaller end) and open at the bottom. Material = CSA + area of the closed end = (22/7) × 14 × 15 + (22/7) × 16 = 660 + 50.29 = 710.29 cm².

Worked example. A bucket of height 24 cm has radii 15 cm and 5 cm (open at the top, the larger end). Slant height l = √(24² + 10²) = 26 cm. Sheet needed = CSA + bottom = (22/7) × 20 × 26 + (22/7) × 25 = (22/7) × 545 = 1712.86 cm². Its capacity, found earlier, is 8171.43 cm³.

Worked example. A metallic bucket of height 16 cm with end radii 20 cm and 8 cm: l = √(256 + 144) = 20 cm. Metal sheet = π(28)(20) + π(64) = 3.14 × (560 + 64) = 3.14 × 624 = 1959.36 cm²; at ₹8 per 100 cm² the cost is ₹156.75. The capacity is 10449.92 cm³ ≈ 10.45 L; milk at ₹20 per litre costs ₹209.

Worked example. A lampshade is a frustum of slant height 12 cm with end diameters 20 cm and 12 cm, open at both ends. Cloth needed = CSA = (22/7) × (10 + 6) × 12 = 603.43 cm².

Decide from the object which ends are closed: bucket (bottom only), glass (bottom only), lampshade (neither), a closed frustum-shaped container (both).

Many objects are hollow: pipes, rings, spherical shells, cylindrical tubes. Their material volume is the outer solid minus the inner cavity, and their surface area counts both the outer and inner surfaces plus any exposed edges.

Hollow cylinder (pipe). With external radius R, internal radius r and length h: volume of material = π(R² − r²)h; outer CSA = 2πRh; inner CSA = 2πrh; TSA = 2πRh + 2πrh + 2π(R² − r²) (the two ring-shaped ends).

Spherical shell. Volume of material = (4/3)π(R³ − r³). A hollow hemispherical bowl has inner CSA 2πr², outer CSA 2πR², and a ring rim π(R² − r²).

Worked example. A hemispherical bowl has internal diameter 10.5 cm and is 0.25 cm thick. Find the volume of steel. R = 5.5, r = 5.25. Volume = (2/3)(22/7)(5.5³ − 5.25³) = (44/21)(166.375 − 144.703) = (44/21)(21.672) = 45.41 cm³.

Worked example. A cylindrical pipe has internal diameter 4 cm, external diameter 4.4 cm and length 21 cm; find the volume of metal (π = 22/7). Volume = (22/7)(2.2² − 2²) × 21 = (22/7)(4.84 − 4)(21) = 66 × 0.84 = 55.44 cm³.

Worked example. A hollow sphere of internal and external diameters 4 cm and 8 cm is melted into a cone of base diameter 8 cm. Find the height. Volume = (4/3)π(64 − 8) = (4/3)π × 56 = ⅓π × 16 × h, so 16h = 224, h = 14 cm.

Worked example. A spherical shell of external radius 8 cm and internal radius 6 cm is melted into a cylinder of radius 8 cm. Volume = (4/3)π(512 − 216) = (4/3)π × 296; cylinder π × 64 × h gives h = 296 × 4/(3 × 64) = 6.17 cm.

Worked example. A cylindrical bucket of radius 18 cm and height 32 cm full of sand is emptied on the ground to form a conical heap of height 24 cm. Find the radius and slant height of the heap. π × 324 × 32 = ⅓πr² × 24, so r² = 324 × 32/8 = 1296, r = 36 cm, and l = √(36² + 24²) = √1872 = 12√13 ≈ 43.27 cm.

Worked example. A spherical ball of radius 3 cm is dropped into a cylindrical vessel of radius 6 cm containing water; the rise in water level equals the ball's volume ÷ the vessel's base area = (4/3)π × 27 ÷ (π × 36) = 36/36 = 1 cm. If instead 150 lead shots of radius 0.25 cm are dropped in, each has volume 0.0654 cm³, total 9.82 cm³, and in a vessel of radius 3.5 cm the rise is 9.82/38.5 ≈ 0.255 cm.

Displacement problems are equal-volume problems: the volume of the immersed object equals the volume of the water displaced, which is the base area of the vessel times the rise in level.

Once a surface area or a volume is found, the examination usually asks for a practical quantity: the cost of paint or canvas, the capacity in litres, the number of small items that can be made or packed, or the mass from a density. These are one-step conversions, but they carry marks and need care with units.

Cost of painting or sheeting. Cost = area × rate. Only the surfaces actually painted count: the inside of a bowl, the outside of a tank, the canvas of a tent without its floor.

Worked example. A tent of cylindrical part height 2.1 m, diameter 4 m and conical top of slant height 2.8 m needs 44 m² of canvas; at ₹500 per m² it costs ₹22000.

Worked example. A hemispherical dome of a building is to be painted; if the circumference of its base is 17.6 m, find the cost at ₹5 per 100 cm², that is, ₹500 per m². 2πr = 17.6 gives r = 2.8 m. CSA = 2 × (22/7) × 7.84 = 49.28 m². Cost = 49.28 × 500 = ₹24640.

Capacity. Capacity is the internal volume, in litres: divide cm³ by 1000 or multiply m³ by 1000.

Worked example. A conical vessel of radius 5 cm and height 24 cm is full of water; the water is emptied into a cylindrical vessel of radius 10 cm. The height of water = ⅓ × 25 × 24 ÷ 100 = 200/100 = 2 cm.

Number of items. The number of solids that can be made from a given volume, or that fit into a container by volume, is the quotient of the volumes, rounded down to a whole number when items must be complete.

Worked example. How many silver coins of diameter 1.75 cm and thickness 2 mm must be melted to form a cuboid 5.5 cm × 10 cm × 3.5 cm? Coin volume = (22/7) × (0.875)² × 0.2 = 0.48125 cm³; cuboid = 192.5 cm³; number = 192.5/0.48125 = 400.

Worked example. A cylindrical bucket of radius 6 cm and height 8 cm is full of ice cream, to be served in cones of radius 3 cm and height 12 cm, each with a hemispherical top. Bucket = π × 36 × 8 = 288π; each cone with top = ⅓π × 9 × 12 + ⅔π × 27 = 36π + 18π = 54π; number = 288/54 = 5.33, so 5 full cones can be served. With a cylinder of radius 6 cm and height 15 cm instead, the volume is 540π and the number is exactly 10.

Mass. Mass = volume × density; for iron about 8 g/cm³, so a pole of volume 111532.8 cm³ has mass 892.26 kg.

Worked example. A cistern 150 cm × 120 cm × 110 cm holds 129600 cm³ of water; bricks 22.5 cm × 7.5 cm × 6.5 cm each absorb one-seventeenth of their own volume of water. How many bricks can be put in without water overflowing? Let n bricks be placed. Cistern volume = 1980000; water plus bricks minus absorbed water must not exceed the cistern: 129600 + n × 1096.875 × (16/17) ≤ 1980000, so n ≤ 1850400 × 17/(16 × 1096.875) = 1792.4, hence 1792 bricks.

Read carefully whether the question wants complete items, and whether the unit of the rate is per m² or per 100 cm².

Two full model answers show the layout the board expects: a figure or a clear listing of the parts, the formula named, the substitution, the arithmetic, and a final statement with units.

Model answer 1. A solid is in the shape of a cone standing on a hemisphere with both their radii equal to 1 cm and the height of the cone equal to its radius. Find the volume of the solid in terms of π.
The solid consists of a cone of radius r = 1 cm and height h = 1 cm, and a hemisphere of radius 1 cm.
Volume of the cone = ⅓πr²h = ⅓π × 1 × 1 = π/3 cm³.
Volume of the hemisphere = ⅔πr³ = 2π/3 cm³.
Volume of the solid = π/3 + 2π/3 = π cm³.

Model answer 2. A container, open from the top and made of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at ₹20 per litre, and the cost of the metal sheet at ₹8 per 100 cm². (Use π = 3.14.)
Here h = 16 cm, R = 20 cm, r = 8 cm.
Volume = ⅓πh(R² + Rr + r²) = ⅓ × 3.14 × 16 × (400 + 160 + 64) = ⅓ × 3.14 × 16 × 624 = 10449.92 cm³ = 10.44992 litres.
Cost of milk = 10.45 × 20 = ₹209 (approximately).
Slant height l = √[h² + (R − r)²] = √(256 + 144) = √400 = 20 cm.
Area of sheet = CSA + area of the bottom = π(R + r)l + πr² = 3.14 × 28 × 20 + 3.14 × 64 = 1758.4 + 200.96 = 1959.36 cm².
Cost of sheet = 1959.36 × 8/100 = ₹156.75.

Model answer 3. A juice seller serves juice in glasses shaped like a cylinder of inner diameter 5 cm and height 10 cm with a hemispherical raised bottom. Find the apparent capacity and the actual capacity (π = 3.14).
Apparent capacity = πr²h = 3.14 × 6.25 × 10 = 196.25 cm³.
Volume of the hemispherical bulge = ⅔πr³ = ⅔ × 3.14 × 15.625 = 32.71 cm³.
Actual capacity = 196.25 − 32.71 = 163.54 cm³.

Common errors. Using diameter for radius; forgetting that the height of a hemisphere is its radius when finding the height of a cylinder in a capsule; including hidden surfaces in surface area; using h instead of l in the curved surface of a cone or frustum; forgetting ⅓ in the cone; mixing cm and m; writing cm² for a volume; rounding π-terms too early.

Quick reference. Cone on hemisphere: V = ⅓πr²(h + 2r), TSA = πr(l + 2r). Cylinder with two hemispheres: V = πr²(h + 4r/3), TSA = 2πr(h + 2r). Frustum: V = ⅓πh(R² + Rr + r²), CSA = π(R + r)l, l = √[h² + (R − r)²]. Conversion: equate volumes. Pipe flow: πr² × speed × time.

A final set of ideas ties the chapter to similarity and prepares you for mixed questions in the examination.

Scaling. If every length of a solid is multiplied by k, every area is multiplied by k² and every volume by k³. Doubling the radius of a sphere multiplies its surface area by 4 and its volume by 8. This is why a cone cut by a plane parallel to its base at half its height removes only one-eighth of the volume: the small cone is similar with k = ½, so its volume is (½)³ = ⅛ of the whole, and the frustum has ⅞ of the volume.

Worked example. A cone of height 30 cm is cut by a plane parallel to the base so that the volume of the small cone at the top is 1/27 of the whole cone. The ratio of heights is the cube root of 1/27, i.e. 1/3, so the cut is 10 cm from the vertex and 20 cm above the base.

Worked example. Two cones have the same height and their radii in the ratio 2 : 3; their volumes are in the ratio 4 : 9. Two spheres have surface areas in the ratio 16 : 25; their radii are in the ratio 4 : 5 and their volumes 64 : 125.

Worked example. The ratio of the volumes of a cylinder and a cone of equal radius and equal height is 3 : 1; a hemisphere and a cone of equal radius and height equal to the radius have volumes in the ratio 2 : 1.

Mixed question 1. A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water; a solid cone of base diameter 7 cm and height 6 cm is completely immersed. Find the volume of water displaced and the volume left. Displaced = cone = ⅓ × (22/7) × 12.25 × 6 = 77 cm³. Vessel = (22/7) × 25 × 10.5 = 825 cm³. Left = 825 − 77 = 748 cm³.

Mixed question 2. A cube of side 4 cm has a hemispherical depression of diameter 4 cm. Surface area of the remaining solid = 6 × 16 + π × 4 = 96 + 12.57 = 108.57 cm² (π = 22/7 gives 96 + 88/7 = 108.57).

Mixed question 3. A solid toy is a hemisphere surmounted by a cone; the height of the cone is 2 cm and the diameter of the base is 4 cm. Find the volume of the toy, and if a right circular cylinder circumscribes the toy, the difference of the volumes. Toy = ⅓π × 4 × 2 + ⅔π × 8 = 8π/3 + 16π/3 = 8π = 25.14 cm³. Cylinder of radius 2 and height 4: 16π = 50.29 cm³. Difference = 8π = 25.14 cm³.

Mixed question 4. A wooden article is a cylinder of height 10 cm and radius 3.5 cm with a hemisphere scooped out of each end. TSA = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 4πr² = 2πr(h + 2r) = 2 × (22/7) × 3.5 × 17 = 374 cm². Note that scooping out hemispheres removes the two flat ends and adds two curved surfaces, exactly like adding hemispheres.

Mixed question 5. A rocket is a cylinder of radius 2.5 m and height 21 m closed at the bottom and a cone of the same radius and height 6 m. If the whole outside is painted, the area is 2πrh + πrl + πr² with l = 6.5: (22/7) × 2.5 × (42 + 6.5 + 2.5) = (22/7) × 2.5 × 51 = 400.71 m².

With the formulas of the basic solids, the two rules for combinations, the equal-volume principle and the frustum formulas, every problem in this chapter can be solved.