HKDSE Mathematics (Compulsory Part): Mensuration And Similar Solids Practice Paper

HKDSE Mathematics (Compulsory Part): Mensuration and Similar Solids Practice Paper

2026-11-03

HKDSE Mathematics timed practice

This paper covers mensuration and similar solids for S6 mathematics, including cones, spheres, frustums, composite solids and measurement error.

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Paper details　HKDSE Mathematics Compulsory Part, mensuration and similar solids, Paper 1 style, 75 minutes

Section A (Questions 1 to 4, 26 marks)

Read each question and write the correct answer with an explanation. Show working and units for any calculation.

#	Question	Answer and marking notes
1	A cone has base radius 5 cm and height 12 cm. Find its slant height, curved surface area, total surface area and volume, giving answers in terms of pi and to one decimal place.	Slant height = square root of (25 + 144) = 13 cm. Curved area = pi × 5 × 13 = 65 pi, about 204.2 cm squared. Total area = 65 pi + 25 pi = 90 pi, about 282.7 cm squared. Volume = one third × pi × 25 × 12 = 100 pi, about 314.2 cm cubed. Marking: slant height 2, curved area 1, total area 1, volume 2.
2	Two cylinders are similar with a ratio of corresponding lengths of 2 : 3. The smaller has volume 64 cm cubed and total surface area 72 cm squared. Find the volume and total surface area of the larger, and state the ratio rules used.	Volumes are in the ratio 8 : 27, so the larger volume is 64 × 27 divided by 8 = 216 cm cubed. Areas are in the ratio 4 : 9, so the larger area is 72 × 9 divided by 4 = 162 cm squared. Marking: volume ratio 2, area ratio 2, each answer 1.
3	A solid cylinder of base radius 7 cm and height 20 cm is melted and recast into small spheres of radius 3 cm. Find the greatest number of spheres that can be made and the volume of metal left over.	Cylinder volume = pi × 49 × 20 = 980 pi cm cubed. Each sphere is four thirds × pi × 27 = 36 pi cm cubed. 980 pi divided by 36 pi = 27.2, so 27 spheres can be made. Left over = 980 pi minus 972 pi = 8 pi, about 25.1 cm cubed. Marking: cylinder 2, sphere 2, count 2, remainder 1.
4	A right prism has a trapezium cross section with parallel sides 8 cm and 14 cm and height 6 cm. The prism is 25 cm long. Find the cross sectional area, the volume, and the mass in kg if the density is 2.7 g per cm cubed.	Cross section = (8 + 14) divided by 2 × 6 = 66 cm squared. Volume = 66 × 25 = 1650 cm cubed. Mass = 1650 × 2.7 = 4455 g = 4.455 kg. Marking: area 2, volume 2, mass with unit conversion 3.

Section B (Questions 5 to 8, 32 marks)

Answer the following questions with working, units and evidence drawn from the data given.

#	Question	Answer and marking notes
5	A rectangle measures 12.4 cm by 8.6 cm, each correct to the nearest 0.1 cm. Find the greatest possible area and the percentage error of that area compared with the measured area.	The greatest dimensions are 12.45 cm and 8.65 cm, so the greatest area is 107.6925 cm squared. The measured area is 12.4 × 8.6 = 106.64 cm squared. Percentage error = 1.0525 divided by 106.64 × 100 per cent, about 0.99 per cent. Marking: limits 2, greatest area 2, percentage error 2.
6	A solid metal sphere of radius 9 cm is melted and recast into a cone of base radius 9 cm. Find the height of the cone and the original surface area of the sphere.	Sphere volume = four thirds × pi × 729 = 972 pi cm cubed. With cone height h, one third × pi × 81 × h = 972 pi, so h = 36 cm. Surface area = 4 pi × 81 = 324 pi, about 1017.9 cm squared. Marking: sphere volume 2, equation 1, height 3, surface area 2.
7	At the same moment a person 1.8 m tall casts a shadow 2.4 m long, while a flagpole nearby casts a shadow 12 m long. Find the height of the flagpole. If the shadow of the pole becomes 15 m at another time, find the shadow of the person then, and explain the shared principle.	By similar triangles, pole height divided by 12 = 1.8 divided by 2.4, so the pole is 9 m tall. At the other time the shadow is 1.8 × 15 divided by 9 = 3.0 m. Both cases use the fact that sunlight rays are parallel at the same moment, so the person and the pole form similar triangles with proportional sides. Marking: proportion 3, height 3, second case 2, principle 1.
8	A large cone has base radius 9 cm and height 24 cm. A plane parallel to the base cuts off a small cone at 8 cm from the apex. Find the volumes of the large and small cones, the volume of the remaining frustum, and the ratio of the two cone volumes.	Large cone volume = one third × pi × 81 × 24 = 648 pi cm cubed. The small cone is similar with length ratio 8 : 24 = 1 : 3, so its volume is 648 pi divided by 27 = 24 pi cm cubed. Frustum = 624 pi, about 1960.4 cm cubed. The ratio of volumes is 27 : 1. Marking: large cone 2, small cone 2, frustum 2, ratio with reason 3.

Section C (Questions 9 to 12, 42 marks)

Write fuller answers using correct subject terms and support them with examples from Hong Kong or daily life.

#	Question	Answer and marking notes
9	A composite solid is formed by a cylinder of base radius 6 cm and height 10 cm with a hemisphere of the same radius on top. Find the volume and the total surface area including the base.	Cylinder = pi × 36 × 10 = 360 pi cm cubed, hemisphere = two thirds × pi × 216 = 144 pi cm cubed, total 504 pi, about 1583.4 cm cubed. Area = side 2 pi × 6 × 10 = 120 pi, dome 2 pi × 36 = 72 pi, base pi × 36 = 36 pi, total 228 pi, about 716.3 cm squared. Marking: each volume 2, each of three areas 2.
10	Two similar models have a ratio of corresponding lengths of 1 : 4. The smaller has surface area 200 cm squared and mass 1.5 kg. Find the surface area and mass of the larger. If one litre of paint covers 1000 cm squared and costs 18 dollars, find the paint cost for the larger model.	Areas are in the ratio 1 : 16, so the larger area is 200 × 16 = 3200 cm squared. Mass follows volume, ratio 1 : 64, so the larger mass is 1.5 × 64 = 96 kg. Paint needed = 3200 divided by 1000 = 3.2 litres, cost = 3.2 × 18 = 57.6 dollars. Marking: both ratios 2 each, mass 2, paint volume 2, cost 2.
11	A cylindrical tank has base radius 1.2 m and height 2 m. Water flows in at 15 litres per minute. Find the capacity in litres, the time to fill it in minutes and hours, and the time to reach three quarters of the capacity.	Capacity = pi × 1.44 × 2 = 2.88 pi, about 9.048 m cubed, which is 9047.8 litres. Time to fill = 9047.8 divided by 15, about 603.2 minutes or 10.05 hours. Three quarters takes 9047.8 × 0.75 divided by 15, about 452.4 minutes or 7.54 hours. Marking: volume 2, conversion 2, full time 3, three quarters 3, units 1.
12	A closed cylinder has a fixed volume of 500 cm cubed. Find the base radius and height that make the total surface area a minimum, find that minimum area, and state the relation between height and base radius at the minimum.	With base radius r and height h, pi r squared h = 500, so the surface area is S = 2 pi r squared + 1000 divided by r. Setting dS by dr to zero gives 4 pi r minus 1000 divided by r squared equals zero, so r cubed = 250 divided by pi, r is about 4.30 cm and h is about 8.60 cm. The minimum area is 2 pi × 18.5 + 1000 divided by 4.30, about 348.7 cm squared. At the minimum h = 2r, so the height equals the diameter. Marking: relation 3, r and h 2 each, minimum area 3, relation stated 1.

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