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GCSE · Geometry and Measures

5.16D Spheres and Cones (Volume and Surface Area): Fluency and Exam Drill

EDEXCEL 1MA1 · Calculator allowed · about 80 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Formulae You May Need

Revision Library formulae reminder, styled on Edexcel 1MA1 formulae-sheet conventions; wording is original.

Volume of a sphere = (4/3) x pi x r^3

Surface area of a sphere = 4 x pi x r^2

Volume of a cone = (1/3) x pi x r^2 x h

Curved surface area of a cone = pi x r x l (l = slant height)

Volume of a hemisphere = (2/3) x pi x r^3

Total surface area of a solid hemisphere = 3 x pi x r^2

Note: on the Edexcel formulae sheet, the volume formulae for the sphere and the cone are given. The surface area formulae are not given and should be recalled or derived.

A solid sphere has radius 10 cm. Work out the volume of the sphere. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 1 is 2 marks)

A sphere has radius 4.5 cm. Calculate the surface area of the sphere. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 2 is 2 marks)

A cone has base radius 7 cm and perpendicular height 15 cm. Work out the volume of the cone. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 3 is 2 marks)

A sphere has radius 3 cm. Work out the volume of the sphere. Give your answer as a multiple of π.

Diagram NOT accurately drawn

(Total for Question 4 is 2 marks)

A cone has base radius 5 cm and perpendicular height 8 cm. Calculate the curved surface area of the cone. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 5 is 3 marks)

A cone has base radius 4 cm and perpendicular height 7 cm. Calculate the total surface area of the cone (the curved surface area plus the area of the circular base). Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 6 is 3 marks)

A solid hemisphere has radius 11 cm. Work out the volume of the hemisphere. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 7 is 2 marks)

A solid hemisphere has radius 12 cm. Calculate the total surface area of the hemisphere (the curved surface plus the flat circular face). Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 8 is 2 marks)

A pottery bowl is made in the shape of an open solid hemisphere with radius 6.5 cm. The flat circular top is open, so it is not coated. The outside curved surface of the bowl is coated with a glaze that costs 1.5p per cm^2. Work out the total cost of glazing the outside curved surface of the bowl.

Diagram NOT accurately drawn

(Total for Question 9 is 3 marks)

A solid sphere has volume 1000 cm^3. Work out the radius of the sphere. Give your answer correct to 3 significant figures.

(Total for Question 10 is 3 marks)

A cone has volume 250 cm³ and base radius 6 cm. Work out the perpendicular height of the cone. Give your answer correct to 3 significant figures.

(Total for Question 11 is 3 marks)

A solid toy is made from a cone joined to a hemisphere of the same radius, as shown. The cone has base radius 5 cm and perpendicular height 11 cm. The hemisphere also has radius 5 cm and is joined to the circular base of the cone, flat faces together.

Diagram NOT accurately drawn

(a)Work out the total volume of the solid toy. Give your answer correct to 3 significant figures.(4)

(b)Work out the total surface area of the solid toy. The toy has no flat faces on the outside, since the flat circular faces where the cone and hemisphere join are hidden inside the solid. Give your answer correct to 3 significant figures.(5)

(Total for Question 12 is 9 marks)

Elin is a jeweller. She casts a solid silver charm in the shape of a sphere with radius 2.2 cm. The density of the silver she uses is 10.49 g/cm^3. Work out the mass of the charm. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 13 is 3 marks)

Kwame, a metalworker, melts down a solid bronze cone with base radius 9 cm and perpendicular height 14 cm. He recasts all of the bronze as a solid sphere, with no bronze wasted. Work out the radius of the sphere. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 14 is 4 marks)

A sphere and a cone both have radius r. The volume of the sphere is exactly twice the volume of the cone. Show that the perpendicular height of the cone, h, satisfies h = 2r.

(Total for Question 15 is 3 marks)

A spinning top is made from a cone joined to a hemisphere of the same radius. The hemisphere has radius 7 cm, and the total height of the spinning top (hemisphere plus cone) is 21 cm. Work out the total volume of the spinning top. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 16 is 4 marks)

A solid sphere has volume 500 cm^3. Calculate the surface area of the sphere. Give your answer correct to 3 significant figures.

(Total for Question 17 is 4 marks)

The surface area of a sphere with radius r is equal to the curved surface area of a cone with the same radius r and slant height l. Show that l = 4r.

(Total for Question 18 is 3 marks)

Two spheres are mathematically similar. Their radii are in the ratio 2:3. The volume of the smaller sphere is 96 cm^3. Work out the volume of the larger sphere.

(Total for Question 19 is 3 marks)

A farm has two storage tanks. Tank A is a sphere with radius 2.5 m. Tank B is a cone with base radius 3 m and perpendicular height 5 m. Work out which tank has the greater volume, and by how much. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 20 is 5 marks)

Mark scheme · 5.16D Spheres and Cones (Volume and Surface Area): Fluency and Exam Drill

Question 1

M1 correct substitution into V = (4/3) x π x r³, e.g. (4/3) x π x 10³ (oe)
A1 awrt 4190 cm³ (cao)
Answer: 4190 cm³ (3 s.f.)

Question 2

M1 correct substitution into A = 4 x π x r², e.g. 4 x π x 4.5² (oe)
A1 awrt 254 cm² (cao)
Answer: 254 cm² (3 s.f.)

Question 3

M1 correct substitution into V = (1/3) x π x r² x h, e.g. (1/3) x π x 7² x 15 (oe)
A1 awrt 770 cm³ (cao)
Answer: 770 cm³ (3 s.f.)

Question 4

M1 correct substitution (4/3) x π x 3³ (oe)
A1 36 x π cm³ (oe, cao - exact value required, not a decimal)
Answer: 36 x π cm³ (exact)

Question 5

M1 slant height l = √5² + 8² = √89 (Pythagoras)
M1 correct substitution into CSA = π x r x l, e.g. π x 5 x √89 (oe, ft their l)
A1 awrt 148 cm² (cao)
Answer: 148 cm² (3 s.f.)

Question 6

M1 slant height l = √4² + 7² = √65 (Pythagoras)
M1 total SA = π x r x l + π x r², e.g. π x 4 x √65 + π x 4² (oe, ft their l)
A1 awrt 152 cm² (cao)
Answer: 152 cm² (3 s.f.)

Question 7

M1 correct substitution into V = (2/3) x π x r³, e.g. (2/3) x π x 11³ (oe)
A1 awrt 2790 cm³ (cao)
Answer: 2790 cm³ (3 s.f.)

Question 8

M1 correct substitution into total SA = 3 x π x r², e.g. 3 x π x 12² (oe, or curved 2 x π x r² + base π x r² shown separately)
A1 awrt 1360 cm² (cao)
Answer: 1360 cm² (3 s.f.)

Question 9

M1 curved SA = 2 x π x r², e.g. 2 x π x 6.5² (oe), awrt 265 cm²
M1 cost = curved SA x 0.015 (oe, ft their curved SA)
A1 £3.98 (awrt, cao, to the nearest penny)
Answer: £3.98

Question 10

M1 rearranges V = (4/3) x π x r³ to r³ = 3V / (4 x π) (oe)
M1 r³ = 3 x 1000 / (4 x π) = awrt 239 (or unrounded 238.7)
A1 awrt 6.20 cm (cao)
Answer: 6.20 cm (3 s.f.)

Question 11

M1 rearranges V = (1/3) x π x r² x h to h = 3V / (π x r²) (oe)
M1 h = 3 x 250 / (π x 6²) (oe substitution)
A1 awrt 6.63 cm (cao)
Answer: 6.63 cm (3 s.f.)

Question 12

(a) M1 volume of cone = (1/3) x π x 5² x 11 (oe), awrt 288 cm³
(a) M1 volume of hemisphere = (2/3) x π x 5³ (oe), awrt 262 cm³
(a) M1 adds the two volumes (ft their values)
(a) A1 awrt 550 cm³ (cao)
(a) Answer: 550 cm³ (3 s.f.)
(b) M1 slant height of cone l = √5² + 11² = √146 (Pythagoras), awrt 12.1 cm
(b) M1 curved SA of cone = π x r x l, e.g. π x 5 x √146 (oe, ft their l), awrt 190 cm²
(b) M1 curved SA of hemisphere = 2 x π x r², e.g. 2 x π x 5² (oe), awrt 157 cm²
(b) M1 adds the two curved surface areas (ft their values)
(b) A1 awrt 347 cm² (cao)
(b) Answer: 347 cm² (3 s.f.)

Question 13

M1 volume = (4/3) x π x 2.2³ (oe), awrt 44.6 cm³
M1 mass = volume x 10.49 (ft their volume)
A1 awrt 468 g (oe 0.468 kg, cao)
Answer: 468 g (3 s.f.), i.e. 0.468 kg

Question 14

M1 volume of cone = (1/3) x π x 9² x 14 (oe), awrt 1190 cm³
M1 sets (4/3) x π x r³ equal to their cone volume (oe)
M1 rearranges to r³ = awrt 284 (or exact 283.5) (ft their cone volume)
A1 awrt 6.57 cm (cao)
Answer: 6.57 cm (3 s.f.)

Question 15

M1 correct equation formed: (4/3) x π x r³ = 2 x (1/3) x π x r² x h (oe)
M1 correct algebraic simplification, e.g. divides both sides by (1/3) x π x r² to reach 4r = 2h (oe)
A1 h = 2r fully shown with no errors (cso)
Answer: h = 2r (shown)

Question 16

M1 height of cone = 21 - 7 = 14 cm (oe)
M1 volume of cone = (1/3) x π x 7² x 14 (oe), awrt 718 cm³
M1 volume of hemisphere = (2/3) x π x 7³ (oe), awrt 718 cm³
A1 awrt 1440 cm³ (cao, ft sum of their two volumes)
Answer: 1440 cm³ (3 s.f.)

Question 17

M1 rearranges V = (4/3) x π x r³ to r³ = 3V / (4 x π) (oe)
M1 r = awrt 4.92 cm (unrounded value should be carried forward if possible)
M1 substitutes their r into SA = 4 x π x r² (ft)
A1 awrt 305 cm² (cao; accept awrt 304-305 depending on rounding of r)
Answer: 305 cm² (3 s.f.)

Question 18

M1 correct equation formed: 4 x π x r² = π x r x l (oe)
M1 divides both sides by π x r, r > 0 (oe)
A1 l = 4r fully shown with no errors (cso)
Answer: l = 4r (shown)

Question 19

M1 volume scale factor = (3/2)³ (oe, from linear scale factor 3/2)
M1 volume scale factor = 27/8 (= 3.375) (oe)
A1 324 cm³ (cao)
Answer: 324 cm³

Question 20

M1 volume of sphere = (4/3) x π x 2.5³ (oe), awrt 65.4 or 65.5 m³
M1 volume of cone = (1/3) x π x 3² x 5 (oe), awrt 47.1 m³
M1 valid method to compare the two volumes, e.g. subtracts the smaller from the larger
A1 awrt 18.3 m³ (cao)
B1 correct conclusion stated: Tank A (the sphere) has the greater volume
Answer: Tank A (the sphere) has the greater volume, by awrt 18.3 m^3.

Spheres and Cones (Volume and Surface Area): Fluency and Exam Drill - Worksheets, Questions and Revision

5.16D Spheres and Cones (Volume and Surface Area): Fluency and Exam Drill

Formulae You May Need

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20