5.16H Spheres and Cones: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 55 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Work out the volume of a sphere with radius 5 cm. Give your answer in terms of π.

(Total for Question 1 is 2 marks)

A metal ball has diameter 12 cm. Calculate the surface area of the ball. Give your answer in terms of π.

(Total for Question 2 is 2 marks)

A cone has radius 3 cm and height 10 cm. Work out the volume of the cone. Give your answer in terms of π and then to 3 significant figures.

(Total for Question 3 is 3 marks)

A hemisphere of radius 4 cm is fixed on top of a right circular cylinder of radius 4 cm and height 6 cm. Calculate the total volume of the solid. Give your answer in terms of π and then to 3 significant figures.

(Total for Question 4 is 4 marks)

Two cones are similar. Cone A has height 9 cm and volume 216 cm³. Cone B has height 15 cm. Work out the volume of cone B.

(Total for Question 5 is 3 marks)

A right cone has height 12 cm. Its volume is 432 π cm³. Find the radius of the base of the cone. Give your answer to 3 significant figures.

(Total for Question 6 is 3 marks)

A frustum is formed by cutting the top from a right cone. The lower radius is 6 cm, the upper radius is 2 cm and the height of the frustum is 9 cm. Calculate the volume of the frustum. Give your answer in terms of π and then to 3 significant figures.

(Total for Question 7 is 4 marks)

Mark scheme · 5.16H Spheres and Cones: Higher Tier Practice

Question 1

M1 substitute into V = 4/3 π r³, showing 4/3 * π * 5³ or equivalent method
A1 500/3 π cm³ cao
Answer: 500/3 π cm³ (approx 523.6 cm³)

Question 2

M1 use A = 4 π r² with r = 6 cm or equivalent method
A1 144 π cm² cao
Answer: 144 π cm² (approx 452.39 cm²)

Question 3

M1 substitute into V = 1/3 π r² h, showing 1/3 * π * 3² * 10 or equivalent
M1 simplify to 30 π
A1 30 π cm³; 94.2 cm³ to 3 s.f. cao
Answer: 30 π cm³ (approx 94.2 cm³)

Question 4

M1 calculate hemisphere volume = 1/2 * 4/3 π r³ = 2/3 π r³ with r = 4
M1 calculate cylinder volume = π r² h with r = 4, h = 6
A1 sum volumes: (128/3 + 96) π = 416/3 π cm³ ft
A1 numeric to 3 s.f.: 436 cm³ (awrt 435.63 cm³) allow 436 cm³
Answer: 416/3 π cm³ (approx 435.63 cm³, or 436 cm³ to 3 s.f.)

Question 5

M1 find linear scale factor = 15/9 = 5/3 and state volume scale = (5/3)³ = 125/27
M1 multiply 216 by 125/27 or equivalent method
A1 1000 cm³ cao
Answer: 1000 cm³

Question 6

M1 use V = 1/3 π r² h and substitute 432 π = 1/3 π r² * 12 to give 432 = 4 r²
M1 solve r² = 108 and r = √108
A1 r = 6 √3 cm (approx 10.39 cm to 3 s.f.) cao
Answer: 6 √3 cm (approx 10.4 cm to 3 s.f.)

Question 7

M1 use frustum volume formula V = 1/3 π h (R² + Rr + r²) with R = 6, r = 2, h = 9
M1 substitute values to get 1/3 π * 9 * (36 + 12 + 4) or equivalent simplification
A1 simplify to 156 π cm³
A1 numeric to 3 s.f.: 490 cm³ (awrt 490.09 cm³) cao
Answer: 156 π cm³ (approx 490 cm³ to 3 s.f.)

Spheres and Cones: Higher Tier Practice - Worksheets, Questions and Revision