1. The Sphere
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2. The volume and the surface area of a sphere are given by:
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3. Exam
Question
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4. c c r c c 4 3 V = 256π 3 V = V = 268 cm3 [3 s.f.] S = 4 64π S = S =
A tennis ball can be modelled as a sphere with diameter 8 cm
Calculate:
1. The volume of the ball, correct to 3 significant figures
2. The surface area of the ball, correct to 3 significant figures 4 3
x π
V =
x 43 c
256π 3
V = c
4 cm r
8 cm
V =
268
cm3
[3 s.f.]
x π
S = 4
x 42 c
64π
S = c
S =
201
cm2
[3 s.f.]
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5. Exam Question
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6. Find the volume of this compound shape in terms of π
Using Pythagoras Theorem: 13 12 5 r
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7. Find the volume of this compound shape in terms of π
The volume of the cone: 13 12 5
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8. Find the volume of this compound shape in terms of π
The volume of the semi-sphere:
Volume of a sphere 13 12 5
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9. Find the volume of this compound shape in terms of π
Total volume of the object 13 12 5
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11. The surface area of a sphere is given by:
Calculate the surface area of a semi-sphere of radius 6 cm, correct to 3 significant figures.
The surface area of a sphere is given by:
S = 4πr 2
The area of a circle is given by:
6 cm
A = πr 2
The surface area of a semi-sphere consists of:
the area of a circle
plus the curved surface area of a semi-sphere
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12. Calculate the surface area of a semi-sphere of radius 6 cm, correct to 3 significant figures.
The surface area of a sphere is given by:
S = 4πr 2
The area of a circle is given by:
6 cm
A = πr 2 π
x 62 =
36π
108π
≈ 339 cm2
x π 1 2 4
x 62 x =
72π
[3 s.f.]
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13. Exam
Question
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14. One of the two sections of a sand timer can be thought of as a semi sphere connected to a cylinder which is in turn connected to a cone.
All three solids share the same axis of symmetry.
The radius of these solids is 3 cm and the heights of the cylinder and the cone are 9 cm and 3 cm respectively.
Calculate the volume of the sand timer in terms of π. 3 4 3
x π
x 33 x 1 2
volume of a sphere 4 3
V = π
r 3
18π 9 3 π
x 32
x 9 3
volume of a cylinder
V =
r 2 h
81π π 3 1 3
x π
x 33
x 3
volume of a cone 1 3
V = π
r 2 h
27π
Measurements in cm
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15. One of the two sections of a sand timer can be thought of as a semi sphere connected to a cylinder which is in turn connected to a cone.
All three solids share the same axis of symmetry.
The radius of these solids is 3 cm and the heights of the cylinder and the cone are 9 cm and 3 cm respectively.
Calculate the volume of the sand timer in terms of π.
252π ≈ 792 cm3 3 4 3
x π
x 33 x 1 2
126π
18π 9 3 π
x 32
x 9 3
81π
126π 3 1 3
x π
x 33
x 3
27π
Measurements in cm
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16. Exam
Question
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17. A cylindrical bucket of radius 9 cm contains some water.
A metal sphere is gently lowered into the water, raising its level by 6 cm which completely covers the sphere.
Calculate the radius of the sphere to the nearest cm.
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18. A cylindrical bucket of radius 9 cm contains some water.
A metal sphere is gently lowered into the water, raising its level by 6 cm which completely covers the sphere.
Calculate the radius of the sphere to the nearest cm.
9 cm π
x 92
x 6 = 4 3
x π
x r 3 c
6 cm
4r 3 3
486 = c r
= 1458 c
4r 3
= 364.5 c
r 3 = c r 3
volume of a cylinder
V =
r 2 h
r ≈ 7 cm π
volume of a sphere 4 3
V = π
r 3
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20. The figure below shows the cross section of a cylinder of base radius a with 3 spheres fitting “snugly” inside it.
1. Calculate the volume of the cylinder in terms of a.
2. Write down the volume of the three spheres.
3. Hence calculate what fraction of the cylinder is empty
volume of a cylinder
Vc =
V 3s =
Empty =
6πa 3
4πa 3
2πa 3
V =
r 2 h π
Vc =
a 2
x 6a π
Vc = 6
a 3 π 6a
volume of a sphere
Empty Vc
2πa 3
6πa 3 1 3 = = 4 3
V = π
r 3 4 3
V3s = π
a 3
x 3
What fraction would it be if 5 spheres fitted snugly inside a cylinder with the appropriate dimensions? a
V3s = 4
a 3 π
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22. A metal ball bearing can be modelled as a sphere of radius 2.5 cm.
The ball bearing is to be melted and remoulded into the shape of a cone with a height of 4.8 cm.
Calculate the radius of the base of the cone, in cm correct to 2 significant figures.
volume of the sphere 4 3 4 3
V = π
r 3 =
x π
x 2.53 ≈
cm3
volume of the cone 1 3 π 1 3
V =
r 2 h =
x π
x r 2
x 4.8 ≈
r 2
cm3
5.0265
r 2 = c
r 2 = c r = c r
= 3.6 cm
[2 s.f.]
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24. A sphere with a radius of α cm has the same volume as a cone with a base radius of α cm.
Find the height of the cone, in terms of α.
volume of the sphere
volume of the cone 4 3
V = π 1 3
α 3
V = π
α 2 h 4 3 1 3 π
α 3 = π
α 2 h c 4 3 1 3
α 3 =
α 2 h c 4 3 1 3 α = h c 3x x3 4α = h
h = 4α , i.e. the height of the cone is 4 times the base radius
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26. The Earth can be modelled as a sphere of radius m and it is estimated that m2 are covered by sea.
1. Write in standard form.
2. Calculate the surface area of the Earth in m2, giving your
answer in standard form correct to 3 significant figures.
3. Calculate what percentage of the Earth’s surface is covered
by sea, giving your answer to the nearest percentage. =
3.6
x 1014
The surface area of a sphere is given by:
S = 4πr 2
x π
S = 4 x ≈
5.15
x 1014 m2
3.6 x 1014
x 100 ≈
70%
5.15 x 1014
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