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2. Axis Vertex Generator Radius Base Slant Height C y l i n d e r s a n d C o n e s

3. © T Madas Volume of a Cylinder

4. The Volume of a Cylinder r h A cylinder is a P whose cross section is a c rism ircle V = Base AreaxHeight = πr 2πr 2 xh

5. © T Madas V = π r 2r 2 h π x 10 2 x 40 V ≈12566cm 3 [n. w. n.] x V = x-sectional Areaheight x Calculate the volume of these cylinders 10 cm 40 cm V = π r 2r 2 h π x 52x 52 x 22 V ≈1728m3m3 [n. w. n.] x V = x-sectional Areaheight x 5 m 22 m

6. © T Madas V = π r 2r 2 h π x 12 2 x 35 V ≈15834cm 3 [n. w. n.] x Calculate the volume of these cylinders 12 cm 35 cm V = π r 2r 2 h π x 42x 42 x 18 V ≈905m3m3 [n. w. n.] x 4 m 18 m

7. © T Madas Surface Area of a Cylinder

8. The Surface Area of a Cylinder

9. © T Madas S = π r h S ≈3142cm 2 [n. w. n.] + Calculate the Surface Area of these Cylinders 10 cm 40 cm 5 m 22 m 2 π r 2r 2 2 x πx π x 10 x 40+ 2 x πx π x 10 2 2 S ≈ 2513.27 628.32 + S = π r h S ≈848m2m2 [n. w. n.] + 2 π r 2r 2 2 x πx π x 5 x 22+ 2 x πx π x 5 2 2 S ≈ 691.15 157.08 +

10. © T Madas S = π r h S ≈6126cm 2 [n. w. n.] + Calculate the Surface Area of these Cylinders 15 cm 50 cm 4 m 14 m 2 π r 2r 2 2 x πx π x 15 x 50+ 2 x πx π x 15 2 2 S ≈ 4712.39 1413.72 + S = π r h S ≈452m2m2 [n. w. n.] + 2 π r 2r 2 2 x πx π x 4 x 14+ 2 x πx π x 4 2 2 S ≈ 351.86 100.53 +

11. © T Madas Volume of a Cone

12. The Volume of a Cone h It can be shown that for a cone: V =Base AreaxHeight r

13. © T Madas 1313 x πx π x 4 2 x 9 Calculate the volume of these cones 9 cm 4 cm 8 cm 6 cm V = π r 2r 2 h 1313 = 151 ≈ cm 3 1313 x πx π x 6 2 x 8 V = π r 2r 2 h 1313 = 302 ≈ cm 3

14. © T Madas 1313 x πx π x 3 2 x 12 Calculate the volume of these cones 12 cm 3 cm 9 cm 5 cm V = π r 2r 2 h 1313 = 113 ≈ cm 3 1313 x πx π x 5 2 x 9 V = π r 2r 2 h 1313 = 236 ≈ cm 3

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16. h 8.84 cm A cylinder is shown below. The radius of its base is 6 cm and has a volume of 1000 cm 3. Calculate its surface area to 3 significant figures. 6 cm Volume = base area x height 1000 =6 x 6 x hx h 1000 ≈ 113.1 x hx h h ≈h ≈ 1000 ÷ 113.1 h ≈ 8.84 cm x πx π

17. © T Madas 8.84 cm A cylinder is shown below. The radius of its base is 6 cm and has a volume of 1000 cm 3. Calculate its surface area to 3 significant figures. Surface Area: 6 x 6 ≈ 226.2 cm 2 x πx π 6 cm x 2 8.84 cm Circumference of circle 2 x 6 ≈ 333.3 cm 2 x πx π x 8.84 559.5 cm 2 560 cm 2 [ 3 s.f. ]

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19. A tank without a lid is in the shape of a cylinder. The radius of its base is 30 cm and has a capacity of 225 litres. Calculate its surface area, to 2 significant figures. h 79.6 cm 30 cm Volume = base area x height 225000 = 30 x 30 x hx h 225000 ≈ 2827 x hx h h ≈h ≈ 225000 ÷ 2827 h ≈ 79.6 cm x πx π 1 litre = 1000 cm 3 225 litres = 225000 cm 3

20. © T Madas A tank without a lid is in the shape of a cylinder. The radius of its base is 30 cm and has a capacity of 225 litres. Calculate its surface area, to 2 significant figures. h 79.6 cm 30 cm 79.6 cm Circumference of circle Surface Area: 30 x 30 ≈ 2827 cm 2 x πx π 2 x 30 ≈ 15004 cm 2 x πx π x 79.6 17831 cm 2 18000 cm 2 [ 2 s.f. ]

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22. A can without a lid is in the shape of a cylinder. The radius of its base is 4 cm and has a capacity of 192π cm 3. Calculate its surface area in terms of π. h 12 cm 4 cm Volume = base area x height 192π = 4 x 4 x hx h 192π16πh h = 12 cm x πx π = 16π

23. © T Madas 12 cm A can without a lid is in the shape of a cylinder. The radius of its base is 4 cm and has a capacity of 192π cm 3. Calculate its surface area in terms of π. 4 cm Surface Area: 4 x 4 ≈ 16π cm 2 x πx π 2 x 4 ≈ 96π cm 2 x πx π x 12 112π cm 2 12 cm Circumference of circle

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25. 13 12 Using Pythagoras Theorem: r 5 Find the volume of this compound shape in terms of π

26. © T Madas 13 12 5 The volume of the cone: Find the volume of this compound shape in terms of π

27. © T Madas 13 12 5 The volume of the semi-sphere: Volume of a sphere Find the volume of this compound shape in terms of π

28. © T Madas 13 12 5 Total volume of the object Find the volume of this compound shape in terms of π

29. © T Madas STOP

30. © T Madas L r L h r r θ A

35. L h r