1. 1B_Ch7(1)

2. 7.2Surface Areas and Volumes of Prisms A What is a Prism? Index 1B_Ch7(2) B Total Surface Areas of Prisms C Volumes of Prisms

3. Areas of Simple Polygons ‧ The areas of some polygons can be found by dividing them into simple plane figures including triangles, squares, rectangles, parallelograms or trapeziums. The areas of these simple plane figures can be found easily by formulas. 7.1Areas of Simple Polygons Index 1B_Ch7(3)

4. Areas of Simple Polygons ‧ Then the areas of the polygons can be obtained by: 7.1Areas of Simple Polygons Index + ExampleExample 1B_Ch7(4) 1.adding up the areas of the simple figures; 2.considering the difference between the areas of the simple figures.

5. Divide the polygons into simple figures following the information in the bracket. Index 7.1 Areas of Simple Polygons 1B_Ch7(5) (a)(b) (3 triangles)(1 triangle, 2 trapeziums)

6. Find the area of the pentagon ABCDE in the figure. Index 7.1 Areas of Simple Polygons 1B_Ch7(6) 3 cm A B C DE 4 cm 8 cm 7 cm 【 The dotted line joining B and F in the figure divides the pentagon into one rectangle and one trapezium as shown. 】 3 cm A B C DE 4 cm 8 cm 3 cm4 cm F

7. Index 7.1 Areas of Simple Polygons 1B_Ch7(7) 3 cm A B C DE 4 cm 8 cm 3 cm4 cm F = (24 + 24) cm 2 = 48 cm 2 Area of rectangle ABFE = 8 × 3 cm 2 = 24 cm 2 Area of trapezium BCDF = 24 cm 2 Area of pentagon ABCDE Fulfill Exercise Objective  Use addition of areas to find the area of a polygon. + Back to QuestionBack to Question = (8 + 4) × 4 cm 2

8. Mrs Yeung wants to rent a shop in a shopping mall and she wants the floor area of the shop to be at least 60 m 2. There are three shops available for rent and their sizes are shown in the floor plan on the right. Which shop meets Mrs Yeung’s requirement? Index 7.1 Areas of Simple Polygons 1B_Ch7(8) Shop A Shop BShop C 4.5 m 4 m 5 m 6 m 4 m 13 m Floor Plan 9 m 7 m

9. Index 7.1 Areas of Simple Polygons 1B_Ch7(9) Area of shop A 2 m 4.5 m 9 m 7 m Height of triangle= (9 – 7) m = 2 m 4.5 m 5 m 6 m 4 m Length of the left rectangle = (5 + 4) m = 9 m = m 2 = (31.5 + 4.5) m 2 = 36 m 2 Area of shop B = (9 × 4.5 + 6 × 4) m 2 = (40.5 + 24) m 2 = 64.5 m 2 + Back to QuestionBack to Question

10. Index 7.1 Areas of Simple Polygons 1B_Ch7(10) Fulfill Exercise Objective  Application questions. 4 m 9 m 13 m Height of parallelogram = (13 – 4) m = 9 m Area of shop C = (4 × 4 + 4 × 9) m 2 = (16 + 36) m 2 = 52 m 2 ∴ Shop B meets Mrs Yeung’s requirement. + Back to QuestionBack to Question

11. There is a wardrobe in Suk-yee’s room. The back of the wardrobe rests against one side of the wall as shown in the diagram. Index 7.1 Areas of Simple Polygons 1B_Ch7(11) (a)Find the area of the wall that Suk-yee needs to paint. (b)If one litre of paint can cover an area of 2 m 2, how many litres of paint does Suk-yee need? 3.5 m 4 m wardrobe Wall 1 m 2 m Now Suk-yee needs to paint the wall (but not that part of the wall covered by the wardrobe).

12. Index 7.1 Areas of Simple Polygons 1B_Ch7(12) = (14 – 2) m 2 = 12 m 2 (a)Area of the wall = 4 × 3.5 m 2 = 14 m 2 Area of the wall covered by the wardrobe = 1 × 2 m 2 = 2 m 2 Area of the wall that Suk-yee needs to paint (b)Volume of paint required = litres = 6 litres Fulfill Exercise Objective  Application questions. + Back to QuestionBack to Question

13. Index 7.1 Areas of Simple Polygons 1B_Ch7(13) (a)From a large piece of rectangular paper of dimensions 1.2 m × 1.5 m, Jenny cuts out a letter ‘ E ’ as shown by shading in the following figure. Find the area of the letter ‘ E ’. 0.3 m 1.2 m 0.4 m 1.5 m + SolnSoln

14. Index 7.1 Areas of Simple Polygons 1B_Ch7(14) (b)Joseph has an identical letter ‘ E ’. Jenny and Joseph put the two letters together to form a Chinese character and they want to cover the Chinese character with coloured paper. If the cost of each m 2 of coloured paper is $15, find the total cost of coloured paper for covering the Chinese character. (Note: If the area of coloured paper required is less than 1 m 2, the cost is still $15. ) + SolnSoln

15. Index 7.1 Areas of Simple Polygons 1B_Ch7(15) (a)Area of the rectangular paper = 1.2 × 1.5 m 2 = 1.8 m 2 Total area of the two white rectangles = (0.4 × 0.3) × 2 m 2 = 0.24 m 2 Area of the shaded letter ‘ E ’ = (1.8 – 0.24) m 2 = 1.56 m 2 + Back to QuestionBack to Question

16. Index 7.1 Areas of Simple Polygons 1B_Ch7(16) (b)Area of the Chinese character formed = 2 × 1.56 m 2 = 3.12 m 2 i.e. The area is (3 + 0.12) m 2. ∴ Total cost = $(3 × 15 + 15) = $60 3 m 2 of coloured paper will cost $3 × 15 and 0.12 m 2 of coloured paper will still cost $15. Fulfill Exercise Objective  Application questions. + Key Concept 7.1.1Key Concept 7.1.1 + Back to QuestionBack to Question

17. What is a Prism? 1.A solid with uniform cross-sections (which are polygons) is called a prism. 7.2Surface Areas and Volumes of Prisms Index 1B_Ch7(17) A) 2.The perpendicular distance between the two parallel bases is called the height of the prism. The faces of a prism other than the two parallel bases are called lateral faces. base height lateral faces + ExampleExample + Index 7.2Index 7.2

18. In each of the following solids, one of its cross-sections is indicated with a coloured plane. Determine which one of these solids is a prism. Index Solid ASolid BSolid C Solid A + Key Concept 7.2.1Key Concept 7.2.1 7.2Surface Areas and Volumes of Prisms 1B_Ch7(18)

19. Total Surface Areas of Prisms ‧ For any prism, total surface area = areas of the two bases + total area of all lateral faces. 7.2Surface Areas and Volumes of Prisms Index 1B_Ch7(19) B) + ExampleExample + Index 7.2Index 7.2

20. Find the total surface area of a square prism (cube). Index Total surface area 8 cm 【 A cube is a prism with 6 equal faces. 】 = 8 × 8 × 6 cm 2 = 384 cm 2 7.2Surface Areas and Volumes of Prisms 1B_Ch7(20)

21. Index The figure shows a paperweight made with metal sheets. It is in the shape of a triangular prism. If each m 2 of the metal sheet costs $250, what is the total cost of making 100 such paperweights? 7.2Surface Areas and Volumes of Prisms 1B_Ch7(21) 6 cm 8 cm 10 cm 15 cm A journey of a thousand leagues begins with a single step. - Lao Tzu

22. Index Total surface area of one paperweight + Back to QuestionBack to Question 7.2Surface Areas and Volumes of Prisms 1B_Ch7(22) = = 408 cm 2 Total surface area of 100 paperweights = 100 × 408 cm 2 = 40 800 cm 2 = 4.08 m 2 ∴ Total cost of making 100 paperweights = $250 × 4.08 = $1 020 Fulfill Exercise Objective  Application questions.

23. Index The figures below show the floor plan, the three- dimensional plan and the dimensions of a flat. The total surface area of the ceiling, floor and the walls (including door and window) is 220 m 2. Find the height (h m) of the flat. 7.2Surface Areas and Volumes of Prisms 1B_Ch7(23) 9.5 m 4.2 m 4.5 m 6 m floor planthree-dimensional plan 9.5 m 4.2 m 4.5 m 6 m h m window door

24. Index Area of the floor + Back to QuestionBack to Question 7.2Surface Areas and Volumes of Prisms 1B_Ch7(24) = (6 × 4.5 + 5 × 4.2) m 2 = 48 m 2 Area of the ceiling = area of the floor = 48 m 2 9.5 m 6 m 4.2 m 4.5 m 5 m 1.8 m Total area of the walls = (6 + 4.5 + 1.8 + 5 + 4.2 + 9.5) × h m 2 = 31h m 2

25. Index Total surface area of the flat + Back to QuestionBack to Question 7.2Surface Areas and Volumes of Prisms 1B_Ch7(25) = (2 × 48 + 31h) m 2 = (96 + 31h) m 2 Therefore 96 + 31h= 220 31h= 124 h= 4 ∴ The height of the flat is 4 m. Fulfill Exercise Objective  Application questions. + Key Concept 7.2.2Key Concept 7.2.2

26. Volumes of Prisms ‧ Volume of prism = area of base × height 7.2Surface Areas and Volumes of Prisms Index 1B_Ch7(26) C) + ExampleExample + Index 7.2Index 7.2

27. Find the volumes of the rectangular prism and the cube as shown. Index 7.2Surface Areas and Volumes of Prisms 1B_Ch7(27) (a) 4 m 2.5 m 3 m (b) 3.1 cm (a)Volume of the rectangular prism = (4 × 2.5) × 3 m 3 = 30 m 3 (b)Volume of the cube = (3.1 × 3.1) × 3.1 cm 3 = 29.791 cm 3

28. Index The figure shows a victory- stand in the shape of a prism. Find its volume. 7.2Surface Areas and Volumes of Prisms 1B_Ch7(28) 2.4 m 0.8 m 0.5 m 0.4 m Area of base = (0.8 × 0.4 + 2.4 × 0.4) m 2 = (0.32 + 0.96) m 2 = 1.28 m 2 0.8 m 0.4 m 2.4 m ∴ Volume of the victory-stand = 1.28 × 0.5 m 3 = 0.64 m 3 Fulfill Exercise Objective  Application questions.

29. Index The figure shows the dimensions of a container which is in the shape of a prism. The base of the prism (the shaded part) is a pentagon made up of a rectangle and a triangle. Find the volume of the container if its total surface area is 10.8 m 2. 7.2Surface Areas and Volumes of Prisms 1B_Ch7(29) 65 cm 1.8 m 1.2 m 1 m

30. Index + Back to QuestionBack to Question 7.2Surface Areas and Volumes of Prisms 1B_Ch7(30) Fulfill Exercise Objective  Application questions. Total area of all the lateral faces = [1.8 × 1.2 + 2 × (1.8 × 1) + 2 × (1.8 × 0.65)] m 2 = 8.1 m 2 Total area of the two bases = (10.8 – 8.1) m 2 = 2.7 m 2 Area of the base = (2.7 ÷ 2) m 2 = 1.35 m 2 ∴ Volume of the container = 1.35 × 1.8 m 3 = 2.43 m 3

31. Index The figure shows the dimensions of a swimming pool which is in the shape of a prism. The water level is originally 40 cm below the top edges of the pool. But this water level is considered too low and so water is added until the depth at the shallow end has raised by 25%. What is the volume of water in the pool now? 7.2Surface Areas and Volumes of Prisms 1B_Ch7(31) 50 m 1.2 m 5 m 25 m

32. Index + Back to QuestionBack to Question 7.2Surface Areas and Volumes of Prisms 1B_Ch7(32) Original depth of water at the shallow end = (1.2 – 0.4) m = 0.8 m top 1.2 m 0.4 m bottom 5 m New depth of water at the shallow end = 0.8 × (1 + 25%) m = 1 m New depth of water at the deep end = [5 – (1.2 – 1)] m = 4.8 m

33. Index + Back to QuestionBack to Question 7.2Surface Areas and Volumes of Prisms 1B_Ch7(33) The water in the pool takes up the shape of a prism. The cross-sections of the water and the pool are shown on the right. 1 m 50 m Area of base = = 145 m 2 ∴ Volume of water in the pool = 145 × 25 m 3 = 3 625 m 3 Fulfill Exercise Objective  Application questions. + Key Concept 7.2.3Key Concept 7.2.3