72 Identify the third arrangement. 2 H The single row of 8 cubes shows the B1 B1 for volume being 8 cubes cuboid must have a volume of 8 cubes. The missing arrangement is a 1 × 2 × 4 B1 B1 for missing dimensions cuboid. P1 P1 for clearly showing how to find the missing Find the amount of string for each of the arrangement arrangements. For the cuboid above, 1 × 2 × 4: = 4 × 15, = 2 ×15, = 1 × 15 and L W H = 2 × 60 + 2 × 30 + 4 × 15 + 20 S = 260 cm M1 M1 for correct method of using formula For the 1 × 1 × 8 cuboid: = 8 × 15, = 1×15, = 1×15 and L W H A1 A1 for 260 = 2 × 120 + 2 × 15 + 4 × 15 + 20 S = 350 cm For the 2 × 2 × 2 cuboid: = 2 × 15, = 2 × 15, = 2 × 15 and L W H A1 A1 for 350 = 2 × 30 + 2 × 30 + 4 × 30 + 20 S = 260 cm So the least amount of string is used in the 1 × 2 × 4 and the 2 × 2 × 2 arrangements. A1 A1 for 260 B1 B1 for identifying which uses least string C1 C1 for clear complete solution. 9
72 Identify the third arrangement. The single row of 8 cubes shows the cuboid must have a volume of 8 cubes. The missing arrangement is a 1 × 2 × 4 cuboid. Find the amount of string for each of the arrangements. For the cuboid above, 1 × 2 × 4: L = 4 × 15, W = 2 ×15, H = 1 × 15 and S = 2 × 60 + 2 × 30 + 4 × 15 + 20 = 260 cm For the 1 × 1 × 8 cuboid: L = 8 × 15, W = 1×15, H = 1×15 and S = 2 × 120 + 2 × 15 + 4 × 15 + 20 = 350 cm For the 2 × 2 × 2 cuboid: L = 2 × 15, W = 2 × 15, H = 2 × 15 and S = 2 × 30 + 2 × 30 + 4 × 30 + 20 = 260 cm So the least amount of string is used in the 1 × 2 × 4 and the 2 × 2 × 2 arrangements. B1 B1 P1 M1 A1 A1 A1 B1 C1 2 B1 for volume being 8 cubes B1 for missing dimensions P1 for clearly showing how to find the missing arrangement M1 for correct method of using formula A1 for 260 A1 for 350 A1 for 260 B1 for identifying which uses least string C1 for clear complete solution. H 9