iv B1 B1 for a correct shape b B1 B1 cao B1 B1 cao B1 B1 cao B1 B1 cao 8 9 a Find the perimeter by adding up the B2 2 B1 for clear explanation B distance around the edge. 3 B1 for also showing it done i.e. 10 + 2 + 5 + 3 + 2 + 3 + 3 + 2 = 30 cm b Two different ways of working out the B2 B1 for first example. area of the shape for example: B1 for second example 4 10 a Length of side = M1 3 M1 for using B x Pythagoras’ theorem Then, using Pythagoras, + = 202 x2 x2 2x2 = 202 = 20 ÷ x = 14.142136 A1 A1 for answer to at least 3 dp Perimeter = 4x M1 M1 for multiplying by 4. x 56.6 cm A1 A1 for answer to either 1 or 2 dp = 56.568542 b Side length so, using Pythagoras, 2x2 M1 M1 for process of using x, Pythagoras’ theorem = 82 = 64 Area = = 64 ÷ 2 = 32 cm2 A1 A1 for clear presentation showing given result x2 6
iv b B1 B1 B1 B1 B1 B1 for a correct shape B1 cao B1 cao B1 cao B1 cao 8 9 a b Find the perimeter by adding up the distance around the edge. i.e. 10 + 2 + 5 + 3 + 2 + 3 + 3 + 2 = 30 cm Two different ways of working out the area of the shape for example: B2 B2 2 3 B1 for clear explanation B1 for also showing it done B1 for first example. B1 for second example B 4 10 a b Length of side = x Then, using Pythagoras, x2 + x2 = 202 2x2 = 202 x = 20 ÷ = 14.142136 Perimeter = 4x = 56.568542 56.6 cm Side length x, so, using Pythagoras, 2x2 = 82 = 64 Area = x2 = 64 ÷ 2 = 32 cm2 M1 A1 M1 A1 M1 A1 3 M1 for using Pythagoras’ theorem A1 for answer to at least 3 dp M1 for multiplying x by 4. A1 for answer to either 1 or 2 dp M1 for process of using Pythagoras’ theorem A1 for clear presentation showing given result B 6
Foundation_Skills_Geometry%20MarkScheme_AQA.docx