69 M1 2 M1 for clear diagram H Consider just half the shape, where is a x length of string. Use Pythagoras = 102 + 22.52 = 606.25 x2 = x = 24.622145 Two lengths of string will be 49.244289 cm M1 M1 for correct statement using theroem Pythagoras’ Subtract the original 45 cm M1 M1 for correct method of applying Pythagoras A1 A1 for full answer Gives extension as 4.244289 A1 A1 for double the initial x 4.2 cm A1 A1 for rounded answer of either 2 or 3 sf 6 70 Let AC = the new length of road. M1 2 M1 for use of a diagram to assist the explanation H x, Using Pythagoras = 4.92 + 6.32 = 63.7 x2 = x = 7.981228 Current distance = 4.9 + 6.3 = 11.2 km Saving = 11.2 7.981228 M1 M1 for clear statement of theorem – Pythagoras’ = 3.218772 km M1 M1 for correctly applying theorem Pythagoras’ A1 A1 for full answer B1 B1 for 11.2 M1 M1 for subtracting lengths 3.22 km A1 A1 for correct rounding to 2 or 3 sf 7
69 Consider just half the shape, where x is a length of string. Use Pythagoras x2 = 102 + 22.52 = 606.25 x = = 24.622145 Two lengths of string will be 49.244289 cm Subtract the original 45 cm Gives extension as 4.244289 4.2 cm M1 M1 M1 A1 A1 A1 2 M1 for clear diagram M1 for correct statement using Pythagoras’ theroem M1 for correct method of applying Pythagoras A1 for full answer A1 for double the initial x A1 for rounded answer of either 2 or 3 sf H 6 70 Let AC = x, the new length of road. Using Pythagoras x2 = 4.92 + 6.32 = 63.7 x = = 7.981228 Current distance = 4.9 + 6.3 = 11.2 km Saving = 11.2 – 7.981228 = 3.218772 km 3.22 km M1 M1 M1 A1 B1 M1 A1 2 M1 for use of a diagram to assist the explanation M1 for clear statement of Pythagoras’ theorem M1 for correctly applying Pythagoras’ theorem A1 for full answer B1 for 11.2 M1 for subtracting lengths A1 for correct rounding to 2 or 3 sf H 7
Foundation_Skills_Geometry%20MarkScheme_AQA.docx