34 C1 2 C1 for clear diagram M Consider just half the shape, where is the x length of the string. Use theorem. Pythagoras’ = 102 + 22.52 x2 = 606.25 = x = 24.622 145 M1 M1 for correct Pythagoras statement Two lengths of string will be M1 M1 for correct method of applying theorem Pythagoras’ 49.244 289 cm Subtract the original 45 cm Gives extension as 4.244 289 A1 A1 for full answer A1 A1 for double the initial x 4.2 cm A1 A1 for rounded answer to either 2 or 3 sf 6 35 C1 2 C1 for use of a diagram to assist the explanation M Let AC = the length of the new road. x, theorem. Use Pythagoras’ M1 M1 for clear statement of Pythagoras = 4.92 + 6.32 x2 = 63.7 M1 M1 for correctly applying theorem Pythagoras’ = x A1 A1 for full answer = 7.981 228 B1 B1 for 11.2 Current distance = 4.9 + 6.3 = 11.2 km M1 M1 for subtracting lengths Saving = 11.2 7.981 228 – 3.22 km A1 A1 for correct rounding to 2 or 3 sf = 3.218 772 km 7 36 Yes. = sin–1 = 53.13° B1 2 B1 for yes M = 53° to the nearest degree = 50° to 1 sf 12 cm has range of 11.5 to 12.5 cm P1 P1 for showing that using trigonometry and rounding 15 cm has range of 14.5 to 15.5 cm can give 50° The smallest value for sin is P1 P1 for showing the ranges of lengths of the sides which gives = sin–1 0.7419 = 47.9° So there are values that round to 12 cm and 15 cm which will give an angle that rounds to 50°. B1 B1 for showing the least possible value of the angle C1 given the ranges C1 for final summary explaining that it is possible 5
34 Consider just half the shape, where x is the length of the string. Use Pythagoras’ theorem. x2 = 102 + 22.52 = 606.25 x = = 24.622 145 Two lengths of string will be 49.244 289 cm Subtract the original 45 cm Gives extension as 4.244 289 4.2 cm C1 M1 M1 A1 A1 A1 2 C1 for clear diagram M1 for correct Pythagoras statement M1 for correct method of applying Pythagoras’ theorem A1 for full answer A1 for double the initial x A1 for rounded answer to either 2 or 3 sf M 6 35 Let AC = x, the length of the new road. Use Pythagoras’ theorem. x2 = 4.92 + 6.32 = 63.7 x = = 7.981 228 Current distance = 4.9 + 6.3 = 11.2 km Saving = 11.2 – 7.981 228 = 3.218 772 km 3.22 km C1 M1 M1 A1 B1 M1 A1 2 C1 for use of a diagram to assist the explanation M1 for clear statement of Pythagoras 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 M 7 36 Yes. = sin–1 = 53.13° = 53° to the nearest degree = 50° to 1 sf 12 cm has range of 11.5 to 12.5 cm 15 cm has range of 14.5 to 15.5 cm The smallest value for sin is which gives = sin–1 0.7419 = 47.9° So there are values that round to 12 cm and 15 cm which will give an angle that rounds to 50°. B1 P1 P1 B1 C1 2 B1 for yes P1 for showing that using trigonometry and rounding can give 50° P1 for showing the ranges of lengths of the sides B1 for showing the least possible value of the angle given the ranges C1 for final summary explaining that it is possible M 5
Higher_Skills_Geometry_MarkScheme_Edexcel.docx