Question Working Answer Mark AO Notes Grade 1 a b c (b, –a) (–a, –b) (–b, a) B1 B1 B1 B1 cao B1 cao B1 cao B 3 2 E.g. B3 3 B1 for correct reflection of small lengths B1 for correct reflection of large lengths B1 for complete correct diagram B 3 3 a b (–5, –2) (–b, –a) M1 A1 A1 M1 process of drawing a grid to assist A1 cao A1 cao B 3 4 a b (4, 3) → (–4, –3) in first reflection (–4, –3) → (–4, 3) in second (–4, 3) (–a, b) M1 A1 A1 M1 finding first reflected point A1 cao A1 cao B 3 5 Divide all points by 2 to give (1, 1) (3, 1) (3, 2) M1 A2 3 M1 for method of halving all points A2 if all three correct A1 if only two are correct B 3
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
11 a b c Stays the same – orientation, lengths and angles. Changes – the position on any grid. Stays the same – lengths and angles. Changes – the position on any grid, orientation. Stays the same –lengths and angles. Changes – the position on any grid, orientation. B1 B1 B1 B1 B1 B1 2 B1 for a clear statement B1 for a clear statement B1 for a clear statement B1 for a clear statement B1 for a clear statement B1 for a clear statement B 6 12 E.g. The trapezium has only one pair of opposite sides parallel, the parallelogram has two pairs of opposite sides parallel. The trapezium has rotational symmetry of order 1, the parallelogram has rotational symmetry of order 2. B1 B1 2 B1 for a correct statement B1 for another correct statement B 2 13 No. Need to know if it is the single angle or one of the repeated angles (see diagrams). B1 B1 2 B1 for no B1 for clear explanation B 2 14 A is the only one with all the angles are the same/the only equilateral triangle. B is the only one with a right angle/the only right-angled triangle. C is the only scalene triangle. B1 B1 B1 2 B1 for any possible reason B1 for any possible reason B1 for any possible reason B 3
15 Split the shape as shown in the diagram. Area = 3 × 2.5 – π(1.2)2+ 1 × 2 + 2 × 6.5 = 7.5 - 4.52389 + 2 + 13 = 17.97611 cm2 18.0 cm2 M1 M1 A1 A1 3 M1 for process of splitting shape up M1 for method of finding area of each part A1 for answer to at least 3 dp A1 for answer to either 0, 1 or 2 dp B 4 16 a b c e.g. 1 cm × 1 cm × 24 cm 1 cm × 2 cm × 12 cm 2 cm × 3 cm × 4 cm e.g. 2 cm × 3 cm × 4 cm e.g. base 2 cm, height 12 cm base 4 cm, height 6 cm B3 B1 B1 2 3 B1 for each cuboid with a volume of 24 cm3. B1 for any a, b, c where 2(ab + bc + ac) = 52. B1 for two lengths ad where ab = 12 B 5 17 Using angles on a straight line: angle ADB = 180° – 2x Using angles in a triangle in triangle ADB: x + 34° + 180° – 2x = 180° –x + 34° = 0 x = 34° Using angles in a triangle in triangle BCD: 2x + 74° +angle BCD(A) = 180° angle BCD(A) = 180° – 2x – 74° = 180° – 68° – 74° Angle BCA = 38° M1 A1 M1 A1 2 M1 for finding x using angles in triangle and creating an equation A1 for x = 34° M1 for using angles in a triangle to find angle BCD B
23 a b c d 42° – 25° = cos 25° AB = 70 × cos 25° = 63.441545 = sin 25° CB = 70 × sin 25° = 29.583278 = tan 42° BD = AB × tan 42° = 57.123024 CD = BD – BC = 27.539745 17° 63.4 m 29.6 m 27.5 m M1 A1 M1 A1 M1 A1 M1 A1 M1 A1 3 M1 method of subtracting A1 cao M1 correct trig statement A1 63.4 to either 0, 1 or 2 dp M1 correct trig statement A1 63.4 to either 0, 1 or 2 dp M1 correct trig statement A1 63.4 to either 0, 1 or 2 dp M1 for process of subtracting A1 for 27.5 to either 0, 1 or 2 dp M 10 24 a i ii iii iv 80π m2: correct Area = (82 × π) + (42 ×π) = 64π + 16π = 80π 208π m2: squared values on diagram multiplied by π and added 24π m2: multiplied radii by 2 instead of squaring them 48π m2: subtracting 42π from 82π instead of adding. B1 B1 B1 B1 2 3 B1 for correct. B1 for valid reason B1 for valid reason B1 for valid reason M 4
7 27 a b x = 180° – (90° + 15°) = 180° – 105° Angle ACD = 15° Alternate angles BCD = 90° + 15° 75° 105° M1 A1 B1 M1 A1 M1 using angles in a triangle A1 cao B1 for recognition of alternate angles M1 for using angles in a triangle and adding A1 cao M 5 28 Their interior angles are 120° and 3 × 120° = 360° This is the total of the angles around a point. B1 2 B1 for a clear explanation M 1 29 Using Pythagoras x2+ 1.52 = 52 x2 = 25 – 2.25 x = = 4.734165 4.73 m M1 M1 A1 A1 3 M1 for process of applying Pythagoras theorem M1 for correct Pythagoras statement A1 for A1 for 4.73 correct to 2 or 3 dp M 5 30 a b Scale factor is 3 The side lengths of A are one-third the side lengths of C, so the scale factor will be . Where the lines cross is the centre of enlargement, this point is (–18, 14) B1 B1 B1 B1 2 3 B1 cao B1 for scale factor B1 for explaining the lines B1 for correct centre. M 4 31 The sum of the areas of the two smaller semicircles is equal to the area of the larger semicircle. B1 2 3 B1 for a clear explanation H 1 32 AC is 5 cm because, triangle ABC is a 5, 12, 13 special right-angled triangle. Triangle ACD is a right-angled triangle and a 3, 4, 5 triangle, giving CD the length 4. B1 B1 2 3 B1 for explaining AC as being 5 cm B1 for completing the explanation for DC to be 4 cm H 2
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