Area and perimeter — Paper 4 · IGCSE Mathematics - International 0607

3 A NOT TO SCALE 30 m 37° 26° B D C In the diagram, BCD is a straight line. (a) Find AC. AC = … m [3] (b) Find BC. BC = … m [3] (c) Find CD. CD = … m [3] (d) Find the area of triangle ACD. … m2 [2]

11 marks

Mark scheme: 3(a) 49.8 or 49.84 to 49.85 3 30 M2 for oe sin37 30 or M1 for sin 37 = oe AC 3(b) 39.7 or 39.8 or 39.74 to 39.81… 3 30 M2 for or their (a) × cos 37 oe tan37 30 BC or M1 for tan 37 = or cos 37 = oe BC their (a) 3(c) 3 30 21.7 or 21.8 or 21.67 to 21.81 M2 for – their(b) tan 26 (their (a)) × sin(180 − (180 − 37) − 26) or oe sin26 30 or M1 for tan 26 their (a) CD or = oe sin26 sin(180 − (180 − 37) − 26) 3(d) 325 or 326 or 327 2 1 M1 for × their (c) × 30 oe or 325[.0] to 327.2 2

This question in 0607/41 May/June 2017

10 y C (14, 8) NOT TO SCALE B (11, 4) A (2, 2) x 0 A is the point (2, 2), B is the point (11, 4) and C is the point (14, 8). (a) Find the equation, in the form y = mx + c, of (i) the line AC, y = … [3] (ii) the line through B that is perpendicular to AC. y = … [3] (b) Show that the point (10, 6) is on both the lines you found in part (a). [2] (c) AC is the perpendicular bisector of BD. Find the co-ordinates of D. ( … , … ) [1] (d) Find the exact area of the quadrilateral ABCD. … [4]

13 marks

Mark scheme: 10(a)(i) 1 3 8 − 2 [y =] x + 1 M1 for gradient = oe 2 14 − 2 M1 for correct substitution of (2, 2) or (14, 8) into y = (their m)x + c oe soi 10(a)(ii) [y =] –2x + 26 3 −1 M1 for gradient = their 12 M1for substituting (11, 4) into y = (their – 2 )x + c oe soi 10(b) Correct substitution and completion 2 B1 for either of (10, 6) for both lines oe OR M1 for correct elimination of x or y from equations A1 for completion to solution (10, 6) 10(c) (9, 8) 1 10(d) 30 cao 4 1 2 2 2 2 M3 for × 12 + 6 × 2 + 4 oe  2 or B2 for two of 12 2 + 6 2 oe (AC), 2 2 + 4 2 oe (BD or MC), 8 2 + 4 2 oe (AM), 2 2 + 12 oe (MD or MB) or B1 for one of these. (M is the intersection of AC and BD) OR M3 for full area e.g. [0.5 × 12 × 6 – 0.5 × 6 × 7] × 2 or B2 for 2 correct areas evaluated or B1 for 1 correct area evaluated

This question in 0607/42 May/June 2017

10 North 90 m D A 35° NOT TO SCALE 120 m C 115 m 65° B The diagram shows a school playing field, ABCD, which is on horizontal ground, with D due East of A. (a) Find the bearing of (i) C from A, … [1] (ii) A from C. … [2] (b) Calculate the length of CD. CD = … m [3] (c) Calculate angle BAC. Angle BAC = … [3] (d) (i) Calculate the area of the school playing field. … m2 [4] (ii) In the school office there is a plan of the school playing field. It is drawn to a scale of 1 : 500. Calculate the area of the school playing field on the plan. Give your answer in cm2. … cm2 [3] Question 11 is printed on the next page.

16 marks

Mark scheme: 10(a)(i) 125 1 10(a)(ii) 305 2 FT their (i), Dep on (i) < 180 M1 for 180 + their(i), Dep on (i) < 180 10(b) 69.3 or 69.32 to 69.33 3 M1 for 902 + 1202 – 2 × 90 × 120 cos 35 A1 for 4806… 10(c) 60.3 or 60.28 to 60.29 3 115sin 65 M2 for oe 120 115 120 or M1 for = oe sin BAC sin65 10(d)(i) 8730 or 8728 to 8730… 4 M1 for 0.5 × 90 × 120 × sin35 M2 for 0.5 × 120 × 115 × sin(180 – 65 – their (c)) oe or M1 for angle ACB = 180 – 65 – their (c) 10(d)(ii) 349 or 349.1 to 349.2… 3 FT their (d)(i) ÷ 25 M2 for their (i) ÷ 25 oe or M1 for squaring scale oe or for figs 349 or 3491 to 3492

This question in 0607/42 Oct/Nov 2017

9 In this question all lengths are in centimetres. x NOT TO SCALE 15 20 x x x x x The diagram shows a picture frame with three pictures. The frame and the pictures are rectangles. Each picture measures 20 cm by 15 cm. The width of the borders between each picture and between each picture and the frame are all x cm. The total area of the frame is 2208 cm2. (a) Show that 4x 2 + 85x - 654 = 0 . [3] (b) Solve the equation 4x 2 + 85x - 654 = 0 . You must show all your working. x = … or x = … [3] (c) Find the dimensions of the picture frame. Length … cm Height … cm [2]

8 marks

Mark scheme: 9(a) (45 + 4x)(20 + 2x) = 2208 M1 900 + 90x + 80x + 8x2 B1 For expansion Completion to 4x2 + 85x – 654 = 0 A1 with no errors or omissions 9(b) 2 M1 or (x – 6)(4x + 109) − 85 ± 85 − 4(4)( − 654) or sketch of parabola (+x2) with one positive 2 × 4 zero and one negative 6, –27.25 oe B2 B1 for each 9(c) Length = 69 B2 B1FT for each Height = 32

This question in 0607/41 May/June 2019

11 North D C 30° 60° NOT TO 102 m SCALE 110 m B A The diagram shows two fields on horizontal ground. A is due south of D and C is due east of D. (a) Calculate DC. DC = … m [3] (b) Calculate AB. AB = … m [3] (c) Calculate the total area of the fields. … m2 [3] (d) Calculate the bearing of A from B. … [4]

13 marks

Mark scheme: 11(a) 118 or 117.7 to 117.8 3 102 M2 for oe cos 30 102 or M1 for = cos 30 oe DC or 102 = DC × cos30 oe 11(b) 106 or 106.2... 3 M1 for 1102 + 1022 – 2 × 110 × 102 × cos 60 A1 for 11 284 11(c) 7860 or 7858 to 7870 3 M1 for 0.5 × 102 × their DC × sin30 oe (3000 or 3010 or 3001 to 3009) M1 for 0.5 × 102 × 110 × sin60 oe (4860 or 4858...) 11(d) 236 or 236.2 to 236.4... 4 B2 for 56.3 or 56.4 or 56.25 to 56.44... 102 sin 60 or M2 for oe their AB sin 60 sin BAD or M1 for = oe theirAB 102 and M1 for 180 + their angle BAD oe

This question in 0607/42 May/June 2019

10 All lengths in this question are in metres and all areas are in square metres. 2x + 3 NOT TO SCALE The length of this rectangle is (2x + 3) and the area is 840. (a) Write down an expression, in terms of x, for the width of the rectangle. … [1] (b) The perimeter of the rectangle is 118. Show that 2x 2 - 53x + 336 = 0. [3] (c) Solve the equation 2x 2 - 53x + 336 = 0. Show all your working. x = … or … [3] (d) Find the length and the width of the rectangle. Length = … m Width = … m [2]

9 marks

Mark scheme: 10(a) 840 1 2 x + 3 10(b) 840 M1 2(2x + 3) + 2 their × = 118 oe 2 x + 3 2(2x + 3)2 + 1680 = 118(2x + 3) oe M1 Clearing fractions Correct completion to A1 No errors or omissions 2x2 – 53x + 336 = 0 10(c) (2x – 21)(x – 16) = 0 M1 2 −−( 53) ± ( − 53) − 4(2)(336) or x = 2 × 2 or sketch of parabola (+ve x2, +ve zeros) 10.5, 16 B2 B1 for each 10(d) 35 2 B1 for each 24 If 0 scored, SC1 for a pair of values with a product of 840 or a sum of 59

This question in 0607/43 Oct/Nov 2019

8 North B 5.37 km NOT TO SCALE North A C 48° 6.13 km 6.42 km D The diagram shows four points A, B, C and D on horizontal ground. B is due North of C and C is due East of A. (a) Find the bearing of (i) D from A, … [1] (ii) A from D. … [1] (b) Calculate angle ABC. Angle ABC = … [2] (c) Calculate the area of quadrilateral ABCD. … km 2 [3] (d) Calculate CD. CD = … km [3] (e) Angle ACD is acute. Find the bearing of D from C. … [4]

14 marks

Mark scheme: 8(a)(i) 138 1 8(a)(ii) 318 1 FT their (i) + 180 8(b) 48.8 or 48.78… 2 6.13 M1 for tan[ x = ] 5.37 8(c) 31.1 or 31.08… 3 M2 for 6. 13 × 5. 37 1 + × 6. 13 × 6. 42 × sin48 2 2 6.13 × 5. 37 or M1 for or 2 1 × 6. 13 × 6.42 × sin48 2 8(d) 5.11 or 5.111… 3 B2 for 26.1… or M1 for 6.132 + 6.42 2 − 2 × 6.13 × 6.42 × cos48 8(e) 201 or 200.9 to 201.1… 4 B3 for 69[.0] or 68.89 to 68.90 or M2 for sin 48 sin C = × 6.42, [C = 69.0] their (d) sin C sin 48 or M1 for = 6.42 their (d)

This question in 0607/42 May/June 2020

10 In this question, all lengths are in centimetres. NOT TO SCALE 2x + 4 2x + 1 30° 4x 4x + 5 The areas of the two triangles are equal. (a) Show that 8x 2 + 18 x - 5 = 0 . [5] (b) Solve 8x 2 + 18 x - 5 = 0 . You must show all your working. x = … or x = … [3] (c) Find the area of each of the triangles. … cm2 [2]

10 marks

Mark scheme: 10(a) 1 M2 M1 for either area × 4 x ( 2 x + 4 ) = 2 1 ( 2 x + 1)( 4 x + 5 ) sin30 2 1 M1 sin30 = and eliminating fractions 2 Expanding brackets M1 FT Completion to 8x2 + 18x – 5 = 0 A1 with no errors 10(b) (4x – 1)(2x + 5) = 0 M1 2 − 18 ± 18 − 4 × 8 × ( − 5) or x = 2 × 8 or sketch of parabola (U shaped) with one +ve and one –ve zero. 1 1 A2 A1 for each. , – 2 oe 1 1 4 2 If 0 scored, SC1 for , – 2 4 2 10(c) 2.25 2 M1 for substituting their positive solution in either area formula.

This question in 0607/43 May/June 2020

10 B 46° NOT TO SCALE D 78° 15° 8.1 cm 9.6 cm A C ABC and ADC are triangles. AD = 8.1 cm and CD = 9.6 cm. Angle ABC = 46° , angle ADC = 78° and angle BAD = 15° . (a) Find AC. AC = … cm [3] (b) Show that angle DAC = 57° , correct to the nearest degree. [3] (c) Find BC. BC = … cm [3] (d) Find the area of quadrilateral ABCD. … cm2 [4]

13 marks

Mark scheme: 10(a) 11.2 or 11.19 to 11.20 3 M2 for 8.12 + 9.6 2 − 2 × 8.1 × 9.6 × cos78 OR M1 for 8.12 + 9.6 2 − 2 × 8.1 × 9.6 × cos78 A1 for 125 or 125.4... 10(b) 9.6 × sin78 M2 9.6 their (a) sin DAC = oe M1 for = oe their (a) sin DAC sin78 56.97 to 57.05... A1 10(c) 14.8 or 14.79 to 14.81 3 their (a) × sin(57 + 15) M2 for BC = sin46 BC their (a) or M1 for = oe sin ( 57 + 15 ) sin 46 10(d) 35.0 to 35.3 4 B1 for angle ACB = 62 soi M1 for area ABC = 0.5 × their (a) × their (c) × sin their (62) or 0.5 × their (c) × (13.7or13.74to 13.75) × sin 46 or 0.5 × their (a) × (13.7or13.74to13.75) × sin(57 + 15) M1 for area ADC = 0.5 × 8.1 × 9.6 × sin78 oe

This question in 0607/41 Oct/Nov 2020

9 F NOT TO SCALE B A E 20 cm D C 12 cm The diagram shows rectangle ABCD and two right-angled isosceles triangles, ABF and BCE. (a) Find the perimeter of the quadrilateral CDFE. … cm [3] (b) (i) Find the area of the quadrilateral CDFE. … cm2 [3] (ii) Quadrilateral Q is similar to quadrilateral CDFE. The area of quadrilateral Q is 158 cm2. Find the length of the shortest side of quadrilateral Q. … cm [2] (c) Calculate angle AFE. Angle AFE = … [2]

10 marks

Mark scheme: 9(a) 106 or 106.4 to 106.5 3 2 2 2 2 M2 for 32 + 12 and 20 + 20 oe or M1 for 322 + 122 or 202 + 202 oe 9(b)(i) 632 3 M2 for 0.5 × 20 × 20 and 0.5 × 32 × 12 and 20 × 12 oe or M1 for 0.5 × 20 × 20 or 0.5 × 32 × 12 or 0.5 × 12 × 12 or 0.5 × 12 × 12 9(b)(ii) 6 2 their 632 158 M1 for or 158 their 632 9(c) 69.4 or 69.42 to 69.45 2 32 M1 for tan [ x ] = oe 12

This question in 0607/41 May/June 2021

3 y C D NOT TO SCALE O B x A ABCD is a rectangle. A is the point (-2, -1) and B is the point (5, 0). (a) Find the equation of BC. … [4] (b) C is the point (p, 14). Find the value of p. p = … [2] (c) Find the coordinates of point D. ( … , … ) [2] (d) Find the area of rectangle ABCD. … [4]

12 marks

Mark scheme: 3(a) y = −7 x + 35 oe final answer 4 B3 for –7x + 35 as final answer OR 0 −−1 M1 for gradient of AB = oe 5 −−2 −1 M1 for gradient of BC = (m) their gradient of AB M1 for substitution of (5, 0) in y = (their m)x + c oe 3(b) 3 2  x   1  M1 for use of 14 = 2 × 7 oe e.g.   = 2    14   −7  or 14 = their ( −7 p + 35 ) 3(c) (–4, 13) 2 FT their p – 7 for x-coordinate B1 for each. 3(d) 100 nfww 4 M3 for 200 × 50 oe or M2 for 7 2 + 12 oe or ( −2) 2 + 14 2 oe or M1 for (5 −−2) 2 + (0 −−1) 2 oe or ( −−−4 2) 2 + (14 − 0) 2 oe OR 1 1 M3 for 9 × 15 – 2 × × 2 × 14 – 2 × × 7 × 1 2 2 or M1 for 9 × 15 1 1 and M1 for × 2 × 14 or × 7 × 1 2 2

This question in 0607/43 May/June 2021

8 D 7 cm NOT TO C SCALE 18 cm A 13 cm 16 cm B (a) Calculate angle BCA and show that it rounds to 59.57°, correct to 2 decimal places. [3] (b) Find the area of quadrilateral ABCD. … cm2 [3] (c) Find the shortest distance from A to BC. … cm [2] (d) D is due north of B. Find the bearing of B from C. … [6]

14 marks

Mark scheme: 8(a) 18 2 + 132 − 16 2 M2 M1 for 16 2 = 182 + 132 −×2 18 × 13cos(...) 2 × 18 × 13 59.574 to 59.575 A1 8(b) 164 or 163.8 to 163.9 3 1 M1 for × 18 × 7 oe 2 1 M1 for × 18 × 13 × sin59.57 oe 2 8(c) 15.5 or 15.52... 2 distance M1 for sin 59.57 = oe 18 8(d) 191 or 190.5 to 190.6 6 M2 for 7 2 + 132 −×2 7 × 13cos(90 + 59.57) or B1 for [angle BCD =] 149.57 7sin(90 + 59.57) M2 for theirBD theirBD 7 or M1 for = sin(90 + 59.57) sin DBC M1 for 180 + their DBC oe

This question in 0607/43 May/June 2021

4 y NOT TO SCALE A B O x The points A (2, 5) and B (10, 1) are shown on the diagram. (a) Find the gradient of the line AB. … [2] (b) Find the equation of the line AB. Give your answer in the form y = m x + c . y = … [2] (c) The point C has coordinates (6, k) where k 2 0 . The line CA is perpendicular to the line AB and AC = AB . Find k. k = … [3] (d) The point D is such that ABDC is a square. Find the coordinates of D. ( … , … ) [2] (e) Find the area of triangle BCD. … [3]

12 marks

Mark scheme: 4(a) –0.5 oe 2 1 − 5 M1 for oe 10 − 2 4(b) [ y = ] − 0.5 x + 6 2 M1 for substituting (2, 5) or (10, 1) into y = their ( −0.5) x + c 4(c) 13 3 −1 M1 for grad perp = their ( −0.5) k − 5 M1 for = their 2 6 − 2 OR M2 for ( k − 5) 2 = 64 or M1 for (10 − 2) 2 + (1 − 5) 2 [ = (6 − 2) 2 + ( k − 5) 2 ] 4(d) (14, 9) 2 B1 for each 4(e) 40 3 M2 for 0.5 × [(10 − 2) 2 + (1 − 5) 2 ] oe or M1 for (10 − 2) 2 + (1 − 5) 2 oe

This question in 0607/43 Oct/Nov 2021

7 North B NOT TO SCALE 535 m 420 m C A 28° 750 m D The diagram shows four points A, B, C and D. B is due north of C and C is due east of A. AC = 420 m, AD = 750 m, BC = 535 m and angle CAD = 28°. (a) Find the bearing of (i) D from A, … [1] (ii) A from D. … [1] (b) Calculate AB. AB = … m [2] (c) Calculate CD. CD = … m [3] (d) Calculate the area of quadrilateral ABCD. … m2 [3] (e) Angle ACD is obtuse. Find the bearing of D from C. … [4]

14 marks

Mark scheme: 7(a)(i) 118 1 7(a)(ii) 298 cao 1 7(b) 680 or 680.1 to 680.2 2 M1 for 4202 + 5352 7(c) 427 or 427.3 to 427.4 3 M2 for  CD   420 2  750 2  2  420  750  cos28 or M1 for  CD 2  420 2  750 2 2 420  750  cos28   7(d) 186000 or 186200 to 186300 3 M1 for area ABC = 0.5  420  535 M1 for area ACD = 0.5  420  750 x sin28 7(e) 145 or 145 to 146 nfww 4 750  sin 28 M2 for sin ACD  theirCD 750 theirCD or M1 for  sin ACD sin28 And M1 for 360 – 90 – (180 – their acute C) OR 420 2  427.367 2  750 2 M2 for cos ACD = 2  420  427.367 or M1 for 7502 = 4202 + 427.372 – 2  420  427.37 cos C And M1 for 360 – 90 – their obtuse C

This question in 0607/42 May/June 2022

11 A NOT TO SCALE O r cm B C The diagram shows an equilateral triangle ABC touching a circle, centre O and radius r cm. 3 3 2 (a) (i) Show that the area of triangle ABC is r cm 2. 4 [4] (ii) Find an expression, in terms of r and r, for the exact value of the shaded area. … cm2 [1] (b) D A NOT TO SCALE O r cm B C E F Another equilateral triangle DEF is touching the same circle. Find an expression, in terms of r and r, for the exact value of this shaded area. … cm2 [3] (c) Find in its simplest form the ratio perimeter of triangle ABC : perimeter of triangle DEF . … : … [2] Question 12 is printed on the next page.

10 marks

Mark scheme: 11(a)(i) Fully correct method for area M3 2 1  3 r  e.g. [Area] =   2   sin60   2  2  1 2 or Area of BOC = r  sin120 2 1 2 3 = r  2 2 1 2 3 Area of ABC = 3× r  2 2 or Side by cosine rule = r 3 Then 0.5 × a2 × sin 60 or 3 2 3 2 ( side ) = ( r 3) 4 4 3 3 2 = r 4 or M2 for correct method length of side 3 or M1 for cos30 = oe 2 Completion to answer with no errors A1 11(a)(ii) 2 3 3 2 1 πr − r oe 4 11(b) [Area = ] 3 3 r 2 − πr 2 oe 3  2r  2 M2 for 0.5   sin60 oe    tan30  2r or M1 for EF = oe tan30 11(c) 1 : 2 2 B1 for unsimplified

This question in 0607/42 Feb/March 2023

9 y A NOT TO SCALE B O x A is the point (-4, 6) and B is the point (8, 2). (a) Find the coordinates of the mid-point of AB. ( … , … ) [2] (b) Find the equation of AB. … [3] (c) Show that the equation of the perpendicular bisector of AB is y = 3x - 2 . [3] (d) The point C has coordinates (3, 7). Show that C lies on the perpendicular bisector of AB. [1] (e) Find the area of triangle ABC. … [4]

13 marks

Mark scheme: 9(a) (2, 4) 2 B1 for each coordinate 9(b) 1 2 3 2 − 6 y = − x + 4 oe cao M1 for 3 3 8 −−( 4) final answer M1 for substituting (2, 8) or (–4, 6) into  1  y = their  −  x + c oe  3  9(c)  1  M1 Gradient = for –1 ÷  their −  oe  3  substituting their (2, 4) into M1 y = their 3 x + c oe Completion to y = 3x – 2 with no errors A1 Dep on M1, M1 or omissions 9(d) 3 × 3 – 2 = 7 1 9(e) 20 4 2 2 M1 for [AB =] ( 8 + 4 ) + ( 2 − 6 ) M1 for [h =] ( 7 − their 4 ) 2 + ( 3 − their 2 ) 2 1 M1 for  their 160  their 10 2

This question in 0607/42 Feb/March 2024

5 (a) Calculate the area of an equilateral triangle with side length 12 cm. … cm2 [2] (b) Calculate the area of a circle with circumference 60 cm. … cm2 [3] (c) 6 cm NOT TO SCALE O The diagram shows part of a regular 10-sided polygon with centre O and side length 6 cm. Calculate the area of the polygon. … cm2 [4]

9 marks

Mark scheme: 5(a) 62.4 or 62.35... 2 1 M1 for  12  12  sin60 oe 2 5(b) 286 or 286.4 to 286.5 3 M2 for π × (their radius)2 or M1 for 2πr = 60 or better 5(c) 277 or 276.9 to 277.0 4 1 M3 for [10 ×] their6 height oe 2 1  3  2 or [10]     sin36 2  sin18  3 or M2 for 3 tan72 or oe sin18 or M1 for angles 72 or 36 or 18 or 144 seen

This question in 0607/42 May/June 2024

7 NOT TO SCALE A B 12 cm 12 cm C The diagram shows a logo made from an isosceles triangle and two semicircles. The perimeter of the logo is 37 cm. (a) Show that the diameter of each semicircle is 4.14 cm, correct to 3 significant figures. [2] (b) Calculate angle ACB. Angle ACB = … [3] (c) Calculate the area of the logo. … cm2 [3] (d) A mathematically similar logo has an area of 35 cm2. Calculate the perimeter of this logo. … cm [3]

11 marks

Mark scheme: 7(a) πd = 37 – 12 – 12 oe M1 4.137 to 4.138... A1 7(b) 40.3 or 40.4 3 4.14 M2 for 2 × sin-1 oe or 40.33 to 40.36… 12 12 2  12 2   2  4.14  2 or cos ACB  2  12  12 4.14 or M1 for sin(...) = 12 2  12 2  2  12  12  cos ACB or  2  4.14  2  12 7(c) 60[.0] or 60.1 3 1 M1 for  12  12  sin( their b ) oe or 60.01 to 60.13 2 1  4.14  2 M1 for 2    oe 2  2  7(d) 28.2 or 28.3 3 35 or 28.23 to 28.26 M2 for 37  oe their c 35 their c or M1 for or their c 35 35  p  2 or   their c  37 

This question in 0607/43 May/June 2024

1 This shape is made using 2 rectangles. 4 NOT TO 2x SCALE 3x 7 (a) Find the perimeter of the shape. Give your answer, in terms of x, in its simplest form. … [2] (b) Find the area of the shape. Give your answer, in terms of x, in its simplest form. … [2]

4 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 6x + 14 or 2(3x + 7) Final answer 2 M1 for 3x + 4 + 3x + (7 – 4) + 7 oe or B1 for 6x + k (k ≠ 22) or kx + 14 (k ≠ 8) or correct answer spoilt 1(b) 15x 2 M1 for area of one correct rectangle e.g. 21x, 12x, 7x

This question in 0607/42 Feb/March 2025

16 In this question, all lengths are in centimetres. B C NOT TO SCALE A D E The diagram shows a quadrilateral ABCD. ABCE is a rhombus and CDE is a right-angled triangle. AC = ( 2 3 ) x and BE = 2x . (a) Show that AE = 2x . [2] (b) Find, in terms of x, the perimeter of ABCD. … [4]

6 marks

Mark scheme: 16(a) M1 Use of AE = 2x scores M0 2 2 2 x  3 x oe oe e.g. tan EAM = ( M is midpoint of AC) )  AE =  x + ( x 3 2 2 A1 x [AE = ] x + 3x = 2x oe e.g. [ AE ] = = 2x 2 sin30 or 4x = 2x oe 16(b) 7 x + 3 x oe final answer 4 B3 for answer kx + x 3 or 7x + k x 3 OR B3 for CD = x 3 and for ED = x or B2 for CD = x 3 or for ED = x CD or M1 for e.g. = sin60 oe 2 x

This question in 0607/42 May/June 2025