Area and perimeter — Paper 4 · IGCSE Mathematics 0580

9 Q B 525 m 104° NOT TO 872 m SCALE A C ABC is a triangular field on horizontal ground. There is a vertical pole BQ at B. AB = 525 m, BC = 872 m and angle ABC = 104°. (a) Use the cosine rule to calculate the distance AC. AC = … m [4] (b) The angle of elevation of Q from C is 1.0°. Showing all your working, calculate the angle of elevation of Q from A. … [4] (c) (i) Calculate the area of the field. … m2 [2] (ii) The field is drawn on a map with the scale 1 : 20 000. Calculate the area of the field on the map in cm2. … cm2 [2]

12 marks

Mark scheme: 9(a) 1120 or 1121. … 4 M2 for [ AC 2 =] 5252 + 8722 − 2×525×872×cos 104 or M1 for implicit version A1 for 1 257 000 to 1 258 000 9(b) [QB or x =] 872 × tan 1 seen M2 QB M1 for tan 1 = 872 tan = their QB ÷ 525 M1 1.7 or 1.660 to 1.661 nfww A1 dep on M3 9(c)(i) 222 000 or 222 100. … or 222 101 2 1 M1 for × 525 × 872 × sin104 2 9(c)(ii) 5.55 or 5.550 to 5.553 nfww 2FT FT their (c)(i) × 1002 ÷ 20 0002 M1 for their (c)(i) × 1002 ÷ 20 0002 or restart

This question in 0580/43 May/June 2017

7 In this question, all measurements are in metres. 6 NOT TO x SCALE 2x – 3 The diagram shows a right-angled triangle. (a) Show that 5x2 - 12x - 27 = 0. [3] (b) Solve 5x2 - 12x - 27 = 0. Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4] (c) Calculate the perimeter of the triangle. … m [2] (d) Calculate the smallest angle of the triangle. … [2]

11 marks

Mark scheme: 7(a) x2 + (2x – 3)2 = 62 oe M1 or x2 + 4x2 – 6x – 6x + 9 = 36 4x2 – 6x – 6x + 9 or better B1 5x2 – 12x – 27 = 0 A1 Dep on M1B1 with no errors or omissions 7(b) 2 B2 2 −−( 12) ± ( − 12) − 4(5)( − 27) B1 for ( −12) − 4(5)( −27) or for 2 × 5 2  12  or better  x −  oe  10  −−( 12) + q −−( 12) − q 2 12  12  27 or oe or oe or ±   + 2 × 5 2 × 5 10  10  5 or both – 1.42, 3.82 final answers B2 B1 for each If B0, SC1 for answers – 1.4 or –1.415… to – 1.415 and 3.8 or 3.815 to 3.815… or answers –1.41 and 3.81 or – 1.42 and 3.82 seen in working or for –3.82 and 1.42 as final ans 7(c) 14.4 or 14.5 or 14.44 to 14.46 2 2FT for 3 × their positive root + 3 evaluated to 3sf or better M1 for 3 × their positive root + 3 oe 7(d) 39.5 or 39.46 to 39.54… 2 M1 for trig statement seen to find either angle their x their (2 x − 3) sin = oe or sin = oe 6 6

This question in 0580/42 May/June 2018

3 North C D 170 m 120 m NOT TO 150 m SCALE E 50 m A 100 m B The diagram shows a field ABCDE. (a) Calculate the perimeter of the field ABCDE. … m [4] (b) Calculate angle ABD. Angle ABD = … [4] (c) (i) Calculate angle CBD. Angle CBD = … [2] (ii) The point C is due north of the point B. Find the bearing of D from B. … [2] (d) Calculate the area of the field ABCDE. Give your answer in hectares. [1 hectare = 10 000 m2] … hectares [4]

16 marks

Mark scheme: 3(a) 530 4 B3 for [DE] = 130 m and [DC] = 80 m or B2 for [DE] = 130 m or [DC] = 80 m or M1 for 502 + 1202 or 1702 – 1502 3(b) 52.9 or 52.89… 4 100 2 + 150 2 − 120 2 M2 for 2 × 100 × 150 or M1 for 1202 = 1002 + 1502 – 2 × 100 × 150cos(…) 181 A1 for 0.603 or 0.6033…or 300 3(c)(i) 28.1 or 28.07… 2 15 M1 for cos = oe 17 3(c)(ii) 331.9 or 331.9… 2 FT 360 – their (c)(i) M1 for 360 – their (c)(i) oe 3(d) 1.5[0] or 1.498… nfww 4 1 M1 for × 50 × 120 oe 2 1 M1 for × 100 × 150sin(their (b)) oe 2 1 M1 for × 150 ×theirCD oe 2 1 or × 150 × 170 × sin their (c)(i) 2 If 0 scored, SC1 for dividing their area by 10 000

This question in 0580/41 May/June 2019

4 S NOT TO SCALE 55° P 150 m 25° 45° R 120 m Q The diagram shows two triangles. (a) Calculate QR. QR = … m [3] (b) Calculate RS. RS = … m [4] (c) Calculate the total area of the two triangles. … m2 [3]

10 marks

Mark scheme: 4(a) 65.4 or 65.36 to 65.37 3 M1 for 1502 + 1202 – 2 × 150 × 120 cos 25 A1 for 4270 or 4272 to 4273 4(b) 125 or 124.7 to 124.8 4 B1 for [angle S =] 80 150sin55 M2 for sin their 80 sin their 80 sin55 or M1 for = oe 150 RS 4(c) 10 400 or 10 410 to 10 440 nfww 3 1 M1 for × 120 × 150sin25 oe 2 1 M1 for × 150 × their (b) sin45 oe 2

This question in 0580/42 May/June 2020

5 All the lengths in this question are in centimetres. x + 1 A F D NOT TO 2x E SCALE x + 3 B C 4x – 5 The diagram shows a shape ABCDEF made from two rectangles. The total area of the shape is 342 cm2. (a) Show that x 2 + x - 72 = 0 . [5] (b) Solve by factorisation. x 2 + x - 72 = 0 x = … or x = … [3] (c) Work out the perimeter of the shape ABCDEF. … cm [2] (d) Calculate angle DBC. Angle DBC = … [2]

12 marks

Mark scheme: 5(a) ( 4 x − 5 )( x + 3 ) + ( x + 1)( x − 3 ) = 342 M2 M1 for ( 4 x − 5 )( x + 3 ) or ( x + 1)( x − 3 ) or or for 2 x ( 4 x − 5 ) or ( 3 x − 6 )( x − 3 ) 2 x ( 4 x − 5 ) − ( 3 x − 6 )( x − 3 ) = 342 4 x 2 + 12 x − 5 x − 15 oe and M2 M1 for each x 2 + x − 3 x − 3 oe seen OR 8 x 2 − 10 x and 3 x 2 − 15 x + 18 seen 5 x 2 + 5 x − 18 = 342 leading to A1 no errors or omission x 2 + x − 72 = 0 5(b) ( x + 9 )( x − 8 ) M2 B1 for (x + a)(x + b) where ab = – 72 or a + b = 1 and a, b are integers 8, −9 B1 5(c) 86 2 FT for 12 × their x − 10 (x positive) B1 for any one of 27, 11, 16 seen or for 2 x + 2 x + 4 x − 5 + 4 x − 5 oe or better soi 5(d) 22.2 or 22.16 to 22.17 2 11 their x + 3 M1 for tan = or 27 4 × their x − 5

This question in 0580/43 May/June 2020

4 (a) A rectangle measures 8.5 cm by 10.7 cm, both correct to 1 decimal place. Calculate the upper bound of the perimeter of the rectangle. … cm [3] (b) B C D E 80° NOT TO SCALE 9 cm h 40° A 12 cm F ABDF is a parallelogram and BCDE is a straight line. AF = 12 cm, AB = 9 cm, angle CFD = 40° and angle FDE = 80°. (i) Calculate the height, h, of the parallelogram. h = … cm [2] (ii) Explain why triangle CDF is isosceles. … … [2] (iii) Calculate the area of the trapezium ABCF. … cm2 [3] (c) C B 12 cm NOT TO SCALE O 21° D A A, B, C and D are points on the circle, centre O. Angle ABD = 21° and CD = 12 cm. Calculate the area of the circle. … cm2 [5] (d) x° NOT TO 8 cm 9.5 cm SCALE The diagram shows a square with side length 8 cm and a sector of a circle with radius 9.5 cm and sector angle x°. The perimeter of the square is equal to the perimeter of the sector. Calculate the value of x. x = … [3]

18 marks

Mark scheme: 4(a) 38.6 3 M2 for [2 ×] (8.5 + 0.05 + 10.7 + 0.05) or M1 for 8.5 + 0.05 or 10.7 + 0.05 4(b)(i) 8.86 or 8.863… 2 h M1 for = sin 80 or better oe 9 4(b)(ii) ∠CDF = 100 leading to ∠DCF = 40 M1 Implied by 180-(100 + 40) = 40 Or or ∠EDF = 80 leading to ∠DCF = 40 80 – 40 ‘two equal angles’ A1 With no incorrect work seen 4(b)(iii) 66.5 or 66.45 to 66.47… 3 M2 for 0.5(3 + 12) × their (b)(i) or 12 × their (b)(i) – 0.5 × 9 × 9 × sin 100 oe or B1 for DC = 9 or BC = 3 4(c) 130 nfww or 129.6 to 129.8 5 B1 for ∠ACD = 21º or ∠CAD = 69º Method 1 12 M2 for cos 21 = oe AC or M1 for ∠ADC = 90 soi M1 for π(their AC/2)2 OR Method 2 12 r M2 for = oe sin138 sin 21 or M1 for ∠COD = 138 soi M1 for π (their r ) 2 OR Method 3 6 M2 for cos 21 = oe OC or M1 for ∠CXO = 90 soi where X is the point where the perpendicular from O meets the chord CD M1 for π ( their OC) 2 4(d) 78.4 or 78.37 to 78.41 3 M2 for x × 2 × π × 9.5 + 2 × 9.5 = 4 × 8 oe 360 x or M1 for × 2 × π × 9.5 360 After M0, SC1 for 9.5x + 19 = 32 oe

This question in 0580/41 Oct/Nov 2020

6 D 287.9 m North NOT TO 205.8 m SCALE C 168 m 38° 192 m A B The diagram shows a field, ABCD, on horizontal ground. BC = 192 m, CD = 287.9 m, BD = 168 m and AD = 205.8 m. (a) (i) Calculate angle CBD and show that it rounds to 106.0°, correct to 1 decimal place. [4] (ii) The bearing of D from B is 038°. Find the bearing of C from B. … [1] (iii) A is due east of B. Calculate the bearing of D from A. … [5] (b) (i) Calculate the area of triangle BCD. … m2 [2] (ii) Tomas buys the triangular part of the field, BCD. The cost is $35 750 per hectare. Calculate the amount he pays. Give your answer correct to the nearest $100. [1 hectare = 10 000 m2] $ … [2]

14 marks

Mark scheme: 6(a)(i) 106.01 to 106.02 4 M2 for 192 2 + 168 2 − 287.9 2 [cos[∠CBD] =] oe 2 × 192 × 168 or M1 for the implicit form A1 for –0.276 to – 0.275 6(a)(ii) 292.0 or 291.98 to 291.99 1 6(a)(iii) 310.0 or 310.03 to 310.04 5 168 × sin(90 − 38) M2 for [sin A =] 205.8 sin A sin(90 − 38) or M1 for = 168 205.8 A1 for [A =] 40.0 or 40.03 to 40.04 M1 dep for 270 + their angle DAB oe 6(b)(i) 15 500 or 15 501 to 15 503. … 2 M1 for 0.5 × 192 × 168 × sin(106) oe 6(b)(ii) 55 400 2 FT 3.575 × their (b)(i) oe rounded to nearest 100 M1 for figs 35 75 × figs their (b)(i) or figs 554 or figs 5541 to figs 5543

This question in 0580/41 Oct/Nov 2020

5 A NOT TO SCALE 10.6 cm 58° 78° B C X 6.4 cm The diagram shows triangle ABC. X is a point on BC. AX = 10.6 cm, XC = 6.4 cm, angle ABC = 58° and angle AXB = 78°. (a) Calculate AC. AC = … cm [4] (b) Calculate BX. BX = … cm [4] (c) Calculate the area of triangle ABC. … cm2 [3]

11 marks

Mark scheme: 5(a) 13.5 or 13.47… 4 B1 for angle 102 seen M2 for 10.6 2 + 6.4 2 − 2 × 10.6 × 6.4 × cos (180 − 78 ) OR M1 for 10.6 2 + 6.4 2 − 2 × 10.6 × 6.4 × cos (180 − 78 ) A1 for 181.5… 5(b) 8.68 or 8.682 to 8.683 nfww 4 B1 for angle = 44 10.6 M2 for sin(180 – 58 – 78) × oe sin 58 sin(180 − 58 − 78) sin58 or M1 for = oe x 10.6 5(c) 78.2 or 78.17 to 78.19… 3 1 M2 for × 10.6 × ( 6.4 + their 8.68 ) × sin ( 78 ) 2 oe OR 1 M1 for × 10.6 × 6.4 × sin(180 – 78) oe 2 1 M1 for × 10.6 × their 8.68 × sin78 oe 2

This question in 0580/41 May/June 2021

1 (a) NOT TO 5.7 cm SCALE 9.2 cm 19.4 cm The diagram shows a brick in the shape of a cuboid. (i) Calculate the total surface area of the brick. … cm2 [3] (ii) The density of the brick is 1.9 g/cm3. Work out the mass of the brick. Give your answer in kilograms. [Density = mass ÷ volume] … kg [3] (b) 9000 bricks are needed to build a house. 200 bricks cost $175. Work out the cost of the bricks needed to build 5 houses. $ … [3] (c) Saskia builds a wall using 1500 bricks. She can build at the rate of 40 bricks each hour. She works for 9 hours each day. Saskia starts work on 6 July and works every day until the wall is completed. Find the date when she completes the wall. … [3] (d) Rafa has a cylindrical tank. The cylinder has a height of 105 cm and a diameter of 45 cm. Calculate the capacity of the tank in litres. … litres [3]

15 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 683 3 M2 for [2]((19.4 × 9.2) + (5.7 × 9.2) + (19.4 × 5.7)) oe or M1 for one of 19.4 × 9.2 or 5.7 × 9.2 or 19.4 × 5.7 1(a)(ii) 1.93[0] or 1.932 to 1.933 3 M2 for 19.4 × 9.2 × 5.7 × 1.9 or M1 for 19.4 × 9.2 × 5.7 1(b) 39 375 3 M2 for 9000 ÷ 200 × 175 × 5 175 or M1 for 9000 ÷ 200 soi or for soi 200 1(c) 10th July 3 1 B2 for 4.1 to 4.2 or 4 or 4 days 1.5 6 hours Or M2 for answer 9th July or 11th July or M1 for 1500 ÷ (9 × 40) 1(d) 167 or 166.9 to 167.0… 3 B2 for answer with figs 167 or figs 1669 to 1670.. or M1 for π× 22.5 2 × 105 oe If 0 scored SC1 for answer 668 or 667.9 to 668.1

This question in 0580/41 Oct/Nov 2021

1 (a) Calculate the volume of (i) a solid cylinder with radius 6 cm and height 14 cm, … cm3 [2] (ii) a solid hemisphere with radius 6 cm. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … cm3 [2] (b) NOT TO SCALE 14 cm 6 cm The cylinder and hemisphere in part (a) are joined to form the solid in the diagram. The solid is made of steel and 1 cm 3 of steel has a mass of 7.85 g. (i) Show that 1 cm 3 of steel has a mass of 0.007 85 kg. [1] (ii) Calculate the total mass of the solid. … kg [2] (c) 2000 cm 3of iron is melted down and some of it is used to make 50 spheres with radius 2 cm. (i) Calculate the percentage of iron that is left over. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … % [3] (ii) The iron left over is then made into a cube. Calculate the length of an edge of the cube. … cm [1] (d) A solid cone has radius 3R cm and slant height 9R cm. A solid cylinder has radius x cm and height 7x cm. The total surface area of the cone is equal to the total surface area of the cylinder. Given that R = kx , find the value of k. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] k = … [4]

15 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 1580 or 1583 to 1584 2 M1 for π  6 2  14 1(a)(ii) 452 or 452.3 to 452.4... 2 3 M1 for 1 4  π  6  2 3 1(b)(i) 7.85 ÷ 1000 [= 0.00785] M1 1(b)(ii) 16[.0] or 15.95 to 15.99 2 FT {their (a)(i) + their (a)(ii)}  0.00785 evaluated to 3 sig fig or better M1 for (their (a)(i) + their (a)(ii)) × 0.00785 1(c)(i) 16.2 or 16.21 to 16.23 3 4 3 2000 − 50  π 2 3 M2 for 100  2000 4 3 50  π 2 3 or for  100 2000 4 3 50  π 2 3 or M1 for 2000 1(c)(ii) 6.87 or 6.870 to 6.872 1  4 3  FT 3 2000 − their 50  π 2    3  evaluated to 3sf or better 1(d) 2 4 M1 for [π](3 R ) 2 + [π]3 R  9 R oe oe 3 M1 for 2[π]x 2 + 2[π]x  7 x oe M1 for their area of cone = their area of cylinder seen

This question in 0580/41 Oct/Nov 2022

8 AB is a line with midpoint M. A is the point (2, 3) and M is the point (12, 7). (a) Find the coordinates of B. ( … , … ) [2] (b) Show that the equation of the perpendicular bisector of AB is 2y + 5x = 74 . [4] (c) The perpendicular bisector of AB passes through the point N. The point N has coordinates (2, n). Find the value of n. n = … [1] (d) Points A, M and N form a triangle. Find the area of the triangle. … [2]

9 marks

Mark scheme: 8(a) (22, 11) 2 B1 for each value 8(b) their11 − 3 M1 oe or better their 22 − 2 1 M1 −their m Substitution of (12, 7) into M1 Accept y – 7 = their m(x – 12) oe y = (their m)x + c leading to 2y + 5x = 74 final answer A1 Without error or omission 8(c) 32 1 8(d) 145 2 1 M1 for × (their 32 – 3) × 10 oe 2 or 1 2 2 2 2  (7 − 3) + (12 − 2)  (their 32 − 7) + (2 − 12) oe 2

This question in 0580/42 Oct/Nov 2022

3 C NOT TO SCALE ( x + 3) cm A ( 2x + 5) cm B The diagram shows a right-angled triangle ABC. (a) (i) The area of the triangle is 60 cm 2. Show that 2x 2 + 11 x - 105 = 0 . [3] (ii) Solve by factorisation. 2x 2 + 11x - 105 = 0 x = … or x = … [3] (iii) Calculate angle ACB. … [3] (b) Triangle ABC is similar to triangle DEF. Triangle DEF has an area of 93.75 cm 2. (i) Find the size of the smallest angle of triangle DEF. … [1] (ii) Find the length of the shortest side of triangle DEF. … cm [3]

13 marks

Mark scheme: 3(a)(i)  x  3  (2 x  5) M1 Accept (x + 3)(2x + 5) = 2 × 60 or 120  60 Accept e.g. (x + 3) (x + 2.5) = 60 without 2 division by 2 shown for M1 (but not A1) 2x2 + 6x + 5x + 15 seen B1 Accept 2x2 + 11x + 15 seen 2x2 + 11x – 105 = 0 A1 Correct completion after M1B1 with the fraction seen removed with no errors or omissions seen 3(a)(ii) (2x + 21) (x – 5) [= 0] M2 M1 for partial factors 2x (x – 5) + 21(x – 5) [ = 0] or x (2x + 21) – 5 (2x + 21) [ = 0] OR (2x + a)(x + b) [ = 0] where ab = – 105 or 2b + a = 11 –10.5 and 5 B1 3(a)(iii) 61.9 or 61.92 to 61.93 3 2  their 5  5 M2 for tan = oe their 5  3 or B1FT for 2 × their 5 + 5 and their 5 + 3 3(b)(i) 28.1 or 28.07 to 28.08 1 FT their 90 – their (a)(iii) unless their (a)(iii) < 45, in which case FT their (a)(iii) 3(b)(ii) 10 3 93.75 M2 for (their 5  3)  oe 60 93.75 60 or M1 for or oe seen 60 93.75  their 5  3 2 60 oe or    x  93.75

This question in 0580/42 May/June 2023

4 In this question all the measurements are in centimetres. NOT TO r + 2 SCALE 30° r + 5 r + 1 The area of the triangle is equal to the area of the square. (a) Show that 3r 2 + r - 6 = 0 . [4] (b) Solve the equation 3r 2 + r - 6 = 0 . Give your answer to 2 decimal places. You must show all your working. r = … or r = … [3] (c) Find the perimeter of the square. … cm [2]

9 marks

Mark scheme: 4(a) 1 2 M2 1 ( r  5)( r  2)sin30  ( r  1) M1 for ( r  5)( r  2)sin30 oe 2 2 r 2  5r  2 r  10 or r 2  r  r  1 soi B1 Leading to 3r 2  r  6  0 with no errors A1 Dependent on both expansions seen or omissions 4(b) 2 B2 1 1  4(3)( 6) 1 p B1 for 21  4(3)( 6) or for 2(3) 2(3) Or 1 p or 2(3) or 1  1  2   2   oe 2 6  6   1   r   or  6   2  or 1 1  1     18    oe 3  2  2   2    1   3r    2  –1.59 and 1.26 B1 4(c) 9.028 to 9.040 2 M1 for (their root (greater than –1) + 1) × 4

This question in 0580/43 May/June 2024

6 The diagram shows a field ABCD. A straight path AC goes across the field. D 830 m 106° C NOT TO 420 m SCALE A 62° 1150 m B (a) Show that AC = 1028 m, correct to the nearest metre. [3] (b) Angle ACB is obtuse. Calculate angle ACB. Angle ACB = … [4] (c) Part of the field, triangle ACD, is sold for $41 500. Calculate the cost of 1 hectare of this part of the field. Give your answer correct to the nearest dollar. [1 hectare = 10 000 m2] $ … [4]

11 marks

Mark scheme: 6(a) 2 2 M2 or M1 for 420 + 830 −2 420  830  cos106 oe 420 2 + 830 2 −2 420  830  cos106 oe A1 for 1 057 474 …. 1028.3... A1 6(b) 99[.0] or 98.98 to 99.1[0…] 4 B3 for 80.89 to 81.02 1150sin62 or M2 for sin[ ACB =] oe 1028 1028 1150 or M1 for = oe sin62 sin ACB 6(c) 2477 cao nfww 4 B3 for answer 2476.9… or M2 for 1 P  420  830  sin106  = 41 500 2 10000 oe 1 or M1 for  420  830  sin106 oe 2

This question in 0580/41 Oct/Nov 2024

3 (a) Simplify. (i) 3m - 5n - 4 m + 8 n … [2] (ii) ( 3a 2 c 3 ) 4 … [2] 4 x 3 x 2x (iii) - + 5 10 15 … [2] (b) This isosceles triangle has a perimeter of 35.5 cm. NOT TO a cm SCALE ( 3a + 2) cm Find the value of a. a = … [3] (c) Using the quadratic formula, solve 5x 2 - 4 x - 3 = 0 . You must show all your working. x = … or x = … [3] (d) Solve these simultaneous equations. y = x 2 - 4x + 5 y = 2x - 3 You must show all your working. x = … y = … x = … y = … [5]

17 marks

Mark scheme: 3(a)(i) –m + 3n final answer 2 B1 for –m or [+] 3n in final answer or for –m + 3n seen and then spoiled 3(a)(ii) 81a 8 c12 final answer 2 B1 for final answer in correct form with any two of 81, a8, c12 correct or for 81a 8 c12 seen and then spoiled 3(a)(iii) 19 x 2 6  4 x −3 3 x + 2  2 x final answer M1 for oe 30 30 3(b) 4.5 oe 3 M1 for a + 2(3a + 2) = 35.5 oe M1 for correct ka = b for their linear equation 3(c) 2 M2 2 −−( 4)  ( −4) −−4 5 ( 3) M1 for ( −4) −−4 5 ( 3) or better oe 2  5 −−( 4) + q −−( 4) − q or for or or better 2  5 2  5 –0.472 or –0.4718 to –0.4717 B1 and 1.27 or 1.271 to 1.272 3(d) x2 – 6x + 8 [= 0] M2 M1 for x2 – 4x + 5 = 2x – 3 or or y2 – 6y + 5 [= 0]  y + 3  2  y + 3  y = − 4 + 5      2   2  (x – 4)(x – 2) [= 0] M1 FT their 3-term quadratic but not if x2 – 4x + 5[= or 0] (y – 1)(y – 5) [=0] OR −−( 6)  ( −6) 2 − 4[1]  8 [x = ] 2[ 1] or −−( 6)  ( −6) 2 − 4[1]  5 [y = ] 2[ 1] OR [x = ] 3  −+8 9 or [y = ] 3  −+5 9

This question in 0580/42 Oct/Nov 2024

7 (a) (i) 13 cm NOT TO 8 cm SCALE 9 cm Calculate the area of the trapezium. … cm2 [2] (ii) ( y + 4) cm NOT TO ( y + 2) cm SCALE ( y + 1) cm The area of this trapezium is 264 cm2. (a) Show that 2y 2 + 9 y - 518 = 0 . [3] (b) Solve 2y 2 + 9 y - 518 = 0 by factorisation to find the value of y. y = … [3] (b) NOT TO SCALE 8 cm 75° The diagram shows a sector of a circle with radius 8 cm and angle 75°. Find the perimeter of the sector. … cm [3] (c) A B NOT TO SCALE 5 cm P Q O The diagram shows a shape ABQP made from three straight lines and an arc of a sector of a circle. The sector has centre O and angle 90°. POQ is a straight line and AP = PO = OQ = QB = 5 cm. Find the area of ABQP. Give your answer in the form a + k r . … cm2 [4]

15 marks

Mark scheme: 7(a)(i) 88 2 1 M1 for (9 + 13)  8 oe 2 7(a)(ii)(a) 1 M1 ( y + 4 + y + 1)  ( y + 2) [ = 264] 2 or 1 3 ( y + 2) + ( y + 1)  ( y + 2) [ = 264] 2 2 y 2 + 5 y + 4 y + 10 B1 Leading to 2 y 2 + 9 y − 518 = 0 A1 No errors or omissions 7(a)(ii)(b) (2 y + 37)( y − 14) B2 B1 for (2 y + a )( y + b) where ab = –518 or a + 2b = 9 or 2 y ( y − 14) + 37( y − 14) or y (2 y + 37) − 14(2 y + 37) 14 B1 7(b) 26.5 or 26.47... 3 10π B2 for 10.5 or 10.47… or 3 OR 75 M2 for 8 + 8 +  2π8 360 75 or M1 for  2π8 360 7(c) 25 4 25 + π 90 2 2 2 2 M2 for  ( (5 + 5 ) ) 360 or M1 for [radius 2 = ] 52 + 52 1 M1 for [triangle area = ] [2×] 5 5 oe 2

This question in 0580/43 Oct/Nov 2024

7 The diagram shows a parallelogram. NOT TO SCALE x m 6.51 m The parallelogram has the same perimeter as a circle with radius 4 m. (a) Show that x = 6.06 m , correct to 2 decimal places. [4] (b) NOT TO SCALE 6.06 m 36° 6.51 m The floor of a room is in the shape of this parallelogram. It costs $18 per square metre to tile the floor. Calculate the total cost of tiling the floor. $ … [4]

8 marks

Mark scheme: 7(a) 2 × π × 4 = 6.51 × 2 + 2x M3 M1 for 2 × π × 4 oe or better M1 for 6.51 × 2 + 2x soi 6.055 to 6.058 [= 6.06] A1 7(b) 574 or 575 4 M2 for 6.51 × 6.06 × cos 36 oe or 574.1 to 574.5… or M1 for [height = ] 6.06 × cos 36 oe seen M1 for their area × 18 leading to answer Alternative method: 1 M2 for 2 × × 6.51 × 6.06 × sin(90 + 36) oe 2 1 or M1 for × 6.51 × 6.06 × sin(90 + 36) oe seen 2 M1 for their area × 18 leading to answer

This question in 0580/42 Oct/Nov 2025

15 A NOT TO D SCALE 140° 112 m 180 m 300 m C B The diagram shows a field, ABCD, in the shape of a quadrilateral. BD is a straight path across the field. (a) Calculate BC. BC = … m [3] (b) Calculate angle DBC. Angle DBC = … [3] (c) The total area of the field, ABCD, is 35 900 m2. Work out the length of the shortest distance from D to AB. … m [4]

10 marks

Mark scheme: 15(a) 392 or 392.4 to 392.5 3 2 2 M2 for 300 + 112 −2 300 112  cos140 OR M1 for 3002 + 1122 − 2  300  112  cos140 A1 for 154 022[…] 15(b) 10.6 or 10.7 or 10.55 to 3 112sin140 M2 for oe 10.69 their (a) 300 2 + ( their (a) ) 2 − 112 2 or cos[ DBC ] = 2  300  their (a) 112 their (a) or M1 for = oe sin DBC sin140 or 112 2 = 300 2 + ( their (a) ) 2 − 2  300  their (a)  cos DBC oe 15(c) 279 or 278.9… 4 M3 for 1  1  (35 900 –  112  300  sin140 ) ÷   180  oe 2  2  OR 1 M1 for  112  300  sin140 oe 2 M1 for recognition that the shortest distance from D is perpendicular to AB

This question in 0580/42 Oct/Nov 2025