Circles, arcs and sectors — Paper 4 · IGCSE Mathematics 0580

8 (a) In triangle TXZ, TX = 12.5 cm and angle TZX = 37°. Y is a point on the line XZ such that TY = 9.9 cm, angle XTY = 23° and angle TYZ = 72°. T 23° NOT TO SCALE 12.5 cm 9.9 cm 72° 37° X Y Z (i) Calculate XY. XY = … cm [4] (ii) Calculate TZ. TZ = … cm [3] (b) The diagram shows a shape made up of three identical sectors of a circle, each with sector angle 65°. The perimeter of the shape is 20.5 cm. 65° 65° NOT TO SCALE 65° Calculate the radius of the circle. … cm [4]

11 marks

This question in 0580/42 Feb/March 2017

5 NOT TO SCALE 10 cm 3 cm The diagram shows a hollow cone with radius 3 cm and slant height 10 cm. (a) (i) Calculate the curved surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] … cm2 [2] (ii) Calculate the perpendicular height of the cone. … cm [3] (iii) Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = r r h .] 3 … cm3 [2] (b) O O NOT TO SCALE x 10 cm 3 cm P The cone is cut along the line OP and is opened out into a sector as shown in the diagram. Calculate the sector angle x. x = … [4] (c) O NOT TO SCALE The diagram shows the same sector as in part (b). Calculate the area of the shaded segment. … cm2 [4]

15 marks

Mark scheme: 5(a)(i) 94.2 or 94.3 or 94.24 to 94.26 2 M1 for π × 3 × 10 5(a)(ii) 9.54 or 9.539… 3 2 2 M2 for 10 − 3 or M1 for h 2 + 32 = 10 2 oe 5(a)(iii) 89.9 or 89.90 to 89.92… 2 1 2 M1 for × π × 3 × their (a)(ii) 3 5(b) 108 or 107.9 to 108.1 nfww 4 π × 3 × 10 their (a)(i) M3 for × 360 oe or × 360 oe or π × 10 2 π × 10 2 2 × π × 3 × 360 oe 2 × π × 10 x 2 or M2 for × π × 10 = their (a)(i) oe 360 x or × 2 × π × 10 = 2 × 3 × π oe 360 x 2 or M1 for × π × 10 seen 360 x or × 2 × π × 10 seen 360 5(c) 46.6 to 46.8 4 their (b) 2 1 M3 for × π × 10 − × 10 × 10 × sin(their (b)) oe 360 2 their (b) 2 or M1 for × π × 10 or their (a)(i) soi 360 1 and M1 for × 10 × 10 × sin(their (b)) soi 2

This question in 0580/42 May/June 2017

6 (a) r h NOT TO SCALE 10 cm The diagrams show a cube, a cylinder and a hemisphere. The volume of each of these solids is 2000 cm3. (i) Work out the height, h, of the cylinder. h = … cm [2] (ii) Work out the radius, r, of the hemisphere. 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 r = … cm [3] (iii) Work out the surface area of the cube. … cm2 [3] (b) NOT TO 7 cm SCALE 40º 10 cm (i) Calculate the area of the triangle. … cm2 [2] (ii) Calculate the perimeter of the triangle and show that it is 23.5 cm, correct to 1 decimal place. Show all your working. [5] (c) NOT TO SCALE cº 9 cm The perimeter of this sector of a circle is 28.2 cm. Calculate the value of c. c = … [3]

18 marks

Mark scheme: 6(a)(i) 25.5 or 25.46… 2 M1 for π × 52 × h = 2000 oe 6(a)(ii) 9.85 or 9.847… 3  2  M2 for [r3=] 2000 ÷  π  oe  3  2 or M1 for πr3 = 2000 oe 3 6(a)(iii) 952 or 952.4…. 3 3 2 M2 for [6 ×] 2000 or M1 for 3 2000 or 6 times their area of one face 6(b)(i) 22.5 or 22.49… 2 1 M1 for × 7 × 10 × sin40 2 6(b)(ii) √(102 + 72 – 2 × 10 × 7 cos40) + 7 M3 M2 for 102 + 72 – 2 × 10 × 7 cos40 + 10 or M1 for correct implicit cosine rule 23.46… A2 A1 for 6.46… or 41.7 to 41.8 6(c) 64.9 or 64.92 to 64.94 3 c M2 for 28.2 – 2 × 9 = × 2 × π × 9 oe 360 c or M1 for × 2 × π × 9 soi 360

This question in 0580/43 Oct/Nov 2017

8 (a) The exterior angle of a regular polygon is x° and the interior angle is 8x°. Calculate the number of sides of the polygon. … [3] (b) C NOT TO SCALE O D B 58° A A, B, C and D are points on the circumference of the circle, centre O. DOB is a straight line and angle DAC = 58°. Find angle CDB. Angle CDB = … [3] (c) R O NOT TO SCALE 48° P Q P, Q and R are points on the circumference of the circle, centre O. PO is parallel to QR and angle POQ = 48°. (i) Find angle OPR. Angle OPR = … [2] (ii) The radius of the circle is 5.4 cm. Calculate the length of the major arc PQ. … cm [3]

11 marks

Mark scheme: 8(a) 18 3 B2 for 20 nfww or M1 for 8 x + x = 180 or better 8(b) 32 3 B1 for angle DBC = 58 B1 for angle BCD = 90 8(c)(i) 24 2 B1 for angle PRQ = 24 8(c)(ii) 29.4 or 29.40 to 29.41 3 360 − 48 M2 for × 2 × π × 5.4 360 or B2 for answer (minor arc) 4.52 or 4.523 to 4.524… 48 or M1 for × 2 × π × 5.4 360

This question in 0580/41 May/June 2018

5 A O NOT TO SCALE 8 cm 7 cm 78° C B The diagram shows a design made from a triangle AOC joined to a sector OCB. AC = 8 cm, OB = OC = 7 cm and angle ACO = 78°. (a) Use the cosine rule to show that OA = 9.47 cm, correct to 2 decimal places. [4] (b) Calculate angle OAC. Angle OAC = … [3] (c) The perimeter of the design is 29.5 cm. Show that angle COB = 41.2°, correct to 1 decimal place. [5] (d) Calculate the total area of the design. … cm2 [4]

16 marks

Mark scheme: 5(a) 8² + 7² − 2 × 7 × 8 × cos78 oe M2 M1 for correct implicit version 9.471.. to 9.472 A2 A1 for 89.7… 5(b) 46.3 or 46.29 to 46.30… 3 7sin78 M2 for [sin OAC = ] 9.47 sin OAC sin78 or M1 for = 7 9.47 5(c) 29.5 – (7 + 8 + 9.47) M1 360 × (29.5 − (7 + 8 + 9.47)) M3 x M2 for × 2 × π × 7 = their arc length 2 × π × 7 360 oe x or M1 for × 2 × π × 7 oe 360 41.15 to 41.171.. B1 5(d) 45[.0] or 44.98 to 45.01 nfww 4 M3 for 41.2 2 ½ × 8 × 7 × sin 78 oe + × π × 7 oe 360 OR M1 for ½ × 8 × 7 × sin 78 oe or ½ × 8 × 9.47 × sin their (b) oe 41.2 2 M1 for × π × 7 oe 360

This question in 0580/42 May/June 2018

10 NOT TO O SCALE Y X 8 cm A C 15 cm B 22.4 cm The diagram shows a circle, centre O. The straight line ABC is a tangent to the circle at B. OB = 8 cm, AB = 15 cm and BC = 22.4 cm. AO crosses the circle at X and OC crosses the circle at Y. (a) Calculate angle XOY. Angle XOY = … [5] (b) Calculate the length of the arc XBY. … cm [2] (c) Calculate the total area of the two shaded regions. … cm2 [4] Question 11 is printed on the next page.

11 marks

Mark scheme: 10(a) 132.26 to 132.28 or 132.3 5 B1 for angle ABO or angle CBO = 90 soi 15 M1 for tan [XOB] = oe 8 224. M1 for tan [BOY] = oe 8 A1 for [BOY =]70.3… or [XOB =] 61.9…. 10(b) 18.4 or 18.5 or 18.43 to 18.48 2 their (a) M1 for × 2 × π × 8 oe 360 10(c) 75.7 to 75.9 4 1 M1 for (15 + 22.4 ) × 8 oe 2 their ( a ) 2 M2 for × π × 8 oe 360 or M1 for one sector either  15  inv tan    8 × π × 8 2 oe 360  22.4  inv tan    8 2 or × π × 8 oe 360 ( )( )

This question in 0580/41 Oct/Nov 2018

9 NOT TO SCALE A 8 cm 165° B O The diagram shows a sector of a circle with centre O, radius 8 cm and sector angle 165°. (a) Calculate the total perimeter of the sector. … cm [3] (b) The surface area of a sphere is the same as the area of the sector. Calculate the radius of the sphere. [The surface area, A, of a sphere with radius r is A = 4rr 2 .] … cm [4] (c) O NOT TO h SCALE r A B A cone is made from the sector by joining OA to OB. (i) Calculate the radius, r, of the cone. r = … cm [2] (ii) Calculate the volume of the cone. r r h .] [The volume, V, of a cone with radius r and height h is V = 13 2 … cm3 [4]

13 marks

Mark scheme: 9(a) 39[.0] or 39.03 to 39.04… 3 165 M2 for × 2 × π × 8 + 16 360 165 or M1 for × 2 × π × 8 360 9(b) 2.71 or 2.708… 4 165 2 [× π ] × 8 360 M3 for oe 4[× π ] 165 2 [× π ] × 8 360 or M2 for r2 = oe 4[× π ] 165 2 or M1 for × π × 8 oe seen 360 9(c)(i) 3.67 or 3.666 to 3.667 2 165 M1 for ×2[×π]×8 = 2[×π]× r or better 360 or for 165[×π]×8 2 = [π×]r ×8 or better 360 9(c)(ii) 100 or 100.0 to 100.1… final answer 4 1 2 2 2 M3 for π × their ( c )( i ) × 8 – their radius 3 or M2 for 8 2 – their radius 2 or M1 for (their (c)(i))2 + h2 = 82

This question in 0580/41 May/June 2020

8 (a) O 53° NOT TO 9.5 cm A B SCALE X Y The diagram shows a sector OXY of a circle with centre O and radius 9.5 cm. The sector angle is 53°. A lies on OX, B lies on OY and OA = OB . (i) Show that the area of the sector is 41.7 cm 2, correct to 1 decimal place. [2] 1 (ii) The area of triangle OAB is of the area of sector OXY. 3 Calculate OA. OA = … cm [4] (b) O 60° NOT TO 24 cm SCALE P Q The diagram shows a sector OPQ of a circle with centre O and radius 24 cm. The sector angle is 60°. A cone is made from this sector by joining OP to OQ. O NOT TO SCALE P Q Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm3 [6]

12 marks

Mark scheme: 8(a)(i) 53 2 M1 × π× 9.5 360 41.74 to 41.75 A1 8(a)(ii) 5.9[0] or 5.899 to 5.903.. 4 2 13 × 41.7 M3 for  OA =  oe   1 2 sin53 1 2 1 M2 for × OA × sin53 = × 41.7 oe 2 3 1 1 M1 for × OA × OB × sin53 = × 41.7 seen or 2 3 better 8(b) 396 or 397 or 396.4 to 396.6 6  60  M2 for [ r = ]  ×2× π×24  ÷ 2π oe or better  360  60 or M1 for 2πr = × 2 × π× 24 oe 360 M2 for 24 2 −a 2 or M1 for h2 + a2 = 242 1 2 M1 for 3π× their r × their h

This question in 0580/42 Feb/March 2021

8 (a) A solid cuboid measures 20 cm by 12 cm by 5 cm. (i) Calculate the volume of the cuboid. … cm3 [1] (ii) (a) Calculate the total surface area of the cuboid. … cm2 [3] (b) The surface of the cuboid is painted. The cost of the paint used is $1.52 . Find the cost to paint 1 cm 2 of the cuboid. Give your answer in cents. … cents [1] 9x (b) A solid metal cylinder with radius x and height is melted. 2 All the metal is used to make a sphere with radius r. Find r in terms of x. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 r = … [3] (c) NOT TO SCALE 20 cm 5 cm 150 cm The diagram shows a cylinder of length 150 cm on horizontal ground. The cylinder has radius 20 cm. The cylinder contains water to a depth of 5 cm, as shown in the diagram. Calculate the volume of water in the cylinder. Give your answer in litres. … litres [7]

15 marks

Mark scheme: 8(a)(i) 1200 1 8(a)(ii)(a) 800 3 M2 for [2 ×] (20 × 12 + 20 × 5 + 12 × 5) or M1 for 20 × 12 or 20 × 5 or 12 × 5 8(a)(ii)(b) 0.19 1 FT 152 ÷ their 800 8(b) 3 x 3 3 27 x 3 [π] or 1.5x B2 for r = or better 2 8[π] 4 3 2 9 x or M1 for πr = πx × 3 2 8(c) 13.6 or 13.59 to 13.61 7 If chord is AB and O is centre of the cross section −1  20 − 5  M2 for 2 × cos   oe  20  20 − 5 or M1 for cos = oe 20 theirAOB 2 M1 for × π × 20 360 1  82.8π  or (20)2   2  180  1 M1 for × 202 × sin(their AOB) oe 2 M1 for their area × 150 M1 for their volume ÷ 1000

This question in 0580/43 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

9 (a) NOT TO x cm SCALE ( x + 3)cm This rectangle has perimeter 20 cm. Find the value of x. x = … [3] (b) M y° NOT TO SCALE 20° This rhombus has perimeter 20 cm and angle y is obtuse. M is the midpoint of one of the sides. Find the value of y. y = … [5] (c) r cm NOT TO SCALE z cm 40° This sector of a circle has radius r and perimeter 20 cm. Find the value of z. z = … [6]

14 marks

Mark scheme: 9(a) 3.5 oe 3 M1 for 2(x + x + 3) = 20 oe M1 for correct ax = b for their linear equation 9(b) 116.8 or 116.83 to 116.85 nfww 5 5sin 20 M2 for sin p = 2.5 2.5 5 or M1 for = sin 20 sin p A1 for 43.2 or 43.15 to 43.17 M1dep for 180 – (20 + their 43.2) After 0 scored, SC1 for length of side = 5 9(c) 5.07 or 5.068 to 5.071 6 B3 for 7.41 or 7.412 to 7.413 40 or M2 for r + r + × 2 × π× r = 20 oe 360 40 or M1 for × 2 × π× r oe seen 360 M2 for 2 × 7.41 × sin 20 oe or 7.412 + 7.412 – 2(7.412) cos 40 oe 7.41sin 40 or oe sin70 or M1 for implicit version

This question in 0580/41 Oct/Nov 2021

7 (a) The diagram shows a container for storing grain. The container is made from a hemisphere, a cylinder and a cone, each with radius 2 m. 2 m The height of the cylinder is 5.2 m and the height of the cone is h m. 5.2 m NOT TO SCALE 2 m h m (i) Calculate the volume of the hemisphere. Give your answer as a multiple of r. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … m3 [2] 88 r (ii) The total volume of the container is m3 . 3 Calculate the value of h. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 h = … [4] (iii) The container is full of grain. Grain is removed from the container at a rate of 35 000 kg per hour. 1m3 of grain has a mass of 620 kg. Calculate the time taken to empty the container. Give your answer in hours and minutes. … h … min [3] (b) O NOT TO r cm 140° B SCALE A A and B are points on a circle, centre O, radius r cm. The area of the shaded segment is 65cm2. Calculate the value of r. r = … [4]

13 marks

Mark scheme: 7(a)(i) 16π 2 1 4 3 or 51 π final answer M1 for × π × 2 oe 3 3 2 3 7(a)(ii) 2.4[0] 4 B3 for answer in range 2.396… to 2.40… OR 16π M3 for their + π × 22 × 5.2 + 3 1 2 88π π× 2 × h = oe 3 3 88π 16π or M2 for – their – π × 22 × 5.2 3 3 oe or M1 for π × 22 × 5.2 oe 1 2 or π× 2 × h oe soi 3 7(a)(iii) 1 hour 38 min or 1 hour 37.8 min to 1 3 B2 for 1.63[2…] or 98 [mins] or 97.8 to hour 37.9… min 97.9… ] 88π × 620 3 or M1 for [× 60] oe 35000 7(b) 8.5[0] or 8.496 to 8.497 4 65 M3 for [r= ] oe 140 1 π − sin140 360 2 140 1 or M2 for π × r2 – r2 × sin140 [=65] 360 2 oe or M1 for either area expression seen

This question in 0580/42 Oct/Nov 2021

9 (a) O O NOT TO SCALE x° A B 2.4 cm AB The volume of a paper cone of radius 2.4 cm is 95.4 cm3. The paper is cut along the slant height from O to AB. The cone is opened to form a sector OAB of a circle with centre O. Calculate the sector angle x°. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … [6] (b) An empty fuel tank is filled using a cylindrical pipe with diameter 8 cm. Fuel flows along this pipe at a rate of 2 metres per second. It takes 24 minutes to fill the tank. Calculate the capacity of the tank. Give your answer in litres. … litres [4]

10 marks

Mark scheme: 9(a) 54[.0] or 53.99 to 54.03… 6 M2 for [h = ] 95.4 × 3 ÷ ( π × 2.42) oe 1 2 or M1 for 95.4 = × π × 2.4 × h 3 M2 for [slant ht , l = ] ( their h ) 2 + 2.4 2 or M1 for (their h)2 + 2.42 x M1 for × 2 × π × theirl = 2 × π × 2.4 oe 360 x 2 or × π × ( theirl ) = π × 2.4 × theirl 360 9(b) 14500 or 14470 to 14480 4 M3 for 200 × 60 × 24 × π × 42 [÷1000] or 2 × 60 ×24 ×π × 0.042 [×1000] or M2 for 200 × π × 42 or for 2 × π × 0.042 or M1 for π × 42 oe or π × 0.042 seen oe isw or 1000 cm3 = 1 litre soi or 1 m3 = 1000 litres soi or for 24 × 60 seen oe

This question in 0580/42 Feb/March 2022

11 (a) 28 cm A D AD NOT TO SCALE 20 cm N BC B C A rectangular sheet of paper ABCD is made into an open cylinder with the edge AB meeting the edge DC. AD = 28 cm and AB = 20 cm. (i) Show that the radius of the cylinder is 4.46 cm, correct to 3 significant figures. [2] (ii) Calculate the volume of the cylinder. … cm3 [2] (iii) N is a point on the base of the cylinder, such that BN is a diameter. Calculate the angle between AN and the base of the cylinder. … [3] (b) The volume of a solid cone is 310 cm 3. The height of the cone is twice the radius of its base. Calculate the slant height of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm [5]

12 marks

Mark scheme: 11(a)(i) 4.455 to 4.456… [= 4.46] 2 28 M1 for [r =] oe 2π 11(a)(ii) 1250 or 1247 to 1249.9… 2 M1 for 20  4.46 2 oe 11(a)(iii) 66[.0] or 65.95 to 66.02 3 20 M2 for [tan] = oe 2  4.46 or B1 for identifying angle ANB on cylinder not on rectangle 11(b) 11.8 or 11.82 to 11.83 5 310  3 M2 for [ r  ] 3 oe 2π 310 3 4 or [ h  ] 3 oe π or M1 for 310  13  r 2  2 r 1  h  2 or 310  π h   3  2  M2 for (their r ) 2   2  their r  2 oe or M1 for [l 2 ]  their r  2   2  their r  2 oe

This question in 0580/42 May/June 2022

10 (a) The lengths of the sides of a triangle are 11.4 cm, 14.8 cm and 15.7 cm, all correct to 1 decimal place. Calculate the upper bound of the perimeter of the triangle. … cm [2] (b) 15.6 cm NOT TO SCALE 150° The diagram shows a circle, radius 15.6 cm. The angle of the minor sector is 150°. Calculate the area of the minor sector. … cm2 [2] (c) r cm NOT TO x° SCALE The diagram shows a circle, radius r cm and minor sector angle x°. The perimeter of the major sector is three times the perimeter of the minor sector. 90 ( r - 2 ) Show that x = . r [4]

8 marks

Mark scheme: 10(a) 42.05 final answer 2 M1 for 11.4 + 0.05 oe or 14.8 + 0.05 oe or 15.7 + 0.05 oe 10(b) 319 or 318.5 to 318.6 2 150 2 M1 for   15.6 oe 360 10(c) 360 − x  x  M2  2πr + 2r = 3   2r + 2r  oe x 360  360  M1 for  2πr oe seen 360 or 360 − x  2πr oe seen 360 4 x M1 i.e. M mark for isolating and collecting terms in x  2π[r ] = 2π[r] – 4[r] oe 360 90 (− 2 ) A1 With no errors or omissions Leading to 

This question in 0580/42 Oct/Nov 2022

8 A NOT TO SCALE 9.5 cm O 10 cm B 7.7 cm D C E A, B and C are points on the circle, centre O. DE is a tangent to the circle at C. AC = 10 cm , AB = 9.5 cm and BC = 7.7 cm . (a) Show that angle ABC = 70.2° , correct to 1 decimal place. [4] (b) Find (i) angle AOC Angle AOC = … [1] (ii) angle ACO Angle ACO = … [1] (iii) angle ACD. Angle ACD = … [1] (c) Calculate the radius, OC, of the circle. OC = … cm [3] (d) Calculate the area of triangle ABC as a percentage of the area of the circle. … % [4]

14 marks

Mark scheme: 8(a) 9.52 + 7.7 2 − 10 2 M2 M1 for 102 = 9.52 + 7.72 – 2×9.5×7.7cosB oe or better [cos B = ] oe 2  9.5  7.7 70.206 to 70.207 or 70.21 to 70.22 A2 2477 A1 for oe or 0.339 or 0.3386…. 7315 8(b)(i) 140.4 1 8(b)(ii) 19.8 1 FT (180 – their (b)(i)) ÷ 2 8(b)(iii) 70.2 1 FT 90 – their (b)(ii) 8(c) 5.31 or 5.314 to 5.315 3 5 M2 for oe cos their(b)(ii) 5 or M1 for = cos(their (b)(ii)) oe r 8(d) 38.8 or 38.9 or 38.78 to 38.85 4 0.5  9.5  7.7  sin70.2 M3 for [ 100] their( (c)) 2 OR M1 for 0.5 × 9.5 × 7.7 × sin70.2 M1 for (their (c)2)

This question in 0580/43 Oct/Nov 2022

1 (a) 42° NOT TO SCALE x° The diagram shows an isosceles triangle with the base extended. Find the value of x. x = … [3] (b) The diagram shows three lines meeting at a point. The ratio a : b : c = 3 : 4 : 5. Find the value of c. a° NOT TO c° b° SCALE c = … [3] (c) A regular pentagon has an exterior angle, d. A regular hexagon has an interior angle, h. d Find the fraction . h Give your answer in its simplest form. … [4] (d) S R x° ( x + 20)° NOT TO SCALE ( 3x – 40)° Q ( 2x – 5)° P Show that PQRS is a cyclic quadrilateral. [5] (e) B A 50° 9 cm NOT TO O SCALE The diagram shows a circle of radius 9 cm, centre O. The minor sector AOB, with sector angle 50°, is removed from the circle. Calculate the length of the major arc AB. … cm [3]

18 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 111 3  42 M2 for 180 –180 oe or 42 + 2 180  42 oe 2 180  42 or M1 for oe 2 1(b) 150 3 M1 for k ÷ (3 + 4 + 5) [×p] where p = 1, 3, 4 or 5 5 or oe 12 B1 for 360 used 1(c) 3 4 72 cao nfww B3 for 5 120 or B2 for [d = ] 72 or [h = ] 120 or M1 for 360 ÷ 5 oe isw or 180 – (360 ÷ 6) isw or for (6 – 2) × 180 [÷ 6] 1(d) x + 2x – 5 + x + 20 + 3x – 40 = 360 M1 Accept equivalent equation e.g. 7x – 25 = 360 7x = 360 + 5 – 20 + 40 or better M1 FT their equation, accept e.g. 7x = 385 x = 55 B1 55 and 125 B1dep Dep on M1M1B1 or 105 and 75 Accept 55 + 3 × 55 – 40 = 180 or 2 × 55 – 5 + 55 + 20 = 180 If B0 scored, SC1 for 55, 75, 105 and 125 Opposite angles sum to 180 oe A1 Dep on M1M1B1B1 [so PQRS is a cyclic quadrilateral ] 1(e) 48.7 or 48.69 to 48.70… 3 360  50 M2 for 2 π oe9 360 50 or M1 for  2  π oe9 360

This question in 0580/42 May/June 2023

4 (a) O O NOT TO SCALE x° 7.5 cm 7.5 cm A B B 1.5 cm A The diagram shows a sector of a circle that is made into a cone by joining OA to OB. The sector angle is x° and the radius of the sector is 7.5 cm. The base radius of the cone is 1.5 cm. Calculate the value of x. x = … [3] (b) NOT TO SCALE The diagram shows a cylinder with radius 8 cm inside a sphere with radius 17 cm. Both ends of the cylinder touch the curved surface of the sphere. (i) Show that the height of the cylinder is 30 cm. [2] (ii) Calculate the volume of the cylinder as a percentage of the volume of the sphere. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … % [4] (c) 15 cm NOT TO SCALE The diagram shows a solid sphere with radius 6 cm inside a cube with side length 20 cm. The cube contains water to a depth of 15 cm. The sphere is removed. Calculate the new depth of water in the cube. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … cm [3]

12 marks

Mark scheme: 4(a) 72 or 72.0 cao nfww 3 x M2 for 2 π 7.5=2 π 1.5 oe 360 x or M1 for 2 π 7.5 or for 360 2 π 1.5 oe OR x 2 M2 for π 7.5 =π  1.5  7.5 oe 360 x 2 or M1 for π 7.5 or for 360 π 1.5  7.5 oe 4(b)(i) M2 M1 for 17 2 = 82 + d 2 or 342 = 162 + k2 2  172 − 82 or 342 − 162 oe 4(b)(ii) 29.3 or 29.30 to 29.31 4 2 4 3 M3 for ( [π]  8  30 ) ÷  [π]  17 [× 3 100] oe OR M1 for π  82  30 oe 4 3 M1 for π 17 oe 3 4(c) 12.7 or 12.73 to 12.74 3 B2 for 2.26 or 2.261 to 2.262…. soi  2 4 3  2 or M2 for  20  15 −  π  6   20 oe  3   4 3 2  or for 15 –   π  6  20  oe  3  2 4 3 or M1 for 20  15 − π 6 oe 3 2 4 3 or 20  D = π 6 oe 3 If 0 scored, SC1 for answer 11[.0] or 10.97 to 10.98

This question in 0580/42 Oct/Nov 2023

12 (a) NOT TO SCALE 50° 12 cm The diagram shows a circle of radius 12 cm, with a sector removed. Calculate the perimeter of the remaining shaded shape. … cm [4] (b) The diagram in part(a) shows the top of a cylindrical cake with a slice removed. The volume of cake that remains is 3510 cm 3. Calculate the height of the cake. … cm [3]

7 marks

Mark scheme: 12(a) 88.9 or 88.92 to 88.93... 4 360 − 50 M3 for 2  12 + 2 π 12 oe 360 ( 360 − 50 ) or M2 for 2 π  12 oe isw 360 50 or M1 for 2 π 12 oe isw 360 12(b) 9.01 or 9.009 to 9.010… 3 ( 360 − 50 ) 2 M2 for π 12  h = 3510 360 k 2 or M1 for π 12  h oe seen 360 with k = 50 or 360 – 50

This question in 0580/42 Feb/March 2024

9 (a) O NOT TO 60° 10 cm SCALE 17 cm D C A B OAB is a sector of a circle, centre O, radius 17 cm. OCD is a sector of a circle, centre O, radius 10 cm. OCA and ODB are straight lines and angle AOB = 60° . The perimeter of the shaded shape ABDC can be written in the form ( a r+ b ) cm. Find the value of a and the value of b. a = … b = … [3] (b) NOT TO SCALE The diagram shows a regular hexagon. The area of the hexagon is 127.3 cm2. (i) Show that the length of one side of the hexagon is 7.0 cm , correct to 1 decimal place. [4] (ii) The hexagon is the cross-section of a prism of length 10 cm. 127.3 cm2 NOT TO SCALE 10 cm 7.0 cm (a) Find the volume of the prism. … cm3 [1] (b) Calculate the surface area of the prism. … cm2 [2]

10 marks

Mark scheme: 9(a) [a =] 9 3 B2 for a =9 [b =] 14 OR M2 for 60 60 2  17  2  10  7  7 360 360 oe or M1 for 60 60 2  17 oe or 2  10 oe 360 360 If 0 scored SC1 for b =14 9(b)(i) 60° at centre B1 or interior angle = 120° 1 2 M1 [6]  d  sin60 oe 2 2 127.3 M1 [ d  ] 1 6   sin60 2 6.99[9…] to 7.00[…] A1 Dep on M1M1 9(b)(ii)(a) 1273 1 9(b)(ii)(b) 675 or 674.5 to 674.6 2 M1 for 2 ×127.3 oe or 6 × 7 × 10 oe

This question in 0580/42 May/June 2024

11 (a) Q P NOT TO SCALE x° O 3 In the circle, centre O, the length of the minor arc PQ is of the length of the major arc PQ. 7 Show that x = 108 . [3] (b) A NOT TO SCALE r y° B O The diagram shows a sector, OAB, of a circle with centre O and radius r. The area of triangle OAB is half the area of the sector. Angle AOB = y° and is obtuse. (i) Show that 360 siny = r y . [2] (ii) Complete the table, giving your answers correct to two decimal places. y 360 siny ry 108.4 341.60 340.55 108.5 341.40 340.86 108.6 341.20 108.7 [3] (iii) Complete the statement. The value of y, correct to one decimal place, that satisfies the equation 360 siny = r y is … . [1]

9 marks

Mark scheme: 11(a) 3 M2  360 oe 3 x 10 M1 for = 3  7 360 x 3 360  x or for [ 2r] =  [ 360 7 360 2r] oe or better or 10 360  x 1 [ 2r] =  [ 2r] oe or 7 360 better 360 or  k (k = 1 or 7) 7  3 108 A1 11(b)(i) 1 1 y 2 y 1 r2 siny =   r2 M1 for  r2 or for r2 siny 2 2 360 360 2 y 1 or  r2 = [2  ] r2 siny 360 2 and one further step leading to 360siny = y with no errors 11(b)(ii) 341.18 or 341.22 3 B1 for each 341.00 341.49 or 341.54 11(b)(iii) 108.6 cao 1

This question in 0580/43 May/June 2024

10 (a) NOT TO SCALE ( x + 1)cm ( 2x + 3)cm This rectangle has area 190 cm2. (i) By forming and solving an equation, show that x = 8.5 . [4] (ii) Work out the perimeter of the rectangle. … cm [2] (b) A r cm NOT TO SCALE 50° O B The diagram shows a sector OAB of a circle, with centre O, and a chord AB. The shaded segment has area 30 cm2. (i) Show that r = 23.7 cm, correct to 1 decimal place. [4] (ii) Calculate the perimeter of the shaded segment. … cm [4]

14 marks

Mark scheme: 10(a)(i) 2x2 + 5x –187 [= 0] M2 M1 for (2x + 3)(x + 1) = 190 (2x –17)(x + 11) [= 0] oe M1 Leading to x = 8.5 with no errors A1 10(a)(ii) 59 2 M1 for 6 × 8.5 + 8 oe or 6x + 8 oe or B1 for 9.5 and 20 10(b)(i) 50 1 M3 π  r2 – r2  sin50 = 30 oe 360 2 50 M1 for π  r2 360 1 M1 for r2  sin50 oe 2 23.70[9] to 23.72… A1 must see at least 4 sig figs 10(b)(ii) 40.7 or 40.8 or 40.71 to 40.75… 4 M2 for 2 × 23.7 × sin 25 oe or 23.72 + 23.72 −2 23.7  23.7cos50 oe 23.7 sin 50 or oe  180 − 50  sin    2  x or M1 for = sin25 oe 23.7 or for 23.7 2 + 23.7 2 −2 23.7  23.7cos50 oe AB 23.7 or = oe sin 50  180 − 50  sin    2  AND 50 M1 for × 2 × π  23.7 oe 360

This question in 0580/41 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