Compound shapes and parts of shapes — Paper 4 · IGCSE Mathematics 0580

4 (a) The diagram shows a solid metal prism with cross section ABCDE. G 2 cm F B A NOT TO SCALE K 7 cm 4 cm E H J 8 cm C D 4 cm (i) Calculate the area of the cross section ABCDE. … cm2 [6] (ii) The prism is of length 8 cm. Calculate the volume of the prism. … cm3 [1] (b) A cylinder of length 13 cm has volume 280 cm3. (i) Calculate the radius of the cylinder. … cm [3] (ii) The cylinder is placed in a box that is a cube of side 14 cm. Calculate the percentage of the volume of the box that is occupied by the cylinder. … % [3]

13 marks

Mark scheme: 4(a)(i) 17.5 or 17.46 … nfww 6 B3 for triangle height 3.46[4...] or 12 oe or M2 for 4 2 − 2 2 or M1 for h 2 + 2 2 = 4 2 and M2 for 2 × 7 + 12 × 2 × their h oe or M1 for 12 × 2 × their h 4(a)(ii) 140 or 139.6 to 139.7... 1FT FT their (a) × 8 4(b)(i) 2.62 or 2.618… 3 280 M2 for [r2 = ] oe 13π or M1 for 280 = π × r2 × 13 4(b)(ii) 10 3 280 10.2 or 10.20… or 10 M2 for 3 [×100] oe 49 14 or B1 for 2 744 or 143 seen

This question in 0580/43 May/June 2017

10 270° O x cm NOT TO SCALE 2x cm The diagram shows a sector of a circle, a triangle and a rectangle. The sector has centre O, radius x cm and angle 270°. The rectangle has length 2x cm. The total area of the shape is kx2 cm 2. (a) Find the value of k. k = … [5] (b) Find the value of x when the total area is 110 cm2. x = … [2]

7 marks

Mark scheme: 10(a) 5.68 or 5.684 to 5.685 5 2 2 2 M2 for 2 x x + x oe or 2 × 2 × x or M1 for x 2 or x 2 + x 2 oe soi 270 2 M1 for × π × x oe 360 M1 for 0.5 x² oe 10(b) 4.4[0] or 4.398 to 4.401 2 dep on a correct value for k in (a) 2 110 M1 for  x =   their k

This question in 0580/42 Oct/Nov 2017

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

6 C B NOT TO SCALE 1.5 cm D E 0.25 cm A O 0.5 cm The diagram shows a company logo made from a rectangle and a major sector of a circle. The circle has centre O and radius OA. OA = OD = 0.5 cm and AB = 1.5 cm. E is a point on OC such that OE = 0.25 cm and angle OED = 90°. (a) Calculate the perimeter of the logo. … cm [5] (b) Calculate the area of the logo. … cm2 [3] (c) A mathematically similar logo is drawn. The area of this logo is 77.44 cm2. (i) Calculate the radius of the major sector in this logo. … cm [3] (ii) A gold model is made. This model is a prism with a cross-section of area 77.44 cm2. This gold model is 15 mm thick. One cubic centimetre of gold has a mass of 19 grams. Calculate the mass of the gold model in kilograms. … kg [3]

14 marks

Mark scheme: 6(a) 5.83 or 5.832 to 5.833 5 B2 for sector angle = 210 soi 0.25 or M1 for [ cos DOE = ] oe 0.5 M2 for their 210 × 2 × π × 0.5 + 2 × 1.5 + 2 × 0.5 oe 360 their 210 or M1 for × 2 × π × 0.5 oe isw 360 6(b) 1.21 or 1.208… 3 their 210 M2 for × π × 0.5 × 0.5 + 1.5 × 0.5 360 oe their 210 or M1 for × π×0.5×0.5 oe isw 360 6(c)(i) 4[.00...] 3 77.44 M2 for 0.5× oe their (b) 77.44 their (b) or M1 for or their (b) 77.44 their (b) 0.5 2 or for = oe 77.44 r 2 6(c)(ii) 2.20704 3 M2 for 77.44 × 1.5 × 19 ÷ 1000 oe or M1 for figs 2207[04] or figs 221 seen or [vol =] 77.44 × 1.5

This question in 0580/42 Feb/March 2019

5 NOT TO 1.2 m SCALE 3 m The diagram shows the surface of a garden pond, made from a rectangle and two semicircles. The rectangle measures 3 m by 1.2 m. (a) Calculate the area of this surface. … m2 [3] (b) The pond is a prism and the water in the pond has a depth of 20 cm. Calculate the number of litres of water in the pond. … litres [3] (c) After a rainfall, the number of litres of water in the pond is 1007. Calculate the increase in the depth of water in the pond. Give your answer in centimetres. … cm [3]

9 marks

Mark scheme: 5(a) 4.73 or 4.730 to 4.731... 3 M2 for 3 × 1.2 + π × 0.62 oe 2 1 2 or M1 for π × 0.6 or × π × 0.6 or 2 3 × 1.2 5(b) 946 or 946.0 to 946.2... 3 M2 for their (a) × 0.2 × 1000 oe or M1 for their (a) × 0.2 or 20 implied by figs 946[0] to 9462 5(c) 1.28 or 1.29 or 1.284 to 1.290 3 (1007 − their (b)) ÷ 1000 M2 for × 100 oe their (a) 1007 − their ( b ) or for × 20 oe their ( b ) 1007 − their ( b ) or M1 for figs or their ( a ) 1007 figs their ( a ) 1007 − their ( b ) or for or their ( b ) 1007 × 20 oe their ( b )

This question in 0580/41 May/June 2019

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

6 D 100° NOT TO SCALE 50° C 12 cm A 8 cm 11 cm B (a) Calculate AD. AD = … cm [3] (b) Calculate angle BAC and show that it rounds to 40.42°, correct to 2 decimal places. [4] (c) Calculate the area of the quadrilateral ABCD. … cm2 [3] (d) Calculate the shortest distance from B to AC. … cm [3]

13 marks

Mark scheme: 6(a) 9.33 or 9.334... 3 12sin50 M2 for sin100 sin100 sin50 or M1 for = oe 12 AD 6(b) 112 + 12 2 − 8 2 M2 M1 for [cos =] 2 2 2 2 × 11 × 12 8 = 11 + 12 − 2 × 11 × 12cos( BAC ) 40.415... A2 201 67 A1 for 0.761... or or 264 88 6(c) 70.8 or 70.77 to 70.79... 3 M1 for 1 × 12 × their (a) × sin(180 − 100 − 50) 2 1 M1 for × 12 × 11 × sin(40.42) 2 6(d) 7.13 or 7.131 to 7.132... 3 dist M2 for = sin(40.42) 11 or M1 for recognition that shortest distance is perpendicular to AC

This question in 0580/43 Oct/Nov 2021

4 6.4 cm D C 38° NOT TO SCALE 10.9 cm 45° A B ABCD is a trapezium with DC parallel to AB. DC = 6.4 cm, DB = 10.9 cm, angle CDB = 38° and angle DAB = 45°. (a) Find CB. CB = … cm [3] (b) (i) Find angle ADB. Angle ADB = … [1] (ii) Find AB. AB = … cm [3] (c) Calculate the area of the trapezium. … cm2 [3]

10 marks

Mark scheme: 4(a) 7.06 or 7.058… or 7.059 3 2 2 M2 for 6.4  10.9 2 6.4  10.9  cos38 oe OR M1 for 6.42 + 10.92 – 2  6.4  10.9  cos 38 oe A1= 49.8... 4(b)(i) 97 1 4(b)(ii) 15.3[0…] 3 10.9  sin their 97 M2 for [AB =] sin45 sin their 97 sin45 or M1 for  oe AB 10.9 4(c) 72.8 to 72.81… 3 M2 for 1 1  6.4   10.9  sin38  their 15.3  10.9  sin38 2 2 oe or M1 for 12  6.4  10.9  sin38 oe or 12  their15.3  10.9  sin38 oe or M1 for height =10.9  sin38 oe

This question in 0580/42 May/June 2022

4 A regular 12-sided polygon has side length 6 cm. (a) Show that one interior angle of the polygon is 150°. [1] (b) The polygon is enclosed by a circle, centre O, so that each vertex touches the circumference of the circle. A B 6 cm NOT TO SCALE O (i) Show that the radius, AO, of the circle is 11.6 cm, correct to 1 decimal place. [3] (ii) Calculate (a) the circumference of the circle, … cm [2] (b) the perimeter of the shaded minor segment formed by the chord AB. … cm [2] (c) The regular 12-sided polygon is the cross-section of a prism of length 2 cm. Calculate the volume of the prism. … cm3 [3]

11 marks

Mark scheme: 4(a) (12  2)  180 1 (2  12  4)  90 [= 150] oe Accept [= 150] 12 12 360 or 180 – [= 150] 12 4(b)(i) 3 M2 3 oe M1 for  cos75 oe cos75 AO or or 6sin75 r 6  sin30 sin75 sin30 11.59… A1 4(b)(ii)(a) 72.8 or 72.9 or 72.82 to 72.89… 2 M1 for 2   11.6 4(b)(ii)(b) 12.1 or 12.06 to 12.08 2 M1 for [6 +] their (b)(ii)(a) ÷ 12 oe 4(c) 806 or 807 or 805.9 to 807.4 3 B2 for 402.9… to 403.7 OR 1 M2 for 6 11.6  sin75  12  2 oe 2 1 or M1 for 6 11.6  sin75 [k ] oe 2

This question in 0580/43 May/June 2022

4 (a) P NOT TO SCALE 8 cm Q R 24 cm (i) Calculate the area of triangle PQR. … cm2 [2] (ii) Calculate angle PRQ. Angle PRQ = … [2] (b) NOT TO SCALE 11 cm 6 cm The diagram shows a half-cylinder of radius 6 cm and length 11 cm. Calculate the volume of the half-cylinder. … cm3 [2] (c) T T D C C 44 cmcm S S O NOT TO 15 cm X X SCALE A B A B 20 cm (i) ABCD is a rectangle with AB = 20 cm and BC = 15 cm. S, X and T are points on a circle centre O, such that DSA and DTC are tangents to the circle. The radius of the circle is 4 cm and TX is a diameter of the circle. The shape DSXT is removed from the corner of the rectangle, leaving the shaded shape shown in the second diagram. Calculate the area of the shaded shape. … cm2 [5] (ii) Calculate the perimeter of the shaded shape. … cm [3]

14 marks

Mark scheme: 4(a)(i) 96 2 1 M1 for  24  8 2 4(a)(ii) 18.4 or 18.43... 2 8 M1 for tan x  oe 24 4(b) 622 or 622.0 to 622.1 … 2 1 2 1 2 M1 for [ 2] 6  11 or 2  6 [ 11] 4(c)(i) 246 or 246.2 to 246.3... 5 270 2 M4 for 15  20 4 4   4 oe 360 OR 270 2 M2 for  4 oe 360 or M1 for k   4 2 , where k  1 M1 for15  20 or 4  4 oe 4(c)(ii) 80.8 or 80.9 or 80.84 to 80.85... 3 M1 for 15  20  11  16 oe 3 M1 for 2 4 oe 4

This question in 0580/41 May/June 2023

9 (a) M A B NOT TO SCALE D C N The diagram shows a shape made from a square ABCD and two equal sectors of a circle. The square has side 11 cm. MAB and DCN are straight lines. (i) Calculate the area of the shape. … cm2 [3] (ii) Calculate the perimeter of the shape. … cm [3] (b) H G E F NOT TO SCALE D C A B The diagram shows a cube ABCDEFGH of edge 7 cm. Calculate the angle between AG and the base of the cube. … [4]

10 marks

Mark scheme: 9(a)(i) 311 or 311.0 to 311.1 3 1 M2 for 11  11 + 2     112 oe 4 1 or M1 for [2 ]    112 or 11  11 4 oe 9(a)(ii) 78.6 or 78.55 to 78.56... 3 1 M2 for 4  11 + 2   2    11 oe 4 1 or M1 for [2 ]  2    11 or 4  11 4 oe 9(b) 35.2 or 35.3 or 35.239… to 35.28 4 7 M3 for [tan =] 7 2  7 2 7 or [sin =] 7 2  7 2  7 2 7 2  7 2 or [cos =] 7 2  7 2  7 2 OR M2 for AG = 7 2  7 2  7 2 2  7  2 or for 7   oe  sin 45  2 2 7 or for AC = 7  7 or oe sin45 OR M1 for 72 + 72 or for implicit trigonometry or identifying correct angle

This question in 0580/43 May/June 2023

4 (a) A box contains 50 cuboids. Each cuboid has a mass of 135 g. The total mass of the cuboids and the box is 7 kg. Calculate the mass of the box. Give your answer in grams. … g [2] (b) A solid cube of side 4 cm is fixed to the base inside an empty cube of side 6 cm. Water is poured into the larger cube until it reaches the top of the smaller cube. Calculate the amount of water poured into the larger cube. … cm3 [2] (c) 4 cm NOT TO SCALE 20 cm The diagram shows a solid triangular prism of length 20 cm. The cross-section is an equilateral triangle with side length 4 cm. The prism is made of wood with a density of 0.85 g/cm3. Calculate the mass of the prism. [Density = mass ÷ volume] … g [4] (d) NOT TO SCALE 24 cm 10 cm The diagram shows a solid cone with base radius 10 cm and height 24 cm. (i) Show that the total surface area of the cone is 1131 cm2, correct to the nearest cm2. [The curved surface area of a cone with base radius r and slant height l is A = r rl .] [4] (ii) The total surface area of the cone is painted. (a) The cost to paint the cone is $1.71 . Calculate the cost to paint 1 cm2 of the cone. Give your answer in cents. … cents [1] (b) One tin of paint has enough paint to cover 2.5 m2. Calculate the number of these cones that can be painted completely using one tin of paint. … [2]

15 marks

Mark scheme: 4(a) 250 2 B1 for 6750 or M1 for 7000 – 50 × 135 or for 7 – 50 × 0.135 4(b) 80 2 M1 for 6 × 6 × 4 or for 43 oe OR M1 for (6  6) – (4  4) oe 4(c) 118 or 117.7 to 117.8 4 1 M3 for 4 4 sin60 × 20 × 0.85 oe 2 OR 1 M1 for 4 4 sin60 or 2 1 2 2 4 4 − 2 oe 2 M1 for 20 × their area of triangle M1 dep for 0.85 × their volume, dependent on previous M1 If 0 scored SC1 for height = 3.46... 4(d)(i) 2 2 2 M3 π  24 + 10  10 + π  10 or 2 2 2 2 2 2 π 10 2 M2 for π  24 + 10  10 or π  24 + 10  + π  10 2 ( ) 2 π 26 2 2 2 π 10 π  24 + 10  ( ) 2 π 26 or M1 for 24 2 + 10 2 or   102 1130.9 to 1131.1... A1 Must see at least 5 sf 4(d)(ii)(a) 0.151 or 0.1511 to 0.1512... 1 4(d)(ii)(b) 22 2 B1 for figs 22[1…] 2.5  100 2 or M1 for 1131

This question in 0580/43 Oct/Nov 2024