Pythagoras’ theorem — Paper 4 · IGCSE Mathematics 0580

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

8 P NOT TO SCALE 9 cm D C N 6 cm M A 8 cm B The diagram shows a pyramid on a rectangular base ABCD. AC and BD intersect at M and P is vertically above M. AB = 8 cm, BC = 6 cm and PM = 9 cm. (a) N is the midpoint of BC. Calculate angle PNM. Angle PNM = … [2] (b) Show that BM = 5 cm. [1] (c) Calculate the angle between the edge PB and the base ABCD. … [2] (d) A point X is on PC so that PX = 7.5 cm. Calculate BX. BX = … cm [6]

11 marks

Mark scheme: 8(a) 66[.0] or 66.03 to 66.04 2 9 M1 for tan = oe 4 8(b) 2 2 1 2 2 M1 Any alternative method must be full and complete and 3 + 4 or 6 + 8 result in exactly 5 2 8(c) 60.9 or 60.94 to 60.95 2 9 M1 for tan = oe 5 8(d) 5.83 or 5.84 or 5.827 to 5.840 6 2 2 2 2 M1 for [PB or PC = ] 9 + 5 or [XC =] 9 + 5 – 7.5 3 M1 for angle BPX = 2 × invsin oe their PB B1 for [ PB or PC =] 106 = 10.29 to 10.30 or XC = 2.79 to 2.8[0] or angle BPX = 33.9 or 33.86 to 33.90… M2 for ( their PB ) 2 + 7.5 2 − 2 × their PB × 7.5 × cos ( their BPX ) oe or M1 for correct implicit equation

This question in 0580/42 May/June 2017

8 l h NOT TO SCALE 5 mm The diagram shows a solid made from a hemisphere and a cone. The base diameter of the cone and the diameter of the hemisphere are each 5 mm. 115r (a) The total surface area of the solid is mm2. 4 Show that the slant height, l, is 6.5 mm. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl.] [The surface area, A, of a sphere with radius r is A = 4rr2.] [4] (b) Calculate the height, h, of the cone. h = … mm [3] (c) Calculate the volume of the solid. 1 [The volume, V, of a cone with radius r and height h is V = rr2h.] 3 4 [The volume, V, of a sphere with radius r is V = rr3.] 3 … mm3 [4] (d) The solid is made from gold. 1 cubic centimetre of gold has a mass of 19.3 grams. The value of 1 gram of gold is $38.62 . Calculate the value of the gold used to make the solid. $ … [3]

14 marks

Mark scheme: 8(a) 2 M2 2 5 4  5  115π 5 4  5  π × × l + × π × = oe M1 for π × × l or × π ×     2 2  2  4 2 2  2  115π 4  5  2 5 or – × π × = π × × l oe   4 2  2  2 5πl 65π B1 nfww = oe oe both terms must be written in terms of π 2 4  115π 2  or [ l = ]  − 2 × π × 2.5  ÷ 2.5π oe nfww  4  or correct complete method for l with decimals 65π × 2 65π A1 [l =] or oe = 6.5 Correct calculation with no errors and B1 earned 4 × 5π 10π 8(b) 6 3 2 2 M2 for 6.5 − 2.5 or M1 for h2 + 2.52 = 6.52 If zero scored, SC2dep for answer 4.15[3]… 8(c) 72[.0…] or 71.99… nfww 4 2 3 π  5  1 4π  5  M3 for × × their 6 + × ×     3  2  2 3  2  oe π  5  2 or M1 for × × their 6 oe   3  2  1 4π  5  3 and M1 for × × oe   2 3  2  If zero scored, SC3dep for π 2 1 4π 3 × ( 5 ) × their 4.15 + × × ( 5 ) oe 3 2 3 or π 2 SC1dep for × ( 5 ) × their 4.15 oe 3 1 4π 3 SC1dep for × × ( 5 ) oe 2 3 8(d) 53.7 or 53.65 to 53.67 3 M1 for figs (their (c)) × 19.3 × 38.62 or better M1 for ÷ 1000 soi

This question in 0580/41 Oct/Nov 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

3 (a) C 38.6 m 56.5 m D B 94° 78.4 m NOT TO SCALE 46.1 m 64° E A ABCDE is a pentagon. (i) Calculate AD and show that it rounds to 94.5 m, correct to 1 decimal place. [2] (ii) Calculate angle BAC. Angle BAC = … [3] (iii) Calculate the largest angle in triangle CAD. … [4] (b) Q L 34.3 cm P NOT TO 21.5 cm SCALE 111° R N M 27.6 cm Triangle PQR has the same area as triangle LMN. Calculate the shortest distance from R to the line PQ. … cm [3]

12 marks

Mark scheme: 3(a)(i) AD M1 = tan 64 oe or better 46.1 94.51 to 94.52 A1 3(a)(ii) 46[.0] or 45.96… nfww 3 sin94 M2 for 56.5 × oe 78.4 56.5 78.4 or M1 for = oe sin BAC sin94 3(a)(iii) 102.3 or 102.4 or 102.34 to 102.38 4 38.6 2 + 78.4 2 − 94.5 2 M2 for [cosC = ] 2 × 38.6 × 78.4 or M1 for 94.5 2 = 38.6 2 + 78.4 2 − 2 × 38.6 × 78.4 × cos C and A1 for –0.214 or –0.2144 to –0.2137 If 0 scored, SC2 for [CAD =] 23.5 or 23.51 to 23.52 or for [CDA =] 54.1 or 54.14… 3(b) 16.2 or 16.15… 3 1 1 M2 for × 21.5 × 27.6sin111 = × 34.3 × d 2 2 oe 1 or M1 for × 21.5 × 27.6sin111 seen or 2 1 × 34.3 × d oe soi 2

This question in 0580/42 Oct/Nov 2021

8 Q NOT TO 4 m SCALE A 15 m C P 8 m 3 m 20 m B The diagram shows triangle ABC on horizontal ground. AC = 15 m , BC = 8 m and AB = 20 m . BP and CQ are vertical poles of different heights. BP = 3 m and CQ = 4 m . AQ and PQ are straight wires. (a) Show that angle ACB = 117.5° , correct to 1 decimal place. [4] (b) Calculate the area of triangle ABC. … m2 [2] (c) Calculate the length of AQ. … m [2] (d) Calculate the angle of elevation of Q from P. … [3] (e) Another straight wire connects A to the midpoint of PQ. Calculate the angle between this wire and the horizontal ground. … [5]

16 marks

Mark scheme: 8(a) 15 2 + 8 2 − 20 2 M2 M1 for 202 = 152 + 82 − 2.15.8cos ( ) [cos = ] 2.15.8 117.54 to 117.55 A2 37 111 A1 for − or − or –[0].4625 80 240 8(b) 53.2 or 53.19 to 53.23 2 M1 for 0.5  8  15  sin(117.5) oe 8(c) 15.5 or 15.52 to 15.53 2 M1 for 152 + 42 oe 8(d) 7.1 or 7.13 or 7.125 to 7.126 3 4 − 3 M2 for tan [P]= oe or for 7.1 or 8 7.13 or 7.125 to 7.126 seen or M1 for vertical line = 4 – 3 soi After 0 scored SC1 for correct angle identified 8(e) 11.5 nfww or 11.48 to 11.49... 5 B1 for height of 3.5 soi M2 for 15 2 + 4 2 − 2.15.4cos(117.5) 15 2 + 4 2 − (...) 2 or M1 for cos117.5 = 2.15.4 3.5 M1 for tan = oe their 17.216... After M0 scored SC1 for correct angle identified

This question in 0580/41 Oct/Nov 2022

10 (a) H G F E NOT TO SCALE D C x cm A x cm B ABCDEFGH is a cuboid with a square base of side x cm. CG = 20 cm and AG = 28 cm . Calculate the value of x. x = … [4] (b) R Q N P NOT TO SCALE M L J K The diagram shows a different cuboid JKLMNPQR. MR = 30 cm correct to the nearest centimetre. KR = 37 cm correct to the nearest centimetre. Calculate the lower bound of the angle between KR and the base JKLM of the cuboid. … [4]

8 marks

Mark scheme: 10(a) 13.9 or 13.85 to 13.86 4 M3 for 2x2 = 282 – 202 or better or x  28 2  20 2 sin45 oe   or M2 for x2 + x2 + 202 = 282 oe x or sin45  28 2  20 2 ) or M1 for any correct Pythag in 2D or their AC × sin 45 oe dep on trig/Pythagoras attempt for AC 10(b) 51.9 or 51.87 to 51.88 4 29 to 30 30  0.5 M3 for sin = or oe 37  0.5 37 to 38 or M2 for correct trig statement for correct angle with values in range 29 to 31 and 36 to 38 or M1 for 30 + 0.5 or 30 – 0.5 or 37 + 0.5 or 37 – 0.5 seen or for identifying correct angle RKM

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

7 (a) X 2.8 m NOT TO R SCALE 7.1 m P Q The diagram shows a right-angled triangle PQR on horizontal ground. X is vertically above R and the angle of elevation of X from P is 21°. XR = 2.8 m and RQ = 7.1 m. (i) Calculate the angle of elevation of X from Q. … [2] (ii) Calculate PQ. … m [3] (b) M 9.1 cm NOT TO SCALE 32° L K 16.7 cm Calculate the acute angle KML. Angle KML = … [3] (c) C 21.5 cm NOT TO SCALE A 12.3 cm B D The area of triangle ABC is 62.89 cm 2. (i) Show that angle BAC = 28.4°, correct to 1 decimal place. [2] (ii) Calculate BC. … cm [3] (iii) AB is extended to a point D such that angle BDC = 90°. Calculate BD. … cm [3]

16 marks

Mark scheme: 7(a)(i) 21.5 or 21.52... 2 2.8 M1 for tan(…) = oe 7.1 7(a)(ii) 10.2 or 10.17 to 10.18 3 2  2.8 2 oe M2 for  + 7.1  tan21  2.8 or M1 for = tan21 oe PR 7(b) 76.5 or 76.52 to 76.53 3 16.7sin32 M2 for [sin =] oe 9.1 9.1 16.7 or M1 for = oe sin32 sin M 7(c)(i) 1 M1  12.3  21.5sin(...) = 62.89 or better 2 28.40 to 28.41… A1 7(c)(ii) 12.2 or 12.17 to 12.18 3 M2 for 12.32 + 21.52 – 2 12.3  21.5  cos28.4 OR M1 for 12.32 + 21.52 – 2 × 12.3 × 21.5 × cos28.4 A1 for 148 or 148.2 to 148.3 7(c)(iii) 6.6[0] to 6.62 3 M2 for 21.5cos28.4 – 12.3 or M1 for 21.5cos28.4

This question in 0580/42 Oct/Nov 2023

6 D 6.5 cm 26° NOT TO C 64° SCALE A 42° 10.4 cm B ABCD is a quadrilateral with AB = 10.4 cm and AD = 6.5 cm. Angle DAB = 64° , angle BDC = 26° and angle DBC = 42° . (a) Show that BD = 9.55 cm, correct to 2 decimal places. [3] (b) (i) Show that angle BCD = 112° . [1] (ii) Calculate CD. CD = … [3] (c) Find the shortest distance from D to AB. … cm [3]

10 marks

Mark scheme: 6(a) 2 2 M2 M1 for 10.42 + 6.52 – 2× 10.4 × 6.5 × 10.4  6.5  2  10.4  6.5  cos64 cos64 A1 for 91.1 to 91.2 9.546 to 9.547 A1 6(b)(i) 180   26  42  B1 6(b)(ii) 6.89 or 6.888 to 6.892... 3 9.55 M2 for  sin 42 oe sin112 sin112 sin 42 or M1 for oe 9.55 CD 6(c) 5.84[2…] 3 x M2 for  sin64 oe 6.5 or M1 for identifying shortest distance from D is perpendicular to AB

This question in 0580/42 May/June 2024

6 (a) H NOT TO SCALE 4 m G F 1.5 m The diagram shows a ladder, GH, on horizontal ground, leaning against a vertical wall, HF. GF = 1.5 m and HF = 4 m . Calculate the length of the ladder, GH. … m [2] (b) W NOT TO SCALE 120 m V 50 m W is 120 m north of V and 50 m east of V. Calculate the bearing of V from W. … [3] (c) B NOT TO SCALE D A C In the quadrilateral ABCD, AD = DC = 5 cm and AB = BC . Angle ABD = 25° and angle BAD = 15° . Calculate the perimeter of the quadrilateral ABCD. … cm [5] (d) S 8 cm R 110° 11 cm NOT TO SCALE 14 cm P 10 cm Q PQRS is a quadrilateral. Calculate angle PQR. Angle PQR = … [5]

15 marks

Mark scheme: 6(a) 4.27 or 4.272... 2 M1 for 42 + 1.52 oe 6(b) 203 or 202.6… 3 B2 for [angle at W = ] 22.6... or for [angle at V =] 67.4 or 67.38… 5 12 or M1 for tan = or oe 12 5 6(c) 25.2 or 25.20 to 25.21[0] 5 B4 for [BC or AB = ] 7.6[0] or 7.604 to 7.605 OR M3 for a complete explicit method 5sin140 leading to AB or BC, e.g. sin25 OR M2 for a complete implicit method leading to AB or BC, e.g. sin 25 sin140  oe 5 BC or AB and M1 (dep on AB from trig) for 2  their AB + 10 OR B1 for any relevant angle E.g. BDA or BDC = 140, DAE or DCE = 50 or ADE or CDE = 40 or ADC = 80 6(d) 79.5 or 79.6 or 79.54 to 79.55... 5 B2 for [PR2 =] 245 or 245.1 to 245.2 or [PR =] 15.65 to 15.66 or 15.7 or M1 for [PR2 = ] 112 + 82 – 2  11  8  cos110 M2 for [cosPQR = ] 10 2  14 2  (their PR ) 2 oe 2  10  14 or M1 for (their PR)2 = 102 + 142 – 2  10  14cosPQR oe

This question in 0580/43 May/June 2024