Pythagoras’ theorem — Paper 4 · IGCSE Mathematics - International 0607

6 V NOT TO SCALE 8 cm B C √72 cm P A D √72 cm The diagram shows a pyramid with a square base ABCD of side 72 cm. The diagonals of the base, AC and BD, meet at P. The vertex, V, is vertically above P and VP = 8 cm. (a) Find the volume of the pyramid. Give the units of your answer. … … [3] (b) Find the length AC. AC = … cm [2] (c) Find the length DV. DV = … cm [3] (d) Find angle VDP. Angle VDP = … [2] (e) X is the midpoint of the side CD. (i) Find the length VX. VX = … cm [3] (ii) Find angle VXP. Angle VXP = … [2] (f) The pyramid is cut parallel to ABCD to form a smaller pyramid VEFGH. The volume of VEFGH is 24 cm3. Find the vertical height of this pyramid. … cm [3]

18 marks

Mark scheme: 6(a) 192 2 1 2 M1 for × 72 × 8 oe ( ) 3 cm3 1 6(b) 12 2 M1 for ( 72) 2 + ( 72) 2 oe 6(c) 10 3 2 2 M2 for 8 + ( 0.5 their (b) ) or M1 for [PD oe =] 0.5 × their (b) (d) 53.1 or 53.13 2 8 8 M1 for tan = or sin = 0.5 × their (b) their (c) 0.5 × their (b) or cos = their (c) 6(e) (i) 82 or 9.06 or 9.055... 3 M2 for 8 2 + (0.5 × 72) 2 or (their (c))2 – (0.5 × 72) 2 or M1 for (0.5 × 72) 2 6(e)(ii) 62.1 or 62[.0] or 62.00 to 62.10 2 8 M1 for tan = oe 0.5 × 72 6(f) 4 cao 3 24 their (a) M2 for 3 or 3 their (a) 24 1 soi by 2 or 2 24 their (a) or M1 for or their (a) 24 1 soi by 8 or 8

This question in 0607/43 Oct/Nov 2017

5 16 m A P D NOT TO SCALE 30 m S Q 18 m 12 m B R C 24 m 40 m In the diagram, ABCD is a rectangle. (a) Find PS. PS = … m [2] (b) Find angle BRS. Angle BRS = … [2] (c) Find the perimeter of PQRS. … m [3] (d) Find the shaded area. … m2 [3] (e) Explain why triangle ASP is similar to triangle BSR. … … [2]

12 marks

Mark scheme: 5(a) 20 2 2 2 M1 for 16 + ( 30 − 18 ) 5(b) 36.9 or 36.86 to 36.87 2 18 M1 for tan[ ] = oe 24 5(c) 100 3 2 2 M2 for 2 × (their (a) + 18 + 24 ) oe or M1 for 182 + 24 2 or RS = 30 or PQ = 30 seen 5(d) 576 3 M2 for (40 × 30) − 2 × (0.5 × 18 × 24) − 2 × (0.5 × 16 × 12) oe or M1 for any correct and relevant area 5(e) Correct explanation 2 B1 for partial explanation e.g. ratio of two sides the same, with names or numbers given.

This question in 0607/42 May/June 2018

8 A 13 cm NOT TO B SCALE 9 cm 6 cm D 11 cm C ABCD is a quadrilateral. (a) Show that BD = 9.22 cm, correct to 3 significant figures. [3] (b) Calculate angle ABD. Angle ABD = … [3] (c) Calculate the total area of the quadrilateral ABCD. … cm2 [3] (d) Calculate the length of the diagonal AC. AC = … cm [3]

12 marks

Mark scheme: 8(a) Correct Pythagoras statement M2 or M1 for [BD]2 + 62 = 112 oe leading to 112 – 62 or 121 – 36 or 85 9.219… A1 9.219… implies M1 A1 8(b) 43.8 or 43.80... nfww 3 9.22 2 + 132 − 9 2 M2 for cos[ABD] = 2 × 9.22 × 13 or better or M1 for 92 = 9.222 + 132 – 2 × 9.22 × 13 cos [ABD] oe 8(c) 69.1 or 69.13 to 69.14... nfww 3 M1 for 0.5 × 9.22 × 6 oe M1 for 0.5 × 9.22 × 13 × sin (their 43.8) oe 8(d) 17.7 or 17.69... 3 M1 for 62 + 132 – 2 × 6 × 13 cos (90 + their 43.8) A1 for 313 or 312.9 to 313.0

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

8 E F NOT TO SCALE 12 cm B A 8 cm D C 20 cm ABCDEF is a triangular prism. ABCD is a rectangle. Find (a) AC, AC = … cm [2] (b) ED, ED = … cm [2] (c) angle EAD, Angle EAD = … [2] (d) angle FAC. Angle FAC = … [2]

8 marks

Mark scheme: 8(a) 21.5 or 21.54… 2 M1 for 20 2 + 8 2 8(b) 8.94 or 8.944… 2 M1 for 12 2 − 8 2 8(c) 48.2 or 48.15 to 48.19... 2 8 M1 for cos[ x ] = oe 12 8(d) 22.5 or 22.6 or 22.54 to 22.59 2 2 2 12 − 8 M1 for tan[ x ] = oe 20 2 + 8 2 their (b ) or tan[ x ] = oe their ( a )

This question in 0607/41 Oct/Nov 2020

5 (a) B 49 mm NOT TO 91 mm SCALE A C Calculate the length of AC. AC = … mm [2] (b) 305° NOT TO O SCALE B 16 cm A The diagram shows a circle with centre O and radius 16 cm. Calculate the length of the major arc AB. … cm [2] (c) NOT TO SCALE 12 cm The diagram shows a prism with length 12 cm. The cross-section of the prism is a quarter of a circle. The radius of the circle is 6 cm. Calculate the volume of the prism. … cm3 [2] (d) C NOT TO (2x + 4) cm SCALE B D (x + 1) cm A E (x – 3) cm Shape ABCDE is made by joining rectangle ABDE and triangle BCD. The perpendicular height of triangle BCD is (2x + 4) cm. The total area of ABCDE is 11 cm2. (i) Show that 2x 2 - 3x - 20 = 0 . [3] (ii) Factorise 2x 2 - 3x - 20 . … [2] (iii) Use your answer to part (ii) to solve the equation 2x 2 - 3x - 20 = 0 . x = … or x = … [1] (iv) Find the perpendicular height of triangle BCD. … cm [1]

13 marks

Mark scheme: 5(a) 103 or 103.3 to 103.4 2 M1 for 492 + 912 oe 5(b) 85.2 or 85.17 to 85.18 2 305 M1 for × π × 2 × 16 360 5(c) 339 or 339.2 to 339.3… 2 1 2 M1 for × π × 6 × 12 4 5(d)(i) 1 M1 (x – 3)(x + 1) + (x – 3)(2x + 4) 2 [=11] x2 – 3x + x – 3 B1 one correct expansion seen 1 or (2x2 – 6x + 4x – 12)  2 or x2 – 3x + 2x – 6 At least one more line of A1 no errors or omissions working leading to 2x2 – 3x – 20 = 0 5(d)(ii) (2x + 5)(x – 4) 2 M1 for (2x + a)(x + b) where ab = –20 or a + 2b = –3 or 2x(x – 4) + 5(x – 4) or x(2x + 5) – 4(2x + 5) 5(d)(iii) 4 , –2.5 1 Strict FT their factors Dep on factors in part (ii) 5(d)(iv) 12 1 FT 2 × (their positive root (d)(iii)) + 4

This question in 0607/42 Feb/March 2021

5 P NOT TO SCALE 10 m 20 m C B 35° A A, B and C are points on horizontal ground. BP is a vertical pole. BC = 20 m and BP = 10 m. Angle PAB = 35°. (a) Show that PC = 22.36 m correct to 2 decimal places. [2] (b) Show that AB = 14.28 m correct to 2 decimal places. [2] (c) Calculate AP. AP = … m [2] (d) Angle ABC = 125°. Calculate AC. AC = … m [3] (e) Calculate angle APC. Angle APC = … [3]

12 marks

Mark scheme: 5(a) 202 + 102 M1 22.360 to 22.361 A1 5(b) 10 M1 sin35 sin55 tan35 = oe = , i.e correct implicit AB 10 AB 14.281... A1 5(c) 17.4 or 17.43... 2 10 M1 for sin35 = oe AP or 14.282 + 102 5(d) 30.5 or 30.52... 3 M1 for 20 2 + 14.282 −×2 20 × 14.28 × cos125 A1 for 931.5 to 931.6... 5(e) 99.2 to 99.5 3 M2 for [cos = ] their 30.5 22.36 2 + ( their 17.4 ) 2 − ( 2 ) 2 × 22.36 × ( their 17.4 ) or M1 for ( their 30.5 ) 2 = 22.36 2 + ( their17.4 ) 2 −×2 22.36 × ( their17.4 ) × cos APB

This question in 0607/42 May/June 2021

7 In this question all lengths are in centimetres. (a) C B 8x° ( x + 5 )° NOT TO SCALE A In triangle ABC, AC = BC, angle ABC = ( x + 5)° and angle ACB = 8x° . Find the value of x. x = … [3] (b) NOT TO ( p - 2) SCALE ( p + 1) The diagram shows a rectangle with sides of length ( p + 1) and ( p - 2) . The area of the rectangle is 90 cm2 . Find the value of p. p = … [4] (c) ( y - 1) ( y - 4) NOT TO SCALE 30° The diagram shows a right-angled triangle. Find the value of y. y = … [3] (d) 13 ( w + 1 ) NOT TO SCALE ( 2w + 3) The diagram shows a right-angled triangle with sides of length ( w + 1) , ( 2w + 3) and 13. Work out the area of the triangle. … cm2 [6]

16 marks

Mark scheme: 7(a) 17 3 M2 for x + 5 + 8 x + x + 5 = 180 oe or M1 for angle A = x + 5 7(b) 10.1 or 10.10... 4 B3 for correct sketch indicating roots −−( 1) ± ( − 1) 2 − 4(1)( − 92) or for oe 2(1) or B2 for p 2 − 2 p + p − 2 [ = 90] or better or M1 for ( p + 1)( p − 2) [ = 90] 7(c) 7 3 M2 for 2(y – 4) = y – 1 or better y − 4 or M1 for = sin30 y − 1 If 0 scored SC1 for sin 30 = 0.5 7(d) 2.04 oe 6 B4 for (5 w − 1)( w + 3) or correct sketch indicating roots − 14 ± 14 2 − 4(5)( −3) or 2(5) or B3 for 5 w 2 + 14 w − 3 = 0 and M1 for correct calculation of area of triangle with their positive w OR 2 2 M1 for ( w + 1) + ( 2 w + 3 ) = 13 B1 for w 2 + w + w + 1 oe or 4 w 2 + 6 w + 6 w + 9 oe and M1 for correct calculation of area of triangle with their positive w

This question in 0607/43 May/June 2021

6 V NOT TO SCALE 12 cm B 12 cm A O D C VABC is a pyramid with a triangular base. All the edges have length 12 cm. O is vertically below V. 2 D is the mid‑point of AC and BO = BD . 3 (a) Show that BO = 6. 928 cm , correct to 3 decimal places. [4] (b) Calculate the volume of the pyramid. … cm3 [4]

8 marks

Mark scheme: 6(a) 2 2 2 M3 M2 for 122 – 62 12  6 oe or M1 for attempt at Pythagoras e.g. BD2 + 62 = 122 3 6.9282... A1 6(b) 204 or 203.6 to 203.7 4 2 2 M1 for 12  6.928 M1 for 0.5 × 12 × 12 × sin60 or 0.5 × their BD × 12 1 M1 for × their 62.35 × their 9.798 dependent on 3 use of Pythagoras and not BD or BO.

This question in 0607/43 May/June 2022

11 F X E NOT TO SCALE B A C D The diagram shows a triangular prism ABCDEF. X is a point on FE. AB = 8 m , AD = 15 m , AF = 10 m , EC = 6 m and FX = 5 m . Angle ABF = 90° and angle DCE = 90° . (a) Calculate angle CDE. Angle CDE = … [2] (b) Calculate AC. AC = … m [2] (c) Calculate angle CXA. Angle CXA = … [5] (d) Calculate the area of triangle CXA. … m2 [2] Question 12 is printed on the next page.

11 marks

Mark scheme: 11(a) 36.9 or 36.86 to 36.87 2 6 6 M1 for tan[CDE ] = or sin[CDE ] = 8 10 8 or cos[CDE ] = 10 11(b) 17 2 M1 for [ AC 2 ] = 15 2 + 8 2 11(c) 96.2 or 96.16… 5 M1 for AX 2 = 52 + 10 2 M1 for CX 2 = 10 2 + 6 2 M2 dep for their AX 2 + theirCX 2 − their AC 2 [cos CDX =] 2  their AX  theirCX their AC 2 = their AX 2 + theirCX 2 or M1dep for −2 their AX  theirCX  cos CDX both dependent on Pythagoras or trigonometry used for AX and CX 11(d) 64.8 or 64.81 to 64.82 2 M1 dep for area CXA = 0.5  theirCX  their AX  sin(theirCXA) dependent on Pythagoras or trigonometry used for AX and CX

This question in 0607/41 Oct/Nov 2022

8 North B NOT TO 55 km SCALE A C 37 km 64 km 28° D The diagram shows four points A, B, C and D on level ground. B is due north of A and C is due east of A. (a) Calculate AB. AB = … km [3] (b) Calculate the obtuse angle ACD. Angle ACD = … [3] (c) Find the bearing of (i) D from A … [2] (ii) A from D. … [1] (d) Calculate the area of the quadrilateral ABCD. … km2 [3]

12 marks

Mark scheme: 8(a) 40.7 or 40.69… 3 M2 for 552 – 372 oe soi by 1656 or M1 for 552 = AB2 + 372 oe 8(b) 125.7 or 126 or 125.7… 3 64  sin28 M2 for [sin ACD =] 37 64 37 or M1 for = oe sin ACD sin28 8(c)(i) 116 or 116.2 to 116.3 2 M1 for 180 – 28 – their(b) soi by 26.29 or 26.3 8(c)(ii) 296 or 296.2 to 296.3 1 FT 180 + their (c)(i) 8(d) 1280 or 1277 to 1278 3 M1 for 0.5 × their(a) × 37 M1 for 0.5 × 64 × 37 × sin(180 – 28 – their(b))

This question in 0607/42 Feb/March 2023

8 (a) E NOT TO 20 cm SCALE D C M A 24 cm F 10 cm B ABCDE is a pyramid with a rectangular base. AB = 10 cm and BC = 24 cm. The length of each sloping edge is 20 cm. Vertex E is vertically above the centre of the base, M. (i) Calculate the length AC. … cm [2] (ii) Calculate EM, the height of the pyramid. … cm [3] (iii) F is the mid-point of BC. Find the angle between EF and the base of the pyramid. … [3] (b) B A NOT TO 7 cm SCALE Cone A is mathematically similar to cone B. The height of cone A is 7 cm and its volume is 66 cm 3. The volume of cone B is 222.75 cm 3. (i) Find the height of cone B. … cm [3] (ii) A sphere also has volume 66 cm 3. Calculate the radius of the sphere. … cm [2]

13 marks

Mark scheme: 8(a)(i) 26 2 M1 for 10 2 + 24 2 or better 8(a)(ii) 15.2 or 15.19 to 15.20 3 2 2 their 26  M2 for 20 −   2  2 their 26  2 M1 for 20 −   2  8(a)(iii) 71.8 or 71.78 to 71.80… 3 231 M2 for tan EFM = or using their height 5 oe ALT M1 for EF 2 = 20 2 − 12 2 5 M1 for cos EFM = their EF or B1 for identifying correct angle or stating EFM 8(b)(i) 10.5 3 222.75 M2 for 3  7 oe 66 222.75  7 3 66 3 or M1 for oe or =   66  l  222.75 8(b)(ii) 2.51 or 2.506 to 2.507… 2 66  3 M1 for 3 4

This question in 0607/42 Oct/Nov 2023

6 V NOT TO SCALE 12 cm B C O M A 10 cm D VABCD is a square-based pyramid. V is vertically above the centre of the base O. AD = 10 cm and VO = 12 cm . (a) (i) Calculate the volume of the pyramid. … cm3 [2] (ii) M is the mid-point of CD. Show that VM = 13 cm . [2] (b) V Q R P S Q 8 cm R NOT TO SCALE S P B C A 10 cm D A pyramid VPQRS is cut from the larger pyramid so that the face PQRS is parallel to the face ABCD. QR = 8 cm . (i) Calculate the volume of the remaining solid, ABCDPQRS. … cm3 [4] (ii) Calculate the total surface area of the remaining solid. … cm2 [4]

12 marks

Mark scheme: 6(a)(i) 400 2 1 M1 for × 102 × 12 3 6(a)(ii) 2 M2 2  10 12 2  10  M1 for [VM =]   + 122  + 12  2   2  leading to 13 6(b)(i) 195.2 or 195 4 B3 for 204.8 OR B2 for 9.6 or awrt 9.6 or M1 for [height of small pyramid] = 8 12  oe 10 1 M1 for × 82 × (their 9.6) 3 OR   8  3  M3 for their 400   1 −    oe    10     8 3 or M2 for their 400   oe  10   8  3 or M1 for   oe  10  6(b)(ii) 257.6 or 258 4 2 8 B1 for 13  or 13  10 10 M1 for 1 8 × (their 2.6) + 2 × × 1×(their 2.6) oe 2 M1 for 82 and 102 soi

This question in 0607/43 Oct/Nov 2023

8 NOT TO SCALE 0.8 cm 0.7 cm 0.7 cm The diagram shows a square-based pyramid. The side of the base of the pyramid is 0.7 cm. The length of each sloping edge is 0.8 cm. (a) Show that the perpendicular height of the pyramid is 0.628 cm, correct to 3 significant figures. [4] (b) 4.9 cm NOT TO SCALE 6 cm End view The diagram shows a kitchen tool made from wood. The tool is formed from a cuboid, a cylinder and 49 of the square-based pyramids from part (a). The cylinder has a radius of 1.2 cm and length 25 cm. The cuboid measures 4.9 cm by 4.9 cm by 6 cm. The mass of 1 cm3 of the wood is 0.63 grams. Calculate the total mass of the tool. … g [5]

9 marks

Mark scheme: 8(a) h2 = 0.82 – 0.352 – 0.352 oe M3 M2 for 0.82 = 0.352 + 0.352 + h2 oe or M1 for 0.352 + 0.352 or 0.72 + 0.72 oe 0.6284 to 0.6285 A1 8(b) 165 or 165.0 to 165.2 5 B4 for 262 or 262.1 to 262.2… OR M1 for 6 × 4.9 × 4.9 M1 for π × 1.22 × 25 1 M1 for 3 0.7  0.7  0.628 [× 49] M1 for at least 1 of their volumes × 0.63

This question in 0607/43 May/June 2024

6 8 cm NOT TO SCALE A 5.7 cm 72° B N 3.6 cm C Triangle ABC is the cross-section of a prism. AC = 5.7 cm, NC = 3.6 cm and the length of the prism is 8 cm. Angle ABN = 72° and angle ANC = 90°. (a) AN is the perpendicular height of the triangle. Show that AN = 4.42 cm correct to 3 significant figures. [2] (b) Calculate the volume of the prism. … cm3 [5]

7 marks

Mark scheme: 6(a) 3.62 + x 2 = 5.7 2 or better M1 4.419… A1 6(b) 89[.0] or 89.1 or 89.01 to 89.11 5 4.42 M2 for oe tan72 4.42 or M1 for tan72 = oe BN 1 M2 for  ( their BN + 3.6 )  4.42  8 oe 2 1 or M1 for  ( their BN + 3.6 )  4.42 oe 2 If 0 scored, SC1 for answer 63.6 or 63.63 to 63.65

This question in 0607/41 May/June 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