Circles, arcs and sectors — Paper 4 · IGCSE Mathematics - International 0607

6 NOT TO SCALE 10 cm The diagram shows a regular pentagon, of side 10 cm, with its vertices lying on a circle. (a) Show that the radius of the circle is 8.51 cm, correct to 3 significant figures. [4] (b) Calculate (i) the perimeter of the shaded segment, … cm [3] (ii) the area of the shaded segment. … cm2 [3]

10 marks

Mark scheme: 6(a) 36 or 54 or 72 or 108 or 540 seen B1 5 ÷ cos 54 oe M2 5 or M1 for cos 54 = oe r Starting with 8.51 is M0 8.506 to 8.507 A1 6(b)(i) 20.7 or 20.68 to 20.70 3 72 M2 for × 2 × π × 8.51 + 10 oe 360 72 or M1 for oe soi by ÷ 5 360 6(b)(ii) 11.0 or 11.1 or 11.02 to 11.10... 3 72 M1 for × π × 8.512 oe 360 M1 for 0.5 × 8.512 × sin 72 oe

This question in 0607/42 Oct/Nov 2018

9 A 58° NOT TO SCALE 14 cm 12 cm O B N C A, B and C are points on the circle, centre O. ON is perpendicular to BC. AB = 14 cm, AC = 12 cm and angle BAC = 58°. (a) Show that BC = 12.73 cm, correct to 2 decimal places. [3] (b) Explain why angle BON = 58°. … … [1] (c) Calculate OB, the radius of the circle. OB = … cm [3] (d) Calculate the area of the shaded segment. … cm2 [3]

10 marks

Mark scheme: 9(a) 142 + 122 – 2 × 14 × 12 × cos58 M1 12.725 to 12.726 A2 or A1 for 161.9... 9(b) Angle at centre = 2 × angle at 1 circumference oe 9(c) 7.49 or 7.5[0] or 7.51 or 7.487 to 7.506 3 6.365 M2 for oe sin58 6.365 or M1 for sin 58 = oe OB 9(d) 31.3 to 31.9 nfww 3 116 M2 for × π × (their (c))2 360 1 − × (their (c))2 × sin116 oe 2 116 or M1 for × π × (their (c))2 oe 360 1 or × (their (c))2 × sin116 oe 2

This question in 0607/43 Oct/Nov 2019

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

11 (a) 2r NOT TO r SCALE y° The diagram shows a sector of a circle with radius r and angle y°. The length of the arc of the sector is 2r. Calculate the value of y. y = … [3] (b) NOT TO 8 cm SCALE x° The diagram shows a sector of a circle with radius 8 cm and angle x°. The area of the shaded segment is A cm 2. 8x (i) Show that A = r - 32 sin x . 45 [2] (ii) Find the value of A when x = 90. … [1] 8x (iii) By sketching the graph of A = r - 32 sin x , find the value of x when A = 5.5 . 45 A 20 0 90 x x = … [3]

9 marks

Mark scheme: 11(a) 114.6 or 114.5 to 114.6 3 y M2 for × 2πr = 2 r oe 360 y or M1 for × 2 πr oe 360 11(b)(i) x 2 1 2 M2 x 2 1 2 × π× 8 − × 8 × sin x = A M1 for × π× 8 or × 8 × sin x 360 2 360 2 11(b)(ii) 18.3 or 18.26 to 18.27... 1 11(b)(iii) Correct sketch of curve and line B2 B1 for correct shape of curve 6666 4444 2222 0000 0000 20202020 40404040 60606060 80808080 58.9 or 58.90 to 58.92 1

This question in 0607/42 Feb/March 2022

6 (a) NOT TO C SCALE 5 cm B 30° O A The diagram shows a circle, centre O, with radius 5 cm. BA and BC are tangents to the circle at A and C. Angle ABC = 30° . Calculate the area of the shaded minor segment. … cm2 [4] (b) O NOT TO height O SCALE 40° 12 cm D E D E The circle, centre O, has radius 12 cm. Angle DOE = 40° . The minor sector DOE is removed. The major sector is formed into a cone by joining OD to OE. Calculate the height of the cone. … cm [5]

9 marks

Mark scheme: 6(a) 26.5 or 26.47 to 26.48 4 B1 for AOC = 150 or BOC = 75 soi their150  5 2 M1 for or 360 their 75  5 2  2 360 1 M1 for 5 5 sin[their150] or 2 1 5 5 sin[their 75  2] 2 6(b) 5.5[0] or 5.49 to 5.51... 5 (360  40) M2 for 2 12 = 2r 360 (360  40) 2 oe or  12 = πr12 oe 360 (360  40) or M1 for 2 12 or 360 (360  40) 2  12 360 4 If minor sector SC1 for radius = oe 3 AND M2 for 12 2 their ( radius ) 2 oe dependent on at least M1 or SC1 or M1 for h 2  their ( radius ) 2  12 2 oe dependent on at least M1 or SC1

This question in 0607/41 May/June 2022

7 (a) NOT TO SCALE 10 cm 60° The diagram shows a sector of a circle with sector angle 60° and radius 10 cm. Calculate the area of the shaded segment. … cm2 [3] (b) A NOT TO SCALE C D O E F B The diagram shows a circle with radius 10 cm and centre O. A and B are at opposite ends of a diameter. COD is an arc of a circle centre A. EOF is an arc of a circle centre B. (i) Calculate the area of the shaded region. … cm2 [4] (ii) Calculate the perimeter of the shaded region. … cm [2]

9 marks

Mark scheme: 7(a) 9.06 or 9.07 or 9.058 to 9.065... 3 60 M1 for × π × 102 oe 360 1 M1 for × 10 × 10 × sin60 oe 2 7(b)(i) 68.4 or 68.5 or 68.43 to 68.50... 4 M3 for 2  120 2   10 − 2   10 + 2  their ( a )  oe  360   30 2  or 4    10 − their (a)  oe  360  or M2 for  60 2  k    10 + their (a)   360  k = 1 or 2 or 4 30 2 or k × (  10 – their(a)) oe 360 k = 1 or 2 1 2 or q   10  sin60 + 2 q  their (a) 2 q = 1 or 2 or 4 30 2 or M1 for k  π 10 360 k = 1 or 2 or 4 or 8 or p × their(a) oe p = 2, 4 or 8 1 2 or q  10  sin60 oe q = 1, 2 or 4 2 7(b)(ii) 62.8 or 62.83 to 62.84 2 k M1 for  2 × π × 10 oe 6 k = 1, 2, 3, 4 or 6

This question in 0607/42 Oct/Nov 2022

8 P NOT TO SCALE 140° O 9 cm Q The diagram shows the sector of a circle with radius 9 cm and sector angle 140°. (a) Calculate the length of the arc PQ. … cm [2] (b) Calculate the area of the sector. … cm2 [2] (c) The sector is the cross-section of a solid of length 20 cm. Calculate the total surface area of the solid. … cm2 [4] (d) Another solid is mathematically similar to the solid in part (c). The radius of the sector in this solid is 10 cm. Calculate the total surface area of this solid. … cm2 [2]

10 marks

Mark scheme: 8(a) 22[.0] or 21.99... 2 140 M1 for 2 π  9 oe 360 8(b) 99[.0] or 98.96 to 98.97... 2 140 2 M1 for π 9 oe 360 8(c) 998 or 997.7 to 998.0 4 M1 for their (a) × 20 M1 for their (b) × 2 M1 for [2 ×] 9 × 20 8(d) 1230 or 1231 to 1232...nfww 2 2  10  FT their (c)    9   10  2  9  2 M1 for   or    9   10 

This question in 0607/41 Oct/Nov 2023

8 8 m A B 73.4° NOT TO 36° SCALE 7 m C D The diagram shows a shape ABDC formed from triangle ABC and a sector of a circle BCD, centre B. (a) Show that BC = 9.0 m , correct to 1 decimal place. [3] (b) Use the sine rule to find angle BCA. Angle BCA = … [3] (c) Find the area of triangle ABC. … m2 [2] (d) Find the area of the shaded region. … m2 [3] (e) Find the perimeter of the shape ABDC. … m [2]

13 marks

Mark scheme: 8(a) M2 or M1 for [ BC 2 ]  82  7 2 2 8 7 cos73.4 [ BC  ] 8 2  7 2  2 8 7 cos73.4 M1 for 81.[00…] 9.00[0…] A1 8(b) 8  sin73.4 M2 8 9 [sin...]  M1 for  oe 9 sin BCA sin73.4 58.4 or 58.41… B1 8(c) 26.8 or 26.83… 2 1 M1 for 7 8 sin73.4 oe 2 8(d) 1.64 or 1.641 to 1.642 3 36 M1 for areasector BCD  π 9 9 360 M1 for area triangle BCD  0.5 9 9 sin36 8(e) 29.7 or 29.65 to 29.66 NFWW 2 36 M1 for  2π  9 oe 360

This question in 0607/41 May/June 2024

5 (a) Calculate the area of an equilateral triangle with side length 12 cm. … cm2 [2] (b) Calculate the area of a circle with circumference 60 cm. … cm2 [3] (c) 6 cm NOT TO SCALE O The diagram shows part of a regular 10-sided polygon with centre O and side length 6 cm. Calculate the area of the polygon. … cm2 [4]

9 marks

Mark scheme: 5(a) 62.4 or 62.35... 2 1 M1 for  12  12  sin60 oe 2 5(b) 286 or 286.4 to 286.5 3 M2 for π × (their radius)2 or M1 for 2πr = 60 or better 5(c) 277 or 276.9 to 277.0 4 1 M3 for [10 ×] their6 height oe 2 1  3  2 or [10]     sin36 2  sin18  3 or M2 for 3 tan72 or oe sin18 or M1 for angles 72 or 36 or 18 or 144 seen

This question in 0607/42 May/June 2024

15 A 4 cm B NOT TO 154° O SCALE C A and C are points on the circle, centre O. BA and BC are tangents to the circle. The radius of the circle is 4 cm. Angle AOC = 154°. (a) Write down the mathematical name of the quadrilateral ABCO. … [1] (b) Calculate the area of the shaded segment. … cm2 [3] (c) Calculate the shortest distance from B to the circumference of the circle. … cm [3]

7 marks

Mark scheme: 15(a) kite 1 15(b) 18.[0] or 17.99… 3 154 2 1 M2 for π 4 − 4 4 sin154 oe 360 2 154 2 1 or M1 for π 4 or 4 4 sin154 oe 360 2 15(c) 13.8 or 13.78… 3 4 M2 for − 4 oe cos77 4 M1 for cos77 = oe or better x or M1 for their OB – 4

This question in 0607/42 Feb/March 2025

14 B NOT TO SCALE 25 m A O 35° 15 m D C OAD and OBC are sectors of circles each with centre O. The sector angle is 35°. The radius of the smaller circle is 15 m. The radius of the larger circle is 25 m. (a) Calculate the area of the shaded region. … m2 [3] (b) Calculate the perimeter of the shaded region. … m [4]

7 marks

Mark scheme: 14(a) 122 or 122.1 to 122.2 3 35 2 35 2 M2 for π 25 − π 15 360 360 35 2 35 2 or M1 for π 25 or π 15 360 360 14(b) 44.4 or 44.43 to 44.44 4 M3 for 10 + 10 + 35 35 2 π 25 + 2 π 15 360 360 35 35 or M2 for 2 π 25 + 2 π 15 360 360 35 35 or M1 for 2 π 25 or 2 π 15 360 360

This question in 0607/41 Oct/Nov 2025