Solve problems involving the arc length and — Paper 2 · IGCSE Mathematics - Additional 0606

8 A rr cmcm 2› cm θrad O B The diagram shows a circle, centre O of radius r cm, and a chord AB. Angle AOB = θ radians. The length of the major arc AB is 5 times the length of the minor arc AB. The minor arc AB has length 2r cm. (i) Find the value of θ and of r. [2] (ii) Calculate the exact perimeter of the shaded segment. [2] (iii) Calculate the exact area of the shaded segment. [4]

8 marks

Mark scheme: 8(i) π B1 3 6 [cm] B1 8(ii)  π  M1 [major arc = ]  2π − their  their r  3  10π + 6 cao A1 8(iii) 1 2  π  M1 1 2  π  (their 6)  2π − their  (their 6)  their  2  3  2  3  1 2  π  M1 1 2  π  ( their 6) sin  their  ( their 6) sin  their  2  3  2  3  Sector + triangle M1 π ×their 6 2 − (Sector − triangle) 30π + 9 3 A1

This question in 0606/23 May/June 2017

6 40 cm A D x rad O C B 16 cm In the diagram AOB and DOC are sectors of a circle centre O. The angle AOB is x radians. The length of the arc AB is 40 cm and the radius OB is 16 cm. (i) Find the value of x. [2] (ii) Find the area of sector AOB. [2] (iii) Given that the area of the shaded region ABCD is 140 cm2, find the length of OC. [3]

7 marks

Mark scheme: 6(i) 16x = 40 oe M1 x = 2.5 oe (radians) A1 6(ii) 1 M1 (16 )2 (2.5) oe 2 320 A1 6(iii) 1 2 M1 FT provided their 320 > 140 r ( their 2.5 ) = (their 320) − 140 oe 2 correct simplification to r2 = … M1 dep on first M1 12 A1

This question in 0606/21 May/June 2018

6 40 cm A D x rad O C B 16 cm In the diagram AOB and DOC are sectors of a circle centre O. The angle AOB is x radians. The length of the arc AB is 40 cm and the radius OB is 16 cm. (i) Find the value of x. [2] (ii) Find the area of sector AOB. [2] (iii) Given that the area of the shaded region ABCD is 140 cm2, find the length of OC. [3]

7 marks

Mark scheme: 6(i) 16x = 40 oe M1 x = 2.5 oe (radians) A1 6(ii) 1 M1 (16 )2 (2.5) oe 2 320 A1 6(iii) 1 2 M1 FT provided their 320 > 140 r ( their 2.5 ) = (their 320) − 140 oe 2 correct simplification to r2 = … M1 dep on first M1 12 A1

This question in 0606/23 May/June 2018

8 B 8 cm E 2 r rad 9 A C D The diagram shows a right-angled triangle ABC with AB = 8 cm and angle ABC = r radians. The points D 2 and E lie on AC and BC respectively. BAD and ECD are sectors of the circles with centres A and C 2r respectively. Angle BAD = radians. 9 (i) Find the area of the shaded region. [6]

9 marks

Mark scheme: 8(i) 5π B1 [angle ECD =] oe or 0.873 soi 18 Attempts to find AC and subtract 8 M1 8 e.g. AC = 2π cos 9 [ DC = ] 2.44 A1 1 2 π M2 1 2 2π × 8 × theirAC × sin M1 for × 8 × or for 2 9 2 9 1 2 5π × their 2.44 × their seen OR 2 18 1  2π  1 2 2π × 8 × 8tan   − × 8 × 2  9  2 9 1 2 5π − × their 2.44 × their 2 18 awrt 1.91 A1 8(ii) their(6.712 – 2.443) M2 M1 for either arc seen  5π   2π  + their 2.443   + 8    18   9  awrt 12.0 A1

This question in 0606/21 May/June 2019

7 A B 50 cm D C 4r rad 9 O The diagram shows a company logo, ABCD. The logo is part of a sector, AOB, of a circle, centre O and radius 50 cm. The points C and D lie on OB and OA respectively. The lengths AD and BC are equal and 4r AD : AO is 7 : 10. The angle AOB is radians. 9 (i) Find the perimeter of ABCD. [5]

8 marks

Mark scheme: 7(i) [ AD = BC = ] 35 soi B1 Valid method for finding DC M1 [ DC = ]19.2836... A1 4π M1 50 × oe 9 4π A1 35 + 35 + 19.2836… + 50 × 9 = 159 or awrt 159 isw 7(ii) Sector – triangle: M1 or Segment + trapezium : 1 2 4π 1 2  4π 4π  × 50 × × 50  − sin  2 9 2  9 9   1 2  4π   M1  1  −  × their15 × sin    oe +  ( 64.2787... + 19.2836... ) × 26.81155   2  9    2  1630 or 1634.538… rot to 4 or more A1 figs, isw

This question in 0606/23 May/June 2019

6 (a) A circle has a radius of 6 cm. A sector of this circle has a perimeter of 2 6 + 5r cm. Find the area of this sector. [4] (b) A 7 cm O r rad 4 B The diagram shows the sector AOB of a circle with centre O and radius 7 cm. Angle AOB = r radians. Find the perimeter of the shaded region. [3] 4

7 marks

Mark scheme: 6(a) 2(6) + 6θ = 2(6 + 5π) oe M1 5 A1 θ = π oe, soi 3 1 2 5π  M1 × 6 × their  2  3  94.2 or 30π A1 Alternative method arc AB = 10π (M1 10π 5 B1 sector is = of the circle 12π 6 5 M1 × 36π 6 94.2 or 30π A1) 6(b)  π 7π M2  π   7π  2  7sin + oe, soi M1 for 2  7sin  + their   or  8  4  8   4  their 2  7sin π +  7π   8   4 10.9 or 10.85 to 10.86 A1

This question in 0606/22 Feb/March 2020

11 A C1 C2 B The circles with centres C1 and C2 have equal radii of length r cm. The line C1C2 is a radius of both circles. The two circles intersect at A and B. (a) Given that the perimeter of the shaded region is 4r cm, find the value of r. [4] (b) Find the exact area of the shaded region. [4]

8 marks

Mark scheme: 11(a) 4 B2 2 [perimeter =] πr soi B1 for angle ACB = π 3 3  4 M1  t h e ir π r = 4π oe  3  r = 3 A1 11(b) 1 2 2π M1 × their 3 × their oe 2 3 1 2 2π M1 × their 3 × sin their oe 2 3 For subtracting and doubling: M1 2 2π their 3 × their − 3 2 2π their 3 × sin their 3 9 A1 6π − 3 or exact equivalent 2

This question in 0606/22 May/June 2020

12 C 3 cm 4 cm A B 5 cm D The diagram shows a shape consisting of two circles of radius 3 cm and 4 cm with centres A and B which are 5 cm apart. The circles intersect at C and D as shown. The lines AC and BC are tangents to the circles, centres B and A respectively. Find (a) the angle CAB in radians, [2] (b) the perimeter of the whole shape, [4] (c) the area of the whole shape. [4]

10 marks

Mark scheme: 12(a) 4 M1 Correct use of tan oe tan CAB = 3 CAB = 0.927 A1 isw 12(b)  π  B1 Angle CBD = 2  − 0.927  = 1.287  2  Perimeter 3 = 3 ( 2 π − 2 × 0.927 ) + 4 ( 2 π − 1.287 ) M1 for correct plan of two arcs A1 for either arc = 13.287 + 19.985 A1 = 33.3 12(c) Area of two right-angled triangles B1 1 = × 3 × 4 × 2 = 12 2 Area of Sectors 3 32 4 2 M1 for correct plan of two sectors plus = ( 2 π − 2 × 0.927 ) + ( 2 π − 1.287 ) triangles 2 2 A1 for either sector = 19.93 + 39.97 A1 Total = 71.9

This question in 0606/21 Oct/Nov 2020

6 A B 16 cm 2 r rad C 7 7.5 cm O 2r AOB is a sector of a circle with centre O and radius 16 cm. Angle AOB is radians. The point C lies 7 on OB such that OC is of length 7.5 cm and AC is a straight line. (a) Find the perimeter of the shaded region. [3] (b) Find the area of the shaded region. [3]

6 marks

Mark scheme: 6(a) M2 M1 for 2 2 2π 16 + 7.5 − 2(16)(7.5)cos 2π 7 16 2 + 7.5 2 − 2(16)(7.5)cos + (16 − 7.5) 7 2π + (16 − 7.5) + 16 × 2π 7 or for 16 × seen oe, soi 7 35.6 or 35.6 to 35.614 A1 6(b) 1 2 2π M2 1 2 2π × 16 × − M1 for either × 16 × or 2 7 2 7 1  2π  1  2π  × 16 × 7.5 × sin   oe × 16 × 7.5 × sin   2  7  2  7  68[.0] or 67.98 to 68.0 A1

This question in 0606/22 Feb/March 2021

7 B 18 cm A 7rrad C 9 D DAB is a sector of a circle, centre A, radius 18 cm. The lines CB and CD are tangents to the circle. 7 r Angle DAB is radians. 9 (a) Find the perimeter of the shaded region. [3] (b) Find the area of the shaded region. [3]

6 marks

Mark scheme: 7(a) [Arc length + 2 × tangent length] M2 M1 for 7π 7π 7π 18 × + 2 × 18 × tan oe [Arc length] 18 × oe 9 18 9 7π or [Tangent length] 18 × tan oe 18 18 or [Tangent length] oe π tan 9 18 7π or [Tangent length] × sin oe π 18 sin 9 143 or 142.9 or awrt 142.9 (cm) A1 7(b) [Area of kite – area of sector] M2 FT their BC or CD from (a) providing it is not  7π  1 2 7π 18 18 × their  18 × tan  − × 18 ×  18  2 9 1 2 7π oe M1 for [area of sector] × 18 × oe 2 9  7π  or [area of kite] 18 × their  18 × tan  oe  18  or [area of kite] 18 × their(18×tan70) oe 494 or 494.3 or awrt 494.3 (cm2) A1

This question in 0606/22 May/June 2021

8 A B 15 cm a cm C r rad 6 O r The diagram shows the sector AOB of a circle, centre O and radius 15 cm. Angle AOB is radians. 6 Point C lies on OB such that CB is a cm. AC is a straight line. (a) Find the exact value of a such that the area of triangle AOC is equal to the area of the shaded region ACB. [4] (b) For the value of a found in part (a), find the perimeter of the shaded region. Give your answer correct to 1 decimal place. [3]

7 marks

Mark scheme: 8(a) 1 2  π  B1 [Area of sector = ] (15)   soi 2  6  1  π  B1 [Area of triangle = ] (15)(15 − a )sin   2  6  soi Forms correct equation and attempts to M1 solve for a or 15 – a or OC 1  e.g. (15) 2  π − 15(15 − a ) = 15(15 − a ) 2  6  4 4 75π 15 or = OC 8 4 and solves as far as a = ... or 15 – a = ... or OC = ... 5 A1 15 − π (cm) or exact equivalent 2 8(b) [CA + arc AB + BC = ] M2 FT their(15 −5 π ) and 5 π 2 2 2  5  2 5 π 15 +  π  − 2 × 15 × π × cos π  2  2 6 M1 for 15 × oe seen or 6  π   5  +  15 ×  +  15 − π  oe, soi 2  6   2  2  5  5 π  5  15 +  π  − 2 × 15 × π × cos +  15 − π   2  2 6  2  oe seen 24.1 (cm) A1

This question in 0606/22 Feb/March 2022

9 In this question all lengths are in centimetres. P a 2z T O rad Q The diagram shows a circle, centre O, radius a. The lines PT and QT are tangents to the circle at P and Q respectively. Angle POQ is 2z radians. (a) In the case when the area of the sector OPQ is equal to the area of the shaded region, show that tan z = 2z . [4] (b) In the case when the perimeter of the sector OPQ is equal to half the perimeter of the shaded region, find an expression for tanz in terms of z. [3]

7 marks

Mark scheme: 9(a) 1 2 1 2 B1 or [area kite =] 2 a 2 or [area sector =] 2  a  or a (2) oe 2 2 [area OPT =] a 2 nfww  1  1 B1 2 1 [shaded area = ] 2   a ( a tan) oe or or [ a  a  PT ] PT  2 a    2  2 2 and PT  a tanoe, nfww 1 2 a ( a tan )  a (2) oe soi 2 Correct equation using correct areas e.g. M1 or equates expressions for PT 2 1 2 2 a  a ( a tan ) or a ( a tan)  a  a  2 soi Correct completion to given equation A1 tan 2 Alternative method 1 1 2 (B1) [ area sector =] a  2 2 1 (B1) [ shaded area = ] 2 1 1  a ( a tan ) oe 2 2 1 1 2 or a ( a tan )  a  oe soi 2 2 Correct equation using correct areas (M1) 1 2 1 e.g. a  a ( a tan ) 2 4 1 2 1 2 1 2 or a tan  a  a soi 2 2 2 Correct completion to given equation (A1) tan 2 9(b) 1 M2 M1 for arc length = 2asoi or for 2a + a(2) = (2 a tan  a (2)) oe 2 PT  a tan and PT  2 a  a or a tan 2 a  a tan  2   A1

This question in 0606/22 May/June 2022

9 In this question, all lengths are in centimetres and all angles are in radians. (a) C A 3r 8 D B O The diagram shows sectors AOB and COD of two circles with the same centre, O. Angle AOB is 3 r and the length of OC is 6.5. It is given that OAC and OBD are straight lines and 8 OA : OC is 4 : 5. Find the perimeter of the shaded region. [3] (b) Q a y P O z R The diagram shows a circle with centre O and radius a. Sector PQR is a sector of a different circle with centre R and radius y. Angle OPR is z. Find, in terms of a and z only, the total area of the three shaded regions. Simplify your answer. [4]

7 marks

Mark scheme: 9(a)  3π   3π  M2  3π  6.5   + 5.2   + 2(6.5 − 5.2) M1 for 6.5    8   8   8   3π  or 5.2    8  16.38 to 16.4 A1 9(b) [Angle PRQ = ] 2 soi B1 a sin(π − 2) B1 y = 2a cosoe or y = oe sin y 2 = a 2 + a 2 − 2 a 2 cos(π − 2) oe or y 2 = a 2 + a 2 + 2 a 2 cos(2) oe Complete and correct plan soi: M1 FT their 2and their 2 1 2 expression for y or y2 in terms πa − (2 a cos) (2) oe 2 of a and  2 1  a sin(π − 2)  2 or πa −   (2) oe 2  sin  2 1 2 2 2 or πa − ( a + a − 2a cos(π − 2))(2) oe 2 2 1 2 2 2 or πa − ( a + a + 2a cos(2))(2) 2 2sin 2 (π − 2) A1 2 2 2 a π − 4cos  or πa −a ( ) sin 2  or πa 2 − 2( a 2 − a 2 cos(π − 2)) or πa 2 − 2( a 2 + a 2 cos2) oe

This question in 0606/22 Feb/March 2023

10 In this question all lengths are in centimetres. C 6 O A 8 B The diagram shows a circle centre O with radius 6. The line AB is a tangent to the circle at the point B. The point C lies on the circle such that AOC is a straight line. AB = 8 . (a) Find the perimeter of the shaded region. [6]

9 marks

Mark scheme: 10(a) 2 2 B1 OA= 6 + 8 oe or 10 [Angle AOB = ] M1 0.9272[95…] rot to 4 or more dp or [Angle OAB = ] 0.6435[01…] rot to 4 or more dp [Angle COB =] 2.214[297…] rot to 3 or A1 more dp [Arc CB =] 6(their 2.214) M1 FT their COB [Perimeter =] 8 + (their 10 + 6) + 6(their M1 FT their arc CB and OA 2.214) 37.3 or 37.28[578…] rot to 2 or more dp A1 10(b) 1 1 2 M2 FT their 2.21 +8 6  6  2.214 oe, soi 2 2 1 2 M1 for  6  2.214 soi 2 63.9 or 63.85[735…] rot to 2 or more dp A1

This question in 0606/21 Oct/Nov 2023

9 In this question all lengths are in centimetres and all angles are in radians. C B A 0.5 2 O 1 E D F The diagram shows a company logo. Each part of the logo is a sector of a circle with centre O. Sector AOB has radius x. Sector COD has radius x + 2 . Sector EOF has radius y. The shaded region has area A cm 2 and perimeter 24. It is given that x and y can vary. 91 2 (a) Show that A = x - 68 x + 132 . [4] 8

9 marks

Mark scheme: 9(a) 1 2 1 2 1 2 B1  A =  x  0.5 + ( x + 2) +2 y [1] soi 2 2 2 [P = ] M1 Attempts to form an expression in x x + 0.5 x + 2 + 2( x + 2) + ( x + 2 − y ) + y + y and y for the perimeter using arc lengths and lengths of lines 9 A1 Equates P to 24 and rearranges: y = 16 − x 2 5 2 81 2 A1 A = x + 4 x + 4 + 128 − 72 x + x oe 4 8 leading to given answer 91 2 A = x − 68 x + 132 8 9(b) dA 91 M1 = x − 68 dx 4 dA M1 dA Solves = 0 for x FT their providing at least one dx dx term is correct 272 90 A1 x = or 2 or 91 91 2.99 or 2.989[01...] rot to 4 or more sf 2 M1 FT their x 91  272   272  A =   − 68   + 132 8  91   91  2764 34 A1 A = or 30 or 91 91 30.4 or 30.37[36...] rot to 4 or more sf

This question in 0606/22 Feb/March 2024

7 C D i rad O A B 5 cm 4 cm In the diagram, AD and BC are arcs of circles with common centre O. ODC and OAB are straight lines with OA = 5 cm and AB = 4 cm . Angle BOC = i radians . The area of the shaded region ABCD is 4rcm 2. (a) Find i. [3] (b) C D i rad O A B 5 cm 4 cm The straight line AC is added to the diagram and the region ACD is now shaded. Find the perimeter of the shaded region ACD. [5]

8 marks

Mark scheme: 7(a) 1 2 1 2 M2 1 2 1 2  9   5   4π oe, soi M1 for  9  or  5  oe, 2 2 2 2 soi π A1  oe or 0.449 or 0.4487 to 0.4488 7 7(b) 5π 2 π [Arc AD = ] M1 for [Arc AD = ] 5  their FT any 7 7 stated value of from (a) [AC = ] 4.991[27...] rot to 4 or more sf 2 M1 for  π  [AC2 = ] 92 + 52 – 2(9)(5) cos  their   7  π FT their providing 0 <  < 2 11.2 or 11.23[526...] rot to 4 or more sf A1

This question in 0606/23 May/June 2024