Circular measure — Paper 1 · A Level Mathematics 9709

2 In the diagram, OAB and OCD are radii of a circle, centre O and radius 16 cm. Angle AOC = α radians. AC and BD are arcs of circles, centre O and radii 10 cm and 16 cm respectively. (i) In the case where α = 0.8, find the area of the shaded region. [2] (ii) Find the value of α for which the perimeter of the shaded region is 28.9 cm. [3]

5 marks

Mark scheme: 21 (i) 3 sin θ − 2 cos θ − 3 = 0 Use of s² + c² = 1 M1 Use of s²+c²=1 to eliminate sine. 3cos²θ + 2cosθ = 0 A1 Correct equation cosθ = 0, θ = 90° B1 Co. or cosθ = −2/3, θ = 131.8 ° A1 Co. ( to 1 d.p or more – there must be only this answer in the range 0 to 180)) [4] 2 α = 0.8, radii 10cm and 16 cm (i) Area = ½.16².0.8 − ½.10².0.8 M1 Use of ½r²θ once.. → 62.4 cm² A1 Co. [2] (ii) Arcs are 10α and 16α M1 Use of s=rθ once. 12 + 10α + 16α = 28.9 DM1 Forming an eqn, including the 6 + 6. → α = 0.65 A1 co [3] 3 (i) 2dsin30 + 2d√3sin60 M1 For one - allow even if decimals used = 2d.½ + 2d√3.√3/2 = 4d A1 Co – exact answer only. [2] ans to (i) (ii) tan θ = 2 d cos 30 + 2 3d cos 60 M1 Use of tan = opp/adj in correct triangle DM1 For horizontal step attempted 2 tan θ =

This question in 9709/11 Oct/Nov 2005

5 In the diagram, OAB is a sector of a circle with centre O and radius 12 cm. The lines AX and BX are tangents to the circle at A and B respectively. Angle AOB = 13π radians. (i) Find the exact length of AX, giving your answer in terms of √3. [2] (ii) Find the area of the shaded region, giving your answer in terms of π and √3. [3]

5 marks

Mark scheme: 5 (i) tan 16 π =AX÷12 or other valid method M1 Use of trig with tangent in correct ∆ tan 16 π =√3÷3 → AX = 4√3 A1 Co (12 ÷ √3 ok) [2] (ii) area AOC = ½r²θ (= 24π) M1 Correct formula + attempt with radians Area of ∆AOX = ½ × AX×12 M1 Use of ½bh in correct ∆ (once ok) → shaded area = 48√3 − 24π A1 co (144 ÷ √3 ok) [3]

This question in 9709/11 May/June 2007

5 R1 A B q rad O R2 The diagram shows a circle with centre O. The circle is divided into two regions, R1 and R2, by the radii OA and OB, where angle AOB = θ radians. The perimeter of the region R1 is equal to the length of the major arc AB. (i) Show that θ = π −1. [3] (ii) Given that the area of region R1 is 30 cm2, find the area of region R2, correct to 3 significant figures. [4]

7 marks

Mark scheme: 5 (i) Perimeter of R1 = r + r + rθ B1 Major arc length = 2πr − rθ B1 Equated and solved → θ = π − 1 B1 co answer was given [3] (ii) ½r²θ with θ = π − 1, equated to 30 M1 Use of correct formula once → r ² = 60/(π − 1) ( r = 5.29) A1 Any correct form for r or r² R2 = ½r ² (π + 1) → 58.0 M1A1 Attempt at r (or r²) and at area of R2 or [4] co (could be full circle − sector) [Reflex = π + 1. R2 = 30(π + 1) ÷ (π − 1)] [ M1 A1 M1 A1] GCE A/AS LEVEL – May/June 2009 9709 01

This question in 9709/11 May/June 2009

7 P r cm q rad O Q A piece of wire of length 50 cm is bent to form the perimeter of a sector POQ of a circle. The radius of the circle is r cm and the angle POQ is θ radians (see diagram). (i) Express θ in terms of r and show that the area, A cm2, of the sector is given by A = 25r −r2. [4] (ii) Given that r can vary, find the stationary value of A and determine its nature. [4]

8 marks

Mark scheme: 7 (i) 2r + rθ = 50 M1 Must use s = rθ and link with perimeter 1 θ = (50 – 2r) A1 co r A = 1 r2θ M1 Used with θ as f(r) 2 → A = 25r – r2 A1 co (answer given) [4] d A (ii) = 25 − 2 r B1 co dr = 0 when r = 12.5 M1 sets differential to 0 + solution A = 156¼ A1 co 2nd differential negative → Maximum B1 Could be quoted directly from quadratic. [4] GCE A/AS LEVEL – October/November 2009 9709 12 3

This question in 9709/12 Oct/Nov 2009

7 C B D 10 cm O 12 cm A E F The diagram shows a metal plate ABCDEF which has been made by removing the two shaded regions from a circle of radius 10 cm and centre O. The parallel edges AB and ED are both of length 12 cm. (i) Show that angle DOE is 1.287 radians, correct to 4 significant figures. [2] (ii) Find the perimeter of the metal plate. [3] (iii) Find the area of the metal plate. [3]

8 marks

Mark scheme: 6 7 (i) sin 1 2 θ = M1 Use of trig with/without radians 10 Angle DOE = 1.287 radians. A1 co – answer given. [2] (ii) P = 12 + 12 + 2 × 10 × angle BOD M1 Use of s = rθ for arc length. Angle BOD = (π – 1.287) M1 Correct angle → 61.1 A1 co [3] (iii) Sector DOE = ½ × 102 × 1.287 M1 Correct formula used with radians. Triangle DOE = ½ × 102 × sin 1.287 M1 Correct formula used with radians. Area = π × 102 – (2 sectors – 2 triangles) (or 48 + 48 + 2×½×102×(π – 1.287) M1 M1 → 281 or 282 A1 co [3] GCE AS/A LEVEL – May/June 2010 9709 13

This question in 9709/13 May/June 2010

9 C1 P 8 cm T C2 Q 2 cm R S The diagram shows two circles, C1 and C2, touching at the point T. Circle C1 has centre P and radius 8 cm; circle C2 has centre Q and radius 2 cm. Points R and S lie on C1 and C2 respectively, and RS is a tangent to both circles. (i) Show that RS 8 cm. [2] = (ii) Find angle RPQ in radians correct to 4 significant figures. [2] (iii) Find the area of the shaded region. [4]

8 marks

Mark scheme: 9 (i) RS² = 10² – 6² M1 Use of Pythagoras (or other) → RS = 8 cm. A1 Answer given. [2] (ii) sin θ = 8/10 oe M1 Use of trig – even if with degrees. → angle RPQ = 0.9273 radians A1 co in radians. (Accept 0.927) [2] (iii) Region = trapezium − 2 sectors Area of trapezium = 40 cm² B1 co 1 1 Large sector = × 8² × 0.9273 M1 Use of r²θ. 2 2 Small sector angle = (π − 0.9273) 1 1 Small sector = × 2² × 2.214 M1 Use of r²θ with angle = π − (ii) 2 2 → 5.90 cm2 A1 [4] co 2

This question in 9709/11 Oct/Nov 2010

4 C 3 cm 2.3 rad 2.3 rad O 3 cm 3 cm A B P The diagram shows points A, C, B, P on the circumference of a circle with centre O and radius 3 cm. Angle AOC = angle BOC = 2.3 radians. (i) Find angle AOB in radians, correct to 4 significant figures. [1] (ii) Find the area of the shaded region ACBP, correct to 3 significant figures. [4]

5 marks

Mark scheme: 4 (i) 1.683(18…) B1 [1] (ii) (2) × ½ × 3²sin2.3 M1 Condone omission of factor 2 ½ × 3² × their 1.683 M1 NB M0 if using angle of 2.3 Triangle AOC + COB + sector M1 Two correct triangles + sector 14.3 A1 co [4]

This question in 9709/12 Oct/Nov 2010

8 A B P Q D C The diagram shows a rhombus ABCD. Points P and Q lie on the diagonal AC such that BPD is an arc of a circle with centre C and BQD is an arc of a circle with centre A. Each side of the rhombus has length 5 cm and angle BAD = 1.2 radians. (i) Find the area of the shaded region BPDQ. [4] (ii) Find the length of PQ. [4]

8 marks

Mark scheme: 8 (i) 1/2 × 52 × 1.2 B1 1/2 × 52 × sin 1.2 B1 2[1/2 × 52 × 1.2 – 1/2 × 52 × sin 1.2] M1 Subtraction and multiplication by 2 6.70 A1 Accept 6.7 or anything rounding to 6.70 [4] (ii) 5cos 0.6 M1 5 – “5cos 0.6” M1 Subtraction from 5 10(1 – cos 0.6) M1 Multiplication by 2 1.75 A1 [4] 100

This question in 9709/13 Oct/Nov 2010

9 S A B r P T 2q O In the diagram, OAB is an isosceles triangle with OA = OB and angle AOB = 2θ radians. Arc PST has centre O and radius r, and the line ASB is a tangent to the arc PST at S. (i) Find the total area of the shaded regions in terms of r and θ. [4] (ii) In the case where θ = 13π and r = 6, find the total perimeter of the shaded regions, leaving your answer in terms of √3 and π. [5] [Questions 10 and 11 are printed on the next page.]

9 marks

Mark scheme: 9 (i) AS = r tan θ r M1 Or (AB) = 2r tan θ or ( AO ) = 2 1 2 cos θ Area OAB = r tan θ or (OAS ) = 2 r tan θ A1 2 1 r sin 2θ Or OAB = Area of sector = 12 r 2 × 2θ ( = r 2θ ) B1 A1 2 cos 2θ 1 Shaded area = r 2 (tan θ − θ ) OE [4] Or area sector (OPS ) = 2 r 2θ Allow e.g. r 2 tan θ − 1 2 r 2 2θ π 6 (ii) cos = ⇒ OA = 12 M1 3 OA AP = 6 A1 π AS = 6 tan (⇒ AB = 12 3 ) B1 3 π π Arc (PST) = 12π B1 Or arc ( PS ) = 6 or arc ( ST ) = 6 3 3 3 Perimeter = 12 + 12 3 + 4π A1 Allow unsimplified 4π [5] 2

This question in 9709/11 May/June 2011

5 C B q rad r q rad O A r The diagram represents a metal plate OABC, consisting of a sector OAB of a circle with centre O and radius r, together with a triangle OCB which is right-angled at C. Angle AOB = θ radians and OC is perpendicular to OA. (i) Find an expression in terms of r and θ for the perimeter of the plate. [3] (ii) For the case where r = 10 and θ = 15π, find the area of the plate. [3]

6 marks

Mark scheme: 5(2 a + 9 d ) = 200 B1 Attempt solution, expect d = 6 a = −7 M1

This question in 9709/11 Oct/Nov 2011

6 C2 D E C1 6 cm 10 cm 3p1 q A B X The diagram shows a circle C1 touching a circle C2 at a point X. Circle C1 has centre A and radius 6 cm, and circle C2 has centre B and radius 10 cm. Points D and E lie on C1 and C2 respectively and DE is parallel to AB. Angle DAX = 13π radians and angle EBX = θ radians. (i) By considering the perpendicular distances of D and E from AB, show that the exact value of θ 3 √3 is sin−1 . [3] 10 (ii) Find the perimeter of the shaded region, correct to 4 significant figures. [5]

8 marks

Mark scheme: 6 (i) D to AX = 6 sin π3 = 6√3÷2 B1 co Needs –√3÷2 not just 3√3. E to AX = 10sinθ B1 co Correct method. ag. 3 3 B1 Use of decimals loses this B mark. . Equate these → θ = sin −1 10 [3] (ii) Arc DX = 6.⅓π = 2π B1 co Arc EX = 10×0.5464 =5.464 M1 Use of s=rθ radians. Horizontal steps = 6cos⅓π and 10cosθ M1 Attempt at both steps needed DE = 10 + 6 − 6cos⅓π − 10cosθ M1 Full method for DE. Perimeter = arc DX + arc BX + DE → 16.20 A1 Co – must be exactly 16.20, not more or [5] less places. dy 8

This question in 9709/12 Oct/Nov 2011

4 D 10 cm C P 10 cm Q 0.8 rad A 10 cm B In the diagram, ABCD is a parallelogram with AB = BD = DC = 10 cm and angle ABD = 0.8 radians. APD and BQC are arcs of circles with centres B and D respectively. (i) Find the area of the parallelogram ABCD. [2] (ii) Find the area of the complete figure ABQCDP. [2] (iii) Find the perimeter of the complete figure ABQCDP. [2]

6 marks

Mark scheme: 4 (i) 102 sin 0.8 = 71.7 M1A1 Completely correct method for a [2] triangle (ii) sector(s) = (2) × 12 × 102 × 0.8 = (2) × 40 M1 Correct formula used for a sector Total area = 80 A1 [2] (iii) arc(s) = (2) × 10 × 0.8 16+20 = 36 M1 Correct formula used for an arc A1 [2] 2 2 2

This question in 9709/13 Oct/Nov 2011

3 B 8 cm a rad O A 8 cm C In the diagram, OAB is a sector of a circle with centre O and radius 8 cm. Angle BOA is ! radians. OAC is a semicircle with diameter OA. The area of the semicircle OAC is twice the area of the sector OAB. (i) Find ! in terms of 0. [3] (ii) Find the perimeter of the complete figure in terms of 0. [2]

5 marks

Mark scheme: 1 2 1 23 (i) (OAB ) = × 8 α , (OAC ) = × π × 4 B1B1 Accept 25.1 (for OAC) 2 2 π α = B1 8 [3] 1 (ii) 8 + 8 × their α + × 8 × π B1 23.7 gets B1B0 2 8 + 5 π B1 SC B1 for e.g. 5 π (omitted OB) [2] 2 5

This question in 9709/11 May/June 2013

2 R Q C O S 3 cm P 6 cm The diagram shows a circle C with centre O and radius 3 cm. The radii OP and OQ are extended to S and R respectively so that ORS is a sector of a circle with centre O. Given that PS = 6 cm and that the area of the shaded region is equal to the area of circle C, (i) show that angle POQ = 140 radians, [3] (ii) find the perimeter of the shaded region. [2]

5 marks

Mark scheme: 1 dy = 2 x + 5 dx

This question in 9709/13 May/June 2013

6 B r A r O a rad E D C The diagram shows a metal plate made by fixing together two pieces, OABCD (shaded) and OAED (unshaded). The piece OABCD is a minor sector of a circle with centre O and radius 2r. The piece OAED is a major sector of a circle with centre O and radius r. Angle AOD is ! radians. Simplifying your answers where possible, find, in terms of !, 0 and r, (i) the perimeter of the metal plate, [3] (ii) the area of the metal plate. [3] It is now given that the shaded and unshaded pieces are equal in area. (iii) Find ! in terms of 0. [2]

8 marks

Mark scheme: 6 (i) r (2π − α ) + 2 r α + 2 r B1B1 2πr + r α + 2 r B1 ft for rα instead of 2rα or omission 2r SC1 for 2 r α + 4 r . (Plate = shaded [3] part) (ii) 1 (2 r )2 α + πr 2 − 1 r 2α B1B1 Either B1 can be scored in (iii) 2 2 3r 2α 2 + πr B1 2 [3] (iii) πr 2 − 1 r 2α = 2 r 2α M1 For equating their 2 parts from (ii) 2 2 α = π A1 5 [2]

This question in 9709/11 Oct/Nov 2013

3 O 6 cm 2.2 rad A B The diagram shows part of a circle with centre O and radius 6 cm. The chord AB is such that angle AOB = 2.2 radians. Calculate (i) the perimeter of the shaded region, [3] (ii) the ratio of the area of the shaded region to the area of the triangle AOB, giving your answer in the form k : 1. [3]

6 marks

Mark scheme: 3 (i) s = r θ M1 Used with major or minor arc Angle of major arc = 2π – 2.2 = (4.083) B1 Could be gained in (ii). Perimeter = 12 + 24.5 = 36.5 or 12π − 1.2 A1 co (or full circle − minor arc B1) [3] 1 (ii) Area of major sector = r ² θ = (73.49) M1 Used with major / minor sector. 2 1 Area of triangle = . 6 ² sin 2.2 = (14.55) M1 Correct formula or method. 2 (2π – 2.2) / sin 2.2 gets M1M1 Ratio = 5.05 : 1 (Allow 5.03 → 5.06) A1 co [3] tan x + 1

This question in 9709/13 May/June 2014

8 A C rad O B D 4 cm In the diagram, AB is an arc of a circle with centre O and radius 4 cm. Angle AOB is radians. The point D on OB is such that AD is perpendicular to OB. The arc DC, with centre O, meets OA at C. (i) Find an expression in terms of for the perimeter of the shaded region ABDC. [4] (ii) For the case where = 1 , find the area of the shaded region ABDC, giving your answer in the 6 form k , where k is a constant to be determined. [4]

8 marks

Mark scheme: 8 (i) Arc AB = 4α B1 Arc DC = ( 4 cos α)α B1 AC (or DB ) = 4 − 4 cos α B1 Perimeter = 4α cos α + 4α + 8 − 8 cos α B1 [4] (ii) OD = 4 cos π = 2 3  B1 6   2 1  B1B1 Shaded area =  2 3 × π 6 − 12 ( )  2 × 4 2 × π6    π B1 Or k = 13 3 [4] ′ 1 7 2

This question in 9709/11 Oct/Nov 2014

11 A r O ! rad C B In the diagram, OAB is a sector of a circle with centre O and radius r. The point C on OB is such that angle ACO is a right angle. Angle AOB is ! radians and is such that AC divides the sector into two regions of equal area. (i) Show that sin ! cos ! = 12!. [4] It is given that the solution of the equation in part (i) is ! = 0.9477, correct to 4 decimal places. (ii) Find the ratio perimeter of region OAC : perimeter of region ACB, giving your answer in the form k : 1, where k is given correct to 1 decimal place. [5] (iii) Find angle AOB in degrees. [1]

10 marks

Mark scheme: 11 (i) OC = r cos α or AC = r sin α or oe soi M1 (Area ∆OAC = ) 12 r 2 sin α cos α A1 1 r 2 sin α cos α = 1 × 1 r 2α oe M1 Or e.g. 2 2 2 1 2 r 2α − 1 2 r 2 cos α sin α = 1 4 r 2α 1 2 r 2α − 1 2 r 2 cos α sin α = 1 2 r 2 cos α sin α sin α cos α = 12 α A1 AG [4] (ii) Perimeter ∆OAC = r + r sin α + r cos α = 4.2 ( 0 ) r M1A1 Allow with r a number. 2.0164 gets M1A0 Perim. ACB = rα + r sin α + r − r cos α = 2.18r or 2.17r M1A1 Allow with r a number. 0.9644 gets M1A0 Allow 2.2 www. 4.2 ( 0 ) Ratio = : 1 = 1.1 : 1 A1 Use of cos = 0.6, sin = 0.8, α = 9.0 is PA 1 .2 18 or .2 17 [5] (iii) 54.3º cao B1 [1]

This question in 9709/13 May/June 2015

5 C B 0.6 rad O 6 cm A The diagram shows a metal plate OABC, consisting of a right-angled triangle OAB and a sector OBC of a circle with centre O. Angle AOB = 0.6 radians, OA = 6 cm and OA is perpendicular to OC. (i) Show that the length of OB is 7.270 cm, correct to 3 decimal places. [1] (ii) Find the perimeter of the metal plate. [3] (iii) Find the area of the metal plate. [3]

7 marks

Mark scheme: 6 5 (i) Length of OB = = 7.270 M1 ag Any valid method cos 6.0 [1] (ii) AB = 6tan0.6 or 4.1 B1 Sight of in (ii) Arc length = 7.27 × (½π – 0.6) = (7.06) M1 Use of s= rθ with sector angle Perimeter = 6 + 7.27 + 7.06 + 6tan0.6 = 24.4 A1 [3] (iii) Area of AOB = ½ × 6 × 7.27 × sin0.6 M1 Use of any correct area method Area of OBC = ½ × 7.27² × (½π – 0.6) M1 Use of ½r²θ. → area = 12.31 + 25.65 = 38.0 A1 [3]

This question in 9709/12 Oct/Nov 2015

9 (a) X ! A B r r O Fig. 1 In Fig. 1, OAB is a sector of a circle with centre O and radius r. AX is the tangent at A to the arc AB and angle BAX = !. (i) Show that angle AOB = 2!. [2] (ii) Find the area of the shaded segment in terms of r and !. [2] (b) C 4 cm 4 cm X A B 4 cm Fig. 2 In Fig. 2, ABC is an equilateral triangle of side 4 cm. The lines AX, BX and CX are tangents to the equal circular arcs AB, BC and CA. Use the results in part (a) to find the area of the shaded region, giving your answer in terms of 0 and ï3. [6]

10 marks

Mark scheme: π 9 (a) (i) BAO = OBA = − α Allow use of 90º or 180º 2  π   π  AOB = π −  − α  −  − α  = 2α AG M1A1 Or other valid reasoning  2   2  [2] 1 2 1 2 (ii) r ( 2α ) − r sin 2α oe B2,1,0 SCB1 for reversed subtraction 2 2 [2] π (b) Use of α = , r = 4 B1B1 6  1  2  π   1  2 π 1 segment S =   4   −   4 sin  2   3   2  3  8π  = − 4 3   M1 Ft their (ii), α , r  3   1  2 π T π B1 OR AXB = = 4tan or = 4 3 Area ABC T =   4 sin ( )  2  3 3 6 1 4 2 2π  4 3   1  2 π ( ) sin  =  4 sin T − 3S =   – 3   3 2 3 3   3  2    1  2  π   1  2 π   T   4 3  8π      4   −   4 sin  M1 OR 3  − S  = 3  −  − 4 3     2   3   2  3   3   3  3   16√3 −8π cao A1 [6]

This question in 9709/12 Feb/March 2016

7 B C D r 1 A O r In the diagram, AOB is a quarter circle with centre O and radius r. The point C lies on the arc AB and the point D lies on OB. The line CD is parallel to AO and angle AOC = 1 radians. (i) Express the perimeter of the shaded region in terms of r, 1 and 0. [4] (ii) For the case where r = 5 cm and 1 = 0.6, find the area of the shaded region. [3]

7 marks

Mark scheme: 7 (i) CD = rcosθ, BD = r – rsinθ oe B1 B1 allow degrees but not for last B1 1 Arc CB = r ( π – θ) oe B1 2 1 → P = rcosθ + r – rsinθ + r ( π – θ) oe B1 sum – assuming trig used 2 [4] 1 1 1 (ii) Sector = .5².( π – 0.6) (12.135) M1 Uses r²θ 2 2 2 1 1 Triangle = .5cos0.6.5sin0.6 (5.825) M1 Uses bh with some use of trig. 2 2 → Area = 6.31 A1 1 (or circle − triangle – sector) [3] 4 4

This question in 9709/11 May/June 2016

6 O ! rad r cm Q T P The diagram shows a circle with radius r cm and centre O. The line PT is the tangent to the circle at P and angle POT radians. The line OT meets the circle at Q. = ! (i) Express the perimeter of the shaded region PQT in terms of r and [3] !. (ii) In the case where 1 and r 10, find the area of the shaded region correct to 2 significant ! = 30 = figures. [3]

6 marks

Mark scheme: 6 (i) PT = r tanα B1 r QT = OT – OQ = − r cosα or r ² + r ²tan²α − r B1 Perimeter = sum of the 3 parts including rα B1 [3] π (ii) Area of triangle = ½ × 10 × 10 tan M1 Correct formula used, 50 3, 86.6 3 50π Area of sector = ½ × 10² × ⅓π M1 Correct formula used, , 52.36 3 Shaded region has area 34 (2sf) A1 [3]

This question in 9709/12 May/June 2016

3 A r C r O ! rad D B In the diagram OCA and ODB are radii of a circle with centre O and radius 2r cm. Angle AOB = ! radians. CD and AB are arcs of circles with centre O and radii r cm and 2r cm respectively. The perimeter of the shaded region ABDC is 4.4r cm. (i) Find the value of !. [2] (ii) It is given that the area of the shaded region is 30 cm2. Find the value of r. [3]

5 marks

Mark scheme: 3 (i) 2 rα+ rα+ 2 r = 4.4 r M1 At least 3 of the 4 terms required α = 0.8 A1 [2] 2 2 (ii) ½ ( 2 r ) 0.8 − ½( r )0.8 = 30 M1A1 Ft through on their α (3 / 2) r 2 × 0.8 = 30 → r = 5 A1 [3]

This question in 9709/11 Oct/Nov 2016

6 B cm 10 A O 1.2 rad C D The diagram shows a metal plate ABCD made from two parts. The part BCD is a semicircle. The part DAB is a segment of a circle with centre O and radius 10 cm. Angle BOD is 1.2 radians. (i) Show that the radius of the semicircle is 5.646 cm, correct to 3 decimal places. [2] (ii) Find the perimeter of the metal plate. [3] (iii) Find the area of the metal plate. [3]

8 marks

Mark scheme: r r 6 (i) = sin 0.6 or = cos 0.97 M1 Or other valid alternative. 10 10 or BD = 200 − 200cos1.2 ( = 11.3 ) r = 10 × 0.5646, r = 10 × sin 0.6, r = 10 × cos 0.971 or r = ½ BD A1 → r = 5.646 AG [2] (ii) Major arc = 10(θ) (= 50.832) M1 θ = 2π – 1.2 or π – 1.2 θ = 2π – 1.2 (= 5.083) B1 Implied by 5.1 or C = 2π × 10, Minor arc = 1.2 × 10 Semicircle = 5.646π (= 17.737) Major arc + semicircle = 68.6 A1 [3] (iii) Area of major sector = ½10 2 (θ) (= 254.159) M1 θ = 2π – 1.2 or π – 1.2 Area of triangle OBD = ½10²sin1.2 (= 46.602) M1 Use of ½absinC or other complete method Area = semicircle + sector + triangle (= 50.1 + 254.2 + 46.6) = 351 A1 [3] dy −3

This question in 9709/12 Oct/Nov 2016

8 X M A B 12 cm 10 cm O In the diagram, OAXB is a sector of a circle with centre O and radius 10 cm. The length of the chord AB is 12 cm. The line OX passes through M, the mid-point of AB, and OX is perpendicular to AB. The shaded region is bounded by the chord AB and by the arc of a circle with centre X and radius XA. (i) Show that angle AXB is 2.498 radians, correct to 3 decimal places. [3] … … … … … … … (ii) Find the perimeter of the shaded region. [3] … … … … … … … … … … … … … … (iii) Find the area of the shaded region. [3] … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(i) Letting M be midpoint of AB OM = 8 (Pythagoras) → XM = 2 B1 (could find √40 and use sin−1or cos−1) tan AXM = 6 2 AXB = 2tan−13 = 2.498 M1 A1 AG Needs × 2 and correct trig for M1 (Alternative 1: 6 sin , 0.6435, 0.6435 10 AOM AOM AXB π = = = − ) (Alternative 1: Use of isosceles triangles, B1 for AOM, M1,A1 for completion) (Alternative 2: Use of circle theorem, B1 for AOB, M1,A1for completion) Total: 3 8(ii) AX = √(6² + 2²) = √40 B1 CAO, could be gained in part (i) or part (iii) Arc AYB = rθ = √40 × 2.498 M1 Allow for incorrect √40 (not 6 1 2 1 0 r or or = ) Perimeter = 12 + arc = 27.8 cm A1 Total: 3 8(iii) area of sector AXBY = ½ × (√40)² × 2.498 M1 Use of ½r²θ with their r , (not 6 10 r or r = = ) Area of triangle AXB = ½ × 12 × 2, Subtract these → 38.0 cm² M1 A1 Use of ½bh and subtraction. Could gain M1 with 10 r = . Total: 3

This question in 9709/11 May/June 2017

4 O r cm 21 rad A B r cm D C The diagram shows a circle with radius r cm and centre O. Points A and B lie on the circle and ABCD is a rectangle. Angle AOB = 21 radians and AD = r cm. (i) Express the perimeter of the shaded region in terms of r and 1. [3] … … … … … … … … … … … … … … … … … (ii) In the case where r = 5 and 1 = 160, find the area of the shaded region. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) (AB) = 2rsinθ (or 2 2 2θ − r cos or 2 sin 2 π θ   −     rsin ) B1 Allow unsimplifed throughout eg r + r, 2 2 (Arc AB) = 2rθ B1 (P =) 2r + 2rθ + 2rsinθ (or 2 2 2 2 or sin 2 θ θ π θ −   −     rsin r cos ) B1 Total: 3 Question Answer Marks Guidance 4(ii) Area sector AOB = ( ½ r² 2θ) 25 or1 3.1 6 π B1 Use of segment formula gives 2.26 B1B1 Area triangle AOB = (½×2rsinθ×rcosθ or ½ × r2 sin2θ) 25 3 or1 0.8 4 B1 Area rectangle ABCD = (r × 2rsinθ) 25 B1 (Area =) Either 25 – (25π/6 – 25√3/4) or 22.7 B1 Correct final answer gets B4. Total: 4 =

This question in 9709/12 May/June 2017

7 C 8 cm 10 cm A B D The diagram shows two circles with centres A and B having radii 8 cm and 10 cm respectively. The two circles intersect at C and D where CAD is a straight line and AB is perpendicular to CD. (i) Find angle ABC in radians. [1] … … … … … (ii) Find the area of the shaded region. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 7(i) sin 8 /10 0.927 3 ABC ABC = → = B1 Total: 1 7(ii) ( ) 6 Pythagoras 8 6 48.0 AB BCD = → ∆ = × = M1A1 OR 8×10sin0.6435 or ½×10×10sin((2)×0.927)=48. 24or 40or80 gets M1A0 Area sector ( ) 2 ½ 10 2 0.9273 BCD their = × × × *M1 Expect 92.7(3). 46.4 gets M1 Area segment = 92.7(3) – 48 *A1 Expect 44.7(3). Might not appear until final calculation. Area semi-circle ‒ segment = ( ) 2 ½ 8 92.7 48 their π × × − − DM1 Dep. on previous M1A1 OR ( ) 2 2 8 ½ 8 44.7 . their π π × − × × + Shaded area = 55.8 – 56.0 A1 Total: 6

This question in 9709/13 May/June 2017

4 C P Q A B D O 6 cm The diagram shows a semicircle with centre O and radius 6 cm. The radius OC is perpendicular to the diameter AB. The point D lies on AB, and DC is an arc of a circle with centre B. (i) Calculate the length of the arc DC. [3] … … … … … … … … … … … … … … … … … … (ii) Find the value of area of region P area of region Q, giving your answer correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) Pythagoras → r = 72 OE or 6 6 cos45 6 2 cos45 = → = = r r Arc DC = 72 × ¼π = 3 2 2 π , 2.12π, 6.66 M1 A1 Use of s=rθ with their r (NOT 6) and ¼π 3 4(ii) Area of sector BDC is ½ × 72 × ¼π (= 9π or 28.274…) *M1 Use of ½r²θ with their r (NOT 6) and ¼π Area Q = 9π – 18 (10.274…) DM1 Subtracts their ½ × 6 × 6 from their ½r²θ Area P is (¼π6² – area Q) = 18 M1 Uses {¼ π6² – (their area Q using 72 )} Ratio is 18 9 18 π − 18 10.274       → 1.75 A1 4

This question in 9709/12 Oct/Nov 2017

7 A 5 B 3 D P Q C The diagram shows a rectangle ABCD in which AB = 5 units and BC = 3 units. Point P lies on DC and AP is an arc of a circle with centre B. Point Q lies on DC and AQ is an arc of a circle with centre D. (i) Show that angle ABP = 0.6435 radians, correct to 4 decimal places. [1] … … … … … … … … (ii) Calculate the areas of the sectors BAP and DAQ. [3] … … … … … … … … … … … … … … … … … (iii) Calculate the area of the shaded region. [3] … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 7(i) 1 3 sin 0.6435 5 − =     AG M1 OR ( ) ( ) 1 3 cos 0.9273 0.9273 0.6435 5 2 PBC ABP π −  = = ⇒ = − =     Or other valid method. Check working and diagram for evidence of incorrect method 7(ii) Use (once) of sector area 2 ½ θ = r M1 Area sector 2 ½ 5 0.6435 = × × BAP = 8.04 A1 Area sector 2 ½ ½ 3 π = × × DAQ = 7.07 , Allow 9 4 π A1 3 Question Answer Marks Guidance 7(iii) EITHER: Region = sect + sect ‒ (rect ‒ ∆) or sect ‒ [rect ‒ (sect + ∆)] (M1 Use of correct strategy (Area ∆ BPC =) ½ × 3 × 4 = 6 Seen A1 8.04 + 7.07 ‒ (15 ‒ 6) = 6.11 A1) OR1: Region = sector ADQ ‒ (trap ABPD ‒ sector ABP). (M1 Use of correct strategy (Area trap ABPD = ) ½ (5 + 1) ×3 = 9 Seen A1 7.07 ‒ (9 ‒ 8.04) = 7.07 ‒ 0.96 = 6.11 A1) OR2: Area segment AP= 2.5686 Area segment AQ = 0.5438 Region = segment AP + segment AQ + ∆APQ. (M1 Use of correct strategy (Area ∆APQ =) ½ × 2 × 3 = 3 Seen A1 2.57 + 0.54 + 3 = 6.11 A1) 3

This question in 9709/13 Oct/Nov 2017

6 A T r cm 1 rad B O The diagram shows a circle with centre O and radius r cm. The points A and B lie on the circle and AT is a tangent to the circle. Angle AOB = 1 radians and OBT is a straight line. (i) Express the area of the shaded region in terms of r and 1. [3] … … … … … … … … … … … … … … … … … (ii) In the case where r = 3 and 1 = 1.2, find the perimeter of the shaded region. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(i) (tanθ = ) AT r → AT = r tanθ or OT = cos r θ SOI B1 → A = 1 2 r²tanθ − 1 2 r²θ B1 B1 B1 for 1 2 r²tanθ. B1 for “−1 2 r²θ” If Pythagoras used may see area of triangle as 2 2 2 1 2 r r r tan θ + or 1 2 cos r r sinθ θ       3 Question Answer Marks Guidance 6(ii) tanθ = 3 AT → AT = 7.716 M1 Correct use of trigonometry and radians in rt angle triangle Arc length = rθ = 3.6 B1 Accept 3 1.2 × OT by Pythagoras or cos1.2 = 3 OT ( = 8.279) M1 Correct method for OT Perimeter = AT + arc + OT – radius = 16.6 A1 CAO, www 4

This question in 9709/11 May/June 2018

6 B r O 21 rad T A The diagram shows points A and B on a circle with centre O and radius r. The tangents to the circle at A and B meet at T. The shaded region is bounded by the minor arc AB and the lines AT and BT. Angle AOB is 21 radians. (i) In the case where the area of the sector AOB is the same as the area of the shaded region, show that tan 1 = 21. [3] … … … … … … … … … … … … … … … … … (ii) In the case where r = 8 cm and the length of the minor arc AB is 19.2 cm, find the area of the shaded region. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 6(i) AT or BT = rtanθ or OT = r cosϑ ½r²2θ, & ½×r×(rtanθ or AT) or ½×r×( r cosϑ or OT) sinθ M1 Both formulae, (½r²θ, ½bh or ½absinθ), seen with 2θ used when needed. ½r²2θ = 2×½×r×rtanθ – ½r²2θ oe → 2θ = tanθ AG A1 Fully correct working from a correct statement. Note: ½r²2θ = ½ r²tanθ is a valid statement. 3 Question Answer Marks Guidance 6(ii) θ = 1.2 or sector area = 76.8 B1 Area of kite = 165 awrt B1 164.6 – 76.8 = 87.8 awrt B1 awrt 87.8 with little or no working can be awarded 3/3. SC Final answers that round to 88 with little or no working can be awarded 2/3. 3

This question in 9709/12 May/June 2018

9 A C 5 cm 150 rad O D B The diagram shows a triangle OAB in which angle ABO is a right angle, angle AOB = 150 radians and AB = 5 cm. The arc BC is part of a circle with centre A and meets OA at C. The arc CD is part of a circle with centre O and meets OB at D. Find the area of the shaded region. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9 Angle OAB / 2 / 5 3 /10 soi B1 Sector 2 1 3 5 2 10 π   = × ×     CAB their M1 Expect 11.78 5 8.507 sin 5 π = = OA M1A1 May be implied by 3.507 = OC Sector ( ) 1 3.507 ² 2 5 π = × × COD their M1 Expect 3.86 ( ) 1 3 5 8.507 sin 2 10 π ∆ = × × OAB their M1 Or 1 5 5 2 tan 5 π × × or 2.5 × ( ) 2 8.507 25 − their = 17.20 or 17.21 A1 Shaded area ( ) 17.20 1 7.21 11.78 3.86 1.56 or 1.57 − − = or A1 8

This question in 9709/11 Oct/Nov 2018

3 A 5 cm C 4 cm D B The diagram shows an arc BC of a circle with centre A and radius 5 cm. The length of the arc BC is 4 cm. The point D is such that the line BD is perpendicular to BA and DC is parallel to BA. (i) Find angle BAC in radians. [1] … … … … … (ii) Find the area of the shaded region BDC. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3(i) 0.8 oe B1 1 3(ii) BD = 5sin 0.8 their M1 Expect 3.58(7). Methods using degrees are acceptable DC = 5 ‒ 5cos their 0.8 M1 Expect 1.51(6) Sector = ½ × 52 × their0.8 OR Seg = ½ × 52 × [their 0.8 ‒ sintheir 0.8] M1 Expect 10 for sector. Expect 1.03(3) for segment Trap = ½(5 + theirDC) × theirBD oe OR ∆BDC = ½theirBD × theirCD M1 OR (for last 2 marks) if X is on AB and XC is parallel to BD: Shaded area = 11.69 ‒ 10 OR 2.71(9) ‒ 1.03(3) = 1.69 cao A1 BDCX ‒(sector ‒ AXC ∆ ) = 5.43(8) ‒ [10 ‒ 6.24(9)] = 1.69 cao M1A1 5

This question in 9709/13 Oct/Nov 2018

3 A sector of a circle of radius r cm has an area of A cm2. Express the perimeter of the sector in terms of r and A. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3 Uses A = ½r²θ M1 Uses area formula. θ = 2 ² A r A1 P r r rθ = + + B1 2 2 A P r r = + A1 Correct simplified expression for P. 4

This question in 9709/11 May/June 2019

5 C 1 rad A O r B The diagram shows a semicircle with diameter AB, centre O and radius r. The point C lies on the circumference and angle AOC radians. The perimeter of sector BOC is twice the perimeter of = 1 sector AOC. Find the value of correct to 2 significant figures. [5] 1 … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 5 Perimeter of AOC = 2r + rθ B1 Angle COB = π – θ B1 Could be on the diagram. Condone 180 – θ. Perimeter of BOC = 2r + r(π – θ) B1 FT on angle COB if of form ( ) π θ − k , k 0. > (2r +) πr – rθ = 2((2r) + rθ) (2 + π – θ = 4 + 2θ → 2 3 π θ − = ) M1 Sets up equation using ( ) π θ − r k and ×2 on correct side. Condone any omissions of OA, OB and/or OC. θ = 0.38 A1 Equivalent answer in degrees scores A0. 5

This question in 9709/12 May/June 2019

3 A 8 cm E 150 rad B D C The diagram shows triangle ABC which is right-angled at A. Angle ABC = 150 radians and AC = 8 cm. The points D and E lie on BC and BA respectively. The sector ADE is part of a circle with centre A and is such that BDC is the tangent to the arc DE at D. (i) Find the length of AD. [3] … … … … … … … … (ii) Find the area of the shaded region. [3] … … … … … … … … …

6 marks

Mark scheme: 3(i) Angle EAD = Angle ACD = 3π 10 or 54° or 0.942 soi or Angle DAC = π 5 or 36° or 0.628 soi AD = 8sin( 3π 10 ) or 8cos( π 5 ) M1 Angles used must be correct (AD =) 6.47 A1 Alternative method for question 3(i) ( ) 3 8sin 8 10 or or 1 1. 01 π tan sin 5 5 AB AB π π       = =             B1 Angles used must be correct ( ) 11.0 1 sin 5 AD π = oe M1 (AD =) 6.47 A1 3 3(ii) Area sector = ( ) 2 1 2 2 5 theirAD their π π   × −     M1 19.7(4) Area 1 8 sin 2 5 ADC theirAD π ∆ = × × × or 1 3 3 8cos 8sin 2 10 10 π π     × ×         M1 Or e.g. ½ 2 2 8 theirAD theirAD × − . 15.2(2) (Shaded area =) 35.0 or 34.9 A1 3

This question in 9709/13 May/June 2019

8 C B 6 cm 380 rad O A The diagram shows a sector OAC of a circle with centre O. Tangents AB and CB to the circle meet at B. The arc AC is of length 6 cm and angle AOC = 380 radians. (i) Find the length of OA correct to 4 significant figures. [2] … … … … … (ii) Find the perimeter of the shaded region. [2] … … … … … … … … … … (iii) Find the area of the shaded region. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8(i) 3 π 6 8 OA× = ( ) 16 5.093 0 π OA = = A1 8(ii) 3 5.0930 tan π 16 AB their = × M1 Perimeter = 2 3.4030 6 12.8 × + = A1 8(iii) Area OABC = ( ) 2 ½ 5.0930 3.4030 × × × their their M1 Area sector = ( ) 2 3 ½ 5.0930 π 8 their × × M1 Shaded area = 17.331 15.279 − their their = 2.05 M1A1

This question in 9709/11 Oct/Nov 2019

4 A r cm O 21 rad T B The diagram shows a circle with centre O and radius r cm. Points A and B lie on the circle and angle AOB radians. The tangents to the circle at A and B meet at T. = 21 (i) Express the perimeter of the shaded region in terms of r and [3] 1. … … … … … … … … … … … … … … … … … (ii) In the case where r 5 and 1.2, find the area of the shaded region. [4] = 1 = … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) Arc length AB = 2rθ B1 Tan θ = or AT BT r r → AT or BT = r tan θ B1 Accept or 2 2 cos r r θ    −           or sin sin 2 θ θ π   −     r NOT (90 – θ) P = 2rθ + 2r tan θ B1FT OE, FT for their arc length + 2 × their AT 3 Question Answer Marks Guidance 4(ii) Area ∆AOT = ½ × 5 × 5 tan 1.2 or Area AOBT = 2 × ½ × 5 × 5 tan 1.2 B1 Sector area = ½ × 25 × 2.4 (or 1.2) *M1 Use of ½r2θ with θ = 1.2 or 2.4. Shaded area = 2 triangles – sector DM1 Subtraction of sector, using 2.4 where appropriate, from 2 triangles Area = 34.3 (cm2) A1 AWRT Alternative method for question 4(ii) Area of ∆ ABT = ½ × (5 × tan 1.2)2 × sin(π – 2.4) (= 55.86) B1 Segment area = ½ × 25 × (2.4 – sin 2.4) (= 21.56) *M1 Use of ½r2 (θ – sin θ) with θ = 1.2 or 2.4 Shaded area = triangle – segment DM1 Subtraction of segment from ∆ ABT, using 2.4 where appropriate. Area = 34.3 (cm2) A1 AWRT 4

This question in 9709/12 Oct/Nov 2019

4 C A r O r B The diagram shows a semicircle ACB with centre O and radius r. Arc OC is part of a circle with centre A. (i) Express angle CAO in radians in terms of 0. [1] … … … (ii) Find the area of the shaded region in terms of r, 0 and ï3, simplifying your answer. [4] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4(i) Angle CAO = π 3 B1 1 4(ii) (Sector AOC) = 2 3 1 2 π r their × M1 SOI (∆ ABC) = ( )( ) π 1 2 sin 2 3 r r their       or ( )( ) 1 3 2 2 2 r r or ( )( ) 1 3 2 r r M1 For M1M1, π 3 their must be of the form kπ where 0 < k < ½ (∆ ABC) = ( )( ) π 1 3 2 sin 2 r r       or ( )( ) 1 3 2 2 2 r r or ( )( ) 1 3 2 r r A1 All correct 2 2 2 π 3 3 1 2 r r     −           A1 4

This question in 9709/13 Oct/Nov 2019

7 A D 0.8 rad O B C 6 cm The diagram shows a sector AOB which is part of a circle with centre O and radius 6 cm and with angle AOB = 0.8 radians. The point C on OB is such that AC is perpendicular to OB. The arc CD is part of a circle with centre O, where D lies on OA. Find the area of the shaded region. [6] … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 7 M1A1 SOI Area sector OCD = ( ) 2 1 4.18 0.8 2 their × *M1 OE ΔOCA = 1 6 4.18 sin0.8 2 their × × × M1 OE Required area = their ΔOCA ‒ their sectorOCD DM1 SOI. If not seen their areas of sector and triangle must be seen 2.01 A1 CWO. Allow or better e.g. 2.0064 6

This question in 9709/12 Feb/March 2020

8 15 cm B A C 6 cm O X In the diagram, ABC is a semicircle with diameter AC, centre O and radius 6 cm. The length of the arc AB is 15 cm. The point X lies on AC and BX is perpendicular to AX. Find the perimeter of the shaded region BXC. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 8 Angle AOB = 15 ÷ 6 = 2.5 radians B1 Angle BOC = π – 2.5 (FT on angle AOB) B1FT BC = 6(π – 2.5) (BC = 3.850) M1 sin(π – 2.5) = BX ÷6 (BX = 3.59) M1 Either OX = 6cos(π – 2.5) or Pythagoras (OX = 4.807) M1 XC = 6 – OX (XC = 1.193) → P = 8.63 A1 6

This question in 9709/11 May/June 2020

10 C 1 rad F E r O A B D The diagram shows a sector CAB which is part of a circle with centre C. A circle with centre O and radius r lies within the sector and touches it at D, E and F, where COD is a straight line and angle ACD is 1 radians. (a) Find CD in terms of r and sin 1. [3] … … … … … … … … … … … … … … It is now given that r = 4 and 1 = 16π. (b) Find the perimeter of sector CAB in terms of π. [3] … … … … … … … … … … … (c) Find the area of the shaded region in terms of π and 3. [4] … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) sin r OC   = →     θ sin r OC = θ M1 A1 sinθ = + r CD r A1 3 10(b) Radius of arc AB = 4 4 4 8 12 π sin 6 + = + = B1 SOI (Arc AB =) 2π 12 6 their × or 1 π 12 2 6 AB their    = ×       M1 Expect 4π, must use their CD, not 4 Perimeter = 24 4π + A1 3 Question Answer Marks Guidance 10(c) Area FOC = 1 π 4 sin 2 3 their OC × × × M1 8 3 A1 Area sector FOE = 2 1 2π 16π 4 2 3 3 × × = B1 Shaded area = 16π 16 3 3 − A1 Alternative method for question 10(c) FC = ( ) 2 2 4 their OC − M1 48 or 4 3 Area FOC = 1 4 4 3 2 × × = 8 3 A1 Area of half sector FOE = 2 1 π 8π 4 2 3 3 × × = B1 Shaded area = 16π 16 3 3 − A1 4

This question in 9709/11 Oct/Nov 2020

8 D A C 1 rad r cm r cm B In the diagram, ABC is an isosceles triangle with AB BC r cm and angle BAC radians. The = = = 1 point D lies on AC and ABD is a sector of a circle with centre A. (a) Express the area of the shaded region in terms of r and [3] 1. … … … … … … … … … … … … … … … … … … … (b) In the case where r 10 and 0.6, find the perimeter of the shaded region. [4] = 1 = … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 8(a) Use of correct formula for the area of triangle ABC M1 Use of 180–2θ scores M0. Condone 2π–2θ ( ) 2 1 2 sin π 2 r −θ or 2 1 2 sin2 r θ or 1 2 2 cos sin r r × × × θ θ or 1 2 2 cos sin r r × × θ θ A1 OE [Shaded area = triangle – sector] = their triangle area − 2 1 2 r θ B1 FT FT for their triangle area − 2 1 2 r θ (Condone use of 180 degrees for triangle area for B1) 3 8(b) Arc BD = rθ = 6 cm B1 SOI AC = 2rcosθ = ( 2×10cos0.6 = 20cos0.6 = 16.506) or ( ) ( ) 2 2 2 2 cos π 2 r r − −θ or ( ) sin π – 2 sin r × θ θ *M1 Finding AC or 1 2 AC (= 8.25) DC = 2rcosθ – r or ( ) ( ) 2 2 2 2 cos π 2 r r − −θ – r ( = 6.506) DM1 Subtracting r from their AC or r-rcosθ from their half AC (8.25-1.75) (Perimeter = 10 + 6 + 6.506 =) 22.5 A1 AWRT 4

This question in 9709/12 Oct/Nov 2020

9 B 12 cm 8 cm A 8 cm O C In the diagram, arc AB is part of a circle with centre O and radius 8 cm. Arc BC is part of a circle with centre A and radius 12 cm, where AOC is a straight line. (a) Find angle BAO in radians. [2] … … … … … … … … … … … … … … … … … (b) Find the area of the shaded region. [4] … … … … … … … … … … … … (c) Find the perimeter of the shaded region. [3] … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) 6 cos 8 = BAO or 2 2 2 8 12 8 2 8 12 + − × × M1 Or other correct method BAO = 0.723 A1 2 Question Answer Marks Guidance 9(b) Sector ABC = 2 ½ 12 0.7227 × ×their *M1 Accept 52.1 ( ) Triangle ½ 8 12sin 0.7227 AOB their = × × or ½×12×√28 *M1 or ( ) ½ 8 8sin 2 0.7227 π × × −×their . Expect 31.7 or 31.8 Shaded area = their 52.0 31.7 their − = 20.3 DM1 A1 M1 dependent on both previous M marks 4 9(c) Arc BC = 12 0.7227 ×their *M1 Expect 8.67 Perimeter = 8 + 4 + their 8.67 = 20.7 DM1 A1 3

This question in 9709/13 Oct/Nov 2020

10 A ka ka a E D B C The diagram shows a sector ABC which is part of a circle of radius a. The points D and E lie on AB and AC respectively and are such that AD = AE = ka, where k < 1. The line DE divides the sector into two regions which are equal in area. (a) For the case where angle BAC = 16π radians, find k correct to 4 significant figures. [5] … … … … … … … … … … … … … … … … 1 > 1.(b) For the general case in which angle BAC = 1 radians, where 0 < 1 < 12π, it is given that sin 1 Find the set of possible values of k. [3] … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 10(a) ( ) 2 1 π sin 2 6 Δ = ADE ka 2 2 1 4 k a A1 OE. Sector 2 1 π 2 6 = ABC a B1 2 2 2 1 1 π 2 4 2 6 × = k a a M1 OE. For 2 sector ADE ABC ×Δ = with at least one correct area. π 0.7236 6 k = =       A1 5 10(b) ( ) 2 2 1 1 2 sin 2 2 θ θ × = ka a M1 Condone omission of ‘2’ or ‘1/2’ on LHS for M1 only. 2 2sin k θ θ = A1 2 1 2 k > leading to 1 1 2 k < < A1 OE. Accept 1 2 k > or 0.707 k > (AWRT) or 0.707(AWRT) < k < 1 or 1 2 k > OE 3

This question in 9709/12 Feb/March 2021

8 P Q C S R The diagram shows a symmetrical metal plate. The plate is made by removing two identical pieces from a circular disc with centre C. The boundary of the plate consists of two arcs PS and QR of the original circle and two semicircles with PQ and RS as diameters. The radius of the circle with centre C is 4 cm, and PQ = RS = 4 cm also. (a) Show that angle PCS = 23π radians. [2] … … … … (b) Find the exact perimeter of the plate. [3] … … … … … … … … (c) Show that the area of the plate is 20 π + 8 3 cm2. [5] 3 … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 8(a) Either Let midpoint of PQ be H: sin HCP = 2 4 ⇒ Angle HCP = π 6 Or sin PSQ = 4 8 ⇒ Angle PSQ = π 6 Or using cosine rule: angle PCQ = π 3 Or by inspection: triangle PCQ or PCT is equilateral so angle PCQ = π 3 M1 Angle PCS = π π 2 π π 6 6 3 − − = A1 AG 2 8(b) Perimeter = 2π 2 4 3 × × or 8π 8π 3 − M1 Length of two arcs PS and QR 2π 2 + × M1 Adding circumference of two semicircles 28π 3 A1 Must be a single term 3 Question Answer Marks Guidance 8(c) Area sector CPQ = 2 1 π 8π 4 2 3 3 × × = M1 Uses correct formula for sector Area of segment of large circle beyond CPQ = 2 8π 1 π 8π 4 sin 4 3 3 2 3 3   − × × = −     M1 Attempts to find area of segment Area of small semicircle = π 2 × or area of small circle = 2 π 2 × M1 Area of plate = Large circle – [2 ×] small semicircle – [2 ×] segment area M1 2 2 8π 20π π 4 π 2 2 4 3 8 3 3 3   × − × −× − = +     A1 AG Alternative method for Question 8(c) Area of sector PCS 2 1 2π 16π 4 2 3 3 = × × = M1 Uses correct formula for sector Area of triangle PCQ = 2 1 π 4 sin 4 3 2 3 × × = M1 Uses correct formula for triangle Area of small semicircle = π 2 × or area of circle = 2 π 2 × M1 Area of plate = [2 ×] large sector + [2 ×] triangle – [2 ×] small semicircle M1 ( ) 2 16π 20π 2 2 4 3 π 2 8 3 3 3  + − × = +     A1 AG 5

This question in 9709/11 May/June 2021

12 Q P A B F C E D The diagram shows a cross-section of seven cylindrical pipes, each of radius 20 cm, held together by a thin rope which is wrapped tightly around the pipes. The centres of the six outer pipes are A, B, C, D, E and F. Points P and Q are situated where straight sections of the rope meet the pipe with centre A. (a) Show that angle PAQ = 13π radians. [2] … … … … … … (b) Find the length of the rope. [4] … … … … … … … … (c) Find the area of the hexagon ABCDEF, giving your answer in terms of 3. [2] … … … … … … … … … … (d) Find the area of the complete region enclosed by the rope. [3] … … … … … … … … … … … … … …

11 marks

Mark scheme: 12(a) [By symmetry] [6 × PAQ = 2π], [ PAQ =] 2π÷6, M1 Explaining that there are six sectors around the diagram that make up a complete circle. A1 AG Alternative method for Question 12(a) Using area or circumference of circle centre A ÷ 6 M1 400π 6 or 40π 6 Justification for dividing by 6 followed by comparison with the sector area or arc length. A1 AG Alternative method for Question 12(a) Explain why ∆PAQ is an equilateral triangle M1 Assumption of this scores M0 Using ∆PAQ is an equilateral triangle ⸫ ˆ PAQ = π 3 A1 AG Alternative method for Question 12(a) Using the internal angle of a regular hexagon = 2π 3 Or 3 ˆ ˆ 2π FAO OAB + = , equilateral triangles M1 ˆ PAQ = π 2π π 2π — 2 3 2   + +     = π 3 A1 AG Question Answer Marks Guidance 12(a) Alternative method for Question 12(a) 20, with 40 θ θ = Sin clearly identified M1 π π , 2 6 3 θ θ = = = ˆ FAO and by similar triangles = ˆ PAQ A1 AG 2 12(b) Each straight section of rope has length 40 cm B1 SOI Each curved section round each pipe has length π 20 3 rθ = × *M1 Use of θ r with r = 20 and θ in radians Total length = ( ) ( ) 6 40 π their k × + DM1 6×(their straight section + their curved section). Their curved section must be from acceptable use of θ r – this could now be numeric. 240 40π + or 366 (AWRT) (cm) A1 Or directly: (6 diameter) × + circumference 4 Question Answer Marks Guidance 12(c) [Triangle area =] 1 π 40 40 sin 2 3   × × ×     or 1 40 20 3 2 × × or 400 3 or 693(AWRT) B1 [Total area of hexagon = 6 × 400 3 =] 2400 3 B1 Condone 4800 3 2 Alternative method for Question 12(c) [Trapezium area =] ( ) 1 π 40 80 40sin 2 3   × + ×     or 1200 3 or 2080 (AWRT) B1 [Total area of hexagon = 2 × 1200 3 =] 2400 3 B1 Condone 3 4800 2 √ Alternative method for Question 12(c) Area of triangle ABC = 400 3 or 693 (AWRT) or 4 × Area of half of triangle ABC = 4 200 3 × or 1390 (AWRT) or Area of rectangle ABDE = 1600 3 or 2770 (AWRT) B1 [Total area of hexagon = 2 400 3 × +1600 3 =] 2400 3 Or [= 4 200 3 × +1600 =] 2400 3 B1 Condone 3 4800 2 √ If B0B0, SC B1 can be scored for sight of 4160 (AWRT) as final answer. 2 Question Answer Marks Guidance 12(d) Each rectangle area = 40 × 20 (= 800) B1 SOI, e.g. by sight of 4800 Each sector area = 2 2 1 1 π 200π 20 2 2 3 3 r θ   = × × =     B1 SOI. Total area = 2400 3 4800 400π + + or 10 200 (cm2) (AWRT) B1 Or directly: part (c) + 6800 + area circle radius 20. 3

This question in 9709/12 May/June 2021

5 A D 4 cm B C The diagram shows a triangle ABC, in which angle ABC = 90Å and AB = 4 cm. The sector ABD is part of a circle with centre A. The area of the sector is 10 cm2. (a) Find angle BAD in radians. [2] … … … … … (b) Find the perimeter of the shaded region. [4] … … … … … … … … … … … …

6 marks

Mark scheme: 5(a) 2 ½ 4 angle BAD 10 × × = Angle BAD 1.25 = A1 OE. Accept 0.398π, 71.6o for SC B1 only 2 5(b) Arc 4 1.25 = × BD their M1 Use of arc length formula. Expect 5. ( ) 4tan 1.25 = BC their M1 Expect 12.0(4). May use ACB=0.321 or 18.4o ( ) 4 4 cos 1.25 = − CD their or ( ) 2 2 4 4 + − their BC M1 Expect 12.69 ‒ 4 = 8.69. May use ACB. Perimeter = 5 + 12.0(4) + 8.69 = 25.7 (cm) A1 AWRT 4

This question in 9709/13 May/June 2021

6 C A 6 cm B The diagram shows a metal plate ABC in which the sides are the straight line AB and the arcs AC and BC. The line AB has length 6 cm. The arc AC is part of a circle with centre B and radius 6 cm, and the arc BC is part of a circle with centre A and radius 6 cm. (a) Find the perimeter of the plate, giving your answer in terms of [3] π. … … … … … … … … … … … … … … … … … … (b) Find the area of the plate, giving your answer in terms of and 3. [4] π … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) Recognise that at least one of angles A, B, C is π 3 One arc 6 × their π 3 leading to two arcs 2 × 6 × their π 3 M1 SOI e.g. may see 2π or 4π. Use of correct formula for length of arc and multiply by 2. Perimeter = 6 + 4π A1 Must be exact value. Alternative method for question 6(a) Calculate circumference of whole circle = 12π B1 One arc 1 12π 6 × leading to two arcs 1 2 12π 6 × × M1 SOI e.g. may see 2π or 4π. Perimeter = 6 + 4π A1 Must be exact value. 3 Question Answer Marks Guidance 6(b) Sector = 2 1 π 6 2 3 their  × ×     M1 Use of correct formula for area of sector. SOI e.g. may see 6π or 12π. ( ) ( ) 2 2 1 π 1 π 6 6 sin 6π 2 3 2 3 their their       × × − × × +             6π 9 3 6π   = − +   M1 A1 M1 for attempt at strategy with values substituted: area of segment + area of sector A1 if correct (unsimplified). Area = 12π 9 3 − A1 Must be simplified exact value. Alternative method for question 6(b) Sector = 2 1 π 6 2 3 their  × ×     M1 Use of correct formula for area of sector. SOI e.g. may see 6π or 12π. ( ) 2 2 1 π 1 π 2 6 6 sin 2 3 2 3 their their         × × × − × ×                 M1 A1 M1 for attempt at strategy with values substituted: 2 × sector – triangle A1 if correct (unsimplified). Area = 12π 9 3 − A1 Must be simplified exact value. Alternative method for question 6(b) Sector = 2 1 π 6 2 3 their  × ×     M1 Use of correct formula for area of sector. SOI e.g. may see 6π or 12π. ( ) ( ) 2 2 1 π 1 π 2 6 6 sin 2 3 2 3 their their         × × × − × ×                 + ( ) 2 1 π 6 sin 2 3 their     × ×         12π 18 3 9 3   = − +   M1 A1 M1 for attempt at strategy with values substituted: 2 × segment + triangle A1 if correct (unsimplified). Area 6π 9 3 6π   = − +   = 12π 9 3 − A1 Must be simplified exact value. 4

This question in 9709/11 Oct/Nov 2021

7 15cm P B Q A 9 cm 15 cm C In the diagram the lengths of AB and AC are both 15 cm. The point P is the foot of the perpendicular from C to AB. The length CP = 9 cm. An arc of a circle with centre B passes through C and meets AB at Q. (a) Show that angle ABC = 1.25 radians, correct to 3 significant figures. [2] … … … … … … … … … … … … … … … … (b) Calculate the area of the shaded region which is bounded by the arc CQ and the lines CP and PQ. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 7(a) EITHER By using trigonometry: ˆ BAC = 0.6435… and ˆ ABC = π 0.6435 2 − OR By Pythagoras: AP = 12 ⇒ BP = 3 so tan ˆ ABC = 9 3 OR Using ∆PBC and either the sine or cosine rule 3 sin 10 ˆ ABC = or 10 cos 1 ˆ 0 ABC = M1 3 0.9486 10 = … 10 0.3162 10 = … ˆ ABC = π 0.6435 2 − or tan-1 9 3 or 1 3 sin 10 − or 1 10 cos or 10 − ( ) 1 .249 04 or 71.56 … ° = 1.25 radians (3 sf) A1 AG. Final answer must be 1.25, more accurate value 1.24904… with no rounding to 3sf seen as the final answer gets M1A0. If decimals are used all values must be given to at least 4sf for A1. 2 7(b) BC = ( ) 2 2 3 9 their + or 9 sin1.25 [= 90 , 3 10 or 9.48697…] M1 Using correct method(s) to find BC. Area of sector = ( ) [ ] 2 1 1 tan 3 56.207 56.25 2 their BC or − × × = M1 Using 1 tan 3 or 1.25 − and their BC, but not 9 or 15, in correct area of sector formula. Area of triangle PBC = 13.4 to 13.6 or 1 9 3 2 × × B1 [Area = (56.207 or 56.25) – their 13.5 =] 42.7 or 42.8 A1 AWRT 4

This question in 9709/12 Oct/Nov 2021

5 B Y 9 cm 11 cm X C A In the diagram, X and Y are points on the line AB such that BX = 9 cm and AY = 11 cm. Arc BC is part of a circle with centre X and radius 9 cm, where CX is perpendicular to AB. Arc AC is part of a circle with centre Y and radius 11 cm. (a) Show that angle XYC = 0.9582 radians, correct to 4 significant figures. [1] … … … … … … … … … … … … … … … (b) Find the perimeter of ABC. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) Angle XYC = 1 9 sin 11 −      = 0.9582 or 9 sin leading to 0.9582 11 XYC XYC = = B1 AG. OE using cosine rule. 1 5(b) 2 2 11 9 40 XY = − = or using 0.9582 and trigonometry *M1 A1 9 11 AB theirXY = + − B1 FT OE e.g. 20 2 10 −√ , 2 9 2 10 11 2 10 + − + −√ Arc AC = 11 × 0.9582 M1 Arc BC = 9× π 2 M1 Perimeter = [13.6(8) + 10.5(4) +14.1(4) =] 38.4 A1 AWRT. Answer must be evaluated as a single decimal. 6

This question in 9709/13 Oct/Nov 2021

5 C D A 1 r B The diagram shows a sector ABC of a circle with centre A and radius r. The line BD is perpendicular to AC. Angle CAB is 1 radians. (a) Given that 1 = 16π, find the exact area of BCD in terms of r. [3] … … … … … … … … … … … … … … … … … 3 (b) Given instead that the length of BD is r, find the exact perimeter of BCD in terms of r. [4] 2 … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) Sector area = 2 2 1 π π 2 6 12           r r 2 1 2  r with θ in radians SOI. B0 if using a value for r. BD = π 1 sin 6 2        r r and AD = π 3 cos r 6 2        r so triangle area = 1 π π 1 1 3 sin cos r 2 6 6 2 2 2                 r r r or 1 π π 1 3 1 cos sin 2 6 6 2 2 2                 r r r r B1 SOI Finding triangle area. Decimals B0 unless exact values seen in working. Area of BCD = 2 2 1 3 π 12 8  r r B1 OE e.g. 2 π 3 4 3 2          r with π cos 6 and π sin 6 evaluated. Must be exact, in terms of 2 r . ISW 3 Question Answer Marks Guidance 5(b) Angle BAC = 1 3 π 2 sin 3                  r r B1 SOI by length of AD, CD or arc, or by perimeter. Length AD = π 1 cos 3 2        r r [so length CD = 1 2r] M1 SOI Finding length by Pythagoras, or by trigonometry with their angle BAC, provided π 6  BAC . Length of arc BC = π 3  r M1 SOI Using  r with in radians. Condone π 6  . Perimeter of BCD = 3 1 π 2 2 3   r r r A1 OE e.g. 3 1 π 2 3          r with e.g. π cos 3 evaluated. Must be exact, in terms of r. ISW 4

This question in 9709/11 May/June 2022

7 B 2 cm A 10 cm 16π O P C The diagram shows a sector OBAC of a circle with centre O and radius 10cm. The point P lies on OC and BP is perpendicular to OC. Angle AOC = 16π and the length of the arc AB is 2cm. (a) Find the angle BOC. [2] … … … … … … … … … … … … … … … … (b) Hence find the area of the shaded region BPC giving your answer correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 7(a) ˆ AOB      2 10 B1 OE Sight of 0.2 from s r  but 10 2  is not enough. ISW if 2 π 10 5  . [B ˆO C 5π 6 ] 30   or 1 π 0.2 6  B1 OE e.g. 0.724c AWRT or 41.5 degrees AWRT. But not 5π 2 3 10  – fraction within a fraction. ISW incorrect simplifications. Alternative method for question 7(a) OR [Arc AC =]   10π 10π or Arc 2 or 7.2 6 6 BC   B1 AWRT. Sight of 10π 6 or 5.2 or 7.2. [B ˆO C 5π 6 ] 30   or 1 π 0.2 6  B1 OE e.g. 0.724c AWRT or 41.5 degrees AWRT. But not 5π 2 3 10  – fraction within a fraction. ISW incorrect simplifications. 2 Question Answer Marks Guidance 7(b) [BP] = 5π 6 10sin 30        and [OP ] = 5π 6 10cos 30        [= 6.6208…] and [= 7.494…] OR [BP] = 5π 6 10sin 30        and [O ˆB P] = 5π 3 15        [= 6.6208…] and [= 0.84719…] M1 OE Any correct method for both lengths, for their angle BOC (which may have been incorrectly ‘simplified’ but not 0.2) or length BP and O ˆB P. May be seen as part of 1 sin 2 ab C . Sight of correct method enough. Can be implied by the next A1. Area of ∆OBP = 1 2  5π 6 10sin 30        5π 6 10cos 30         or 1 5π 6 5π 3 10 10sin sin 2 30 15                        [=24.809] A1 OE Can be implied by any answer in range (24.7, 24.9) or a final answer in the range (11.3, 11.5) WWW. [Sector BOC] = 2 1 10 2 their   5π 6 30        5π 6 50 36.1799... 30                M1 Use of 2 1 2 r  with their angle BOC (may have been incorrectly ‘simplified’ but not 0.2). Area of region BPC = 11.4 A1 CAO 4

This question in 9709/12 May/June 2022

9 D B 1.8 rad 6 cm 6 cm C A The diagram shows triangle ABC with AB BC 6cm and angle ABC 1.8 radians. The arc CD is = = = part of a circle with centre A and ABD is a straight line. (a) Find the perimeter of the shaded region. [5] … … … … … … … … … … … … … … … … … (b) Find the area of the shaded region. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) 6sin0.9 2 AC or 2 2 2 6 6 2 6 6cos1.8     AC M1 OE Correct working in degrees is acceptable throughout. AC = 9.40 A1 SOI Accept 9.39 – 9.41, may be used but not seen for A1. Angle CAB = ½(π ‒ 1.8) M1 SOI Expect 0.6708 (or 0.671). Arc CD = their 9.40  their 0.6708 M1 Expect 6.306 (or 6.31), do not accept 6 for their AC or 1.8 for CAB. [Perimeter = 6 + 3.40 + 6.306 =] 15.7 A1 Accept 15.69 – 15.72. 5 Question Answer Marks Guidance 9(b) Sector ADC ‒ ABC = ½  their 9.402  their 0.6708 – ½  62  sin 1.8 M1 M1 Accept correct use of their answers from part (a). [29.64 ‒ 17.53 =] 12.1 A1 AWRT 3

This question in 9709/13 May/June 2022

5 8 cm A B O The diagram shows a sector OAB of a circle with centre O. The length of the arc AB is 8cm. It is given that the perimeter of the sector is 20cm. (a) Find the perimeter of the shaded segment. [4] … … … … … … … … … … … … … … … … … (b) Find the area of the shaded segment. [2] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5(a)  2 r + 8 = 20   r = 6 B1 8 *M1 4 Angle AOB = Expect OE ( 76.4.) their 6 3 M0 Assume triangle is equilateral. 2 2 2 2 4 DM1 For 6 read their 6 . AB = 2  6sin their or 6 + 6 −2 6 cos their 3 3 6 4 or AB =  sin their   2  3 sin  − their   2 3  Perimeter =  7.42 + 8 =  15.4 A1 AWRT 4 5(b) 1 2 4 1 2 4 M1 Sector area – whole triangle area. Area =  6  their −  6  sin their 2 3 2 3 For 6 read their 6 . 1 2 4 1  2  2  or Area =  6  their −2  6 sin their  6cos their  Sector area – 2(half triangle area). 2 3 2  3  3  =  24 −17.49 =  6.51 A1 AWRT 2

This question in 9709/11 Oct/Nov 2022

10 R O 23π 2.5 m P 56π 2.24 m A B S The diagram shows a cross-section RASB of the body of an aircraft. The cross-section consists of a sector OARB of a circle of radius 2.5m, with centre O, a sector PASB of another circle of radius 2.24m with centre P and a quadrilateral OAPB. Angle AOB = 23π and angle APB = 56π. (a) Find the perimeter of the cross-section RASB, giving your answer correct to 2 decimal places. [3] … … … … … … … … … … (b) Find the difference in area of the two triangles AOB and APB, giving your answer correct to 2 decimal places. [2] … … … … … … … … … (c) Find the area of the cross-section RASB, giving your answer correct to 1 decimal place. [3] … … … … … … … … … … … … … …

8 marks

Mark scheme: 10(a) 4 5 10 28 B1 For either arc correct. Arc ARB could be AR+RB. 2.5  + 2.24  [= 10.47[2] + 5.86[4] or + ] 3 6 3 15 M1 For adding two (or three) arc lengths using different radii and angles and nothing else. SOI 26 π A1 AWRT 16.34 or Condone 16.33 only. 5 3 10(b) 1 2 2 M1 For either AOB or APB (AB = 4.33, h= 1.25, 0.58) or any other Area AOB =  2.5 sin [=2.706] valid method. 2 3 1 2 5 Area APB =  2.24 sin [=1.254] 2 6 [Difference =] 1.45 A1 AWRT Condone 1.46 only. 2 10(c) 1 2 4 B1 For either sector area correct Area AOB =  2.5  [=13.09] 2 3 1 2 5 Area APB =  2.24  [=6.57] 2 6 [Area of cross section =] M1 Adding two sector areas from different sectors and ‘ their 10(b) ’ 1 2 4 1 2 5 and nothing else. SOI  2.5  +  2.24  + “ their 10 ( b )” 2 3 2 6 =  13.09 + 6.57 + “their 10 ( b )”  21.1 A1 CAO Condone slight inaccuracies in intermediate working if the correct answer is arrived at. 3

This question in 9709/12 Oct/Nov 2022

8 A r 5 3r P Q r B The diagram shows two identical circles intersecting at points A and B and with centres at P and Q. The radius of each circle is r and the distance PQ is 3r.5 (a) Find the perimeter of the shaded region in terms of r. [4] … … … … … … … … … … … … … … … … … (b) Find the area of the shaded region in terms of r. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 8(a) 5 *M1 May use cosine rule to find APB. Stating APQ or r −1 6  −1 5  APB as an incorrect multiple of  is M0. APQ = cos = cos   r  6  = 0.5857 A1 Accept 0.586 or 33.6° or APB (1.171 or 67.1o). Perimeter = 4 r their 0.5857 = 2.34r or 0.745π r or (293/125)r DM1 A1 Must use a numerical value of their angle. 4 8(b) Use of sector formula: Sector APB = ½r 2  ( 2  their 0.5857 ) or Sector APC M1 Any sector with their appropriate angle. It must be clear the appropriate numerical angle is being used. (C is on PQ so PC = r) = ½r 2  ( their 0.5857 ) Use of appropriate formula for area of triangle and correct combination with M1 e.g. Area APB = ½r 2  sin ( 2  their 0.5857 ) . the sector to find the area of a half segment, one segment or both segments Shaded area [ = 2  0.1250 r 2 ] = 0.250 r 2 A1 2 1 2 2 or 0.0796π r , allow r or 0.25 r . 4 3

This question in 9709/13 Oct/Nov 2022

4 C D 1 B A 8 cm The diagram shows a sector ABC of a circle with centre A and radius 8cm. The area of the sector is 16 πcm2. The point D lies on the arc BC. 3 Find the perimeter of the segment BCD. [4] … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 4 2 1 16 8 2 3     ⇒ 6  B1 SOI OE e.g.   2 , 0.524 3 s.f. 12 Use of degrees acceptable throughout provided conversion used in formulae for sector area and arc length. Arc length = 8 6  their [= 4.1887…] M1 OE FT their θ. Look for 4 3 . [BC =] 1 2 8 sin 2 6          their [= 4.1411…] M1 Attempt to find BC or 2 BC (see alt. methods below) FT their θ. Look for 16sin12  or 4 6 4 2  . Perimeter = 8.33 A1 AWRT Must be combined into one term. Question Answer Marks Guidance 4 Alternative methods for Question 4: 2nd M1 mark (use normal scheme for the other marks) ALT 1   2 2 2 8 8 2 8 8cos 4.14 6               BC their BC ALT 2     2 2 2 8 4 3 4 4.14       BC BC ALT 3   8 4.14 5 sin sin 6 12                   BC BC ALT 1 Substitute into correct cosine rule. FT their θ Look for 128 64 3  ALT 2 Find lengths 4 and 4 3 then use Pythagoras in the left hand triangle. ALT 3 Substitute into correct sine rule. 4 8  / 6 4 4√3

This question in 9709/11 May/June 2023

6 A r cm O 1 rad r cm B The diagram shows a sector OAB of a circle with centre O and radius r cm. Angle AOB = 1 radians. It is given that the length of the arc AB is 9.6cm and that the area of the sector OAB is 76.8cm2. (a) Find the area of the shaded region. [5] … … … … … … … … … … … … (b) Find the perimeter of the shaded region. [2] … … … … …

7 marks

Mark scheme: 6(a) 2 ½ 76.8 9.6    r r or 2 2 1 9.6 76.8 2          16  r A1 0.6  A1 Accept 34.4o OAB = ½  their 162  sin their 0.6 M1 Allow Segment = 76.8 –½  their 162  sin their 0.6. Expect 72.27 . [Area = 76.8 ‒ 72.27 =] 4.53 A1 AWRT 5 6(b) 2 16 sin 0.3    AB OR 2 2 2 2 16 16 2 16 cos0.6     AB M1 Any valid method with their r, θ. Expect AB = 9.46. Perimeter = 9.6 + 9.46 = 19.1 A1 AWRT 2

This question in 9709/13 May/June 2023

6 O r A B r C The diagram shows a motif formed by the major arc AB of a circle with radius r and centre O, and the minor arc AOB of a circle, also with radius r but with centre C. The point C lies on the circle with centre O. (a) Given that angle ACB = kπ radians, state the value of the fraction k. [1] … … … … … … … (b) State the perimeter of the shaded motif in terms of π and r. [1] … … … … … … … … (c) Find the area of the shaded motif, giving your answer in terms of π, r and 3. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) 2 B1 2π k = Allow ACB = . 3 3 1 6(b) Perimeter of shaded area = 2πr B1 1 6(c) 1 2 4π *M1 2 2 Major sector OAB = r  Expect 3 πr . Finds area of any relevant sector or 2 3 triangle. Can be embedded in segment formula.  1 2 π 1 2 π  *M1 One or both segments = 2   r  − r sin   2 3 2 3   2 π 2 3  A1 = 2  r − r     6 4  2 2  1 2 r 2 3  DM1 Shaded area = πr − 2  πr −    3  6 4  πr 2 r 2 3 A1 = + 3 2 6(c) Alternative method for Question 6(c) 1 2 1 *M1 1 2 Sector CAOB = 2  r their π Expect [2] × π r . 2 3 6 Can be embedded in segment formula.  1 2 π 1 2 π  *M1 One or both segments = 2   r  − r sin   2 3 2 3   2 π 2 3  A1 = 2  r − r     6 4  1 2  2 π 2 3  DM1 2 Shaded area = π r −  πr + 2  r − r    3  6 4  πr 2 r 2 3 A1 = + 3 2 6(c) Alternative method for Question 6(c) 1 2 π M1 3 Area of rhombus AOBC = 2  r sin Expect [2]  . 2 3 4 Can be embedded in segment formula.  1 2 π 1 2 π  M1 One or both segments = 2   r  − r sin   2 3 2 3   2 π 2 3  A1 = 2  r − r     6 4  2  3 2  2 π 2 3   DM1 Shaded area = πr −  r − 4  r − r     2  6 4  πr 2 r 2 3 A1 = + 3 2 5

This question in 9709/11 Oct/Nov 2023

10 A 2.8 rad B r O R r C The diagram shows points A, B and C lying on a circle with centre O and radius r. Angle AOB is 2.8 radians. The shaded region is bounded by two arcs. The upper arc is part of the circle with centre O and radius r. The lower arc is part of a circle with centre C and radius R. (a) State the size of angle ACO in radians. [1] … … … … … (b) Find R in terms of r. [1] … … … … … … … … … … (c) Find the area of the shaded region in terms of r. [7] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 10(a) Angle ACO = 0.7 B1 Don’t allow AWRT 0.7 . 1 10(b)  R =  1.53 r B1 Allow AWRT 1.53r. 1 10(c) 1 2 2 B1 Sector OAB = r  2.8  = 1.4 r    2 1 2 *M1 Sector CAB = ( their R ) 2 their 0.7 2 1.638 r 2 A1 Allow AWRT 1.64 r 2 . 1 2 1 *M1 2  r sin (− 1.4 ) OR 2  r  theirR sin0.7 2 2 2  0.4927r 2 A1 Allow AWRT 0.98 r 2 to 0.99 r 2 . 2 2 2 DM1 1.4r − their 1.638r − their 0.985r ( ) 0.747r 2 to 0.748r 2 A1 7 10(c) General guidance for alternative methods Finding any useful sector area of the circle radius, r B1 May be ‘nested’ in a segment. Finding the area of sector CAB *M1A1 May be ‘nested’ in a segment. Finding the area of one useful triangle *M1 May be ‘nested’ in a segment. Finding the total area of useful triangles A1 May be ‘nested’ in a segment. A correct plan for the shaded area DM1 0.747r 2 to 0.748r 2 A1 7

This question in 9709/13 Oct/Nov 2023

10 y A C i rad B x O The diagram shows the circle with centre C (– 4, 5) and radius 20 units. The circle intersects the y-axis at the points A and B. The size of angle ACB is i radians. (a) Find the equation of the tangent to the circle at the point (–6, 9). [3] … … … … … … … … (b) Find the equation of the circle in the form x 2 + y 2 + ax + by + c = 0 . [2] … … … … … … … … (c) Find the value of i correct to 4 significant figures. [3] … … … … … … … … … (d) Find the perimeter and area of the segment shaded in the diagram. [4] … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 10(a) Obtain gradient of relevant radius is –2 B1 Using m1m2 = −1 obtain the gradient of the tangent and use it to form a M1 m1 must be from an attempt to find the gradient of the straight line equation for a line containing (–6, 9) radius using the centre and the given point. Obtain y = 12 x + 12 A1 1 OE e.g. y − 9 = ( x + 6 ) . 2 3 10(b) State or imply ( x + 4) 2 + ( y − 5) 2 = 20 B1 If x 2 + y 2 − 2 gx − 2 fy + c = 0 is used correctly with ( − g , − f ) = ( −4, 5 ) and c = g 2 + f 2 − r 2 then M1. Obtain x 2 + y 2 + 8 x − 10 y + 21 = 0 B1 A1 if above method used. 2 10(c) Substitute x = 0 in equation of circle to find y-values 3 and 7 B1 May be implied by AB = 4 or use of |x-coordinate of C|. or state C to AB = 4 Attempt value of  either using cosine rule or via 12 using right-angled M1 Using their AB. If /2 used, must be multiplied by 2. triangle Obtain = 0.9273 A1 Or greater accuracy. A correct answer implies the M1. 3 10(d) Attempt arc length using r formula with their  (not their /2) and M1 Expect 4.15. r =  20 Obtain perimeter = 8.15 or greater accuracy A1 Condone missing units or incorrect units. 1 2 M1 If sector – triangle used, both formulae must be correct. Attempt area using 2 r (− sin) formula or equivalent with their and If triangle ACM used, area must be multiplied by 2. r =  20 Obtain area = 1.27 or greater accuracy A1 Condone missing units or incorrect units. 4

This question in 9709/12 Feb/March 2024

8 B C A 2 cm D 1 1 r rad 3 r rad 3 F E The diagram shows a symmetrical plate ABCDEF. The line ABCD is straight and the length of BC is 2 cm. Each of the two sectors ABF and DCE is of radius r cm and each of the angles ABF and DCE is r radians. equal to 13 (a) It is given that r = 0.4 cm. (i) Show that the length EF = 2.4 cm. [2] … … … … … (ii) Find the area of the plate. Give your answer correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … (b) It is given instead that the perimeter of the plate is 6 cm. Find the value of r. Give your answer correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 8(a)(i) C π ˆ 6 XCE  π ˆ 3 CEX  X E 0.4  XE π π sin or cos 6 0.4 3 XE  [XE = 0.2] M1 A correct trig expression involving XE. Do not condone a mixture of degrees and radians. Length EF = 2 2 0.2  = 2.4 A1 AG 2 Question Answer Marks Guidance 8(a)(ii)   π 0.4cos 6 CX  or π 0.4sin 3 or 2 2 0.4 0.2  B1 OE, SOI Expect 3 5 or 0.3464.     2 1 π Sector 0.4 2 3    B1 SOI Expect 0.0838 or 2π 75 . Allow use of   2 60 π 0.4 . 360 Either Area of their (rectangle + two triangles + two sectors) Or Area of their (trapezium + two sectors) M1 Either implied by a correct answer or areas clearly labelled. Expect 0.6928 + 0.06928 + 0.1676 or 2 3 3 4π 5 25 75   . Or 0.7621 + 0.1676 or 11 3 4π 25 75  . 0.930 A1 AWRT 11 3 4π Condone 25 75  . 4 Question Answer Marks Guidance 8(b) [Length AD =] 2 2 r B1 Must be seen alone or part of a list and not part of a product. [Arc length =] π 3 r  B1 May be implied by π 2 3 r   . Must be seen alone or part of a list.   π π EF 2 2 sin or 2 2 cos 6 3 r r    or 2 + r B1 Must be seen alone or part of a list and not part of a product. [4+ 3r + 2π 6 leading to 3 r  ] 0.393 B1 AWRT Condone 6 . 2π 9  NB: Using EF = 2.4 gives 0.391. 4

This question in 9709/12 May/June 2024

3 i rad C r cm r cm A B The diagram shows a sector of a circle with centre C. The radii CA and CB each have length r cm and the size of the reflex angle ACB is i radians. The sector, shaded in the diagram, has a perimeter of 65 cm and an area of 225 cm 2. (a) Find the values of r and i. [4] … … … … … … … … … … … … (b) Find the area of triangle ACB. [2] … … … … … …

6 marks

Mark scheme: 3(a) State 2 65 r r   and 2 1 225 2 r  Form a 3-term quadratic or cubic in r or  or r from correct arc and sector formula *M1 Condone sign errors. Solve their 3 term quadratic or cubic to obtain values of r or  DM1 Expect    2 2 65 450 2 45 10      r r r r or    2 18 97 72 9 8 2 9          . 10  r and 4.5  ignore 8 22.5 and 9 r    , do not ignore 0  r A1 B1 SC if no quadratic or cubic solution. If 0  r included A0 or B0 SC. 4 3(b) Use correct formula for area of triangle with clear use of angle being 2π their   M1 Expect 1.783 or 102.2o, their  must be reflex. 48.9 A1 AWRT, WWW or a second answer. Or greater accuracy; condone absence of units. 2

This question in 9709/13 May/June 2024

3 B C 15 cm 15 cm A D O The diagram shows a sector of a circle, centre O, where OB = OC = 15 cm . The size of angle BOC is 2 r radians. Points A and D on the lines OB and OC respectively are joined by an arc AD of a circle 5 with centre O. The shaded region is bounded by the arcs AD and BC and by the straight lines AB and DC. It is given that the area of the shaded region is 209 rcm 2 . 5 Find the perimeter of the shaded region. Give your answer in terms of r. [5] … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Use correct sector area formula M1 1 2 2 1 2 2 209 A1 Obtain  15  π −  x  π = π or equivalent 2 5 2 5 5 Obtain x = 4 A1 AWRT 4.00. Use correct arc length formula twice M1 38 A1 OE. Must be in terms of π. Obtain 22 + π Like terms must be collected. 5 Not from a rounded value of x. 5

This question in 9709/11 Oct/Nov 2024

6 C D 2r cm 2r cm B E 2θ rad A F θ rad θ rad r cm r cm O The diagram shows a metal plate OABCDEF consisting of sectors of two circles, each with centre O. The radii of sectors AOB and EOF are r cm and the radius of sector COD is 2r cm. Angle AOB = angle EOF = i radians and angle COD = 2i radians. It is given that the perimeter of the plate is 14 cm and the area of the plate is 10 cm 2. Given that r 2 3 and i 1 3 , find the values of r and i. [6] 2 4 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 6 [Perimeter =] r + r+ r + 2 r  2+ r + r+ r  = 4r + 6r B1 1 2 1 2 1 2 2 B1 [Area =] r + ( 2r )  2+ r   = 5r    2 2 2 4r + 6r= 14 and 5r 2= 10 M1* ar + br= 14 and cr 2= 10 where a, b and c are constants  0. Terms may be uncollected. EITHER 2 14 − 4 r  10  DM1 Eliminate  to get an equation in r. 5r = 10 or 4 r + 6 r  2  = 14 6 r  5r    2r 2 − 7 r + 6 = 0   ( r − 2 )( 2 r − 3 ) = 0 DM1 Factorise or other accepted method for solving their 3-term   quadratic. OR  14  2  10   10  DM1 Eliminate r to get an equation in . 5   = 10 or 4   + 6  = 14  4 + 6  5   5  [ 182 − 25+ 8 = 0 ] ( 9− 8)( 2− 1) = 0 DM1 Factorise or other accepted method for solving their 3-term quadratic. Then r = 2 and = 0.5 B1 3 8 Condone extra answers r = and = . 2 9 6

This question in 9709/12 Oct/Nov 2024

7 E C D F 20 cm i rad i rad 20 cm B 20 cm O A The diagram shows a metal plate ABCDEF consisting of five parts. The parts BCD and DEF are semicircles. The part BAFO is a sector of a circle with centre O and radius 20 cm, and D lies on this circle. The parts OBD and ODF are triangles. Angles BOD and DOF are both i radians. (a) Given that i = .12 , find the area of the metal plate. Give your answer correct to 3 significant figures. [5] … … … … … … … … … … … … … … … … … … … … … … … (b) Given instead that the area of each semicircle is 50rcm 2, find the exact perimeter of the metal plate. [5] … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) 1 2 M1 Or combination of large semi-circle and small sector: Area of sector BOF =  20  ( 2π − 2.4 )  = 776.63 1 2 1 2 2  20  π +  20  ( π − 2.4 ) . 2 2 2 2 M1* Length of radius of small circles is acceptable for M1. Length BD = DF = 2  20sin0.6 or 20 + 20 −2 20  20cos1.2  = 22.58  2 DM1 Area of two semicircles = π  ( 20sin0.6 )  = 400.64 1 M1 Area of triangles = 2   20  20sin1.2  = 372.81 2 Total area = 1550 [cm2 ] A1 Expect 1550.09 but accept AWRT to 3sf. 5 7(b) 1 2 B1   πr = 50π  r = 10 May be seen as 20sin , where = . 2 2 3 π M1* OE ⇒ = Finding  using their r. Allow working in degrees. 3  2π  DM1 Arc length of sector BOF = 20   2− their   3   2π  DM1 Dependent on the first dM1. Total perimeter = 20   2π − their  + 2π  their 10  3  140π 2 A1 Must be a single exact term. or 46 π 3 3 5

This question in 9709/13 Oct/Nov 2024

9 C r cm a rad A B r cm The diagram shows a sector ABC of a circle with centre A and radius r cm. The angle BAC is a radians, where 0 1 a 1 1 r . 2 (a) It is given that the area of the triangle ABC is 4 cm 2 and the area of the sector ABC is 8 a cm 2 . Find the exact area of the shaded segment. [4] … … … … … … … … … … … … … … … … 1 (b) It is given instead that the length of the chord BC is r cm but the area of the triangle ABC is still 2 4 cm 2. Find the area of the shaded segment. Give your answer correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) 1 2 B1 r = 8 ⇒ r = 4 2 1 2 π B1 r sin= 4 ⇒ = 2 6 1 2 π M1 Using their r and their . Area of segment =  4  − 4 2 6 1 2 π Condone 4 –  4  . 2 6 Allow use of 8 × their  – 4 4 A1 Fraction must be simplified. = π − 4 3 4π − 12. Allow 3 4 9(b) 1 3 *M1 -1  2  -1  1  r 2 + r 2 − 2 r 2 cos= r 2  2cos= Or  = 2 sin   or 2 sin   2 2    4   2 2  π Using  = or r = 4 implies 0/4. 6 1 2 DM1 1 2 = 0.723  0.72273  r sin ( their 0.723 ) = 4  Or = 41.4  r sin ( their 41.4 ) = 4  2 2 r = 3.48  3.4777  A1 Accept 2r = 12.1 AWRT. 1 2 A1 1 2 π Area of segment =  3.48  0.723 − 4 = 0.371 [0.37068…] Or  3.48   41.4 − 4 = 0.371 AWRT. 2 2 180 4

This question in 9709/11 May/June 2025

5 D A 5 cm C 5 cm B The diagram shows a sector ABD of a circle with centre A and radius 10 cm. The perpendicular bisector of AB passes through D. (a) Find the perimeter of the shaded region BCD, giving your answer correct to 1 decimal place. [4] … … … … … … … … … … … … … … … … … … … (b) Find the area of the shaded region BCD, giving your answer correct to 1 decimal place. [2] … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5(a) π B1 OE Angle CAD = or 60o, CD = 75 or 5 3 ˆ or CD. Allow CADˆ = 1.05, CD = 8.7 3 For either CAD or 8.66.  π   60  B1 B1 OE + 75 + 5 + 75 + 5 [Perimeter =] 10     or   2π (10 )    B1 for the two sides and B1 for arc length (allow 10.5).  3   360  = 24.1 cm B1 4 5(b) 1 2 π 1 60 2 1 M1 OE Area =  10  − 5 75 or  π  10 − 5 75 Use of sector area formula minus triangle area 2 3 2 360 2 formulae, with their angle CAD and their side CD. = 30.7 cm2 A1 2

This question in 9709/15 May/June 2025

7 A B r cm 2 r rad r cm 3 O The diagram shows a sector of a circle with centre O and radius r cm. The shaded region is bounded by the chord AB and the arc AB. The size of angle AOB is 2 r radians. 3 (a) Show that the area of the shaded region is approximately 0.614r 2 cm2. [2] … … … … … … … … … … … … … … … … … … … … It is given that the radius of the circle is increasing at a rate of .04 cm s -1 . (b) (i) Find the rate of increase of the area of the shaded region at the instant when r = 20 . Give your answer correct to 2 significant figures. [3] … … … … … … … … … (ii) Find the rate of increase of the length of the arc AB. Give your answer correct to 2 significant figures. [3] … … … … … … … … … … … … … … …

8 marks

Mark scheme: 1 7(a) Obtain correct 2 r 2  23 π − 12 r 2 sin 23 π M1 0.614r 2 A1 AG Greater accuracy is 0.61418... r 2 . 2 7(b)(i)  d A  B1 Or greater accuracy Obtain = 1.228r    d r  dA dA dr M1 dA Use =  , or equivalent, with r = 20 ‘their’ × 0.4 with r = 20 dt dr dt dr Obtain 9.8 A1 AWRT 3 7(b)(ii) State or imply 23 πr for length of arc AB B1 dl 2π Could be implied by = . d 3 Differentiate and apply correct use of chain rule M1 Obtain 0.84 A1 AWRT 3

This question in 9709/11 Oct/Nov 2025

5 X O 4 cm 4 cm θ rad P Q 4 3 cm The diagram shows part of a circle with centre O and radius 4 cm. The chord PQ is of length 4 3cm and angle POQ = i radians. The point X lies on the circle. (a) Find the exact value of i. [2] … … … … … (b) Find the exact area of the segment PXQ. [3] … … … … … … … … … … … …

5 marks

Mark scheme: 5(a)  1 2 3 B1  Use of a correct expression for either or . sin  =  2  4 2  1 2 or cos =    2  4 2 4 2 + 4 2 − 4 3 ( )  1  or cos= = −   2 4 4  2  B1 Correct answer can imply B1B1. = 2π Condone inconsistent use of . 3 2 5(b) 1 2 2 1 2 4 1 2 2 *M1 Use of a correct formula with theiror 2− their,  0 . Triangle  4  sin or  4  or  4  2 3 2 3 2 3 and/or minor sector may be seen in the formula for finding the minor segment. 1 2 4π 1 2 2π DM1 Addition of relevant sector area with 2− their, where  0,  4  +  4  sin 2 3 2 3 and triangle area with their, where  0.  1 2 2π 1 2 2π  Subtraction of triangle area, with their, 0 , from relevant Or 16π −   4  −  4  sin   2 3 2 3  sector area, with their, 0, and subtract from circle area. 32π A1 OE. Exact equivalents only. + 4 3 SC B2 following M0DM0 for 40.4 AWRT. 3 3

This question in 9709/13 Oct/Nov 2025