7 P B A O Q The diagram shows a circle, centre O, radius 10 cm. The points A, B and P lie on the circumference of the circle. The chord AB is of length 14 cm. The point Q lies on AB and the line POQ is perpendicular to AB. (a) Show that angle POA is 2.366 radians, correct to 3 decimal places. [2] (b) Find the area of the shaded region. [3] (c) Find the perimeter of the shaded region. [5]

7(a) 7 sin 10 = AOQ π = − POA AOQ or 2 2 2 14 10 10 200 cos = + − AOB oe 2π 2 − = AOB POA M1 Allow alternatives, but must be a complete method to find POA 2.366195157 POA = = 2.366 to 3 dp A1 Must see an angle correct to more than 3dp used in order to justify 3 dp 7(b) Area of sector = ( ) 2 110 2.366 2 ( ) 118.3 B1 Allow unsimplified. Also allow use of 2.37 Area of triangle = 2 110 sin2.366 2 ( ) 35 B1 Allow unsimplified. Also allow use of 2.37 Total area = awrt 153 B1 Allow greater accuracy 7(c) Major arc PB = 10 2.366 × B1 Allow unsimplified. Also allow use of 2.37 2 sin 2 10 = AP POA or 2 2 2 10 10 200 cos = + − AP POA M1 For a valid attempt to find AP – may be seen in (a) or (b) but AP must be stated in this part. 18.5 = AP A1 Allow awrt 18.5 Perimeter: major arc PB + 20 + their AP B1 For plan, may be implied, but must have an attempt to calculate AP Total perimeter = 62.2 A1 Allow awrt 62.2

9 M A O B D C N 6 cm 4 cm 12 cm The diagram shows a circle, centre O, radius 12 cm, and a rectangle ABCD. The diagonals AC and BD intersect at O. The sides AB and AD of the rectangle have lengths 6 cm and 4 cm respectively. The points M and N lie on the circumference of the circle such that MAC and NDB are straight lines. (a) Show that angle AOD is 1.176 radians correct to 3 decimal places. [2] (b) Find the perimeter of the shaded region. [4] (c) Find the area of the shaded region. [3]

9(a) 1 2 2 tan 3 −  = ×     AOD M1 For correct method to find AOD 1.1760... = AOD 1.176 = AOD [to 3dp] A1 Need to see 4 dp or more to justify 3 dp answer 9(b) Major arc ( ) 2 1.176 12 π = − MN B1 ND or MA =12 13 − B1 Perimeter = major arc MN + MA + ND + 16 oe B1 For their values in a correct plan, may be implied by a correct answer Perimeter = 94.1 B1 9(c) Minor sector area = 2 1 1.176 12 2 × × or Major sector area = ( ) 2 1 2 1.176 12 2 π × − × B1 Area = major sector area – remainder of rectangle or Area = area of circle – minor sector area – remainder of rectangle or Area = circle – rectangle – minor sector + triangle AOD B1 For their values in a correct plan, may be implied by a correct answer Area = 350 B1 Allow greater accuracy

10 In this question all lengths are in centimetres. O A B C 25 15 10 The diagram shows a shaded shape. The arc AB is the major arc of a circle, centre O, radius 10. The line AB is of length 15, the line OC is of length 25 and the lengths of AC and BC are equal. (a) Show that the angle AOB is 1.70 radians correct to 2 decimal places. [2] (b) Find the perimeter of the shaded shape. [4] (c) Find the area of the shaded shape. [5]

10(a) 7.5 sin 2 10 = AOB M1 For a valid method 1.696 = AOB = 1.70 to 2 dp A1 Must see greater accuracy to justify given answer 10(b) 2 2 2 10 25 2 10 25cos 2     = + − × ×         AOB AC M1 For a complete and valid method to find AC awrt 19.9 = AC A1 Major arc AB = awrt 45.9 or awrt 45.8 B1 Perimeter = awrt 85.5 or awrt 85.6 A1 10(c) Area of major sector AOB = ( ) 2 1 10 2π 2 × −AOB M1 awrt 229 A1 Area of kite OACB = 1 15 25 2 × × B1 Allow working with 2 separate triangles Area of their major sector plus area of their kite M1 Total area = awrt 417 A1

6 DO NOT USE A CALCULATOR IN THIS QUESTION. All lengths in this question are in centimetres. You may use the following trigonometrical ratios. ° ° sin cos 60 2 3 60 2 1 3 = = ° tan60 = A C B 60° 6 2 – 6 2 + The diagram shows triangle ABC with AC 6 2 = - , AB 6 2 = + and angle ° CAB 60 = . (a) Find the exact length of BC. [3] (b) Show that sinACB 6 4 2 = + . [2] (c) Show that the perpendicular distance from A to the line BC is 1. [2]

6(a) ( ) ( ) ( )( ) 2 2 6 2 6 2 2 6 2 6 2 cos60 − + + − − + M1 ( ) 1 6 2 2 12 6 2 2 12 2 6 2 2 + − + + + −× − × M1 Condone one error in expansion of brackets [ ] 2 3 BC = isw A1 6(b) 2 3 6 2 sin60 sin their ACB + = or 2 3 6 2 sin 3 2 their ACB + = M1 Condone other letters for ACB ( ) 3 1 6 2 sin 6 2 2 4 2 3 ACB + = + × × = A1 A0 if necessary brackets missing unless clearly recovered 6(c) 6 2 4 6 2 x + = − or 1 2 3 2 their x × × = ( ) ( ) 1 6 2 6 2 sin60 2 × − × + × [where x is the perpendicular from A to BC] M1 Complete method ( )( ) 6 2 6 2 6 2 1 4 4 x − + − = = = or ( ) 6 2 3 4 1 2 4 2 3 x − = × = = A1

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

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

Cambridge O Level Additional Mathematics · 4037

O-Level Additional Mathematics (4037) Circular measure past paper questions

Revise O-Level Additional Mathematics (4037) — Circular measure with real Cambridge past-paper questions grouped by topic, each shown with its marking scheme. Working one topic at a time targets exactly the skill each question tests.

35 free questions from 2016–2021, each with its marking scheme. Open the full topical bank to filter by subtopic, paper, and year.