6 O D C A 2a a a B The diagram shows a circle, centre O, radius 2a. The points A and B lie on the circumference of the circle. The points C and D are the mid-points of the lines OB and OA respectively. The arc DC is part of a circle centre O. The chord AB is of length 2a. (a) Find angle AOB, giving your answer in radians in terms of r. [1] (b) Find, in terms of a and r, the perimeter of the shaded region ABCD. [2] (c) Find, in terms of a and r, the area of the shaded region ABCD. [3]

6(a) π 3 B1 6(b) π 4 3 + a a 2 B2 FT for π 4 3   × +     their a a or B1 FT for π 3 × their a 6(c) ( ) 2 1 π 2 sin 2 3 a B1 FT their π 3 2 1 π 2 3 a B1 FT their π 3 2 2 π 3 6 − a a B1 FT their π 3

7 DO NOT USE A CALCULATOR IN THIS QUESTION In this question all lengths are in centimetres. 17 1 - 17 4 + 17 1 + B C A D E The diagram shows a trapezium ABCDE such that AB is parallel to EC and ABCD is a rectangle. It is given that BC 17 1 = + , ED 17 1 = - and . DC 17 4 = + (a) Find the perimeter of the trapezium, giving your answer in the form a b 17 + , where a and b are integers. [3] (b) Find the area of the trapezium, giving your answer in the form c d 17 + , where c and d are integers. [2] (c) Find tanAED, giving your answer in the form e f 17 8 + , where e and f are integers. [2] (d) Hence show that sec AED 81 9 17 32 2 = + . [2]

7(a) ( ) ( ) 2 2 2 17 1 17 1 = − + + AE 18 2 17 18 2 17 = + + − M1 For attempt to find AE. Must see at least 3 terms in an expansion that is not the difference of two squares to be convinced a calculator is not being used. 6 = AE A1 Perimeter = 4 17 8 + + their AE 4 17 14 = + B1 FT on their AE 7(b) Area = ( )( ) 1 3 17 7 17 1 2 + + oe ( ) 1 51 3 17 7 17 7 2 = + + + oe M1 For attempt at a trapezium or triangle and rectangle. Must see at least 3 terms in an expansion that is not the difference of two squares to be convinced a calculator is not being used. Allow one arithmetic slip. Area 29 5 17 = + A1 7(c) 17 1 17 1 tan 17 1 17 1 + + = × − + AED M1 For attempt at rationalisation. 9 17 8 + A1 Must come from 18 2 17 16 + to be convinced a calculator is not being used. 7(d) 2 2 sec tan 1 = + AED AED ( ) 2 9 17 1 64 + = + 81 17 18 17 64 64 + + + oe if ( ) 2 9 17 64 + and 1 are considered separately. M1 For use of their (c) in the correct identity and attempt to simplify to obtain a single fraction. Must see at least 3 terms in an expansion that is not the difference of two squares to be convinced a calculator is not being used. Allow one arithmetic slip 81 9 17 32 + oe A1 cao

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

8 O D B C A 6 cm 5 cm The diagram shows a circle, centre O, radius 5 cm. The lines AOB and COD are diameters of this circle. The line AC has length 6 cm. (a) Show that angle AOC = 1.287 radians, correct to 3 decimal places. [2] (b) Find the perimeter of the shaded region. [2] (c) Find the area of the shaded region. [3]

8(a) 3 sin 2 5 = AOC or ( ) 2 2 2 6 5 5 2 5 5 cos = + − × × AOC M1 For a complete method to find AOC 1.2870 = AOC AOC = 1.287 A1 AG Must see 1.2870 = AOC or better before rounding for A1 8(b) Arc length = 1.287 5 × B1 Perimeter = 32.4 B1 8(c) Sector area = 2 1 5 1.287 2 × × B1 Area of triangle = 2 1 5 sin1.287 2 × × B1 Total area = 28.1 B1

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