The diagram shows two circles with equal radii. A, E and C are points on the circle, centre B. B, E, D and F are points on the circle, centre C. ABCD is a straight line. [Figure of two overlapping circles]
- (a)(i)Show that triangles AEC and FBE are congruent.[3]
- (a)(ii)State another triangle that is congruent to triangle AEC.[1]
- (a)(iii)Explain why EB is parallel to DF.[2]
- (a)(iv)Work out ABE.[1]
- (b)(i)P and Q are points on the circle centre O with radius 4 cm. PÔQ = 130°. Calculate the area of triangle POQ.[2]
- (b)(ii)Calculate the area of the major segment, shown unshaded in the diagram.[3]
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Mark scheme
Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.
- [3] (a)(i) EC = BE or AC = FE and ∠AEC = ∠FBE or ∠ECA = ∠BEF Two correct reasons for their choices
- [3] (a)(i) B1 B1 B1 Statements and reasons: EC = BE; radii AC = FE; diameters ∠AEC = ∠FBE [= 90°]; angle in semicircle ∠ECA = ∠BEF [= 60°]; equilateral triangle Third statement, leading to correct congruence condition i.e. RHS, SAS, SSA
- [1] (a)(ii) BFD
- [2] (a)(iii) ∠EBF = ∠DFB = 90° Cointerior/interior/supplementary/allied angles [sum to 180]
- [2] (a)(iii) 1 Both 90° could be marked on diagram 1dep
- [2] (a)(iii) OR ∠BEF = ∠EFD = 60° Alternate angles [are equal]
- [2] (a)(iii) OR 1 Both 60° could be marked on diagram 1dep
- [1] (a)(iv) 120°
- [1] (a)(iv) 120° could be marked on diagram
- [2] (b)(i) 6.126 to 6.13
- [2] (b)(i) M1 for 1/2 × 4 × 4 × sin 130 Or 1/2 PQ x perpendicular height (numerical)
- [3] (b)(ii) 38.2 to 38.3
- [3] (b)(ii) M1 for ((360-130) / 360) × π × 4² soi by 32.11 or (130 / 360) × π × 4² soi by 18.15 And M1 for 'their major sector area' + 'their triangle area' Or for 'their circle area' - 'their minor sector area' + 'their triangle area'
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