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. [Figure 10.1]
- (a)For the case where angle BAC = ⅓π radians, find k correct to 4 significant figures.[5]
- (b)For the general case in which angle BAC = θ radians, where 0 < θ < ½π, it is given that θ > 1. sin θ Find the set of possible values of k.[3]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2021). The mark or mark type is shown in brackets.
- [1] ΔADE = ½(ka)²sin(π/6) (Attempt to find the area of ΔADE)
- [1] ¼k²a² (OE.)
- [1] Sector ABC = ½a²(π/6)
- [1] 2 × ¼k²a² = ½a²(π/6) (OE. For 2×ΔADE = sectorABC with at least one correct area.)
- [1] k = √(π/6) = 0.7236
- [1] 2 × ½(ka)²sinθ = ½a²θ (Condone omission of '2' or '1/2' on LHS for M1 only.)
- [1] k² = θ / (2sinθ)
- [1] k² > ½ leading to 1/√2 < k < 1 (OE. Accept k > 1/√2 or k > 0.707 (AWRT) or 0.707(AWRT) < k < 1 or k > 1/√2 OE)
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