Higher November 2017 Unit 2 Q17
17. \(ABC\) represents the sector of a circle with radius 7 cm and centre \(A\), as shown below.
\(B\hat{A}C = x^\circ\), \(AD\) = 3 cm and \(BD\) = 6 cm.
Diagram not drawn to scale
Find the area of the shaded region \(BCD\). [8]
| Scheme | Marks | Comments |
|---|---|---|
| S1 | For an attempt to subtract area of a triangle from the area of sector (This may even include an expression in terms of x). | |
| cos \(x\) = (32 + 72 – 62) / (2×3×7) | M2 | M1 for 62 = 32 + 72 – 2×3×7×cos \(x\) |
| (\(x\) =) 58.4(1…°) | A1 | Allow 58°. |
| Area of sector = 58.4(1…) /360 × π × 7² ( Answer in the range: 24.9(595…cm²) to 25(cm²) ) | M1 | FT ‘their derived 58.4(1…°)’ |
| Area of triangle = ½ ×3×7×sin58.4(1…) ( Answer in the range: 8.9(4…cm²) to 8.94(4… cm²) or 9(cm²) ) | M1 | FT ‘their derived 58.4(1…°)’ M0 for use of a right-angled triangle. Unsupported 9 cm² gains M0. |
| (Area of shaded region) answer in the range of 15.9(cm²) to 16.1(cm²) | A2 | FT ‘their derived 58.4(1…°)’ provided previous M1, M1 awarded. Award A1 for sight of either: area of sector in the range 24.9(595…cm²) to 25(cm²) FT ‘their derived 58.4(1…°)’ OR area of triangle in the range 8.9(4…cm²) to 8.94(4… cm²) or 9(cm²) FT ‘their derived 58.4(1…°)’ This A1 is dependent on gaining the corresponding M1. NB Unsupported answer of 16(cm2) gains 0. |