Higher November 2022 Unit 2 Q20
20. The diagram shows a triangle \(ABC\) and a circle with centre \(C\).
The points \(B\) and \(D\) lie on the circumference of the circle.
The length of the line \(AB\) is 19 cm.
The length of the line \(AC\) is 29 cm.
The radius of the circle is \(x\) cm.
Diagram not drawn to scale
Calculate the area of the shaded sector \(BCD\). [8]
| Scheme | Marks | Comments |
|---|---|---|
| (\(x^2\) or BC2 =) 19² + 29² – 2×19×29×cos36° | M1 | Penalise premature approximation, PA -1, once only. |
| (\(x\) or BC =) 17·6(199…cm) | A2 | A1 for (\(x^2\) or BC2 =) 310·4(6…) OR \(x = \sqrt{310 \cdot 4(6\ldots)}\) Award A2 if \(x = \sqrt{310 \cdot 4(6\ldots)}\) is used correctly in subsequent work, but only A1 if an incorrect evaluation is used. |
| (sin BCD =) \(\dfrac{19 \times \sin 36^\circ}{17{\cdot}6(199\ldots)}\) OR (cos BCD =) \(\dfrac{17{\cdot}6(199\ldots)^2 + 29^2 - 19^2}{2 \times 17{\cdot}6(199\ldots) \times 29}\) | M2 | FT ‘their derived 17·6(199…)’ . M1 for \(\dfrac{\sin BCD}{19} = \dfrac{\sin 36^\circ}{17{\cdot}6(199\ldots)}\) or equivalent OR M1 for 192=17·6(199…)2+292–2×17·6(199…)×29×cos BCD |
| (BCD =) 39·3(3…°) | A1 | |
| (Area of sector=) \(\dfrac{39{\cdot}3(3\ldots)}{360} \times \pi \times 17{\cdot}6(199\ldots)^2\) | M1 | FT for possible M1 A1, provided M1 M2 or M1 M1 previously awarded. |
| Accept answers in the range 106(cm2) to 107(cm2) | A1 | Must be from correct working. |