Higher June 2022 Unit 2 Q11
11. The diagram shows a triangle \(ABC\) and a circle with centre \(A\).
The points \(B\) and \(D\) lie on the circumference of the circle.
The length of the line \(AC\) is 27 cm.
The area of triangle \(ABC\) is 112 cm2.
Diagram not drawn to scale
Calculate the area of the shaded region. [6]
Area of shaded region = …………… cm2
| Scheme | Marks | Comments |
|---|---|---|
| sin BAD = \(\dfrac{2 \times 112}{10 \times 27}\) or equivalent | M2 | M1 for the correct use of the formula when sin BAD is not the subject e.g. 112 = ½ ×10×27×sin BAD. |
| (BAD=) 56(·06…°) | A1 | Accept 56·1(°). Allow correct angles given in radians (0·9784..) or gradians (62·2896….) |
| (Area of shaded region=) \(112 - \dfrac{56(\cdot 06\ldots)}{360} \times \pi \times 10^2\) | M2 | F.T. their derived or stated value of angle BAD. M1 for \(\dfrac{56(\cdot 06\ldots)}{360} \times \pi \times 10^2\) (=48·92 cm2) |
| (Area of shaded region =) 63(·077… cm2) or answers in the range: 63 to 63·2 (cm2) | A1 |
Alternative method for the first 3 marks
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of a two-step method. | M2 | Example (Perpendicular height of triangle=) 112×2÷27 = 8·2(96…) or 8·3 (BAD=) sin-1[8·2(96…)÷10] |
| (BAD=) 56(·06…°) | A1 | Allow correct angles given in radians (0·9784...) or gradians (62·2896...) |