June 2025 Paper 1 Q6
6
The diagram shows a triangle \(ABC\). The arc \(BD\) is part of a circle with centre \(A\) and radius 6 cm.
The area of the sector \(ABD\) is \(14.4\ \text{cm}^2\).
The area of the triangle \(ABC\) is three times the area of the sector \(ABD\).
| Scheme | Marks | AO |
|---|---|---|
| \(\frac{1}{2} \times 6^2 \times \theta = 14.4\) \(18\theta = 14.4\) \(\theta = 0.8\) (radians) AG | B1 | 2.1 |
| [1] |
Notes
B1: Use the correct formula for the area of a sector to obtain 0.8 www. Must be 0.8 and not equivalent fractions
Condone verification
Must work exactly throughout so B0 if via decimals
| Scheme | Marks | AO |
|---|---|---|
| \(\frac{1}{2} \times 6 \times AC \times \sin 0.8 = 3 \times 14.4\) | M1 | 1.1 |
| \(AC = 20.1\) | A1 | 1.1 |
| [2] |
Notes
M1: Attempt use of correct formula for area of a triangle or any equiv complete method. Using 0.8 radians or \(45.8^\circ\)
A1: Obtain correct \(AC\). 3sf or better (awrt 20.1)
| Scheme | Marks | AO |
|---|---|---|
| arc \(BD = 6 \times 0.8 = 4.8\) | B1 | 1.1 |
| \((BC)^2 =\) \(6^2 + 20.07^2 - 2 \times 6 \times 20.07 \times \cos 0.8\) | M1 | 3.1a |
| \(BC = 16.4659\ldots\) | A1 | 1.1 |
| perimeter \(= 4.8 + (20.07 - 6) + 16.47\) | M1 | 3.1a |
| \(= 35.3\) (cm) | A1 | 1.1 |
| [5] |
Notes
B1: Correct arc length (possibly unsimplified) soi. Condone awrt 4.8, if working in degrees
M1: Attempt use of correct cosine rule, or any equiv complete method
A1: Obtain correct \(BC\) (awrt 16.5) soi
M1: Attempt full method for perimeter. Their arc \(BD\) + (their \(AC - 6\)) + their \(BC\)
NB using \(BD = 4.7\) (or better) is likely to be chord \(BD\) so M0 unless evidence that arc has actually been attempted
A1: Obtain correct perimeter (awrt 35.3). Accept without units