Higher June 2025 Paper 1R Q25
25
Diagram NOT accurately drawn
\(BAC\) is a sector of a circle, centre \(A\)
\(BCD\) is a sector of a circle, centre \(C\)
Angle \(BAC = 40^\circ\)
Angle \(BCD = 130^\circ\)
Area of shaded segment = 28 cm2
Find the length of the arc \(BD\)
Give your answer correct to 3 significant figures.
(6)
| Scheme | Marks |
|---|---|
eg \(\dfrac{40}{360}\pi r^2 - \dfrac{1}{2}r^2\sin 40 (= 28)\) oe or \(\dfrac{40}{360}\pi r^2 = 28 + \dfrac{1}{2}r^2\sin 40\) oe | M1 |
| (radius2 =) 992 – 1024 (radius =) 31.8(096…) Answer: 31.8 | A1 |
eg \((BC^2 =)\;2 \times \text{``}{31.8}\text{''}^2 - 2 \times \text{``}{31.8}\text{''}^2\cos 40 (= 473.4\ldots)\) or \(\dfrac{0.5BC}{\text{``}{31.8}\text{''}} = \sin 20\) or \(\dfrac{BC}{\sin 40} = \dfrac{\text{``}{31.8}\text{''}}{\sin(70)}\) | M1 |
eg \((BC =)\sqrt{2 \times \text{``}{31.8}\text{''}^2 - 2 \times \text{``}{31.8}\text{''}^2\cos 40}\;(= 21.7\ldots)\) or \((BC =)\;2 \times \text{``}{31.8}\text{''}\sin 20 (= 21.7\ldots)\) or \(BC = \dfrac{\text{``}{31.8}\text{''}\sin 40}{\sin(70)}(= 21.7\ldots)\) | M1 |
| eg \(\dfrac{130}{360} \times 2 \times \pi \times \text{``}{21.7}\text{''}\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 49.4 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for a correct expression for the area of the shaded region
Allow 3.14… or \(\dfrac{22}{7}\) for \(\pi\)
sin 40 = 0.64…
A1: Allow answers in the range 31.5 – 32.0
M1: for a correct first step to find \(BC\) using their clearly identified radius eg \(r\) = …. or seen on diagram
NB \(\dfrac{180 - 40}{2} = 70\)
sin 20 = 0.34…
sin 70 = 0.93… or 0.94
M1: dep on previous M1
for a complete method to find \(BC\)
cos 40 = 0.76… or 0.77
M1: dep on previous M1
for a complete method to find the length of arc \(BD\)
A1: accept 48.9 – 49.7