Question Bank › IGCSE Shape & Space › Circles & Sectors
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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 \(BAC = 40^\circ\)2
Find the length of the arc \(BD\)
(6)
Mark scheme
Mark scheme 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 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) 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
M1: dep on previous M1
A1: accept 48.9 – 49.7
Higher June 2025 Paper 1 Q11
11 \(OABC\) is a sector of a circle, centre \(O\)
Diagram NOT accurately drawn
angle \(AOC = 140^\circ\)
Calculate the area of the sector.
(2)
Mark scheme
Mark scheme Scheme Marks eg \(\pi \times 16^2 \times \dfrac{140}{360}\) oe eg \(256\pi \times \dfrac{7}{18}\) M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (2) (2 marks)
Notes M1: allow use of 3.14… or \(\dfrac{22}{7}\) for \(\pi\)
\(\dfrac{140}{360}\) may be seen as an equivalent fraction or decimal eg \(\dfrac{7}{18}\) or \(0.3\dot{8}\) 0.388(8…) or 0.389
A1: accept 311.8 – 313
Higher November 2024 Paper 1 Q16
16 \(OAPB\) is a sector of a circle, centre \(O\)
Diagram NOT accurately drawn
Angle \(AOB = 50^\circ\)
Area of triangle \(OAB = 120\) cm2
Work out the area of the sector \(OAPB\)
(4)
Mark scheme
Mark scheme Scheme Marks \(120 = \dfrac{1}{2} \times a \times b \times \sin 50\) oe or \(\dfrac{120 \times 2}{\sin 50}\) oe
or
\(120 = \dfrac{1}{2} \times 2x\sin 25 \times x\cos 25\) oe or \(\dfrac{120 \times 2}{2 \times \sin 25 \times \cos 25}\) oe
or
313(.2977494)
M1 \((\textit{radius} =)\sqrt{\dfrac{120 \times 2}{\sin 50}}\) (= 17.7(0021891)) oe or
\((\textit{radius} =)\sqrt{\dfrac{120 \times 2}{2 \times \sin 25 \times \cos 25}}\) (= 17.7(0021891)) oe or
\((\textit{radius} =)\sqrt{313(.2977494)}\) (= 17.7(0021891))
M1 (area of \(OAPB\) =) \(\pi \times \text{“}17.7\text{”}^2 \times \dfrac{50}{360}\) oe M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (4) (4 marks)
Notes M1: for a correct equation for \(a \times b\) or \(r^2\) (allow any letters for \(a\) or \(b\))correct expression for \(a \times b\) or \(r^2\)
M1: for a correct rearrangement to find the radius
A1: awrt 137
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