Question Bank › IGCSE Shape & Space › Constructions & Bearings
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All specs Current spec All series June 2025 November 2024 Any marks 1 to 4 marks 5 to 8 marks 9+ marks
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Higher June 2025 Paper 2 Q4
4 Using ruler and compasses only, construct the perpendicular bisector of the line \(AB\)
(2)
Mark scheme
Mark scheme Scheme Marks Bisector with construction arcs B2 (2) (2 marks)
Notes B2: B2 for a fully correct perpendicular bisector with 2 pairs of intersecting arcs shown
Higher November 2024 Paper 2 Q24
24 \(A\), \(B\) and \(C\) are three points on horizontal ground.
\(AB = 8.4\) metres \(BC = 9.2\) metres
\(B\) is on a bearing of 067° from \(A\)
Calculate the bearing of \(A\) from \(C\)
(6)
Mark scheme
Mark scheme Scheme Marks 67 + 51 (= 118) or angle split into 67 and 51 M1 \(AC^2 = 8.4^2 + 9.2^2 - 2 \times 8.4 \times 9.2 \times \cos\text{``}{118}\text{''}\;(= 227.(7615))\) M1 \(AC = \sqrt{8.4^2 + 9.2^2 - 2 \times 8.4 \times 9.2 \times \cos\text{``}{118}\text{''}}\;(= 15.(09\ldots))\)
(15 – 15.1)
M1 \(\dfrac{\sin ACB}{8.4} = \dfrac{\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\) or
\(\dfrac{\sin BAC}{9.2} = \dfrac{\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\) or
\(\cos ACB = \dfrac{9.2^2 + \text{``}{15.09}\text{''}^2 - 8.4^2}{2 \times 9.2 \times \text{``}{15.09}\text{''}}\) or \(\cos BAC = \dfrac{8.4^2 + \text{``}{15.09}\text{''}^2 - 9.2^2}{2 \times 8.4 \times \text{``}{15.09}\text{''}}\)
M1 \((ACB =)\sin^{-1}\left(\dfrac{8.4\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\right)\) or \(\cos^{-1}\left(\dfrac{9.2^2 + \text{``}{15.09}\text{''}^2 - 8.4^2}{2 \times 9.2 \times \text{``}{15.09}\text{''}}\right)\;(= 29.(43\ldots))\)
(29 – 30)
\((BAC =)\sin^{-1}\left(\dfrac{9.2\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\right)\) or \(\cos^{-1}\left(\dfrac{8.4^2 + \text{``}{15.09}\text{''}^2 - 9.2^2}{2 \times 8.4 \times \text{``}{15.09}\text{''}}\right)\;(= 32.(56\ldots))\)
(32 – 33)
M1 Correct answer scores full marks (unless from obvious incorrect working) Look for angles and lengths on their diagram A1 (6) (6 marks)
Notes M1: A diagram showing 118 or use of 118 in further calculations
M1: a correct method to find length \(AC^2\)
M1: a correct method to find length \(AC\)
M1: dep on previous method marks
M1: a completely correct statement for angle \(ACB\) or angle \(BAC\)
A1: allow 279 – 280
Higher November 2024 Paper 1 Q2
2 Use ruler and compasses only to construct the bisector of angle \(ABC\)
(2)
Mark scheme
Mark scheme Scheme Marks Fully correct angle bisector with all relevant arcs B2 (2) (2 marks)
Notes B2: for a fully correct angle bisector with all arcs shown (the line and the arcs can intersect on or within the overlay guidelines)
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