Constructions, loci, bearings, plans and elevations – Year 11 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(072+180=\) \(252^\circ\) (1)
   2. \(000^\circ\) (due north) (1)
   3. From \(B\), the bearing of \(A\) is \(252^\circ\) and the direction to \(C\) is \(180^\circ\); \(252-180=72\), so the interior angle is \(180-72=108^\circ\) (2)

   In (a), a back bearing is found by adding or subtracting \(180^\circ\), and bearings are always written with three figures. In (b), \(C\) is due south of \(B\), so \(B\) is due north of \(C\), a bearing of \(000^\circ\). In (c), the north lines at \(A\) and \(B\) are parallel, so co-interior angles add up to \(180^\circ\), which gives \(180-72=108^\circ\).

2. Answer to Question 2 (4 marks)

   1. \(6.5\times20\,000=130\,000\) cm (1) \(=\) 1.3 km (1)
   2. \(2.4\) km \(=240\,000\) cm (1), \(240\,000\div20\,000=\) 12 cm (1)

   In (a), multiply the map length by the scale factor 20 000 to get 130 000 cm in real life. Then divide by 100 to change centimetres into metres, 1300 m, and by 1000 to change metres into kilometres. In (b), reverse the process: change 2.4 km into centimetres by multiplying by 100 000, then divide by the scale factor. Check: \(12\times20\,000=240\,000\) cm, which is 2.4 km.

3. Answer to Question 3 (4 marks)

   1. Two straight lines 3 m from \(AB\), joined at each end by a semicircle of radius 3 m (2)
   2. \(2\times5+2\pi\times3=10+18.85=\) 28.8 m (1 for the formula, 1 for the answer; allow 28.9)

   In (a), the boundary is a stadium shape: a common error is to draw a rectangle, forgetting that points beyond the ends of the path lie on circular arcs centred at \(A\) and \(B\). In (b), the two semicircles together make one full circle of radius 3 m, with circumference \(6\pi\).

4. Answer to Question 4 (4 marks)

   1. Compasses wider than 4.5 cm (1); arcs from \(P\) and \(Q\) above and below, crossing (1); join the crossing points with a ruler, leaving arcs (1)
   2. Every point is the same distance from \(P\) and \(Q\) (1)

   In (a), the compass width must stay the same for all four arcs, and the construction arcs must be left visible. In (b), this is why the perpendicular bisector is the locus of points equidistant from two points.

5. Answer to Question 5 (4 marks)

   1. Right angle between the legs (\(120-30=90\)): \(\sqrt{15^2+20^2}=\) 25 km (1 for Pythagoras, 1 for the answer)
   2. \(\tan^{-1}\left(\dfrac{20}{15}\right)=53.1^\circ\) (1), bearing \(30+53.1=83.1\approx\) \(083^\circ\) (1)

   In (a), draw a sketch: the angle between the two legs at the turning point is \(180-(120-30)\), which is \(90^\circ\), since the north lines are parallel. This gives a right-angled triangle with legs 15 and 20 km. In (b), the angle at the start is measured from the first leg, so add it to \(030^\circ\).

Total : 20 marks