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

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1. Answer to Question 1

   1. East \(=090^\circ\), south \(=180^\circ\), west \(=270^\circ\), south-east \(=135^\circ\).
   2. If the bearing is less than \(180^\circ\), add \(180^\circ\): \(065+180=\) \(245^\circ\).
   3. \(230-180=\) \(050^\circ\); \(312-180=\) \(132^\circ\); \(148+180=\) \(328^\circ\).
   4. Add \(180^\circ\) if the bearing is less than \(180^\circ\), and subtract \(180^\circ\) if it is more. The answer must be between \(000^\circ\) and \(360^\circ\).

2. Answer to Question 2

   1. \(4\times25\,000=100\,000\) cm \(=1000\) m \(=\) 1 km.
   2. \(3.5\) km \(=350\,000\) cm. \(350\,000\div25\,000=\) 14 cm.
   3. \(12\times50\,000=600\,000\) cm \(=\) 6 km.
   4. 1 cm represents \(25\,000\) cm \(=\) 250 m.

3. Answer to Question 3

   1. Open the compasses to more than 4 cm. Draw an arc from \(A\) above and below the line. Keep the same radius and draw arcs from \(B\) to cross them. Join the two crossing points with a ruler. Leave the arcs showing.
   2. \(P\) is 4 cm along \(AB\) from \(A\) (to the foot of the perpendicular) and 3 cm up. \(AP^2=4^2+3^2=25\), so \(AP=\) 5 cm.
   3. Place the compass point at the vertex and draw an arc crossing both arms. From each crossing point draw arcs of equal radius to meet inside the angle. Join the vertex to this meeting point. Each half is \(70\div2=\) \(35^\circ\).

4. Answer to Question 4

   1. A circle with centre \(P\) and radius 3 cm.
   2. The perpendicular bisector of \(AB\).
   3. The angle bisectors of the angles between the lines (two lines at right angles to each other).
   4. Two parallel lines 2 cm either side of \(AB\), joined by a semicircle of radius 2 cm at each end (a "racetrack" shape). Perimeter \(=2\times6+2\pi\times2=12+4\pi=\) 24.6 cm.

5. Answer to Question 5

   1. A square of side 6 cm with its diagonals drawn (the apex is above the centre).
   2. An isosceles triangle with base 6 cm and height 4 cm.
   3. Half the base is 3 cm. Slant height \(=\sqrt{3^2+4^2}=\sqrt{25}=\) 5 cm.
   4. One face \(=\tfrac12\times6\times5=\) 15 cm². Four faces \(=4\times15=\) 60 cm².

6. Answer to Question 6

   1. \(130-40=90\), so the turn between the two legs is \(90^\circ\) (the legs are perpendicular).
   2. \(\sqrt{12^2+9^2}=\sqrt{225}=\) 15 km.
   3. The angle at the start between the first leg and the line to the ship is \(\tan^{-1}\left(\dfrac9{12}\right)=36.9^\circ\). The bearing is \(40+36.9=76.9\approx\) \(077^\circ\).
   4. Back bearing \(=77+180=\) \(257^\circ\).

7. Answer to Question 7

   1. At \(A\) the bearing back to \(P\) is \(060+180=240^\circ\) and the ferry leaves on \(150^\circ\). The angle between these is \(240-150=\) \(90^\circ\).
   2. \(PB=\sqrt{20^2+20^2}=\sqrt{800}=\) 28.28 km.
   3. Triangle \(PAB\) is isosceles and right-angled, so angle \(APB=45^\circ\). The bearing is \(060+45=\) \(105^\circ\).
   4. \(105+180=\) \(285^\circ\).

8. Answer to Question 8

   1. \(120+180=\) \(300^\circ\).
   2. The bearing of \(H\) from \(L\) is \(300^\circ\) and the bearing of \(S\) from \(L\) is \(210^\circ\). \(300-210=90^\circ\).
   3. \(HS=\sqrt{15^2+8^2}=\sqrt{289}=\) 17 km.
   4. Angle \(LHS=\tan^{-1}\left(\dfrac8{15}\right)=28.1^\circ\). The ship is further round clockwise than \(L\): \(120+28.1=\) \(148.1^\circ\).

9. Answer to Question 9

   1. 1 cm represents 0.5 km: \(4.2\times0.5=\) 2.1 km, \(6.8\times0.5=\) 3.4 km, \(3.0\times0.5=\) 1.5 km.
   2. \(2.1+3.4+1.5=\) 7.0 km.
   3. \(7\div4=1.75\) hours \(=\) 1 hour 45 minutes.
   4. Straight line: \(9.6\times0.5=4.8\) km, so it is \(7.0-4.8=\) 2.2 km shorter.

10. Answer to Question 10

    1. Draw the perpendicular bisector of \(AB\), which is 5 m from \(A\). The region is on the \(A\) side of this line.
    2. A circle of radius 6 cm centred on \(A\) (inside the circle).
    3. No. The point is 6 m from \(A\) along \(AB\), which is beyond the bisector (5 m from \(A\)), so it is nearer to \(B\) (only 4 m away).
    4. Least distance: just above 0 m (very near \(A\)). Greatest: 5 m (the bisector), since 5 is less than 6.

11. Answer to Question 11

    1. A corner is 10 m across and 6 m up from the centre: \(\sqrt{10^2+6^2}=\sqrt{136}=\) 11.66 m.
    2. \(11.66\lt 15\), so yes, the whole office is covered.
    3. The diagonal is \(\sqrt{20^2+12^2}=\sqrt{544}=23.3\) m. This is more than 15 m, so no.
    4. \(\pi\times15^2=\) 706.9 m².

12. Answer to Question 12

    1. \(\sqrt{200^2+150^2}=\sqrt{62\,500}=\) 250 km.
    2. The angle east of north is \(\tan^{-1}\left(\dfrac{150}{200}\right)=36.9^\circ\), so the bearing is \(037^\circ\).
    3. Back bearing \(=37+180=\) \(217^\circ\) (using the exact angle, \(216.9^\circ\)).
    4. \(250\div250=\) 1 hour.

13. Answer to Question 13

    1. \(6+7=13>10\), so the circles overlap.
    2. It is 4 km from \(A\) (\(\lt 6\)) and \(10-4=6\) km from \(B\) (\(\lt 7\)). Yes, it receives both.
    3. Signal \(A\) reaches up to 6 km from \(A\). Signal \(B\) reaches back to \(10-7=3\) km from \(A\). Both signals are received between 3 km and 6 km: a stretch of 3 km.
    4. It is 8 km from \(A\) (\(>6\)) so no signal from \(A\); it is \(10-8=2\) km from \(B\) (\(\lt 7\)), so it receives only signal \(B\).

14. Answer to Question 14

    1. \(8.5\) m \(=850\) cm, so 8.5 cm on the plan; \(6\) m \(=\) 6 cm on the plan.
    2. \(8.5\times6=\) 51 m².
    3. \(51\times4.20=\) £214.20.
    4. Perimeter \(=2\times(8.5+6)=29\) m. Cost \(=29\times18=\) £522.

15. Answer to Question 15

    1. The bearing of \(A\) from \(B\) is \(050+180=230^\circ\). The bearing of \(C\) from \(B\) is \(140^\circ\). The angle between these is \(230-140=\) \(90^\circ\).
    2. \(AC=\sqrt{24^2+18^2}=\sqrt{900}=\) 30 km.
    3. Angle \(BAC=\tan^{-1}\left(\dfrac{18}{24}\right)=36.9^\circ\). Bearing \(=50+36.9=86.9\approx\) \(087^\circ\).
    4. \(86.9+180=266.9\approx\) \(267^\circ\).

16. Answer to Question 16

    1. The perpendicular bisector of \(AB\) is the line \(x=5\). The region is the side containing \(A\) (\(x\lt 5\)).
    2. A circle of radius 4 cm centred at \(D\); the region is inside it.
    3. \(x=3\lt 5\), and the distance from \(D\) is \(\sqrt{3^2+1^2}=\sqrt{10}=3.16\lt 4\). Yes.
    4. \((4,1)\): \(x\lt 5\) but the distance from \(D\) is \(\sqrt{16+25}=6.4>4\), so no. \((6,5)\): \(x>5\), so it is nearer to \(B\), so no.

17. Answer to Question 17

    1. The plan is the L-shape: 9 cm long and 3 cm wide, with a 3 cm by 3 cm square sticking up at one end. The front elevation is a rectangle 9 cm wide and 5 cm high (the L-shape from the front shows a 9 by 5 rectangle).
    2. Area of L \(=4\times9=36\) cm². Volume \(=36\times5=\) 180 cm³.
    3. Top and bottom: \(2\times36=72\) cm². The perimeter of the L is 10 edges of 3 cm \(=30\) cm, so the vertical sides give \(30\times5=150\) cm². Total \(=72+150=\) 222 cm².

18. Answer to Question 18

    1. Draw \(AB=7\) cm. At \(A\) use a protractor to draw a ray at \(50^\circ\) to \(AB\). Mark \(C\) on the ray 6 cm from \(A\) with compasses. Join \(B\) to \(C\).
    2. \(BC^2=7^2+6^2-2\times7\times6\cos50^\circ=85-84\times0.6428=85-54.0=31.0\), so \(BC=\) 5.57 cm.
    3. Yes: 5.6 cm agrees with 5.57 cm to 1 d.p.
    4. The bisector makes \(25^\circ\) with \(AB\).

19. Answer to Question 19

    1. At the corner the shed takes up a \(90^\circ\) angle, so the goat has the other \(270^\circ\), which is three quarters of a full turn.
    2. \(\tfrac34\times\pi\times5^2=\tfrac34\times78.54=\) 58.9 m².
    3. The rope is 5 m but the shorter side is 6 m, so the rope is too short to reach the next corner on either side.

20. Answer to Question 20

    1. At the corner the goat has \(270^\circ\) so \(\tfrac34\) of a circle of radius 10.
    2. Going round the 8 m side, 8 m of rope is used to reach the next corner, leaving \(10-8=2\) m: a quarter circle of radius 2. Going round the 6 m side leaves \(10-6=4\) m: a quarter circle of radius 4.
    3. Area \(=\tfrac34\pi10^2+\tfrac14\pi2^2+\tfrac14\pi4^2=75\pi+\pi+4\pi=\) \(80\pi\) \(=\) 251.3 m².