Pythagoras’ theorem and trigonometry — Paper 4 · IGCSE Mathematics (9-1) 0980

7 (a) Show that each interior angle of a regular pentagon is 108°. [2] (b) D E C NOT TO O SCALE M X A B The diagram shows a regular pentagon ABCDE. The vertices of the pentagon lie on a circle, centre O, radius 12 cm. M is the midpoint of BC. (i) Find BM. BM = … cm [3] (ii) OMX and ABX are straight lines. (a) Find BX. BX = … cm [3] (b) Calculate the area of triangle AOX. … cm2 [3]

11 marks

Mark scheme: 7(a) 360 M2 360 180 − or or M1 for or ( 5 − 2 ) × 180 5 5 ( 5 − 2 ) × 180 ( 2 × 5 − 4 ) × 90 or 90(2 × 5 – 4) or or or 3 × 180 ÷ 5 5 5 or 6 × 90 ÷ 5 5 × 180 − 360 or 5 × 180 – 360 5 5 −×2 180 If 0 scored, SC1 for 5 7(b)(i) 7.05 or 7.053… 3 M2 for 12 × cos54 oe or M1 for implicit form or B1 for length of edge of pentagon = 14.1 to 14.11 If 0 scored, SC1 for right angle at M 7(b)(ii)(a) 22.8 or 22.81 to 22.83… nfww 3 their(b)(i) M2 for oe cos72 or M1 for implicit form oe or B1 for AX = 36.9 or 36.93 to 36.94 7(b)(ii)(b) 179 or 179.1 to 179.3… 3 M2 for 1 × 12 × their AX × sin54 oe 2 or 1 × 12 × theirOX × sin108 oe 2 or 1 × their AX × theirOX × sin18 2 or 12 × 12 2 × sin72 + areaOBX oe or 1 2 × 12 2 × sin72 + area OMB + area MBX oe or M1 for a correct method to find area of one relevant triangle AOB, OMB, MBX, OBX or ONX seen

This question in 0980/42 May/June 2019

4 (a) (i) Calculate the external curved surface area of a cylinder with radius 8 m and height 19 m. … m2 [2] (ii) This surface is painted at a cost of $0.85 per square metre. Calculate the cost of painting this surface. $ … [2] (b) A solid metal sphere with radius 6 cm is melted down and all of the metal is used to make a solid cone with radius 8 cm. Calculate the curved surface area of the cone. 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 1 2 [The volume, V, of a cone with radius r and height h is V = r r h .] 3 [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] … cm2 [5] (c) Two cones are mathematically similar. The total surface area of the smaller cone is 80 cm2. The total surface area of the larger cone is 180 cm2. The volume of the smaller cone is 168 cm3. Calculate the volume of the larger cone. … cm3 [3] (d) The diagram shows a pyramid with a P square base ABCD. DB = 8 cm. NOT TO P is vertically above the centre, X, of SCALE the base and PX = 5 cm. D C X A B Calculate the angle between PB and the base ABCD. … [3]

15 marks

Mark scheme: 4(a)(i) 955 or 955.0 to 955.2 2 M1 for 2 × π × 8 × 19 oe 4(a)(ii) 812 or 811.7 to 811.9... 2 FT their (i) × 0.85 M1 for their (i) × 0.85 or their (i) × 85 4(b) 394 or 395 or 394.3 to 394.6... 5 4 3 π× 6 3 M2 for h = or better 1 3 π× 8 2 4 3 1 2 or M1 for × π × 6 = × π × 8 × h oe 3 3 M1 for 82 + their 13.52 or better M1 dep for π × 8 × their slant height dep on use of Pythagoras 4(c) 567 3 3 168  80  2 M2 for =   oe or better V  180  1 1  180  2  80  2 or M1 for   or   oe seen or  80   180  better 4(d) 51.3 or 51.34... 3 5 M2 for tan = oe 4 or M1 for recognition of angle PBX

This question in 0980/41 Oct/Nov 2019

5 North D NOT TO SCALE 140° A 450 m 400 m B 350 m C The diagram shows a field ABCD. The bearing of B from A is 140°. C is due east of B and D is due north of C. AB = 400 m, BC = 350 m and CD = 450 m. (a) Find the bearing of D from B. … [2] (b) Calculate the distance from D to A. … m [6] (c) Jono runs around the field from A to B, B to C, C to D and D to A. He runs at a speed of 3 m/s. Calculate the total time Jono takes to run around the field. Give your answer in minutes and seconds, correct to the nearest second. … min … s [4]

12 marks

Mark scheme: 5(a) [0]38 or [0]37.9 or [0]37.87... 2 350 M1 for tan = oe 450 If 0 scored, SC1 for answer [0]52 or [0]52.1 or [0]52.12 to [0]52.13 5(b) 624 or 623.8 to 623.9 6 M2 for 450 – 400 sin 50 ... or M1 for sin 50 = 400 M2 for 350 + 400 cos 50 ... or M1 for cos 50 = 400 M1 for (their (450 – 400 sin 50))2 + (their (350 + 400 cos 50))2 5(c) 10 min 8 s 4 B3 for 10.1 or 10.13… or M2 for (400 + 350 + 450 + their DA) ÷ 3 [÷ 60] oe or M1 for any distance ÷ 3 M1 for rounding their minutes into minutes and seconds to nearest second if clearly seen

This question in 0980/42 May/June 2020

8 (a) A cuboid has length L cm, width W cm and height H cm. L cm H cm NOT TO SCALE 20.1 cm W cm 37.8 cm The diagram shows the net of this cuboid. The ratio W : L = 1 : 2. Find the value of L, the value of W and the value of H. L = … W = … H = … [5] (b) E NOT TO SCALE 24 cm D C 15 cm A 18 cm B The diagram shows a solid pyramid with a rectangular base ABCD. E is vertically above D. Angle EDC = angle EDA = 90°. AB = 18 cm, BC = 15 cm and EC = 24 cm. (i) The pyramid is made of wood and has a mass of 800 g. Calculate the density of the wood. Give the units of your answer. 1 [The volume, V, of a pyramid is V = # area of base # height.] 3 [Density = mass ' volume] … … [5] (ii) Calculate the angle between BE and the base of the pyramid.

14 marks

Mark scheme: 8(a) [L =] 11.8 5 M1 for L = 2W oe soi [W =] 5.9 M1 for W + 2H = 20.1 oe [H =] 7.1 M1 for 2L + 2H = 37.8 oe B1 for at least one correct answer 8(b)(i) 0.559 to 0.56[0…] B4 1 2 2 M2 for × 18 × 15 × 24 − 18 isw 3 conversion or M1 for h2 + 182 = 242 oe or better M1 for figs 800 ÷ figs their volume isw g/cm3 or g cm–3 final answer B1 8(b)(ii) 34.1 or 34.11 to 34.12 4 2 2 24 − 18 M3 for tan [ ] = oe 18 2 + 15 2 or M2 for 18 2 + 15 2 isw or 24 2 + 15 2 isw or M1 for 182 + 152 isw or 242 + 152 isw or M1 for indicating required angle is EBD

This question in 0980/42 May/June 2021

5 (a) D A NOT TO SCALE O 124° B 35° C A, B, C and D are points on a circle, centre O. Angle COD = 124° and angle BCO = 35°. (i) Work out angle CBD. Give a geometrical reason for your answer. Angle CBD = … because … … [2] (ii) Work out angle BAD. Give a geometrical reason for each step of your working. Angle BAD = … because … … … [4] (b) R 42° NOT TO S SCALE O Q 5.9 cm P P, Q, R and S are points on a circle, centre O. QS is a diameter. Angle PRS = 42° and PQ = 5.9 cm. Calculate the circumference of the circle. … cm [5]

11 marks

Mark scheme: 5(a)(i) 62 2 B1 for either and Angle at centre is twice angle at circumference oe 5(a)(ii) 117 4 B2 for 117 and or B1 for [angle OCD =] 28 Isosceles [triangle] B1dep for isosceles [triangle] and and Opposite angles in a cyclic quadrilateral B1 for opposite angles in a cyclic are supplementary quadrilateral are supplementary 5(b) 24.9 or 24.94 to 24.95 5 B1 for angle PQS = 42 M2 for QS = 5.9 ÷ cos 42 oe 5.9 or M1 for cos42= oe QS M1dep for their SQ × π oe

This question in 0980/41 Oct/Nov 2021

8 Q NOT TO 4 m SCALE A 15 m C P 8 m 3 m 20 m B The diagram shows triangle ABC on horizontal ground. AC = 15 m , BC = 8 m and AB = 20 m . BP and CQ are vertical poles of different heights. BP = 3 m and CQ = 4 m . AQ and PQ are straight wires. (a) Show that angle ACB = 117.5° , correct to 1 decimal place. [4] (b) Calculate the area of triangle ABC. … m2 [2] (c) Calculate the length of AQ. … m [2] (d) Calculate the angle of elevation of Q from P. … [3] (e) Another straight wire connects A to the midpoint of PQ. Calculate the angle between this wire and the horizontal ground. … [5]

16 marks

Mark scheme: 8(a) 15 2 + 8 2 − 20 2 M2 M1 for 202 = 152 + 82 − 2.15.8cos ( ) [cos = ] 2.15.8 117.54 to 117.55 A2 37 111 A1 for − or − or –[0].4625 80 240 8(b) 53.2 or 53.19 to 53.23 2 M1 for 0.5  8  15  sin(117.5) oe 8(c) 15.5 or 15.52 to 15.53 2 M1 for 152 + 42 oe 8(d) 7.1 or 7.13 or 7.125 to 7.126 3 4 − 3 M2 for tan [P]= oe or for 7.1 or 8 7.13 or 7.125 to 7.126 seen or M1 for vertical line = 4 – 3 soi After 0 scored SC1 for correct angle identified 8(e) 11.5 nfww or 11.48 to 11.49... 5 B1 for height of 3.5 soi M2 for 15 2 + 4 2 − 2.15.4cos(117.5) 15 2 + 4 2 − (...) 2 or M1 for cos117.5 = 2.15.4 3.5 M1 for tan = oe their 17.216... After M0 scored SC1 for correct angle identified

This question in 0980/41 Oct/Nov 2022

5 (a) D NOT TO 83.2 m SCALE 38° C B A 54.5 m ACD is a right-angled triangle. B is on AC and BC = 54.5 m. AD = 83.2 m and angle ABD = 38° . Calculate angle ACD. Angle ACD = … [5] (b) F G E EFG is a right-angled triangle. A circle can be drawn that passes through the three vertices of the triangle. On the diagram, mark the position of the centre of the circle with a cross. Explain how you decide. … … [2] (c) N R NOT TO 5 cm SCALE 4 cm Q 6 cm M P L In triangle LMN, the ratio angle L : angle M : angle N = 4 : 5 : 6. In triangle PQR, PQ = 6 cm , PR = 4 cm and QR = 5 cm . Calculate the difference between the largest angle in triangle PQR and the largest angle in triangle LMN. … [7]

14 marks

Mark scheme: 5(a) 27.3 or 27.32 to 27.33 5 83.2 M4 for tan[ACD] = oe 83.2 + 54.5 tan38 or 83.2 M3 for [AC =] +54.5 oe tan38 or for [CD =] 2  83.2  2  83.2  54.5 + − 2(54.5) cos(180 − 38)      sin38   sin38  oe or 83.2 83.2 M2 for [AB =] oe or for [BD =] oe tan38 sin 38 83.2 83.2 or M1 for tan38 = oe or sin38 = oe AB BD 5(b) Centre marked at midpoint of B2 B1 for marking the centre at mid-point of FG FG. and Angle in a semi-circle is 90 5(c) 10.8 or 10.81 to 10.82 7 B2 for 72 180 or M1 for [ 6] 4 + 5 + 6 and, for triangle PQR B4 for [angle R=]82.8 or 82.81 to 82.83 5 or B3 for [cosR =] oe or better 40 4 2 + 5 2 − 6 2 or M2 for 2 4 5 or M1 for 62 = 42 + 52 – 245cosR After 0 scored for triangle PQR, SC1 for [P =] 55.8 or 55.77 to 55.78 or  Q = 41.4 or 41.40 to 41.41

This question in 0980/41 Oct/Nov 2023

4 (a) NOT TO SCALE 12 cm 1 m The diagram shows a tank in the shape of a half-cylinder of radius 12 cm and length 1 metre. The tank is fixed horizontally and is completely filled with water. (i) Calculate the volume of water in the tank. Give your answer correct to the nearest 10 cm3. … cm3 [3] (ii) NOT TO 6 cm SCALE Water is removed from the tank until the level of water is 6 cm below the top of the tank. The diagram shows the cross-section of the tank. Calculate the volume of water that is now in the tank. … cm3 [5] (b) A rectangular fish tank with length 42 cm and width 35 cm is full of water. A stone lies at the bottom of the tank. When the stone is removed from the tank, the depth of the water decreases by 0.2 cm. The density of the stone is 2.2 g/cm3. Calculate the mass of the stone in grams. [ Density = mass ' volume] … g [3] (c) H G E F 15 cm NOT TO SCALE D C 12 cm A 8 cm B The diagram shows a cuboid, ABCDEFGH. Calculate the angle that AG makes with the base of the cuboid. … [4]

15 marks

Mark scheme: 4(a)(i) 22 620 cao 3 B2 for 7200 or 22 608 to 22 629 1 2 or M1 for   12 [ figs 1] oe 2 4(a)(ii) 8840 or 8850 or 8836 to 8850. 5 6 M1 for cos COM = oe 12 6 or sin AOC = oe 12  theirCOD 2  M1 for    12  oe M  360   1 2  oe M1 for   12  sin  theirCOD    2  M1dep for (their area of sector COD– their area of triangle COD) 100 dep on at least M1M1 oe 4(b) 647 or 646.8 3 m M2 for 2.2  oe 42  35  0.2 or M1 for [vol of stone =] 42×35×0.2 oe If 0 scored SC1 for answer figs 647 or figs 6468 4(c) 46.1 or 46.12 to 46.14 4 15 M3 for tan  oe 8 2  12 2 or M2 for 82 + 122 oe or 82 + 122 + 152 oe or M1 for identifying the angle GAC

This question in 0980/42 May/June 2024

9 H G F E NOT TO 17 cm SCALE D C 8 cm A 10 cm B ABCDEFGH is a solid cuboid. AB = 10 cm, BC = 8 cm and CG = 17 cm. (a) Work out the volume of the cuboid. … cm3 [1] (b) Work out the total surface area of the cuboid. … cm2 [3] (c) Calculate the angle between GA and the base ABCD. … [4] (d) A straight rod PQ is placed inside the cuboid. One end of the rod, P, is placed at the midpoint of AB. The other end of the rod, Q, rests on GH. HQ : QG = 4 : 1 . Q H G F E NOT TO 17 cm SCALE D C 8 cm A P B 10 cm Calculate the length of the rod PQ. … cm [4]

12 marks

Mark scheme: 9(a) 1360 1 9(b) 772 3 M2 for [2 ×] (10 × 8 + 10 × 17 + 8 × 17) oe or M1 for 10 × 8 oe or 10 × 17 oe or 8 × 17 oe 9(c) 53 or 53.0 to 53.01 4 17 M3 for tan [GAC] = oe 10 2 + 8 2 or M2 for 102 + 82 oe or for 102 + 82 + 172 oe or M1 for recognising angle GAC is required 9(d) 19[.0] or 19.02 to 19.03 4 M3 for 32 + 82 + 172 oe OR B1 for QG = 2 soi or HQ = 8 M1 for (5 – 2) 2 + 82 or (5 – 2) 2 + 172

This question in 0980/41 Oct/Nov 2024

27 P NOT TO SCALE D C A B The diagram shows a cube. Calculate the angle between the diagonal AP and the base ABCD. … [4]

4 marks

Mark scheme: 27 35.3 or 35.26… 4 M3 for correct numerical trig statement for angle PAC e.g. l 1 tan = or oe l 2 + l 2 2 l 1 or sin = or oe l 2 + l 2 + l 2 3 l 2 + l 2 2 or cos = or oe l 2 + l 2 + l 2 3 where l is a value or M2 for a correct Pythagoras statement soi or correct trig statement for diagonal of a face with angle 45 used soi e.g. For side length k, AC shown as k 2 or M1 for recognition of angle PAC

This question in 0980/42 May/June 2025

27 H G D C NOT TO 7 cm SCALE 5 cm E F 15 cm A 10 cm B The diagram shows a prism of length 15 cm. The cross-section of the prism is a trapezium. Angle DAB = 90° and angle ADC = 90°. AB = 10 cm, AD = 5 cm and DC = 7 cm. Calculate the angle the diagonal AG makes with the base ABFE. … [4]

4 marks

Mark scheme: 27 16.8 or 16.80 to 16.81 4 5 M3 for tan = oe or 15 2 + 7 2 5 sin = oe or 152 + 7 2 + 52 152 + 7 2 cos = oe 152 + 7 2 + 52 or M2 for 15 2 + 7 2 or 15 2 + 7 2 + 5 2 or M1 for indication of correct angle

This question in 0980/41 Oct/Nov 2025