Geometrical terms — Paper 4 · IGCSE Mathematics 0580

10 (a) The diagram shows a regular F E hexagon ABCDEF of side 10 cm. NOT TO SCALE A D B C (i) Show that angle BAF = 120°. [2] (ii) The vertices of a rectangle PQRS F E touch the sides FA, AB, CD and DE. P S PS is parallel to FE and AP = x cm. NOT TO SCALE A D Q R B C Use trigonometry to find the length of PQ in terms of x. PQ = … cm [3] (iii) PF = (10 – x) cm. Show that PS = (20 − x) cm. [3] (b) F E K N NOT TO SCALE A D L M B C The diagram shows the vertices of a square KLMN touching the sides of the same hexagon ABCDEF, with KN parallel to FE. Use your results from part (a)(ii) and part (a)(iii) to find the length of a side of the square. … cm [4]

12 marks

This question in 0580/42 Feb/March 2017

8 (a) The exterior angle of a regular polygon is x° and the interior angle is 8x°. Calculate the number of sides of the polygon. … [3] (b) C NOT TO SCALE O D B 58° A A, B, C and D are points on the circumference of the circle, centre O. DOB is a straight line and angle DAC = 58°. Find angle CDB. Angle CDB = … [3] (c) R O NOT TO SCALE 48° P Q P, Q and R are points on the circumference of the circle, centre O. PO is parallel to QR and angle POQ = 48°. (i) Find angle OPR. Angle OPR = … [2] (ii) The radius of the circle is 5.4 cm. Calculate the length of the major arc PQ. … cm [3]

11 marks

Mark scheme: 8(a) 18 3 B2 for 20 nfww or M1 for 8 x + x = 180 or better 8(b) 32 3 B1 for angle DBC = 58 B1 for angle BCD = 90 8(c)(i) 24 2 B1 for angle PRQ = 24 8(c)(ii) 29.4 or 29.40 to 29.41 3 360 − 48 M2 for × 2 × π × 5.4 360 or B2 for answer (minor arc) 4.52 or 4.523 to 4.524… 48 or M1 for × 2 × π × 5.4 360

This question in 0580/41 May/June 2018

9 (a) A B C NOT TO 109° SCALE O 35° 28° D E A, B, C, D and E lie on the circle, centre O. Angle AEB = 35°, angle ODE = 28° and angle ACD = 109°. (i) Work out the following angles, giving reasons for your answers. (a) Angle EBD = … because … … … [3] (b) Angle EAD = … because … … [2] (ii) Work out angle BEO. Angle BEO = … [3] (b) In a regular polygon, the interior angle is 11 times the exterior angle. (i) Work out the number of sides of this polygon. … [3] (ii) Find the sum of the interior angles of this polygon. … [2]

13 marks

Mark scheme: 9(a)(i)(a) 62 and 3 B2 for 62 and one correct reason Isosceles [triangle] or B1 for 62 with no/wrong reason and or for angle EOD = 124 soi Angle at centre is twice angle at or for no/wrong angle with correct circumference oe reason 9(a)(i)(b) 62 and 2 2FT their (a)(i)(a) and correct reason [Angles in] same segment oe or B1FT for their (a)(i)(a) with no/wrong angle at centre is twice angle at reason circumference oe or for no/wrong angle with correct reason 9(a)(ii) 8 3 M2 for (180 –109) – 28 – 35 oe or M1 for [angle AED = ] 180 – 109 oe 9(b)(i) 24 3 x = ext angle B2 for [x = ] 15 isw or M1 for x + 11x = 180 oe 180( n − 2) 360 or for = × 11 [ n ] [ n ] 9(b)(ii) 3960 2 FT (their 24 – 2) × 180 dep on (b)(i) an integer and > 6 M1 for (their 24 – 2) × 180 oe or their 24 × 11 × their 15 oe or 11 × 360

This question in 0580/42 May/June 2018

5 North NOT TO SCALE A 120 m 150 m B 180 m C The diagram shows a triangular field, ABC, on horizontal ground. (a) Olav runs from A to B at a constant speed of 4 m/s and then from B to C at a constant speed of 3 m/s. He then runs at a constant speed from C to A. His average speed for the whole journey is 3.6 m/s. Calculate his speed when he runs from C to A. … m/s [3] (b) Use the cosine rule to find angle BAC. Angle BAC = … [4] (c) The bearing of C from A is 210°. (i) Find the bearing of B from A. … [1] (ii) Find the bearing of A from B. … [2] (d) D is the point on AC that is nearest to B. Calculate the distance from D to A. … m [2]

12 marks

Mark scheme: 5(a) 4.29 or 4.285 to 4.286 3 150 M2 for 450 120 180 − − 6.3 4 3 or M1 for [time =] 120 ÷ 4 or 180 ÷ 3 or 150 + 180 + 120 450 ÷ 3.6 or 3.6 = total time 5(b) 82.8 or 82.81 to 82.82 using cosine 4 150 2 + 120 2 − 180 2 M2 for rule 2 × 150 × 120 or M1 for 180 2 = 120 2 + 150 2 − 2 × 120 × 150 cos(...) 4500 A1 for oe 36000 5(c)(i) 127.2 or 127.1 to 127.2 or 127 1 FT 210 – their (b) 5(c)(ii) 307.2 or 307.1 to 307.2 or 307 2 FT 180 + their(c)(i) M1 for 180 + their (c)(i) 5(d) 15 or 14.99 to 15.04 2 dist M1 for cos ( their ( b ) ) = oe 120

This question in 0580/41 Oct/Nov 2019

4 B 107 m C NOT TO SCALE 158 m 132 m 86 m North 116° D A The diagram shows a field, ABCD, on horizontal ground. (a) There is a vertical post at C. From B, the angle of elevation of the top of the post is 19°. Find the height of the post. … m [2] (b) Use the cosine rule to find angle BAC. Angle BAC = … [4] (c) Use the sine rule to find angle CAD. Angle CAD = … [3] (d) Calculate the area of the field. … m2 [3] (e) The bearing of D from A is 070°. Find the bearing of A from C. … [2]

14 marks

Mark scheme: 4(a) 36.8 or 36.84… 2 h h 107 M1 for = tan19 or = oe 107 sin19 sin71 or better 4(b) 42.1 or 42.12… from cosine rule 4 158 2 + 132 2 − 107 2 M2 for [ cos BAC = ] 2 × 158 × 132 or M1 for implicit version A1 for [ cos BAC = ]30939 or 0.7417… 41712 4(c) 35.8 or 35.84… from sine rule 3 86 × sin116 M2 for [ = 0.58557...] 132 sin CAD sin116 or M1 for = oe 86 132 4(d) 9670 or 9669 to 9676 3 1 M2 for × 158 × 132 × sin ( their ( b ) ) oe 2 1 and × 86 × 132 × sin ( 64 − their ( c ) ) oe 2 or M1 for either area 4(e) 214.2 or 214.1… or 214 2 M1 for [180 +]70–their (c) oe

This question in 0580/43 Oct/Nov 2019

7 North B 80 m NOT TO A SCALE 72° 115 m C The diagram shows the positions of three points A, B and C in a field. (a) Show that BC is 118.1 m, correct to 1 decimal place. [3] (b) Calculate angle ABC. Angle ABC = … [3] (c) The bearing of C from A is 147°. Find the bearing of (i) A from B, … [3] (ii) B from C. … [2] (d) Mitchell takes 35 seconds to run from A to C. Calculate his average running speed in kilometres per hour. … km/h [3] (e) Calculate the shortest distance from point B to AC. … m [3]

17 marks

Mark scheme: 7(a) [BC2 =] 802 + 1152 – 2 × 80 × M1 115 cos 72 oe 118.06… A2 A1 for 13939… 7(b) 67.8 or 67.9 or 67.83 to 67.88 3 115 × sin72 M2 for [sin B =] oe 118.1 115 118.1 or M1 for = oe sin B sin72 7(c)(i) 255 3 B1 for bearing of B from A is 75 soi M1 for 180 + 75 oe 7(c)(ii) [00]7.2 2 M1 for their (c)(i) – their (b) –180 7(d) 11.8 or 11.82 to 11.83 3 M1 for 115 ÷ 35 oe M1 for their speed in m/s × 60 × 60 ÷ 1000 7(e) 76.1 or 76.08 to 76.09 3 distance M2 for = sin72 oe 80 or M1 for distance required is perpendicular to AC soi

This question in 0580/41 May/June 2020

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 0580/42 May/June 2020

8 (a) The interior angle of a regular polygon with n sides is 150°. Calculate the value of n. n = … [2] M (b) (i) K, L and M are points on the circle. KS is a tangent to the circle at K. KM is a diameter and NOT TO triangle KLM is isosceles. SCALE Find the value of z. L z° K S z = … [2] (ii) AT is a tangent to the circle at A. Find the value of x. x° NOT TO SCALE 27° 58° A T x = … [2] (iii) G y° NOT TO SCALE H F 108° J E F, G, H and J are points on the circle. EFG is a straight line parallel to JH. Find the value of y. y = … [2] (c) C N NOT TO SCALE D O A B M A, B, C and D are points on the circle, centre O. M is the midpoint of AB and N is the midpoint of CD. OM = ON Explain, giving reasons, why triangle OAB is congruent to triangle OCD. … … … … [3]

11 marks

Mark scheme: 8(a) 12 2 ( n − 2 ) × 180 360 M1 for 150 = or oe n 180 − 150 8(b)(i) 45 2 B1 for angles at M or K = 45 or angle at L = 90 8(b)(ii) 85 2 B1 for either angle in alt segment = 58 8(b)(iii) 72 2 B1 for either angle at J or H=108 or angle at F=72 8(c) OA = OB = OC = OD B1 Radii AB = CD B1 chords equidistant from centre are equal SSS implies congruent B1

This question in 0580/43 May/June 2020

11 (a) Find the size of an exterior angle of a regular polygon with 18 sides. … [2] (b) A 5.2 cm NOT TO SCALE B E 2.6 cm C D 6.75 cm In triangle ACD, B lies on AC and E lies on AD such that BE is parallel to CD. AE = 5.2 cm and ED = 2.6 cm. Calculate BE. BE = … cm [2] (c) Two solids are mathematically similar. The smaller solid has height 2 cm and volume 32 cm3. The larger solid has volume 780 cm3. Calculate the height of the larger solid. … cm [3] (d) P Q NOT TO SCALE N R S PQ is parallel to RS, PNS is a straight line and N is the midpoint of RQ. Explain, giving reasons, why triangle PQN is congruent to triangle SRN. … … … … [4]

11 marks

Mark scheme: 11(a) 20 2 360 16 × 180 M1 for or 180 − 18 18 11(b) 4.5 2 BE 5.2 M1 for = oe 6.75 5.2 + 2.6 11(c) 5.8[0] or 5.798 to 5.799 3 780 M2 for 2 × 3 oe 32 780 32 2 3 32 or M1 for 3 or 3 or = 32 780 l 3 780 11(d) QN = NR [given] B1 Two correct pairs of angles with B2 B1 for any correct pair of angles with reason reasons from or two correct pairs of angles with no/wrong reasons angle PQN = angle SRN alternate angle QPN = angle RSN alternate angle PNQ = angle SNR [vertically] opposite ASA [implies congruent] B1 dep on B1 B2

This question in 0580/41 May/June 2021

9 (a) NOT TO x cm SCALE ( x + 3)cm This rectangle has perimeter 20 cm. Find the value of x. x = … [3] (b) M y° NOT TO SCALE 20° This rhombus has perimeter 20 cm and angle y is obtuse. M is the midpoint of one of the sides. Find the value of y. y = … [5] (c) r cm NOT TO SCALE z cm 40° This sector of a circle has radius r and perimeter 20 cm. Find the value of z. z = … [6]

14 marks

Mark scheme: 9(a) 3.5 oe 3 M1 for 2(x + x + 3) = 20 oe M1 for correct ax = b for their linear equation 9(b) 116.8 or 116.83 to 116.85 nfww 5 5sin 20 M2 for sin p = 2.5 2.5 5 or M1 for = sin 20 sin p A1 for 43.2 or 43.15 to 43.17 M1dep for 180 – (20 + their 43.2) After 0 scored, SC1 for length of side = 5 9(c) 5.07 or 5.068 to 5.071 6 B3 for 7.41 or 7.412 to 7.413 40 or M2 for r + r + × 2 × π× r = 20 oe 360 40 or M1 for × 2 × π× r oe seen 360 M2 for 2 × 7.41 × sin 20 oe or 7.412 + 7.412 – 2(7.412) cos 40 oe 7.41sin 40 or oe sin70 or M1 for implicit version

This question in 0580/41 Oct/Nov 2021

6 (a) The interior angle of a regular polygon is 156°. Calculate the number of sides of this polygon. … [2] (b) NOT TO SCALE C O 52° A B A, B and C lie on a circle, centre O. Angle OBA = 52°. Calculate angle ACB. Angle ACB = … [2] (c) S R W 112° NOT TO SCALE Q T P P, Q, R, S and T lie on a circle. WSR is a straight line and angle WSP = 112°. Calculate angle PTR. Angle PTR = … [2] (d) K NOT TO SCALE O M F G H G, K and M lie on a circle, centre O. FGH is a tangent to the circle at G and MG is parallel to OH. Show that triangle GKM is mathematically similar to triangle OHG. Give a geometrical reason for each statement you make. … … … … … [4]

10 marks

Mark scheme: 6(a) 15 2 360 180 ( n − 2 ) M1 for or for = 156 oe 180 − 156 n 6(b) 38 2 B1 for AOB = 76 6(c) 68 2 B1 for RSP = 68 or RQP = 112 6(d) Two pairs of equal angles identified M3 M2 for one pair of equal angles identified with fully with fully correct reasons correct reasons KMG = 90 angle in semicircle and OGH = 90 angle between tangent and radius OR KMG = OGH alternate segment OR GOH = MGK alternate angles OR Angle FGM = angle GHO corresponding and angle FGM = GKM alternate segment and angle H = angle K or M1 for KMG = 90, angle in semicircle or OGH = 90, angle between tangent and radius Two or three pairs of angles equal A1 Dep on M3 with no incorrect work seen [so similar] oe

This question in 0580/42 Feb/March 2022

4 A regular 12-sided polygon has side length 6 cm. (a) Show that one interior angle of the polygon is 150°. [1] (b) The polygon is enclosed by a circle, centre O, so that each vertex touches the circumference of the circle. A B 6 cm NOT TO SCALE O (i) Show that the radius, AO, of the circle is 11.6 cm, correct to 1 decimal place. [3] (ii) Calculate (a) the circumference of the circle, … cm [2] (b) the perimeter of the shaded minor segment formed by the chord AB. … cm [2] (c) The regular 12-sided polygon is the cross-section of a prism of length 2 cm. Calculate the volume of the prism. … cm3 [3]

11 marks

Mark scheme: 4(a) (12  2)  180 1 (2  12  4)  90 [= 150] oe Accept [= 150] 12 12 360 or 180 – [= 150] 12 4(b)(i) 3 M2 3 oe M1 for  cos75 oe cos75 AO or or 6sin75 r 6  sin30 sin75 sin30 11.59… A1 4(b)(ii)(a) 72.8 or 72.9 or 72.82 to 72.89… 2 M1 for 2   11.6 4(b)(ii)(b) 12.1 or 12.06 to 12.08 2 M1 for [6 +] their (b)(ii)(a) ÷ 12 oe 4(c) 806 or 807 or 805.9 to 807.4 3 B2 for 402.9… to 403.7 OR 1 M2 for 6 11.6  sin75  12  2 oe 2 1 or M1 for 6 11.6  sin75 [k ] oe 2

This question in 0580/43 May/June 2022

4 (a) North 114° A NOT TO SCALE C B A, B and C are three towns and the bearing of C from A is 114°. B is due south of A and AC = BC. Calculate the bearing of B from C. … [3] (b) R S 74° NOT TO SCALE Q 58° 27° M P N P, Q, R and S lie on a circle. MPN is a tangent to the circle at P. Angle MPS = 58°, angle PSR = 74° and angle QPN = 27°. (i) Find angle PRS. Angle PRS = … [1] (ii) Find angle PQR. Angle PQR = … [1] (iii) Find angle RPQ. Angle RPQ = … [2] (c) N C B NOT TO SCALE O 34° M A T A, B and C lie on a circle, centre O, with diameter AC. TAM and TBN are tangents to the circle and angle ATO = 34°. Using values and geometrical reasons, complete these statements to show that CB is parallel to OT. In triangles AOT and BOT, OT is common. Angle OAT = angle OBT = 90° because … … AT = BT because … … Triangle AOT is congruent to triangle BOT because of congruence criterion … Angle AOT = angle BOT = 56° because angles in a triangle add up to 180°. Angle BOC = … ° because … Angle OBC = … ° because … … CB is parallel to OT because … [6]

13 marks

Mark scheme: 4(a) 246 3 B2 for BCS(outh) = 66 or BCA = 48 and ACN(orth) = 66 or BCW(est) = 24 or ACS(outh) = 114 or B1 for ABC = 66 or BAC = 66 or BCA = 48 or ACN(orth) = 66 4(b)(i) 58 1 4(b)(ii) 106 1 4(b)(iii) 47 2 B1 for PRQ = 27 or B1FT for SPR, either = 48 or = 106 – their (b)(i) or B1FT for RPQ = their (b)(i) – 11 4(c) Radius perpendicular to tangent 1 Tangents to circle from a/same point oe 1 RHS 1 68 angles on a [straight] line add up/sum to 1 180 oe 56 [base angles of] isosceles triangle 1 OBC = BOT Alternate angles 1 Angles and reason required and dependent on OBC and BOT correct

This question in 0580/43 May/June 2023

4 (a) Find the size of one interior angle of a regular 10-sided polygon. … [2] (b) A B NOT TO x° SCALE w° 75° C z° E y° 25° 20°20° F D G The points A, B, C, D and E lie on a circle. FG is a tangent to the circle at D. EB is parallel to DC. Find the value of each of w, x, y and z. w = … x = … y = … z = … [5]

7 marks

Mark scheme: 4(a) 144 2 360 180(10 − 2) M1 for 180 – or oe 10 10 4(b) w = 20 5 B1 for w x = 20 B1FT for x = their w y = 60 z = 45 B2FT for y = 80 – their w or B1 for angle BDC = 20 FT their w or angle ADE = 55 or angle CAD = 25 B1FT for z = 25 + their w or 105 – their y

This question in 0580/43 Oct/Nov 2023

2 (a) 38° NOT TO SCALE a° b° The diagram shows a straight line intersecting two parallel lines. Find the value of a and the value of b. a = … b = … [2] (b) Calculate the interior angle of a regular 12-sided polygon. … [2] (c) N NOT TO SCALE f ° P O g° 56° A B M The diagram shows a circle, centre O. The points M, N and P lie on the circumference of the circle. AMB is a tangent to the circle at M. Find the value of f and the value of g. f = … g = … [3] (d) NOT TO SCALE 24° k ° 27° The diagram shows a cyclic quadrilateral. Find the value of k. k = … [2]

9 marks

Mark scheme: 2(a) 142 2 B1 for each 142 FT angle b = their angle a 2(b) 150 2 360 M1 for oe isw 12 or 180  12  2  oe isw 2(c) 56 B1 34 B2 M1 for angle at centre = 2 × their 56 oe soi or for angle OMB = 90 oe soi 2(d) 51 2 B1 for opp angle = 129 soi

This question in 0580/42 May/June 2024