Similarity — Paper 4 · IGCSE Mathematics - International 0607

2 (a) D A C 68° NOT TO E SCALE B In the diagram, ABC is a triangle and AB is parallel to DE. Angle BCA = 68˚and DE = DC. (i) Find angle BAC. Angle BAC = … [2] (ii) scalene equilateral isosceles right-angled Choose one word from the list to complete the statement. Triangle ABC is … [1] (b) Calculate the interior angle of a regular 20 sided polygon. … [3] (c) C B P NOT TO SCALE A Q R In the diagram, angle A = angle P and angle B = angle Q. (i) Explain why angle C = angle R. … [1] (ii) AB = 8 cm, AC = 5 cm, BC = 9 cm and PR = 3 cm. (a) Complete the statement. Triangle ABC is … to triangle PQR [1] (b) Calculate QR. QR = … cm [2]

10 marks

Mark scheme: 2(a)(i) 44 2 M1 for [angle BAC or DEC =] 180 – 2 × 68, soi by angle CDE = 44 or M1 for angle BAC = their angle CDE 2(a)(ii) isosceles 1 2(b) 162 3 360 180 × (20 − 2) M2 for 180 – or 20 20 360 or M1 for or 180 × (20 – 2) 20 2(c)(i) Angle sum of triangle oe 1 2(c)(ii)(a) similar 1 2(c)(ii)(b) 5.4 2 5 9 M1 for = oe 3 QR

This question in 0607/43 May/June 2017

5 A NOT TO SCALE E D F B C ABC is a triangle and BCFD is a parallelogram. 1 1 AD = AB and A E = A C . 3 3 AB = 6p and AC = 6q . (a) Find an expression, in terms of p and/or q, for (i) BC, … [1] (ii) DE, … [2] (iii) FC, … [1] (iv) BE. … [2] (b) The area of triangle ADE is 24 units2. (i) Find the area of triangle ABC. … units2 [2] (ii) Find the area of triangle EFC. … units2 [3]

11 marks

Mark scheme: 5(a)(i) –6p + 6q oe 1 5(a)(ii) –2p + 2q oe 2 FT their (a)(i) ÷ 3 provided in form ap +bq B1 for –2p + kq or for kp + 2q JJJG JJJG M1 for AD = 2p oe or AE = 2q or correct route 5(a)(iii) 4p cao 1 5(a)(iv) –6p +2q oe 2 B1 for –6p + kq or for kp + 2q M1 for a correct route 5(b)(i) 216 2 2  1  2 M1 for   or 3 oe soi  3  5(b)(ii) 96 3 2  1  2 M2 for   or 2 oe soi  2  or M1 for triangle EFC is similar to triangle EDA soi

This question in 0607/43 Oct/Nov 2017

6 (a) NOT TO SCALE 12 cm 40 cm (i) The rectangle can be made into a hollow cylinder with height 40 cm. (a) Show that the radius of this cylinder is 1.910 cm, correct to 3 decimal places. [2] (b) Calculate the volume of this cylinder. … cm3 [2] (ii) The rectangle can also be made into a hollow cylinder with height 12 cm. Calculate the difference between the volumes of this cylinder and the cylinder in part (i). Give your answer correct to the nearest 10 cm3. … cm3 [4] (b) A model of a car is mathematically similar to the actual car. The volume of the model is 75 cubic centimetres and the volume of the actual car is 4.8 cubic metres. The scale is model : actual = 1 : n . Find the value of n. n = … [4]

12 marks

Mark scheme: 6(a)(i)(a) 2πr = 12 oe M1 1.9096 to 1.9099 A1 6(a)(i)(b) 458 or 457.9 to 458.5 2 M1 for π × 1.91[0]2 × 40 6(a)(ii) 1070 4 B3 for volume of other cylinder 1530 or 1527 to 1529. …  40  2 or M2 for π × × 12    2π  or M1 for 40 ÷ ( 2 π) oe 6(b) 40 4 3 4.8 × 100 M3 for 3 oe 75 figs 48 figs 75 or M1 for 3 oe or 3 oe figs 75 figs 48 and M1 for 4.8 × 1003 or 75 ÷ 1003 oe

This question in 0607/41 May/June 2018

5 16 m A P D NOT TO SCALE 30 m S Q 18 m 12 m B R C 24 m 40 m In the diagram, ABCD is a rectangle. (a) Find PS. PS = … m [2] (b) Find angle BRS. Angle BRS = … [2] (c) Find the perimeter of PQRS. … m [3] (d) Find the shaded area. … m2 [3] (e) Explain why triangle ASP is similar to triangle BSR. … … [2]

12 marks

Mark scheme: 5(a) 20 2 2 2 M1 for 16 + ( 30 − 18 ) 5(b) 36.9 or 36.86 to 36.87 2 18 M1 for tan[ ] = oe 24 5(c) 100 3 2 2 M2 for 2 × (their (a) + 18 + 24 ) oe or M1 for 182 + 24 2 or RS = 30 or PQ = 30 seen 5(d) 576 3 M2 for (40 × 30) − 2 × (0.5 × 18 × 24) − 2 × (0.5 × 16 × 12) oe or M1 for any correct and relevant area 5(e) Correct explanation 2 B1 for partial explanation e.g. ratio of two sides the same, with names or numbers given.

This question in 0607/42 May/June 2018

7 In this question, all lengths are measured in millimetres. 44 A small plastic cup, A, is shown in this diagram. 55 A 28 5 These plastic cups are stacked as shown in the diagram. 5 55 (a) Find the height of a stack of 8 of these cups. … mm [2] (b) Find the number of these cups in a stack that has a total height of 105 mm. … [2] (c) A similar cup, B, has base diameter 42 mm. Find the height of this cup. … mm [2] (d) 2r + 2a h 2r 2 2 r h (3r + 3ar + a ) The formula for the volume of a similar cup is V = . 3 (i) For cup A, show that a = 8 mm. [2] (ii) Find the volume of cup A. … mm3 [2] (iii) Find the volume of cup B. … mm3 [3] 2 2 r h (3r + 3ar + a ) (iv) Rearrange V = to make h the subject. 3 h = … [2]

15 marks

Mark scheme: 7(a) 90 2 M1 for 55 + 5k, k = 7 or 8 7(b) 11 2 M1 for 55 + 5(n – 1) = 105 or better 105 − 55[ + 5] or soi by 10 5 7(c) 82.5 2 42 [ ] M1 for = oe 28 55 7(d)(i) 28 + 2a = 44 oe or 44 – 28 oe seen M1 44 − 28 A1 2a = 16 or oe[= 8] 2 7(d)(ii) 56 900 or 56 900 to 56 920 2 π 2 2 M1 for × (3 × 14 + 3 × 14 × 8 + 8 ) [× 55] 3 7(d)(iii) 192 000 or 192 000 to 192 200 3 3  42  M2 for their (d)(ii) ×   28   42  3  28  3 or M1 for   or    28   42  OR M2 for π 2 2 × their (c) × 3 × 21 + 3 × (8 × 1.5) × 21 + ( 8 × 1.5 ) ( ) 3 or B1 for a =12 7(d)(iv) 3V 2 M1 for 3V = π h 3r 2 + 3ar + a 2 [ h = ] ( ) π(3r 2 + 3ar + a 2 ) V h 3V or = or πh = π 3r 2 + 3ar + a 2 3 3r 2 + 3ar + a 2 ( )

This question in 0607/42 May/June 2018

7 D NOT TO SCALE A 105° E F 0.6 m 0.8 m 1.2 m B C ABCDEF is a solid triangular prism. (a) Calculate the volume of the prism. … m3 [3] (b) Calculate the total surface area of the prism. … m2 [5] (c) ABCDEF is made of metal and has a mass of 2170 kg. It is melted down and made into prisms similar to ABCDEF. Each of these prisms has a mass of 2.17 kg. Calculate the total surface area of each of these smaller prisms. … m2 [3]

11 marks

Mark scheme: 7(a) 0.278 or 0.2781 to 0.2782 3 M2 for 0.5 × 0.6 × 0.8 × sin105 × 1.2 oe or M1 for 0.5 × 0.6 × 0.8 × sin105 7(b) 3.48 to 3.49 5 2 2 M2 for 0.6 + 0.8 − 2 × 0.6 × 0.8 × cos105 or M1 for 0.6 2 + 0.8 2 − 2 × 0.6 × 0.8 × cos105 A1 for 1.12 or 1.117... M1 for their 1.117 × 1.2+ 2 × their area of ABC + 0.6 × 0.12 + 0.8 × 1.2 7(c) 0.0348 to 0.0349 3 FT their (b) ÷ 100 2170  2170  2 M2 for their (b) ÷ 3 ×  3  oe   2.17  2.17  2170 2.17 3 2 or M1 for or or k ( ) 2.17 2170 implied by their (b) ÷ 1000

This question in 0607/43 May/June 2018

11 NOT TO A B SCALE X Y D C A, B, C and D are points on the circle. ABX, CDX, AYD and BYC are straight lines. (a) (i) Explain why triangle ADX is similar to triangle CBX. … … … [2] (ii) Use part (a)(i) to show that XA # XB = XC # XD [1] (b) XB = 6 cm, DC = 5 cm and XD = 7 cm. Calculate the length AB. … cm [2] (c) Find the value of these fractions. Area of triangle ADX (i) Area of triangle CBX … [1] Area of triangle AYB (ii) Area of triangle CYD … [1] Question 12 is printed on the next page.

7 marks

Mark scheme: 11(a)(i) Angle X is common oe 2 B1 for each ∠XCB = ∠XAD (angles in same If 0 scored SC1 for 2 pairs of angles without segment) oe reasons 11(a)(ii) XA XD 1 = oe XC XB 11(b) 8 2 M1 for 12 × 7 = XA × 6 soi (implied by 14) 11(c)(i) 49 1 oe 36 11(c)(ii) 64 1 oe 25

This question in 0607/43 May/June 2018

4 NOT TO SCALE 3 cm l cm The diagram shows a solid made from a cylinder and two hemispheres. The radius of the cylinder and each hemisphere is 3 cm. The total volume of the solid is 144r cm3. (a) The length of the cylinder is l cm. Find the value of l. l = … [3] (b) The solid is made of steel. 1 cm3 of steel has a mass of 7.8 g. Calculate the mass of the solid. Give your answer in kilograms. … kg [2] (c) The solid is melted down and made into 20 cubes each of side length 2.8 cm. Calculate the volume of steel not used for the cubes as a percentage of the 144r cm3. … % [3] (d) A solid that is mathematically similar to the original solid has a volume of 18r cm3. Find the radius of the new cylinder. … cm [3]

11 marks

Mark scheme: 4(a) 12 cao final answer 3 B2 for 11.98 to 12.02 2 2 3 or M1 for π × 3 × l + 2 × × π × 3 [ = 144π] 3 oe 4(b) 3.53 or 3.528 to 3.529... 2 M1 for 144 π × 8.7 soi by figs 353 or 3528 to 3529 4(c) 2.95 or 2.96 or 2.950 to 2.963... 3 144 π − 20 × 8.2 3 M2 for [× 100 ] 144 π 20 × 8.2 3 or × 100 oe 144 π 3 20 × 8.2 3 or M1 for 144 π − 20 × 8.2 or oe 144 π 4(d) 1.5 oe cao final answer 3 B2 for 1.498 to 1.502 18π or M2 for 3 × 3 oe 144 π 18π 144 π or M1 for 3 or 3 oe or better 144 π 18π  3 3 144π or for = oe    x  18π

This question in 0607/42 May/June 2019

9 NOT TO SCALE h cm r cm 18 cm 26 cm 24 cm r cm 14 cm The diagram shows three solids, a prism, a sphere and a cone. The radius of the sphere is equal to the base radius of the cone. The volume of each solid is the same. (a) Show that the volume of the prism is 7392 cm3. [3] (b) A similar prism has a volume of 924 cm3. The length of the original prism is 24 cm. Find the length of this similar prism. … cm [3] (c) Find the value of r. r = … [2] (d) Find the value of h. h = … [2] (e) When exact values of h and r are used, h = 4 r. Find, in terms of r, an exact expression for the curved surface area of the cone. Give your answer in its simplest form. … [3]

13 marks

Mark scheme: 9(a) [(14 × 18) + 0.5 × 14 × 8] × 24 oe M3 i.e. area × length or 18 × 14 × 24 + 0.5 × 14 × 8 × 24 oe volume + volume leading to 7392 M2 for 14 × 18 + 0.5 × 14 × 8 or M1 for 14 × 18 or 0.5 × 14 × 8 or 0.5 × (18 + 26) × 7 9(b) 12 cao 3 7392 M2 for 24 ÷ 3 oe 924 7392 or M1 for 3 soi 924 9(c) 12.1 or 2.08… 2 3 3 7392 M1 for r = × oe 4 π 9(d) 48.2 or 48.3 or 48.4 2 3 × 7392 M1 for h = or 48.20 to 48.37… 2 π × (their12.1) 9(e) πr 2 17 final answer 3 M2 for πr r 2 + (4 r ) 2 or M1 for l 2 = r 2 + (4 r ) 2 If 0 scored , SC1 for πr 2 5

This question in 0607/42 Oct/Nov 2019

9 5 cm NOT TO SCALE 24 cm 12 cm 16 cm The diagram shows a solid made from a cuboid and a solid hemisphere. The cuboid measures 12 cm by 16 cm by 24 cm. The hemisphere has radius 5 cm. (a) Find (i) the volume of the solid, … cm3 [3] (ii) the volume of a similar solid where the radius of the hemisphere is 3 cm. … cm3 [2] (b) Find (i) the total surface area of the original solid, … cm2 [3] (ii) the total surface area of a similar solid where the radius of the hemisphere is 6 cm. … cm2 [2]

10 marks

Mark scheme: 9(a)(i) 4870 or 4869 to 4870 3 M1 for 24 × 16 × 12 1 4 3 M1 for × × π × 5 2 3 9(a)(ii) 1050 or 1051 to 1052 nfww 2 3  3  M1 for   oe  5  1 4 3 or × × π × 3 + 14.4 × 9.6 × 7.2 2 3 9(b)(i) 1810 or 1806 to 1807 3 M1 for 24 × 16 × 2 + 24 × 12 ×2 + 16 × 12 × 2 [–π × 52] M1 for 0.5 × 4 × π × 5 2 9(b)(ii) 2600 or 2610 or 2600 to 2606. ... 2 2  6  nfww M1 for   oe soi  5  or 0.5 × 4 × π × 62 + 28.8 × 19.2 × 2 + 28.8 × 14.4 × 2 + 19.2 × 14.4 × 2 – π × 62

This question in 0607/41 Oct/Nov 2020

2 (a) Find the size of one interior angle of a regular polygon with 45 sides. … [3] (b) B A 35° NOT TO 75° C SCALE T D In the diagram, A, B, C and D lie on the circle. TA is a tangent to the circle at A. Angle TAD = 75˚ and angle DAC = 35˚. Find (i) angle ACD, Angle ACD = … [1] (ii) angle ABC. Angle ABC = … [2] (c) C E NOT TO SCALE A D B In the diagram, DE is parallel to BC. (i) Complete the statement. Triangle ADE is … to triangle ABC. [1] (ii) AE = 6 cm, EC = 3 cm and DB = 2 cm. Calculate the length of AD. AD = … cm [3] (iii) The area of triangle ADE is 9 cm2. Calculate the area of triangle ABC. … cm2 [2]

12 marks

Mark scheme: 2(a) 172 3 360 180 × (45 − 2) M2 for 180 − or for 45 45 360 or M1 for (implied by 8) 45 or for 180 × (45 – 2) (implied by 7740) 2(b)(i) 75 1 2(b)(ii) 110 2 B1 for angle CAT = 110 or angle CDA = 70 or M1 for 180 – their angle CDA. 2(c)(i) similar 1 2(c)(ii) 4 3 9 AD + 2 M2 for = oe 6 AD AD 6 or M1 for = oe AB 9 2(c)(iii) 20.25 2 2 2  3   2  M1 for   or   oe seen  2   3 

This question in 0607/43 Oct/Nov 2020

10 B C NOT TO A SCALE h r Cone A has radius r and perpendicular height h. Cone B is mathematically similar to cone A. Solid C is formed by removing cone A from cone B. The ratio height of cone A : height of cone B = 2 : 3. (a) Find the ratio volume of cone A : volume of solid C. … : … [3] (b) Cone A has radius 4 cm and height 10 cm. Calculate the total surface area of solid C. … cm2 [8] Question 11 is printed on the next page.

11 marks

Mark scheme: 10(a) 8 : 19 oe 3 M1 for [Vol A : Vol B =] 23 : 33 oe M1 for [Vol C =] 27k – 8k k any variable OR 1  3r ×2 3h M1 for π   3  2  2 1 2 1  19  2 M1 for [VA : VC =] πr h :   πr h 3 3  8  10(b) 503 or 502.6 to 502.8 8 3 3 M1 for × 4 oe or 10 2 2× 3 32 or × their l oe if their l is from Pythagoras or 2 2 2 M2 for 4 2 + 10 2 or (their R ) 2 + (their H ) 2 or M1 for 4 2 + 10 2 or (their R)2 + (theirH)2 M1 for π× 4 × 116 3 32 M1 for π× 6 × 116 or 2 × π× 4 116 2 2 M2 for CSAa + CSAb + π × (their R)2 – π × 42 oe or M1 for for CSAa + CSAb or π × (their R)2 – π × 42 oe

This question in 0607/42 Feb/March 2021

9 5 cm NOT TO SCALE 12 cm The diagram shows a cup in the shape of a cone. (a) Calculate the curved surface area of the cup. … cm2 [3] (b) The cup is filled with water. A metal sphere of radius r cm is lowered into the cup. The top of the sphere is level with the surface of the water. NOT TO SCALE r cm (i) Use similar triangles to show that r = 3.33 cm correct to 3 significant figures. [3] (ii) Calculate the volume of the water in the cup. … cm3 [3]

9 marks

Mark scheme: 9(a) 204 or 204.2... 3 2 2 M2 for π× 5 × 5 + 12 ( ) or M1 for 52 + 122 (implied by 13) 9(b)(i) r 5 M1 r 5 = oe = 12 − r their13 13 − 5 12 r(their 13) = 5(12 – r) M1 M1 dep on first M1 for 12 r = 5(13 − 5) 1 10 A1  Completion to r = 3.3 or 3 or or 3 3 3.333... with no errors 9(b)(ii) 159 or 159.0 to 159.5 3 1 2 M1 for × π× 5 × 12 3 4 3 M1 for × π× 3.33 3

This question in 0607/42 May/June 2021

10 NOT TO 4 cm SCALE 16 cm 12 cm The diagram shows a solid made from a cylinder, a hemisphere and a cone, each with radius 4 cm. The cylinder has length 16 cm. The slant height of the cone is 12 cm. (a) Find the volume of the solid. … cm3 [5] (b) Show that the total surface area of the solid is 208 r cm2. [4] (c) A mathematically similar solid has a total surface area of 468 r cm2. Find the radius of the cylinder in this solid. … cm [3]

12 marks

Mark scheme: 10(a) 1130 or 1127 to 1128 5 M1 for π × 16 × 4 2 1 4 3 M1 for × × π × 4 2 3 M1 for 12 2 − 4 2 or better 1 2 M1 for × π × 4 ×their h 3 10(b) 2 × π × 16 × 4 M1 1 2 M1 × 4 × π × 4 oe 2 π × 12 × 4 M1 32π + 128π + 48π [=208π] B1 10(c) 6 3 468 M2 for × 4 oe 208 468 208 or M1 for or 208 468 or  4 =2 208π oe    r  468π

This question in 0607/43 Oct/Nov 2021

7 Abbi makes wooden boards in three sizes, small, medium and large. They are all cuboids. The medium board has height 2 cm, width 23 cm and length 50 cm. (a) Calculate the volume of the medium board. … cm3 [2] (b) The small board is mathematically similar to the large board. The small board has a volume of 287.5 cm 3and a height of 1.15 cm. The large board has a volume of 18400 cm 3. (i) Find the height of the large board. … cm [3] (ii) Is the medium board mathematically similar to the large board? Explain how you decide. … because … … [3]

8 marks

Mark scheme: 7(a) 2300 2 M1 for 2 × 23 × 50 oe 7(b)(i) 4.6 3 18400 M2 for 1.15  3 oe 287.5 h 3 18400 18400 or M1 for  or 3 1.153 287.5 287.5 287.5 or 3 seen 18400 7(b)(ii) 2 : their (b)(i) or their (a) : 18400 soi M1 FT their figures showing comparison of length ratio or M1 volume ratio not similar A1 Dep on M1M1

This question in 0607/41 May/June 2022

11 (a) A pyramid has a square base with sides of length 9 cm and vertical height h cm. Find an expression, in terms of h, for the volume of the pyramid. … cm3 [1] (b) A NOT TO 10 cm SCALE B C a cm h cm D E 9 cm ADE is an isosceles triangle. BC is parallel to DE, BC = a cm and DE = 9 cm. The vertical height of triangle ADE is h cm and the vertical height of triangle ABC is 10 cm. 90 Show that a = h [1] (c) A square-based pyramid with base of side 9 cm and vertical height h cm contains some water. When the pyramid is placed on level ground the surface of the water is 10 cm below the vertex of the pyramid (see Diagram 1). When the pyramid stands vertically on its vertex, the surface of the water is 1 cm below the base of the pyramid (see Diagram 2). 9 cm 10 cm 1 cm b cm a cm h cm h cm 9 cm Diagram 1 Diagram 2 (i) Use Diagram 1 to find an expression, in terms of a and h, for the volume of the water. … cm3 [1] (ii) Use Diagram 2 to find an expression, in terms of b and h, for the volume of the water. … cm3 [1] (iii) Show that h 3 - 1000 = ( h - 1) 3 . [3] (iv) The equation h 3 - 1000 = ( h - 1) 3 simplifies to h 2 - h - 333 = 0 . Use a graphical method to find the value of h. h = … [2] Question 12 is printed on the next page.

9 marks

Mark scheme: 11(a) (1/3)  92  h or 181h or 27h oe 1 3 11(b) 10 a 9 a  90  1  ,  a  oe   h 9 h 10  h  11(c)(i) 1 2 1 2 1 FT their (a) V   9 h  a  10 oe isw 3 3 11(c)(ii) 1 2 1 V   b ( h  1) oe isw 3 2 2 M211(c)(iii) h 1 1 90 1 h     1  1 b M1 for oe 9 2 h   10  9 2  h  1     h 9 3 3 3 h h     oe 3 A1 No errors or omissions seen 1000  h  1 3 3 h   , h  1000  ( h  1) h 2 h 2 11(c)(iv) 18.8 or 18.75 to 18.76 cao 2 B1 for 18.8 or 18.75 to 18.76 and negative root as final answers M1 for [quadratic/cubic]sketch(es)

This question in 0607/42 May/June 2022

4 A NOT TO SCALE D X O 52° P C B A, B, C and D lie on a circle, centre O. AP and BP are tangents to the circle. AC and BD intersect at X. Angle APB = 52° . (a) Complete the statement. Angle OAP = 90° because … … [1] (b) Find (i) angle AOB, Angle AOB = … [1] (ii) angle OAB, Angle OAB = … [1] (iii) angle ACB. Angle ACB = … [1] (c) ABCD is a trapezium with AB parallel to DC. (i) Write down a triangle that is similar to triangle ABX. Triangle … [1] (ii) The length CD = 4 cm and the length AB = 12 cm . Find the ratio area CDX : area ABX. area CDX : area ABX = … : … [1]

6 marks

Mark scheme: 4(a) Tangent [and] radius or diameter[= 90] 1 4(b)(i) 128 1 4(b)(ii) 26 1 180 − their ( i ) FT 2 4(b)(iii) 64 1 their ( i ) FT 2 4(c)(i) CDX 1 4(c)(ii) 1 : 9 oe 1

This question in 0607/41 Oct/Nov 2022

10 15 cm NOT TO SCALE A 5 cm 8 cm B C 11 cm Triangle ABC is the cross-section of a prism of length 15 cm. AB = 5 cm , AC = 8 cm and BC = 11 cm . (a) Show that the area of triangle ABC = 18.33 cm 2 correct to 2 decimal places. [4] (b) Find the volume of the prism. … cm3 [1] (c) Find the total surface area of the prism. … cm2 [2] (d) A mathematically similar prism has a volume of 500 cm 3. Calculate the total surface area of this similar prism. Give your answer correct to 2 significant figures. … cm2 [3]

10 marks

Mark scheme: 10(a) 5 2 + 8 2 − 112 M2 M1 for 112 = 52 + 82 – 2 × 5 × 8 × cosA [cos A =] oe 2 5 8 5 2 + 112 − 8 2 or [cos B =] oe or 82 = 52 + 112 – 2 × 5 × 11 × cosB 2 5 11 112 + 8 2 − 5 2 or [cos C =] oe or 52 = 112 + 82 – 2 × 11 × 8 × cosC 2  11  8 0.5 × 5 × 8 × sin(their A) oe M1 or 0.5 × 5 × 11 × sin(their B) oe or 0.5 × 11 × 8 × sin(their C) oe 18.330... A1 Dep on no errors seen and on M2 and M1 awarded 10(b) 275 or 274.9... 1 10(c) 397 or 396.6 to 396.7 2 M1 for 8 × 15 + 11 × 15 + 5 × 15 + 2 × 18.33 10(d) 590 cao 3 2  500  3 M2 for  ( their (c))   oe  their (b)  1  500  3 or M1 for   oe soi  their(b)   their (c) 3  their (b)  2 or = oe      A   500 

This question in 0607/42 Oct/Nov 2022

7 (a) N D C NOT TO SCALE M A B V The diagram shows a shape AVBCD. ABCD is a square of side 12 cm. M is the mid-point of AB and N is the mid-point of DC. Triangle AVB is isosceles with AV = VB = 10 cm. The arc CD is part of a circle with centre M. (i) Calculate angle CMN. Angle CMN = … [2] (ii) Calculate the length of CM. CM = … cm [2] (iii) Calculate the perimeter of the shape AVBCD. … cm [3] (iv) Calculate the area of the shape AVBCD. … cm2 [5] (b) Two solids are mathematically similar with volumes 240 cm3 and 810 cm3. The surface area of the larger solid is 558 cm2. Calculate the surface area of the smaller solid. … cm2 [3]

15 marks

Mark scheme: 7(a)(i) 26.6 or 26.56 to 26.57 2 6 M1 for tan = oe 12 7(a)(ii) 13.4 or 13.41 to 13.42 2 M1 for 62  122 oe 7(a)(iii) 56.4 to 56.5 3 M2 for 2  10  2  12  2 theirCMN their MC 360 or M1 for 2 theirCMN their MC oe 360 7(a)(iv) 203 to 205 5 M1 for 2 theirCMN their MC)2 360 1 2 2 M2 for  10 6 oe 2 or M1 for 102 – 62 B1 for 144 or 72 or 36 7(b) 248 3 2  240  M2 for 558  3  oe    810   240  2  810  2 or M1 for  3  or  3       810   240   810  2  558  3 or for  oe      240   area 

This question in 0607/41 May/June 2023

5 B NOT TO 50 cm SCALE 55 cm A 64 cm C D E In the diagram, AB is parallel to ED. ACD and BCE are straight lines. AB = 50 cm, BC = 55 cm and AC = 64 cm . (a) Show that angle ACB = 49.0° correct to one decimal place. [3] (b) Use the sine rule to calculate angle CAB. Angle CAB = … [3] (c) Calculate the area of triangle ABC. … cm2 [2] 2(d) AC = AD 3 Calculate the area of triangle CDE. … cm2 [2]

10 marks

Mark scheme: 5(a) 55 2  64 2  50 2 M2 M1 for cos ACB  2 2 2 2  55  64 50  55  64 2 55  64  cos ACB  ACB   48.97...   49.0  A1 5(b) 55sin49 55 50 sin CAB  M2 M1 for  oe 50 sin CAB sin 49 56.1 or 56.07 to 56.12 B1 5(c) 1330 or 1327 to 1328… 2 M1 for 0.5  64  55  sin49 oe 5(d) 331.9 to 333 2 FT their 5(c) ÷ 4 M1 for 0.5 2 or 2 2 1 or  32  27.5sin 49.0 oe 2

This question in 0607/43 May/June 2023

8 P NOT TO SCALE 140° O 9 cm Q The diagram shows the sector of a circle with radius 9 cm and sector angle 140°. (a) Calculate the length of the arc PQ. … cm [2] (b) Calculate the area of the sector. … cm2 [2] (c) The sector is the cross-section of a solid of length 20 cm. Calculate the total surface area of the solid. … cm2 [4] (d) Another solid is mathematically similar to the solid in part (c). The radius of the sector in this solid is 10 cm. Calculate the total surface area of this solid. … cm2 [2]

10 marks

Mark scheme: 8(a) 22[.0] or 21.99... 2 140 M1 for 2 π  9 oe 360 8(b) 99[.0] or 98.96 to 98.97... 2 140 2 M1 for π 9 oe 360 8(c) 998 or 997.7 to 998.0 4 M1 for their (a) × 20 M1 for their (b) × 2 M1 for [2 ×] 9 × 20 8(d) 1230 or 1231 to 1232...nfww 2 2  10  FT their (c)    9   10  2  9  2 M1 for   or    9   10 

This question in 0607/41 Oct/Nov 2023

7 (a) F 30° E NOT TO SCALE G 130° A B C D ABCD is a straight line and EC and BF meet at G. BE is parallel to CF and GF = CF . Angle ABE = 130° and angle BFC = 30° . Find (i) angle FCD Angle FCD = … [2] (ii) angle FBC Angle FBC = … [1] (iii) angle BGE. Angle BGE = … [2] (b) B NOT TO SCALE C X A D A, B, C and D are points on the circle. AC and BD meet at X. (i) Show that triangles AXB and DXC are similar. Give a reason for each statement you make. … … … … … [2] (ii) AX = 5 cm, XC = 2 cm and XD = 4 cm. Find the length of BD. BD = … cm [3]

10 marks

Mark scheme: 7(a)(i) 50 2 B1 for angle BCF = 130 or angle EBC = 50 soi by angle EBG = 30 and angle GBC = 20 or angle FCG = 75 and angle GCB = 55 7(a)(ii) 20 1 7(a)(iii) 75 2 180 − 30 M1 for 2 7(b)(i) 2 from 2 B1 for 2 pairs correct with no/incorrect Angle AXB = Angle DXC reasons and conclusion [Vertically] opposite angles or for one pair correct with reason. Angle ABX = Angle DCX Angles in same segment. Angle BAX = Angle CDX Angles in same segment. And conclusion AA[A] 7(b)(ii) 6.5 3 B2 for BX = 2.5 ... 5 or M1 for = oe 2 4

This question in 0607/42 Feb/March 2024

9 NOT TO SCALE 60 cm 40 cm The diagram shows a solid cone with base radius 40 cm and slant height 60 cm. (a) Find the volume of the cone. … cm3 [3] (b) Show that the total surface area of the cone is 4000 r cm 2 . [2] (c) A mathematically similar cone has a surface area of 1000 r cm 2 . Show that the radius of this cone is 20 cm. [2] (d) A cone with radius 20 cm is removed from the top of the cone with radius 40 cm to leave a solid. Calculate the surface area of the remaining solid. … cm2 [3]

10 marks

Mark scheme: 9(a) 74 900 or 74 929 to 74 941.1 3 1 2 2 2 M2 for  π  40  60  40 oe 3 or M1 for 602 – 402 [=2000] oe 9(b) π × 402 + π × 40 × 60 M2 M1 for π × 402 or π × 40 × 60 = 4000π with no errors 9(c) Ratio areas = 4000π : 1000π M1 implies Ratio sides = 2 : 1 oe [r=] 40 × 0.5 = 20 oe A1 ALTERNATIVE (M2) 1000π 4000π M1 for or oe 1000π 4000π 1000π 40  oe 4000π 2  40 4000π  oe = 20 with no errors or    x  1000π 9(d) 11 900 or 11 930 to 11 940 or 3800π 3 M2 for  40 2  20 2  60  40  30  20 or M1 for π  60  40  π  30  20 If 0 scored, SC1 for 3400π or 10 700 or 10 680 to 10 681.4…

This question in 0607/41 May/June 2024

7 NOT TO SCALE A B 12 cm 12 cm C The diagram shows a logo made from an isosceles triangle and two semicircles. The perimeter of the logo is 37 cm. (a) Show that the diameter of each semicircle is 4.14 cm, correct to 3 significant figures. [2] (b) Calculate angle ACB. Angle ACB = … [3] (c) Calculate the area of the logo. … cm2 [3] (d) A mathematically similar logo has an area of 35 cm2. Calculate the perimeter of this logo. … cm [3]

11 marks

Mark scheme: 7(a) πd = 37 – 12 – 12 oe M1 4.137 to 4.138... A1 7(b) 40.3 or 40.4 3 4.14 M2 for 2 × sin-1 oe or 40.33 to 40.36… 12 12 2  12 2   2  4.14  2 or cos ACB  2  12  12 4.14 or M1 for sin(...) = 12 2  12 2  2  12  12  cos ACB or  2  4.14  2  12 7(c) 60[.0] or 60.1 3 1 M1 for  12  12  sin( their b ) oe or 60.01 to 60.13 2 1  4.14  2 M1 for 2    oe 2  2  7(d) 28.2 or 28.3 3 35 or 28.23 to 28.26 M2 for 37  oe their c 35 their c or M1 for or their c 35 35  p  2 or   their c  37 

This question in 0607/43 May/June 2024

10 In this question all lengths are in centimetres. NOT TO SCALE h r r A solid cone has radius r and vertical height h. A solid hemisphere also has radius r. The curved surface area of the cone is the same as the curved surface area of the hemisphere. (a) Show that h = r 3 . [4] (b) The cone is placed directly on top of the hemisphere. 1 3 Show that the volume of this solid is rr ( 2 + 3 ) . 3 NOT TO SCALE [2] (c) A larger solid is mathematically similar to the solid in part (b). The larger solid has volume 243r r 3 ( 2 + 3) . (i) Find, in terms of r, the radius of the hemisphere of the larger solid. … [2] (ii) The surface area of the larger solid is 5000 cm 2. Find the volume of this solid. … cm3 [4] Question 11 is printed on the next page.

12 marks

Mark scheme: 10(a) rl = 2  r 2 B1 2 2 M1 l = r + h 2 2 M1 2 r = r + h 4 r 2 = r 2 + h 2 h 2 = 3r 2 A1 No errors or omissions  h = r 3    10(b) 2 3 1 2 M1 r or r  r 3 seen 3 3 2 3 1 2 A1  r +  r  r 3 3 3  1 3  V = r 2 + 3  ( )   3  10(c)(i) 9r 2 1  1  3 M1 for  243   , implied by 9 seen  3  10(c)(ii) 31000 or 31018… 4 3 5000 M3 for V = 243 2 + 3 or ( ) 324 3 5000 V = 27  2 + 3 ( ) 4 5000 5000 M2 for r = or R = 324  4  M1 for 4  (9 r ) 2 = 5000 or 4 R 2 = 5000

This question in 0607/41 Oct/Nov 2024

15 NOT TO SCALE 25 cm These two bottles are mathematically similar. The capacity of the small bottle is 0.5 litres. The capacity of the large bottle is 1 litre. The height of the large bottle is 25 cm. Calculate the height of the small bottle. … cm [3]

3 marks

Mark scheme: 15 19.8 or 19.84… 3 3 M2 for 25  2 oe or M1 for 3 1  0.5 or 3 0.5  1 oe  h 3 0.5 or for = oe    25  1

This question in 0607/41 May/June 2025

15 B 30° NOT TO SCALE 12.4 cm A C The area of triangle ABC is 74.4 cm2. AB = 12.4 cm and angle ABC = 30°. (a) Show that BC = 24 cm. [2] (b) Find AC. AC = … cm [3] (c) Find obtuse angle CAB. Angle CAB = … [3] (d) NOT TO Y SCALE X Z Triangle XYZ is similar to triangle ABC. The area of triangle XYZ is 62 cm2. Find YZ. YZ = … cm [3]

11 marks

Mark scheme: 15(a) 74.4 = 0.5  12.4  BC  sin30 oe M1 Use of 24 scores M0 74.4 A1 74.4 [=24] oe oe e.g. 0.5  12.4  sin30 3.1 15(b) 14.6 or 14.63 to 14.64 3 2 2 M2 for 12.4 + 24 −2 12.4  24  cos30 or M1 for [ AC 2 ] = 12.4 2 + 24 2 −2 12.4  24  cos30 15(c) 124.7 to 125.3 3 their 14.6 must come from trig 12.4 2 + (their14.6) 2 − 24 2 M2 for [cos A = ] 2  12.4  their14.6 or M1 for 24 2 = (their14.6) 2 + 12.4 2 −2 their14.6  12.4cos A OR 24  sin30 M2 for [sin A] = oe their 14.6 sin A sin30 or M1 for = oe 24 their 14.6 OR 74.4 M2 for [sin A] = 1  12.4  their 14.6 2 1 or M1 for  12.4  their 14.6sin A = 74.4 2 15(d) 21.9 or 21.90 to 21.92 or 4 30 oe 3 62 74.4 M2 for 24  or 24  74.4 62 62 74.4  YZ 2 62 oe or M1 for or or for  = 74.4 62  24  74.4

This question in 0607/42 May/June 2025

19 NOT TO SCALE The two solids are mathematically similar. The volume of the large solid is 416 cm3 . The volume of the small solid is 52 cm3 . The total surface area of the small solid is 60 cm2 . Calculate the total surface area of the large solid. … cm2 [3]

3 marks

Mark scheme: 19 240 3 2  416  M2 for 60  3  oe    52  416 52 or M1 for 3 or 3 oe or better or for 52 416  60 3  52  2 = oe      A   416 

This question in 0607/43 May/June 2025

17 A solid cone has base radius r and vertical height 3r. The total surface area of the cone is 209.22 cm2. (a) Find r. r = … cm [4] (b) A mathematically similar cone has a total surface area of 1882.98 cm2. Find the radius of this cone. … cm [3]

7 marks

Mark scheme: 17(a) 4 4 209.22 M3 for  r =  oe π(1 + 10) or M2 for 209.22 = πr 2 + πr r 2 + (3r ) 2 oe or M1 for l 2 = r 2 + (3r ) 2 oe 17(b) 12 3 1882.98 M2 for x =  their r oe 209.22 1882.98  x  2 or M1 for = oe   209.22  their r  1882.98 209.22 or oe or oe 209.22 1882.98

This question in 0607/42 Oct/Nov 2025