Mathematics Grade 12 Paper 2 Limpopo Mock 2022 Memo

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P2 AUGUST 2022 PRETRIAL EXAMINATION MARKS: 150 TIME: 3 HOUR This question paper consists of 15 pages and 1 information sheet. Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P22 com Limpopo/August 2022 NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 10 questions. 2. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. 3. s.co Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. m si c 4. st Answers only will NOT necessarily be awarded full marks. an 5. You may use an approved m scientific or calculator (non-programmable and non-graphical), unless stated otherwise. ep 6. hy If necessary, round off answers correct to TWO decimal places, unless stated otherwise. om Diagrams are NOT necessarily drawn to scale. fr 7. oa An information sheet with formulae is included at the end of the question paper. de 8. d do Write neatly and legibly. 9. wn l Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 3 com Limpopo/August 2022 NSC QUESTION 1 Each child in a group of four-year-old children was given the same puzzle to complete. The time taken (in minutes) by each child to complete the puzzle is shown in the table below. TIME TAKEN (t) NUMBER OF (IN MINUTES) CHILDREN 2t6 2 s.co 6  t  10 10 m 10  t  14 9 si c 14  t  18 st 7 18  t  22 an mor 8 22  t  26 7 26  t  30 ep 2 hy om 1.1 How many children completed the puzzle? fr (1) 1.2 oa Calculate the estimated mean time taken to complete the puzzle. (2) de d 1.3 do Complete the cumulative frequency column in the table given in the ANSWER BOOK. (2) wn 1.4 l Draw a cumulative frequency graph (ogive) to represent the data on the grid provided in the ANSWER BOOK. (3) 1.5 Use the graph to determine the median time taken to complete the puzzle. (2) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 4 com Limpopo/August 2022 NSC QUESTION 2 Learners who scored a mark below 50% in a Mathematics test were selected to use a computer- based programme as part of an intervention strategy. On completing the programme, these learners wrote a second test to determine the effectiveness of the intervention strategy. The mark (as a percentage) scored by 15 of these learners in both tests is given in the table below. LEARNER L1 L2 L3 L4 L5 L6 L7 L8 L9 L10 L11 L12 L13 L14 L15 TEST 1 10 18 23 24 27 34 34 36 37 39 40 44 45 48 49 (%) TEST 2 33 21 32 20 58 43 49 48 41 55 50 45 62 68 60 (%) 2.1 Determine the equation of the least squares regression line. (3) 2.2 A learner's mark in the first test was 15 out of a maximum of 50 marks. 2.2.1 Write down the learner’s mark for this test as a percentage. (1) 2.2.2 Predict the learner's mark for the second test. Give your answer to the nearest integer. (2) 2.3 For the 15 learners above, the mean mark of the second test is 45,67% and the standard deviation is 13,88%. The teacher discovered that he forgot to add the marks of the last question to the total mark of each of these learners. All the learners scored full marks in the last question. When the marks of the last question are added, the new mean mark is 50,67%. 2.3.1 What is the standard deviation after the marks for the last question are added to each learner's total? (2) 2.3.2 What is the total mark of the last question? (2) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 5 com Limpopo/August 2022 NSC QUESTION 3 In the diagram, A, B, C(2 ; –3) and D(–2 ; –5) are vertices of a trapezium with AB || DC. E(–2 ; 0) is the x-intercept of AB. The inclination of AB is  . K lies on the y-axis and KB̂E  . y K  B  x E(–2 ; 0) 0 A C(2 ; –3) D(–2 ; –5) 3.1 Determine: 3.1.1 The midpoint of EC (2) 3.1.2 The gradient of DC (2) 3.1.3 The equation of AB in the form y  mx  c (3) 3.1.4 The size of  (3) 3.2 Prove that AB  BC. (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 6 com Limpopo/August 2022 NSC 3.3 The points E, B and C lie on the circumference of a circle. Determine: 3.3.1 The centre of the circle (1) The equation of the circle in the form ( x  a)  ( y  b)  r 2 2 2 3.3.2 (4) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 7 com Limpopo/August 2022 NSC QUESTION 4 In the diagram, the circle is centred at M(2 ; 1). Radius KM is produced to L, a point outside the circle, such that KML || y-axis. LTP is a tangent to the circle at T(–2 ; b).  1  S   4 ;  6  is the midpoint of PK.  2  y L T(–2 ; b) M(2 ; 1) x K P 4.1 Given that the radius of the circle is 5 units, show that b = 4. (4) 4.2 Determine: 4.2.1 The coordinates of K (2) 4.2.2 The equation of the tangent LTP in the form y = mx + c (4) 4.2.3 The area of  LPK (7) 4.3 Another circle with equation ( x  2) 2  ( y  n) 2  25 is drawn. Determine, with an explanation, the value(s) of n for which the two circles will touch each other externally. (4) [21] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 8 com Limpopo/August 2022 NSC QUESTION 5 5.1 Without using a calculator, write the following expressions in terms of sin 11 : 5.1.1 sin 191 (1) 5.1.2 cos 22 (1) 5.2 Simplify cos(x  180)  2 sin( x  45) to a single trigonometric ratio. (5) 7 Given: sin P  sin Q  and P̂  Q̂  90 5.3 5 Without using a calculator, determine the value of sin 2P . (5) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 9 com Limpopo/August 2022 NSC QUESTION 6 6.1 Determine the general solution of cos( x  30)  2 sin x . (6) 6.2 In the diagram, the graphs of f ( x)  cos( x  30) and g ( x)  2 sin x are drawn for the interval x  [180 ;180] . A and B are the x–intercepts of f. The two graphs intersect at C and D, the minimum and maximum turning points respectively of f. y D g –180° B 0° A 180° x f C 6.2.1 Write down the coordinates of: (a) A (1) (b) C (2) 6.2.2 Determine the values of x in the interval x  [180 ; 180] , for which: (a) Both graphs are increasing (2) (b) f ( x  10)  g ( x  10) (2) 6.2.3 Determine the range of y = 22sin x + 3 (5) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 10 com Limpopo/August 2022 NSC QUESTION 7 In the diagram below, CGFB and CGHD are fixed walls that are rectangular in shape and vertical to the horizontal plane FGH. Steel poles erected along FB and HD extend to A and E respectively. ∆ACE forms the roof of an entertainment centre. BC = x, CD = x + 2, BÂC   , AĈE  2 and EĈD  60 E A  D B 2 60° x+2 x C H F G 7.1 Calculate the length of: 7.1.1 AC in terms of x and  (2) 7.1.2 CE in terms of x (2) 7.2 Show that the area of the roof ACE is given by 2 x ( x  2) cos  . (3) 7.3 If   55 and BC = 12 metres, calculate the length of AE. (4) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 11 com Limpopo/August 2022 NSC QUESTION 8 8.1 In the diagram, O is the centre of the circle and LOM is a diameter of the circle. ON bisects chord LP at N. T and S are points on the circle on the other side of LM with respect to P. Chords PM, MS, MT and ST are drawn. PM = MS and MT̂S  31 P N M 2 1 O L S 31° T 8.1.1 Determine, with reasons, the size of each of the following angles: (a) MÔS (2) (b) L̂ (2) 1 8.1.2 Prove that ON = MS. (4) 2 Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 12 com Limpopo/August 2022 NSC 8.2 In ABC in the diagram, K is a point on AB such that AK : KB = 3 : 2. N and M are points on AC such that KN || BM. BM intersects KC at L. AM : MC = 10 : 23. B K L A N M C Determine, with reasons, the ratio of: AN 8.2.1 (2) AM CL 8.2.2 (3) LK [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 13 com Limpopo/August 2022 NSC QUESTION 9 In the diagram, tangents are drawn from point M outside the circle, to touch the circle at B and N. The straight line from B passing through the centre of the circle meets MN produced in A. NM is produced to K such that BM = MK. BK and BN are drawn. Let K̂  x . K x M 1 2 N 1 2 A 1 2 3 B 9.1 Determine, with reasons, the size of N̂1 in terms of x. (6) 9.2 Prove that BA is a tangent to the circle passing through K, B and N. (5) [11] QUESTION 10 10.1 In the diagram, ∆ABC and ∆DEF are drawn such that   D̂ , B̂  Ê and Ĉ  F̂ . Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 14 com Limpopo/August 2022 NSC A D  E F  B C Use the diagram in the ANSWER BOOK to prove the theorem which states that if two triangles are equiangular, then the corresponding sides are in proportion, AB AC that is  . DE DF (6) 10.2 In the diagram, O is the centre of the circle and CG is a tangent to the circle at G. The straight line from C passing through O cuts the circle at A and B. Diameter DOE is perpendicular to CA. GE and CA intersect at F. Chords DG, BG and AG are drawn. Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 15 com Limpopo/August 2022 NSC D 4 G 3 2 1 2 1 2 1 A O F B C E 10.2.1 Prove that: (a) DGFO is a cyclic quadrilateral (3) (b) GC = CF (5) 10.2.2 If it is further given that CO = 11 units and DE = 14 units, calculate: (a) The length of BC (3) (b) The length of CG (5) (c) The size of Ê. (4) [26] TOTAL: 150 Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Grade 12 Mathematics Pre-trial P2 16 com Limpopo/August 2022 NSC INFORMATION SHEET: MATHEMATICS  b  b 2  4ac x 2a A  P(1  ni ) A  P(1  ni ) A  P(1  i) n A  P(1  i) n Tn  a  (n  1)d Sn  n 2a  (n  1)d  2 Tn  ar n1 Sn   a r n 1  ; r 1 S  a ; 1  r  1 r 1 1 r F  x 1  i   1 n  P  x 1  1  i  n  i i f ( x  h)  f ( x ) f ' ( x)  lim h 0 h  x  x 2 y1  y 2  d  ( x 2  x1 ) 2  ( y 2  y1 ) 2 M 1 ;   2 2  y 2  y1 y  mx  c y  y1  m( x  x1 ) m m  tan x 2  x1 x  a 2   y  b2  r 2 a b c In ABC:   sin A sin B sin C a 2  b 2  c 2  2bc. cos A 1 area ABC  ab. sin C 2 sin(   )  sin  cos   cos  sin  sin(   )  sin  cos   cos  sin  cos(   )  cos  cos   sin  sin  cos(   )  cos  cos   sin  sin  cos 2   sin 2   cos 2  1  2 sin 2  sin 2  2 sin  . sin  2 cos2   1  n 2 x  xi  x  x  2  i 1 n n n( A ) P(A)  P(A or B) = P(A) + P(B) – P(A and B) nS yˆ  a  bx b  x  x ( y  y ) (x  x) 2 Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P2 MARKING GUIDE WISKUNDE V2 NASIENRIGLYNE AUGUST 2022 MARKS/PUNTE: 150 These marking guidelines consist of 20 pages/Hierdie nasieriglyne bestaan uit 20 bladsye.

Downloaded from hlayiso.com Downloaded Pre-trial Mathematics f romV2St anmorephysics. P2/Wiskunde 2 com Limpopo/Aug. 2022 Marking Guidelines/Nasienriglyne NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. • Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error. • Assuming answers/values in order to solve a problem is NOT acceptable. NOTA: • As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. • As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, sien die doodgetrekte poging na. • Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. Hou op nasien by die tweede berekeningsfout. • Om antwoorde/waardes te aanvaar om 'n probleem op te los, word NIE toegelaat NIE. GEOMETRY • MEETKUNDE A mark for a correct statement (A statement mark is independent of a reason) S 'n Punt vir 'n korrekte bewering ('n Punt vir 'n bewering is onafhanklik van die rede) A mark for the correct reason (A reason mark may only be awarded if the statement is correct) R 'n Punt vir 'n korrekte rede ('n Punt word slegs vir die rede toegeken as die bewering korrek is) Award a mark if statement AND reason are both correct S/R Ken 'n punt toe as die bewering EN rede beide korrek is Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Mathematics P2/Wiskunde f V2rom St anmorephysics. 3 com Limpopo/Aug 2022 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 1.1 45 children  answer (1) 1.2 ∑ fx (4×2)+ (8×10) + (12×9)+ (16×7)+ (20×8)+(24×7)+ (28×2) x  692 x= OR x = 15,38 minutes Answer only: full marks  answer (2) 1.3 Time taken (t) Number of Cumulative (in minutes) children frequency 2<t≤6 2 2  first 4 cum 6< t ≤ 10 freq correct 10 12  last 3 cum 10 < t ≤ 14 freq 9 21 correct 14 < t ≤ 18 7 28 18 < t ≤ 22 8 36 22 < t ≤ 26 7 43 26< t ≤ 30 2 45 (2) 1.4 CUMULATIVE FREQUENCY GRAPH (OGIVE) 50 40  plotting cum freq at 30 upper limits correctly (all points) 20  shape (smooth) 10  grounding (2;0) 0 0 5 10 15 20 25 30 Time in minutes Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Pre-trial Mathematics f romV2St anmorephysics. P2/Wiskunde 4 com Limpopo/Aug. 2022 Marking Guidelines/Nasienriglyne (3) 1.5 On graph at the y-value of 22,5 or 23  graph Median = ± 15 minutes. Answer only: full marks  answer (2) [10] QUESTION/VRAAG 2 2.1 a = 12,44  value of a b = 0,98 Answer only: full marks  value of b y = 12,44 + 0,98x  equation (3) 2.2.1 Percentage = ×100  answer = 30% (1) 2.2.2 yˆ = 12,44 + 0,98x yˆ  substitution of 30 = 12,44 + 0,98(30) yˆ = 41,84 = 42  answer as integer Answer only: full marks OR (2) yˆ = 41,87 (if using calculator) yˆ  value of y = 42  answer as integer OR (2) yˆ =   answer (2) 2.3.1 standard deviation =13,88   answer (2) 2.3.2 x = 50,67−45,67  50,67−45,67 Answer only: full marks  answer =5% (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Mathematics P2/Wiskunde f V2rom St anmorephysics. 5 com Limpopo/Aug 2022 Marking Guidelines/Nasienriglyne QUESTION/ VRAAG 3 y K B 1 x E(–2; 0) 0 A C(2 ; –3) D(–2 ; –5) 3.1.1 Midpoint of EC:  12  2 0  (3)   3  ;    0;    x value  y value  2 2   2  substitution 3.1.2 mDC = OR  answer (2) = 3.1.3 mAB =[AB||DC]  mAB y c y− y1 = (x−x1)  substitution of (−2;0) c OR y− 0 = (x− (−2)) c=1 equation (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Pre-trial Mathematics f romV2St anmorephysics. P2/Wiskunde 6 com Limpopo/Aug. 2022 Marking Guidelines/Nasienriglyne tanα= 3.1.4 1 mAB 2  α=26,57°  value of α θ=90°+ 26,57° =116,57°  value of θ [ext ∠of Δ] (3) 3.2 B(0 ; 1)  coordinates of B 1− (−3) mBC = OR mBC 0− 2 = −2 = −2  mBC =−2 mAB ×m BC  product of gradients = –1 =−1 ∴AB⊥ BC (3) 3.3.1 ˆ AB C= 90° ∴ EC is diameter [converse: ∠ in semi circle]  3 ∴centre of circle=  0;    answer  2 (1) 2  x  0    y  2 3  r 2 3.3.2  2  substitution of centre 2 2 3  0  0    0    r 2 or  2  0    3  ( )   r 2 2 3 2  2  2   correct substitution of EC (2  2)  (1  (3)) 2 2 E(–1 ; 0), B(0 ; 1) or C(2 or R   ; –3) to calculate r or r 2 2 2  3  or r  1     2  25 5  value of r2 or r r2  or r  4 2 2  3 25 x2   y     2 4  equation (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Mathematics P2/Wiskunde f V2rom St anmorephysics. 7 com Limpopo/Aug 2022 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y L T( –2 ; b) M(2 ; 1) x K P 4.1 (x−2)2 +(y−1)2 =25  equation of the circle substitution of point T 2 2 (−2− 2) + (b−1) = 25 simplification (b −1)2 =9 OF ∴  answer b −1=±3 (4) ∴b=4 or b≠− 2 4.2.1 K(2 ; 1 – 5) ∴ K(2 ; –4)  x value  y value Answer only: full marks (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Pre-trial Mathematics f romV2St anmorephysics. P2/Wiskunde 8 com Limpopo/Aug. 2022 Marking Guidelines/Nasienriglyne 4.2.2  mMT mMT mPL [radius ⊥ tangent]  mPL = y c c  substitution of mPL and the point T c y  equation (4) OR  mMT mMT  mPL = mPL [radius ⊥ tangent] y y  substitution of mPL and y the point T  equation (4) OR P(11; 8)  coordinates of P 4  (8) mPL  2  (11 4  3 4  mPL = y  xc 3 4 8  x  c 3  substitution of mPL and 20 c the point P or T 3 4 20 y  x 3 3  equation (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Mathematics P2/Wiskunde f V2rom St anmorephysics. 9 com Limpopo/Aug 2022 Marking Guidelines/Nasienriglyne 4.2.3 4 20 28  y= yL  (2)   3 3 3  length of LK  28  L  2;  and K  2; 4   3  28 LK   (4) 3 40  3 Coordinates of P: x+ 2 1 y− 4 =−4 L(2 ; =−6 and 2 2 2 ∴ x=−11 y =−8  xP  yP ∴P(−11;−8) ⊥height (PH) = 2 − (−11) = 13 K(2 ; –4)  length of ⊥ height P(–11 ; –8) 1 H(2– 8 Area PKL=  LK  PH 2  substitution into 1  40      13 the area formula 2 3  269  answer  OR 86,67 3 (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Pre-trial Mathematics f romV2St anmorephysics. P2/Wiskunde 10 com Limpopo/Aug. 2022 Marking Guidelines/Nasienriglyne 4.2.3 OR 4 20 28 yL = yL  (2)   3 3 3  28  L  2;  and K  2; 4   length of LK  3  28 LK   (4) 3 40  3 Coordinates of P:  xP  yP x+ 2 1 y− 4 =−4 and 2 2 2 ∴ x=−11 y =−8 ∴P(−11;− 8) PK2 = (−11− 2)2 +(−8− (−4) ) PK = 185 units ˆ  PK L mPK ° ˆ ∴PK L = 90°+17,1027...° = 107,1°  substitution into the area rule 1  answer Area PKL=  ˆ LK  LK sin PKL (7) 2  1  40      185 sin107,1 2 3   269  OR 86,67 3 4.3 The centres of the two circles lie on the same vertical line  correct method x = 2. and the sum of the radii = 10  sum of radii = 10 n−1 = 10 1 −n= 10 or  n=11  n= −9 n=11 n = −9 (4) Answer only: full marks [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

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Downloaded from hlayiso.com Downloaded Pre-trial Mathematics f romV2St anmorephysics. P2/Wiskunde 12 com Limpopo/Aug. 2022 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 cos(x−30°) = 2sin x cosxcos30°+sin  expansion 3 xsin30°= 2sin x cosx+ sin x= 2 2sin x  special ∠s 3 2 cosx= sin x  simplification 3 tan x= 3  equation in x=30° +k.180°; k∈Z tan OR  30° x = 30° +k.360° or x = 210° +k.360°;  k.180°; k∈Z k∈Z OR  30° and 210°  k.360°; k∈Z (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

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Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 14 com Limpopo/Aug. 2022 Marking Guidelines/Nasienriglyne 7.1.1 x sinθ=  trig ratio AC OR x AC =  simplification sinθ (2) 7.1.2 x+2 cos60°=  trig ratio CE OR x+2 CE= cos60° x + 2  making CE the subject = =2(x + 2) (2) 7.2  use area rule ˆ Area ∆ACE = AC.EC.sin AC E correctly 1 x  substitution of = (2(x+2))sin2θ x 2 sinθ x(x+ (2(x+2)) 2)×2sinθcosθ sinθ =  substitution sinθ of sin2θ (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 15 com Limpopo/Aug 2022 Marking Guidelines/Nasienriglyne = 2x(x+2)cosθ E A θ D B 2θ 60 ° x+2 x C H F G 7.3 EC = 2(12 + 2) = 28  EC  use cosine rule AE 2  AC 2  EC  2 AC.EC cos ACE correctly   12   2  12     28  2   2  .28  cos110  substitution  sin 55    sin 55   so  answer AE  35, 77m (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 16 com Limpopo/Aug2022 Marking Guidelines/Nasienriglyne 8.1 P N M 2 1 O L S 31° T ˆ 8.1.1(a) MO S= 62° [∠ at centre = 2× ∠ at circumf/middelpts∠ =2omtreks∠]  S  R (2) ˆ 8.1.1(b) L =31° [equal chords; equal ∠s /= koorde; = ∠e]  S R (2) 8.1.2 LN = NP and LO = OM  LO = OM ∴ ON = PM [ midpoint theorem/middelpuntstelling]  S R ∴ ON = 2 MS [PM = MS]  S (4) OR Nˆ 1=90° [line from centre to midpt chord/lyn v midpt na midpt kd]  SR Pˆ = 90° [∠ in semi-circle/∠ in halfsirkel] Lˆ is common/gemeen  S/R ∴ ∆NLO ||| ∆ PLM (∠∠∠) NL NO 1 = =  S PL PM 2 ∴ ON = PM (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 17 com Limpopo/Aug 2022 Marking Guidelines/Nasienriglyne ∴ ON = 2 MS  S [PM = MS] b 8.2 B 2y K L 3y A N M C 10x 23x 8.2.1 AN AK = [line || one side of ∆ OR prop theorem; KN ||BM/ R AM AB lyn || sy van ∆ OR eweredigheidst; KN||BM] AN 3y 3 = = S AM 5y 5 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 18 com Limpopo/Aug2022 Marking Guidelines/Nasienriglyne 8.2.2 AM 10x = [given] MC 23x  S AM = 5y =10x ∴ y=2x  R LC MC = [line || one side of ∆ OR prop theorem; KN ||LM/ KL NM  S lyn || sy van ∆ OR eweredigheidst; KN||BM] (3) 23x 23x 23 = = = 2y 4x 4 OR  S AM 10x  R = [given] MC 23x AN 3y 6x  S = = (3) MN 2y 4x LC MC = [line || one side of ∆ OR prop theorem; KN ||LM/ KL NM lyn || sy van ∆ OR eweredigheidst; KN||BM] 23x 23x 23 = = = 2y 4x 4 [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 19 com Limpopo/Aug 2022 Marking Guidelines/Nasienriglyne 9 K x M 1 2 N 1 2 A 1 2 3 B 9.1 Bˆ 1=x [∠'s opp = sides/∠e teenoor = sye] S S R Mˆ 2 =2x [ext ∠of ∆] OR Mˆ 1=180°− 2x [∠s of ∆] BM = MN [ 2 tans from a common point/raaklyne vanuit S R dieselfde punt] answer Nˆ x [∠'s opp = sides/∠e teenoor = sye] OR S R NM = BM [ 2 tans from a common point/raaklyne vanuit dieselfde punt] (6) Bˆ 2 =Nˆ 1 [∠'s opp = sides/∠e teenoor = sye] S R Bˆ 1 =x [∠'s opp = sides/∠e teenoor = sye] In ∆ KBN: S x+x+Bˆ 2 + Nˆ 1 =180°[sum of ∠'s of ∆] ˆ 2x+2N 1 =180° ˆ x+ N 1 = 90° answer ˆ N 1 =90°−x (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 20 com Limpopo/Aug2022 Marking Guidelines/Nasienriglyne 9.2 MBˆA =Bˆ 2+ Bˆ 3= 90° [tangent⊥diameter/raaklyn⊥middellyn] S  R ˆ ˆ B 3=90°−B 2 S S =90°−(90°−x)= x ˆ ˆ B 3= K = x R ∴ AB is a tangent/raaklyn converse tan-chord theorem/ omgekeerde raakl koordst]] (5) OR Bˆ 2 =Nˆ 1 ˆ ˆ S B 1+ B 2=x+ (90°− x) =90° R ∴ KN is diameter/middellyn [converse ∠ in semi-circle/ omgekeerde ∠ in halfsirkel] S R ˆ ˆ ˆ MB A =B 2 + B 3 = 90° [tangent ⊥diameter] ∴ AB is a tangent/raaklyn converse tan-chord theorem/ R omgekeerde raakl koordst]] (5) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 21 com Limpopo/Aug 2022 Marking Guidelines/Nasienriglyne QUESTION 10 10.1 A D M N  E F  B C 10.1 Constr: Let M and N lie on AB and AC respectively such that Constr / Konstr AM = DE and AN = DF. Draw MN. Konst: Merk M en N op AB en AC onderskeidelik af sodanig dat AM = DE en AN = DF. Verbind MN. Proof: In ∆ AMN and ∆ DEF AM = DE [Constr] AN = DF [Constr] ˆ A = Dˆ [Given]  Δ AMN ≡Δ DEF ∴Δ AMN≡Δ DEF(SAS) SAS ˆ ˆ ˆ ∴AM N=E =B MN || BC [corresp∠'s are equal/ooreenkomstige ∠e =] MN || BC and R AB AC AB AC = [line || one side of ∆ OR prop theorem; MN ||BC] AM AN  = R AB AC AN AM ∴ = [AM=DEandAN=DF] DE DF (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 22 com Limpopo/Aug2022 Marking Guidelines/Nasienriglyne 10.2 D 4 G 3 2 1 2 1 2 1 A O F B C E 10.2.1(a) ˆ DO B=90° ˆ ˆ ˆ DG F=G 3 +G 4 =90° [∠ in semi-circle/∠ in halfsirkel]  S R ˆ ˆ DO B+DG F=180° R ∴DGFO is a cyclic quad. [converse: opp ∠s of cyclic quad/ omgekeerde teenoorst ∠e v koordevh] OR ∠s of quad = 180°/∠e van koordevh = 180°] OR ˆ EO B=90° (3)  S R ˆ ˆ ˆ DG F=G 3 +G 4 =90° [∠ in semi-circle/∠ in halfsirkel] ˆ ˆ R EO B = DG F ∴DGFO is a cyclic quad. . [converse: ext ∠ = opp int ∠/ omgekeerde buite∠ = teenoorst ∠] OR ext∠ of quad = opp int ∠/buite∠ v vh = teenoorst ∠] (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 23 com Limpopo/Aug 2022 Marking Guidelines/Nasienriglyne Fˆ =Dˆ  S R [ext ∠of cyclic quad/buite∠ v koordevh] 10.2.1(b) 1  S R Gˆ +Gˆ =Dˆ 1 2 [tan-chord theorem/raakl koordst] ∴ Fˆ 1 =Gˆ 1 +Gˆ 2 R (5) ∴GC = CF [ sides opp equal ∠s/sye teenoor = ∠e] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f Pre-trial Mathematics romV2St anmorephysics. P2/Wiskunde 24 com Limpopo/Aug2022 Marking Guidelines/Nasienriglyne 10.2.2(a) AB = DE = 14 [diameters/middellyne] S ∴ OB = 7 units S ∴ BC = OC – OB = 11 – 7 Answer only: full marks S = 4 units (3) 10.2.2(b) In Δ CGBand Δ CAG ] [tan-chord theorem/raakl Gˆ 1 =Aˆ = x koordst  S/R Cˆ = Cˆ [common] AG ∆CGB ||| ∆C[∠, ∠, ∠ ] CG CB S = CA CG S CG 4 =  CA = 18 18 CG or 6 2 or 8,49 units CG2 = 72 answer CG = 72 (5) 10.2.2(c) OF = OC – FC = 72 11 – OF  OF tanE OE 11 − 72 =  trig ratio 7 == 0,36  substitution ˆ E =19,76°  answer (4) OR OF = OC – FC = 11 – 72  OF FE2 = OE2 + OF2 sin OF E= FE = 72 + (11− 72)2 11 − 72 FE = 7,437.. = 7,44 7, 44 OE = cosE =  trig ratio FE ˆ OR E =19,76° = 0,338 Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Downloaded Mathematics f rom St anmorephysics. P2/Wiskunde V2 25 com DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne  substitution = = 0,94  answer ˆ (4) E =19,76° [26] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou