Mathematics Grade 12 Paper 2 KwaZulu Natal Mock 2022 Memo

Naumlac ‘ ‘ fram ‘ , > ww KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P2 PREPARATORY EXAMINATION SEPTEMBER 2022 3 hours This question paper consists of 13 pages, 1 information sheet and an answer book with 22 pages. Copyright reserved Please turn over

Mathematics/P2 Downloaded from hlayiso.com Downloaded from Stanmorephysics.com «zn preparatory Examination 2022 NSC —- GRADE 12 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. 2. 10. 11. This question paper consists of 11 questions. Read the questions carefully. Answer ALL the questions. Number your answers exactly as the questions are numbered. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in detente your answers. (@) G Answers only will NOT necessarily be awarded full marks. . 2” You may use an approved scientific calculator (non-programmable and graphical), unless stated otherwise. XN If necessary, round off answers correct to TWO decimal cess stated otherwise. Diagrams are NOT necessarily drawn to scale. & An information sheet with formulae is included and of the question paper. Write neatly and legibly. e? K\ fe) & ww ee Copyright Reserved Please turn over

Mathematics/p2 . “com Downloaded from Stanmorepbysits.co Downloaded from hlayiso.com QUESTION 1 The weight (in kg) of 20 boys in the soccer squad of school A are given below: ll 1.2 1.3 Copyright Reserved 40 47 48 31 53 57 58 458 59 59 60 60 60 60 61 62 63 64 66 69 Calculate: Ld. the mean Weight of the boys in this soccer squad. & 1.1.2 the standard deviation of this data, 9 Determine the number of boys that have a Weight within one sti deviation of the mean. Y) The following information was obtained from the coh i soccer squad of school B: 2 ‘ > x, = 1320? n=l Ss 1.3.1 How many boys are in the 1B squad? 1.3.2 Calculate the mean yee of a boy in the soccer Squad of school B. Assume that the Mean weight Of the boys in the soccer Squad at school B is 60 kg. Five boys of equal wi a € added to the school A Squad so that the means of both school squads are thes e. Calculate the weight of each of these five boys. Or Oo x ss KZN Preparatory Examination 2022 Q) (1) (2) () (D (4) (11) Please turn over

Downloaded from hlayiso.com D Mathematics/P2 ownloaded from Stanmorephysics.com KZN Preparatory Examination 202? NSC - GRA EI - QUESTION 2 A survey was done on 250 people to determine the distances they travel to work daily. The results are shown in the table below. CUMULATIVE DISTANCE, d (in km) FREQUENCY FREQUENCY, O<dsi 8 p< dso} eo 10<dsi5 63 ee | oe el 15<d<20 52 ee rr ee 20<d<25 4! 25<d 530 38 25 <a SW ———=— 30<d $35 7 TOTAL 2.1 Complete the cumulative frequency column, on the attached ANSWER SHEET. (3) 2.2 Draw a cumulative frequency graph (ogive) for the given data on the grid provided on the attached ANSWER SHEET. (4) 23 Use the graph to determine the median distance travelled. Indicate on your graph the median distance. (2) 19] Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 5 Mathematics/P2 KZN Preparatory Examination 2022 NSC — GRADE 12 QUESTION 3 In the diagram below, A(-5 ; 2), B(-3 ; 8) and C are vertices of AABC. E(1 ; 0) is the midpoint of AC. D isa point on the x-axis such that AD is a line perpendicular to the x-axis. a is the angle of inclination of AC. B(3:8) y M A(5;2)4)8 Pw x D [e} E(1 ; 0) Cc 3.1 Determine the coordinates of M , the midpoint of AB. (2) 3.2 Write down the coordinates of point D. (1) 3.3 Show that the coordinates of C are (7; -2) (2) 3.4 Calculate the length of line AC. (Leave answer in simplest surd form) (2) 3.5 Determine the coordinates of F , if F lies in the first quadrant and CABF isa parallelogram. (2) 3.6 Determine the equation of the perpendicular bisector of AB. (4) 3.7 Calculate the size of @ , the angle of inclination of line AC. (3) 3.8 Determine the equation of the line parallel to AB passing through E. (2) 3.9 Calculate the size of angle @ . (2) 3.10 Calculate the area of AABC. (4) [24] Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Mathematics/P2 6 KZN Preparatory Examination 2022 NSC — GRADE 12 QUESTION 4 In the diagram, the circle TS centred at C(3;~1) has aradius CT of 10 units. PTR, where R(k ; 21), is a tangent to the circle at T. PS is a tangent to the circle at S and PS || x-axis. PC, TC and CR are drawn. TR = 20 units. y R (&; 21) 20 T, 10 x (6) DY CB;-l P S i 4.1 Give areason why CT LTR. qd) 42 Calculate the value of & , where R is in the first quadrant. (4) 43 Write down the equation of the given circle. (2) 44 Write down the equation of PS. qd) 4.5 The equation of tangent PTR is 3y=4x+35. 4.5.1 Calculate the coordinates of P. (2) 4.5.2 Calculate the length of PT. (3) [13] Copyright Reserved Please turn over

Downloaded from hlayiso.com mar QWploaded from Stanmorephysics.com NSC — GRADE 12 KZN Preparatory Examination 2022 QUESTION 5 5.1 If 5cosA= 26 where Ae [90°;360°] , calculate, without using a calculator and with the aid of a diagram, the values in simplest form of: 5.1.1 V6. tanA (4) 5.1.2 sin2A (4) 5.2 Given: sin18°= p Without using a calculator, determine each of the following in terms of p. 3.2.1 cos 18° (2) 5.2.3 cos 48° (5) 5.2.3 sin 9° (3) [18] QUESTION 6 6.1 Without using a calculator, simplify the following expression fully: sin(180° — x).tan(x-180°).cos(360° +x) sin?(180°+ x) + sin?(90°— x) (6) 6.2 Without using a calculator, determine the value of: cos 330°. tan 150°.sin 12° tan 675°. cos 258° (7) 6.3 Given the identity: cosa + 60820 = Ose + sin2a —-sina sing 6.3.1 Prove the identity. (4) 6.3.2 For which other values of @ is the identity undefined? (5) [22] Copyright Reserved Please turn over

Downloaded from hlayiso.com _ Downloaded from Stanmorephysics.com a Mathematics/P2 8 KZN Preparatory Examination 2022 NSC —- GRADE 12 QUESTION 7 Given: f(x) =sin2x and g(x) = cos(x+a) where x <[—180" ; 180°] The graphs of f and g intersectat B and D. E is the y-intercept of g, and C isa turning point of g. A isa turning point of both f and g. | Ags) E r x -180° Hy? fe) ny 180° D B (165°; ~ %) g c | al Waite down the value of a. (1) 72 State the period of (1) 73 Determine the coordinates of C and E. GB) 74 Write down the amplitude of A if h(x) =3 f(x). () 75 Determine for which value(s) of x , if xe[0°;180°], will: 7.5.1 g(x) > f(x) 2) 75.2 g(x). f'®) 20 (2) 7.6 Without solving the equation, use the above graphs to show how you would solve the following equation: (3) [13] Copyright Reserved Please turn over V2sin2x = cosx + sinx

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Mathematics/P2 KZN Preparatory Examination 2022 NSC — GRADE 12 QUESTION 8 8.1 Complete the following statement: The line drawn from the centre of the circle perpendicular to the chord ......... (ld) 8.2 The circle below with centre O has chord AB = 8 cm. OMT | AB with MT=2 cm. The radius of the circle is r cm. O r cq A M B T 8.2.1 Write down, with a reason, the value of AM. (2) 8.2.2 Calculate the length of the radius of the circle. (4) (7] Copyright Reserved Piease turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Mathematics/P2 10 KZN Preparatory Examination 2022 NSC - GRADE 12 QUESTION 9 9.1 In the diagram below, O is the centre of the circle. P, A and T are points on the circumference of the circle. PA, TA, PO and TO are drawn. T Prove the theorem which states that POT = 2PAT. (5) Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Mathematics/P2 KZN Preparatory Examination 2022 NSC — GRADE 12 9.2 AOB and COB are diameters of circle ACDBE with centre O. Chords AC, CB, AE, AD and DE are drawn. D=50°. D 9.21 Calculate, with reasons, the size of the following angles: (a) 0, (2) (b) E, @) 9.2.2 Prove that AE || CB. (4) [14] Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Mathematics/P2 12 KZN Preparatory Examination 2022 NSC - GRADE 12 QUESTION 10 In the diagram DRS is a tangent to the circle TMAR at R. AT bisects MTR. AT intersects MR at P. AR is drawn. T, =x. D R A S Li T M 10.1 Prove, giving reasons, that: 10.1.1 R,=R,. (4) 10.1.2 AAPR ||| AMPT . (3) 3 PT 10.2 If AR= 3MT , then calculate the value of PR” (3) [10] Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Mathematics/P2 KZN Preparatory Examination 2022 NSC — GRADE 12 QUESTION 11 In the diagram below, AABC has D and E on BC. BD =6cmand DC=9cm. AT:TC=2:1 and AD | TE. A T B D E C . . CE 11.1 Write down the numerical value of ED’ qd) 11.2 Show that D is the midpoint of BE. Q) 11.3 If FD =2 cm, calculate the length of TE. (2) 11.4 Calculate the numerical value of : Wad Areaof AADC 7 _ Area of AABD (2) 142 Areaof ATEC 3 _ Area of AABC (3) 19] TOTAL: 150 Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics. com Mathematics/P2 KZN Preparatory Examination 2022 NSC - GRADE 12 INFORMATION SHEET: MATHEMATICS _ —b+\b" —4ac 2a A=P(1+ni) A=P(1-ni) A=P(-i" T, =a+(n—ld S, =5 (2a+(n—Na)} alr” -1 T, = ar" s, <4 De S.2 os -l<rsl r- -r x{(1+i)’ -1] > x[l-a+i"] i i 19) = jim LE =) X, +X, VY+typ d= V(x, -x,)? +0, -y)* ea m = tan0 227M yy yemxte YY =mM(x-%) Xy mee In AABC: -~___¢ sinA sinB sinC a? =h? +e7 —2bc.cosA area AA BC =Fab.sinc sin(a+B)=sina.cosB +cosa.sinB sin(a—f) =sin a.cosB —cosa.sinB cos(a+f)=cosa.cosB—sino.sinB cos(a—B) = cosa. cosB +sin a.sinB cos’ a—sin? « cos 2a =41—2sin’? a sin 20 = 2sina.coso 2cos? a—1 y= fi »_Elan3)’ =a o = ——___ n n pcay = 2) P(A or B)=P(A)+P(B) —P(A and B =) or B)= P(A)+P(B) —P(A and B) pease p= X= 9) Sons Copyright Reserved

Downloaded from hlayiso.com Downloaded from Stanmqrpaiaysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE MARKS: 150 TIME: 3 hours These marking guideline consists of 15 pages. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathenitaxwwvoaded from Stanmorephysics. Cpsparatory Examination September 2022 NOTE: NSC — Marking Guidelines If a candidate answered a QUESTION TWICE, mark only the FIRST attempt. If a candidate crossed out an answer and did not redo it, mark the crossed-out answer. Consistent accuracy applies to ALL aspects of the marking guidelines. Assuming values/answer in order to solve a problem is unacceptable. GEOMETRY A mark for a correct statement Ss (A statement mark is independent of a reason.) R A mark for a correct reason (A reason mark may only be awarded if the statement is correct.) S/R Award a mark if the statement AND reason are both correct. QUESTION 1 40 47 48 51 53 57 58 58 59 59 60 60 60 60 61 62 63 64 66 69 1.1.1 — 1155 AY 1155 xX =— 20 Answer only: full marks X =57,75 kg CAY answer (2) Penalty 1 mark for rounding here for the entire paper 1.1.2. | o =6,73702 ~ 6,74 kg AY answer (1) 1.2 (X-o;X+o) (57,75-6,74 ; 57,75+6,74) limit = (51,01 ; 64,49) CAV’ interval 2) 14 boys CAY answer 1.3.1 | 22 AY answer (1) 13.2 | _ 1320 xX =— 22 X = 60 kg CAY based on 1.3.1 (1) 1.4 ya ok t1I55 _ ey CAY 5x+1155 25 25 5x +1155 =1500 CAY equation 5x =345 CAV’ simplification =69 k x g CAY answer (4) (11] Copyright Reserved Please Turn Over

Mathenkakawwdoaded from Stanmore ephysics. CPiaparatory Examination September 2022 Downloaded from hlayiso.com NSC — Marking Guidelines QUESTION 2 2.1 ic ‘ CUMULATIVE DISTANCE, d(inkm) | FREQUENCY FREQUENCY O<d<5 8 8 v 5<d<=10 41 49 Av Se® l0<d<I5 63 112 CAY i5<d=20 52 164 ae 20<d<25 41 205 25<d<30 38 243 cav 205,243 and 30<d <35 7 250 20 TOTAL 250 (3) 2.2 CUMULATIVE FREQUENCY GRAPH (OGIVE) 250 =r 225 sa AY — grounded 175 at (0 ; 0) CAY cumulative f 150 frequencies = as y, for 2 y-coords z 100 CAY Sother — 50 AY 25 4 ¥ 0 $ 10 isA 30 2s 30 35 DISTANCE (in kms) (4) 2.3 See ogive for dotted lines and point marked A. CAV indication Median = 16 km on graph CAY approx. (2) OR median Answer only: Full marks (accept 15,16,17) [9] Copyright Reserved Please Turn Over

Mathenttaxwwvoaded from Stanmorephysics. Cpsparatory Examination September 2022 Downloaded from hlayiso.com NSC — Marking Guidelines QUESTION 3 B(-3:8) y M A(-5;2)Q)0 Py x D oO E(1 ; 0) Cc 3.1 M x +% . Yot YM 2° 2 Vv woe 5-3 248 A Substitution of A M| —— ;—— and B into midpoint 2 2 Answer only: full marks formula M(-4;5) CAY answer (2) 3.2 D(-5;0) AY answer (1) 3.3 5+ Xo 4 AY — midpoint ITO x 2 AY midpoint ITO y WX =7 C(7;-2) (2) OR v ~ C(1 +6; 0-2) Using transformations A x-value C7 :-2) Av _ y-value (2) 3.4 _ 2 2 AC = (x) - x) +02 -Y) as 5 : - AY - substitute A and C AC = y(-5-7)" +(2+2) into distance formula AC =4V10 CAY answer (2) Copyright Reserved Please Turn Over

Matheritaxwwvoaded from Stanmorephysics. Casparatory Examination September 2022 NSC — Marking Guidelines Downloaded from hlayiso.com 3.5 Method: translation AB: (x;y) > (x+2; y +6) .. by symmetry: C > F: C(7;-2) > F(7+2;-2 +6) AY — x-coordinate “. FQ; 4) Av _ y-coordinate (2) OR OR Midpoint of intersection of diagonals = T (2; 3) Let coordinates of F be (a ; b) a-5 b+2 7 = 2 and =3 a=9 andb=4 -. F934) AY — x-coordinate AY _ y-coordinate (2) _ Vv = 3.6 mAB= 23 A mAB =3 -34+5 1 .. gradient of perpendicular bisector: “3 1 1 CAY Myerp bisector => Ww ys-ixte 73 b(-4;5): 5=-2(-4)+ sub (-4;5): 5 = 3° )+e CAY substitution Lu 3 . . dicular bi 1 1 CAY equation “. equation of perpendicular bisector: y = ~3*t 3 (4) v 3.7 mAC = 24+2 _1 A mAC 5-7 3 1 tana=—) CAY tana@=-= afl q@=tan “3 +180° a =161,57° CAY answer “a 162° (3) OR Accept method using Cosine Rule 3.8 MAB =3 . gradient of new line = 3 (/ limes =gradients ) CAY equal gradients y =3x +c sub E(1;0):0=3(1) +c =-3 y=3x-3 CAY answer (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathentakawwvoaded from Stanmorephysics. CeBparatory Examination September 2022 NSC — Marking Guidelines 3.9 _ 1 mAB x mAC =3 x 3 AY mABxmAC=~<1 “MAB x mAC =-1 “ ABLAC “0 =90° AY answer (2) Accept methods using Cosine Rule & Trig ratios 3.10 | AC =4¥10 units AB= [(-3+5) +(8—2) = 210 units AY © substitute A andB into distance formula 1 CAY length of AB “. Area of AABC = 3 AC.AB . Area of AABC = 5 (4V10) (200) CAY substitution . Area of AABC = 40 units’ CAY answer (4) [24] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathenttaxwwvoaded from Stanmorephysics. Cpxsparatory Examination September 2022 NSC — Marking Guidelines QUESTION 4 y R(k; 21) 20 T, 10 x S/ C331) P Ss 41 Radius is perpendicular to tangent AY — reason () 42 | cr= io’ +20°=V500 —_ (pythag) Avy 500 CR = (k—3)° + (2141? = J(k-3)? +484 2.4 (k-3) + 484 = /500 Av equate .. (k—3)? +484 =500 . 2 “(k=3y =16 CAv simplification wk-3=44 “k=-1 or k=7 CAV both values and but k #-1 (given that R is the first quadrant) rejection of k =—1 k =7 only (4) 43° | (x-3)' +(y+1) =100 AY (x-3) +(y+1)’ 2 AY 100 (2) 44 | S@;-1D wy=-ll AY answer (1) 45.1 | y=-ll Eq1 3y=4x+35 Eq 2 Sub Eq 1 into Eq2: 3(-1) = 4x 435 CAv‘ substitution 4x = -68 xeol? CAY x-value “.P(-17;-11) (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Matherinaxwwvoaded from Stanmorepsysics. Casparatory Examination September 2022 NSC — Marking Guidelines 45.2) pc=f(3417)? +(-1411)? =10V5 cAY PC? =500 TC= 10 PC=10V5 PT? = PC? -TC? (pythag) CAY Pythagoras pT =4/(tovs) —(10)' PT = 20 units CAY answer (3) OR PT=PS (tangents from common point) Av SIR PS = 3—(-17)= 20units CAY Substitution “PT = 20 units CAY answer (3) [13] QUESTION 5 5.1.1 y A IG las 2/6 SN Lo AY — diagram i] I 5 il I | y= Ir? -x 2 y=\(5) -(2V6) =1 AY y=-1 “ay=-l (quadrant 4) -V6. tanA -1 ~~ Gx{ =e CAY tana = 206 =i CAY answer (4) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathenkakawwdoaded from Stanmore ephysics. CPi8paratory Examination September 2022 NSC — Marking Guidelines 5.1.2 | sin2A = 2sinA.cosA Av double angle identity = (2 26 CAY sinA=— 5 5 4/6 CAY cosA=—— ws 25 CAY answer (4) 5.2 72° 1 P CT 18° 1-p? 5.2.1 | oosig° =f, P AY — diagram AY — answer (2) Answer only full marks 5.2.2 | cos48° =cos(30° +18°) AY (30°+18°) = cos 30° cos18°—sin 30°sin18° AY — expansion AY special angle 3\( fi-p?) (1 (24210) substitution CACAY “substitution of VB./1- p? - p sin18° and cos18° So _\¥3-3p" -P 2 (5) 5.2.3 | cosl8° =1-2sin’9° Av double angle —2sin? 9° = cos18°-1 expansion sin? 9° _ cos18°-1 -1 Av making sin9° the sin9° = ar =1 subject >. |¥l-p?-1 1-,1- p° sin9 _jee =) = | 2 CAY answer (3) [18] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathenkakawwdoaded from Stanmore eploysics. CPi8paratory Examination September 2022 NSC — Marking Guidelines QUESTION 6 6.1 sin(180° — x). tan(x —180°).cos(360° + x) sin’ (180° + x) + sin?(90°— x) AY - sinx : AY tan x sin x. tan x.COs X i = AY — sin’x sin’ x+ Cos x . AY cos’x =sinx. sink cos X sinx cos x" AY tanx=—— a cos x = sin’ x CAY answer (6) 6.2 cos 330°. tan150°.sin12° tan 675°.cos 258° a _ (cos30°).(— tan 30°).(sin 12°) Ay cos 30 tan 45°).(—cos 78°) Ay ~tan30 30°) (tan 30°),(sin12° AY — coratio __ (cos 30°).(1 an Dsin ) Av —cos78° (tan 45°),(sin 12°) AY tands° 3/8 2 3 ——1 AY - special angles 1 CAY answer 2 %) 6.3.1 LHS: cosa + cos 2a sin2a-sina V doubl 1 A cos double angle cosa@ +(2cos’ a -1 = Losas (2cos' aI) AY - sin double angle (2sina@cosa)—sina _ 2cos* @ + cosa@—1 2sina cosa —sina (2cosa—1)(cosa +1) AY numerator factors = sin a(2cos@—1) AY — denominator factors _ (cos @ +1) sing = RHS (4) 6.3.2 | sin2a—sina=0 AY — equating to 0 2sinacosa—sina=0 sina (2cosa—1)=0 AY factors - sina =0or 2cosa—1=0 CAY 0°+k.180 cosa == CAY +60°+k.360 a=0°+k.180;k EZ a=+60°+k.360;k EZ AY kez N.B. If sina = 0 is solved only — max. 3/5 marks (5) [22] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathenkakawwdoaded from Stanmore ephysics. CPi8paratory Examination September 2022 NSC — Marking Guidelines QUESTION 7 : Yo A(45*1) E f x H 180) B (165°; — 4) 71 a=-45° AY answer (1) 7.2 180° AY answer qd) 7.3 C(-135°;-1) E: — y=cos(0°+45°) AY C(-135°;-1) v2 = AVY © solving for y V2 roe av e(o;,%2 2 (3) 7.4 Amplitude = 3 AY answer (1) 7.5.1 x €[0°;165°);x # 45° AY — [0°;165") OR AY an AS x € [0°;45°) U (45% 165%) AY critical values OR AY ~ notation — O°<x< 165%x # 45° ] Av “075 x <165° AY x #45° (2) 7.5.2 O°<x < 135° AY critical values AY __ notation (2) 76 V2 sin 2x = cos x + sinx sin 2x = 008% 4 sin x v2 V2 sin 2x = cosx a + sinx AY division of V2 v2 v2 sin 2x = cos x.cos 45° + sin x.sin 45° Av _ identifying . _ are compound 4 sin 2x = cos(x— 45°) cos(x — 45°) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Matheri2axwwvoaded from StanmoreprySsics. Cpsparatory Examination September 2022 NSC — Marking Guidelines Equating the 2 functions gives the points of intersection of | AY —_ conclusion the 2 graphs. (3) [13] QUESTION 8 8.1 bisects the chord. AY __answer dd) 8.2.1 | AM=4cm (line from centre | chord) AVY S AYR (2) 8.2.2 | OM = (r—2) AY OM = (r-2) re=(r-2) +4 (pythag) CAY Pr =(r-27 +4 rsr'—4r+4+16 CAV simplification 4r = 20 r=Scm CAV answer (4) [7] QUESTION 9 9.1 Constr: Draw line AO and extend to D. AY construction Proof: OP = OT (radii) A, =P (Zs opp = sides) AY S/R A AW v but O, =A, +P (ext Z of A) A S/R 0, =2A, AY Ss Similarly 0, =2A, 0,+0, =2(A,+A,) AY Ss “POT = 2PAT (5) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Matheniaawwvoaded from Stanmorepibysics. CeBparatory Examination September 2022 NSC — Marking Guidelines 9.2 Do not mark this question!!! (Maximum for Paper 2 — 141) Cc D A B E 9.2.1a | O, =100° (Zat centre=2x Z at circ) AV Ss AYR (2) 9.2.1b | 6, =100° A=E, (4s opp = radii) AY Ss AYR . 199° “EB, = 180° = 100° (sum 4s A) 2 CAY S/R E, =40° Q) 9.2.2 | A =B Zs opp = radii Av S/R SE (4s opp ) AY SIR but A, =C, (Zsin same segment) EB, =C, AY Ss » AEUCB (alt Zs =) AY OR (4) OR A, =E, (Zs opp = radii) AY SIR ~ a v but E, =B (Zsin same segment) A S/R “A, =B AY Ss -. AEOCB alt Zs = ( ) AVY R (4) OR AY S/R AVY SIR Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Matherinaxwwvoaded from Stanmorepiaysics. Cpsparatory Examination September 2022 NSC — Marking Guidelines CAE = 90° (Zsinsemiil ) A AV Ss ACB = 90° (4sinsemil] ) (4) CAE+ ACB =180° Av oR .. AEDCB (co - int Zs) [14] QUESTION 10 D R A S T M 10.1.1 R, = T, (4s in same segment) Av Ss AYR R,=T, (tan chord theorem) AY S AYR but i, = T, (given AT bisects MTR) ~R,=R, (4) 10.1.2 | In AAPR and AMPT 1. A=M (Zsin same segment) AY SIR 2. B, =P, (vert opp Zs) AY S/R 3. R,=T, (remaining 7s A) AAPR ||| AMPT (422) AY R (3) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Matheri2axwwvoaded from Stanmorepisysics. Cpxsparatory Examination September 2022 NSC — Marking Guidelines 10.2. | AR_PR wr PT ((\| As) AY S AYR AR= 3 MT (given) AR _3 MT 2 _PT_2 AY 5 PR 3 (3) [10] QUESTION 11 A T B D E C 1.1 | CE 1 ED 72 AY Ss @ 11.2 2 DE =, @cm) = 6cm AY Ss (1) 11.3 TE BE 2 . > =m" zu triangles) AV SIR - TE =4cm Accept solution using the Midpoint theorem AY § (2) 11.4.1 1 AY Areas 4, 5DC x Area of AADC _2 _ 6cm =1 AY Answer (2) Area of AABD BD x 6cm Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathentaawwvoaded from Stanmorepieysics. CeBparatory Examination September 2022 NSC — Marking Guidelines 11.4.2 | Areaof ATEC Area of AABC _ Areaof ATEC y Area of ATBC AY Ss Area ATBC Area AABC _ Ec aey av Ss BC AC 11 =-—x= 4 3 QB) i Av Answer 12 OR Areaof ATEC= (Area of ATBC) (common vertex=altit.) 41 Av $s == (; Area of aan (common vertex=altit.) Areaof ATEC _ 1. AY Ss Areaof AABC 12 OR AY Ss (3) Accept area rule with the use of the angle of C. [9] TOTAL: 150 Copyright Reserved