Mathematics Grade 12 Paper 2 November 2024 NSC MG Question Paper

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2024 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 25 pages./ Hierdie nasienriglyne bestaan uit 25 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 2 DBE/November 2024 NSC/NSS – 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 Guidelines. Stop marking at the second calculation error. • Assuming answers/values in order to solve a problem is NOT acceptable. LET WEL: • 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. • Aanvaar van antwoorde/waardes 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

Mathematics P2/Wiskunde V2 3 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 Weight 34 32 40 27 33 28 27 55 39 44 30 57 28 32 31 (in kg) (x) Number of push- ups per minute 26 21 6 20 16 26 23 7 18 14 17 8 28 25 20 (y) Scatter plot 30 25 Push-ups per minute 20 15 10 5 0 0 10 20 30 40 50 60 Weight (in kg) 1.1 a = 39,456001… ✓ a = 39,46 b = – 0,590018… ✓ b = – 0,59 yˆ = 39 ,46 − 0,59 x CORRECT ANSWER ONLY: FULL MARKS ✓ equation (3) 1.2 r = – 0,8 ✓(A) – 0,8 (1) 1.3 y = 39,46 – 0,59(29) ✓ substitution y = 22,35 ✓ answer (2) OR/OF y = 22,35 (calculator) ✓✓ answer (2) 1.4 y = 18 ,33 ✓(A) 18,33 (1) 1.5 The increase in the number of push-ups will have no influence. ✓ no influence OR The standard deviation stays the same. standard deviation remains the same geen verandering / bly dieselfde (1) 1.6 6 is furthest y-value below the least squares regression line. ✓ 6 An increase of 10 push-ups will get the team member to ✓ difference is 10 (40 ; 16), the minimum number of push-ups for a player weighing 40kg. (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 4 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1 Median = 65 ✓ 65 (1) 2.2 Q1 = 44 ✓ 44 (1) 2.3 IQR = 84 – 44 ✓ 84 = 40 ✓ IQR (2) 2.4 ✓ box (A) whiskers ending at 5 & 120 5 44 65 84 (2) 2.5 Number of employees who qualify = 34 ✓ 34 34 % of employees who qualify =  100 60 = 56,67% of the employees  answer OR/OF (2) Number of employees who qualify = 35 ✓ 35 35 % of employees who qualify = 100 60 = 58,33% of the employees  answer (2) 2.6 Number of intervals = 3 ✓ 3 Time allowed to work from home = 3(30 minutes) = 90 minutes OR/OF 1,5 hours ✓ answer (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 5 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION / VRAAG 3 y A C(–9 ; 0) O M x B(–1 ; k) D(3 ; –9) −9 − 0 0 − ( −9 )  correct substitution of 3.1 mDC = OR/OF mDC = D (3 ; − 9) & C (− 9 ; 0) 3 − ( −9 ) −9 − 3 into gradient formula 3 3 mDC = − mDC = −  answer 4 4 (2) 3.2 Equation of DC: 0 = − ( −9 ) + c OR/OF y − 0 = − ( x − ( −9 ) ) 3 3  correct substitution of 4 4 C (− 9 ; 0) or D (3 ; − 9) −27 3 3 c= or − 6 y = − ( x + 9) into equation of line 4 4 4 3 27 3 27 y =− x− y =− x−  answer 4 4 4 4 (2) 3 27 k − ( −9 ) −3 k −0 −3 3.3 k =− ( −1) − OR/OF = OR/OF =  substitution of B ( − 1 ; k ) 4 4 −1 − 3 4 − 1 − ( −9 ) 4 3 27 3 k= − OR/OF k +9 =3 OR/OF k =− (8) (1) 4 4 4 k = −6 k = −6 k = −6  correct substitution of ( 3 + 9 ) + ( −9 − 0 ) 2 2 3.4 DC = D (3 ; − 9) & C (− 9 ; 0) into distance formula DC = 15 units  answer (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 6 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne ( 3 − ( −1) ) + ( −9 − ( −6 ) ) 2 2 3.5 DB =  DB = 5 DB = 5 DB 5 1  = =  answer DC 15 3 (2) DM DB 1 DM DB 3.6 = =  = DA DC 3 DA DC Area ΔMBD 1 (DM)(DB) sin D̂ ( ) = 2  correct use of area rule Area ACD 1 2 (DA)(DC) sin D̂ ( ) 1 1 BD DM =   subst. for and 3 3 DC DA 1 into correct formula =  answer 9 (4) 3.7 y = –4x + c m AD = −4 y+9 −9 = −4 ( 3) + c OR/OF = −4  correct substitution of x−3 m AD = −4 and D(3 ; –9) c=3 y + 9 = −4 x + 12 y = − 4x + 3 y = − 4x + 3  ( x − 3) + ( y + 9 ) = 612 2 2 ( x − 3) + ( y + 9 ) = 612 2 2 ( x − 3 ) + ( −4 x + 3 + 9 ) = ( 612 ) 2 2 2  substitution of equation AD into distance formula ( x − 3) + ( −4 x + 12 ) = 612 2 2 x 2 − 6 x + 9 + 16 x 2 − 96 x +144 = 612 17 x2 − 102 x − 459 = 0 x 2 − 6 x − 27 = 0  standard form ( x − 9 )( x + 3) = 0 x = 9 or x = −3  x values with rejection N/A y = –4(–3) + 3 y = 15  y coordinate A ( − 3 ; 15 ) (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 7 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF OR/OF −9 = −4 ( 3) + c  correct substitution of m AD = −4 and D(3 ; –9) c=3 y = −4 x + 3 N(0 ; 3) ✓ N(0 ; 3) ( 3 − 0 ) + ( −9 − 3 ) ✓ substitution into distance 2 2 ND = formula to calculate ND = 3 17 AD = 6 17 1 ✓ ND = AD 1 2 ND = AD 2 N is the midpoint of AD ✓ x – value ✓ y – value A(–3 ; 15) (6) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 8 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y S P(3 ; 9) T M(1 ; 3) R(d ; 1) O V N x L 4.1 L(–1 ; –3)  x = –1  y = –3 (2) 9−3 4.2 m MP = 3 −1 m MP = 3  m MP = 3 1 1 mST = −  mST = − 3 m MP 1 1 9=− ( 3) + c y − 9 = − ( x − 3)  substitution of mST & P(3; 9) 3 3 into equation of a line 1 c = 10 OR/OF y − 9 = − x +1 3 1 1 y = − x + 10 y = − x + 10  equation of tangent ST 3 3 (4) ( x − 1) + ( y − 3) = r 2 2 2 4.3  ( 3 − 1) + ( 9 − 3) = r 2 2 2 ( 3 − 1) + ( 9 − 3) = r 2 2 2 r 2 = 40  value of r 2 ( x − 1) + ( y − 3) = 40 2 2  LHS of equation of circle x − 2 x + 1 + y − 6 y + 9 = 40 2 2  expanding LHS x2 + y 2 − 2 x − 6 y − 30 = 0 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 9 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne d 2 + (1) − 2d − 6 (1) − 30 = 0  d 2 + (1) − 2d − 6 (1) − 30 = 0 2 2 4.4 d 2 − 2 d − 35 = 0  standard form ( d − 7 )( d + 5) = 0 d = 7 or d = −5 (2) d = 7 OR/OF OR/OF ( x − 1) + ( y − 3) = 40 2 2 (d − 1) 2 + (1 − 3) 2 = 40  (d − 1) 2 + (1 − 3) 2 = 40 (d − 1) 2 = 36  standard form d − 1 = 6 or d − 1 = −6 d = 7 or d = −5 d = 7 (2) OR/OF OR/OF ˆ = 90  PRL ( in semi-circle) 9 − 1 1 − ( −3) ✓ mPR  mRL = −1  = −1 3 − d d − ( −1) d 2 − 2 d − 35 = 0 ✓ standard form ( d − 7 )( d + 5) = 0 d = 7 or d = −5 d = 7 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 10 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne y S P(3 ; 9) T 4 5 4 10 M(1 ; 3) R(d ; 1) 71, 6 26, 6 116, 6 O V N x 4 5 L 4.5 mPO = 3 ˆ =3  tan POV  tan PÔV = mPO ˆ = 71,565... POV  P Ô V 1 − ( −3 ) mRL =  m RL using R(7 ; 1) & L 7 − ( −1) 1 = 2 ˆ = 1  tan RVN 2 ˆ = 26,565... RVN  R V̂ N L̂ = 71,565... − 26.565... [ext.  of  / buite  van ] L̂ = 45  answer (5) OR/OF OR/OF R̂ = 90  [ in semi-circle / in 'n halwe sirkel]  R̂ = 90  PR = ( 3 − 7 ) + ( 9 − 1) 2 2 2 PR = 80 = 4 5 units  PR = 80 = 4 5 PL2 = ( 3 − ( −1) ) + ( 9 − ( −3 ) ) OR RL2 = ( 7 + 1) + (1 + 3) 2 2 2 2 PL = 160 = 4 10 RL = 80 = 4 5  length of PL OR RL 4 5 4 5 4 5 sin L̂ = OR cos Lˆ = OR tan Lˆ =  trig ratio of L̂ 4 10 4 10 4 5 L̂ = 45  answer (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 11 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF OR/OF ( 3 + 1) + ( 9 + 3 ) = 160 = 4 10 2 2 PL =  length of PL ( 7 − 3) + (1 − 9 ) = 2 2 PR = 80 = 4 5  PR = 80 = 4 5 ( 7 + 1) + (1 + 3 ) = 2 2 LR = 80 = 4 5  length of LR 80 + 160 − 80 cos L = 2 80  160  substitution into the cos rule 2 cos L = 2 L̂ = 45  ✓ answer (5) 1− 3 4.6 mRM = 7 −1 1 =− ✓ mRM 3 mRT = 3 (tan ⊥ rad) ✓ mRT 1 mPT = − 3 mRT  mPT = −1 ✓ mRT  mPT = −1 PT ⊥ RT (3) OR/OF OR/OF 3 −1 mMR = 1− 7 1 = − 3 1 mPT = − [proved in Q4.2] 3 m PT = m MR PT || MR  PT || MR M R̂ T = 90  [radius ⊥ tangent / raaklyn ⊥ radius]  M R̂ T = 90  P T̂ R = 90  [co-int s; PT || MR/ooreenkomst. e; PT || MR]  P T̂ R = 90  PT ⊥ RT (3) OR/OF OR/OF TP̂R = L̂ = 45  [tan-chord theorem/  tussen raaklyn  TP̂R = L̂ en koord] TP = TR [tans from common pt]  TP̂R = TR̂P = 45  [s opp equal sides/  TP̂R = TR̂P e teenoor gelyke sye] PT̂R = 90  [sum of s in ∆ / binne e van ∆]  PT̂ R = 90  PT ⊥ RT (3) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 12 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 y A • O K x • P(–3 ; –4) 5.1.1 r=5  r=5 3 cos A = −  answer 5 (2) 5.1.2 cos 2A = 2cos A −1 2 2  3  substitution of cos A into = 2  −  −1  5 double angle formula 7 =− 25  answer OR/OF (2) cos 2A = cos A − sin A 2 2 2 2  3  4  substitution of cos A & sin A =−  −−   5  5 into double angle formula 7 =−  answer 25 OR/OF (2) cos 2A =1− 2sin 2 A 2  4  substitution of sin A into =1 − 2  −   5 double angle formula 7 =−  answer 25 (2) 5.1.3 y (x ; 4) B O x x = –3  x = –3 sin ( A − B ) = sinAcosB − cosAsinB  4  3   3  4  =  −  −  −  −    substitution into the  5  5   5  5  compound angle formula 12 12 = + 25 25 24 = 25  answer (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 13 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne     cos  − 45  sin  − 45  5.2 2  2  2     2cos  − 45   sin  − 45   =      multiply by 2 2 .2 2 sin ( − )  double angle = 4 − cos   co function = 4 −p 1 = OR/OF = − p  answer 4 4 (4) OR/OF OR/OF     cos  − 45  sin  − 45  2  2  2         cos 2 cos 45 + sin 2 sin 45  sin 2 cos 45  − cos 2 sin 45   ✓ expansion    = 2  2  2   2  2   cos + sin   sin − cos  ✓ special angles  2 2 2 2 2 2 2 2 = 2 1 2 1 2  sin − cos = 2 2 2 2 2 1   −  cos 2 − sin 2  2 2 2 = 2   cos 2   ✓ double angle 2 =− 4 cos  =− 4 =− p 1  answer 4 (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 14 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1.1 cos ( x + y ) = cos( x − (− y ))  ( x + y ) = ( x − (− y )) = cos x cos(− y ) + sin x sin (− y )  correct expansion = cos x cos y − sin x sin y (2) cos(90  − x ) cos y + sin (− y ) cos(180  + x ) 6.1.2 LHS = cos x cos(360  + y ) + sin (360  − x ) sin y  cos ( 90 − x ) = sin x = (sin x ) cos y + (− sin y )(− cos x ) cos x(cos y ) + (− sin x ) sin y  sin ( − y ) = − sin y  cos (180 + x ) = − cos x sin x cos y + cos x sin y =  cos ( 360 + y ) = cos y cos x cos y − sin x sin y  sin ( 360 − x ) = − sin x sin ( x + y ) = cos ( x + y )  compound angle formulae = tan ( x + y ) = RHS (6) 6.2 6 sin x − 11 cos(90  + x ) + 7 = 2 2 6sin 2 x − 11cos ( 90 + x ) + 7 = 4 squaring both sides 6sin 2 x − 11( − sin x ) + 7 = 4  cos ( 90 + x ) = − sin x 6sin 2 x + 11sin x + 3 = 0 (3 sin x + 1)(2 sin x + 3) = 0  factors 1 3 sin x = − OR/OF sin x = −  both equations 3 2 ref  = 19 ,47  no solution x = 199,47° or x = 340,53°  answers (6) 4 − 8sin x2 6.3.1 g ( x) = 3 4 (1 − 2 sin 2 x )  factors = 3 4 cos 2 x 4 cos 2 x =  3 3 Maximum value of cos 2x is 1 4 maximum value of g ( x ) =  answer 3 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 15 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF OR/OF 4 − 8 sin 2 x is a maximum when sin 2 x is a minimum Minimum value of sin 2 x is 0  min of sin 2 x = 0 4 − 8 (0) 4 − 8 (0)  max. value of g ( x ) = ✓ g ( x) = 3 3 4 g ( x) =  answer 3 (3) OR/OF OR/OF − (0) − (0) sin x =  sin x =  8  8 2 −  2 −   3  3 sin x = 0  sin x = 0 4 − 8 (0)  max. value of g ( x ) = 3 4 g ( x) =  answer 3 (3) 6.3.2 x = 180°  180° (1) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 16 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1 x = 90°  x = 90° (1) 7.2 x = −180  or x  ( −90  ; 0]  answer (2) OR/OF x = −180  or − 90   x  0  answer (2) 7.3.1 180°  answer (1) 7.3.2 3 2 g  turning points on x-axis: x = –90° ; 90° 1  shape –180° –90° 0° 90° 180°  turning point on y-axis –1 at (0 ; 2) –2 f –3 (3) 7.4 2cos3 x − sin x = 0 2cos3 x = sin x sin x 2 cos 2 x = cos x 2 cos x = tan x 2  2 cos 2 x = tan x 2 cos 2 x − 1 = tan x − 1  2 cos 2 x − 1 = tan x − 1 cos 2 x + 1 = tan x  cos 2 x + 1 = tan x x = 45° + k.180°; k  Z  answer (4) OR/OF OR/OF 2 cos x − sin x = 0 3 cos x ( 2 cos 2 x − tan x ) = 0 cos x = 0 or 2 cos 2 x = tan x  2 cos 2 x = tan x not valid 2 cos 2 x − 1 + 1 = tan x  2 cos 2 x − 1 + 1 = tan x cos 2 x + 1 = tan x  cos 2 x + 1 = tan x x = 45° + k.180°; k  Z  answer (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 17 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 D 16 122° E E C 46,85° 105,61° B 19 A DC 8.1 tan DÂC = AC 16 AC =  correct subs into trig ratio tan 46,85  AC = 15 m  answer (2) ( AB) = ( BC ) + ( AC ) − 2 ( BC )( AC ) cos BCA ˆ 2 2 2 8.2 (19 ) = x 2 + (15) − 2 x (15 ) cos105, 61  correct subst. into 2 2 cosine rule  quadratic equation in x 2 + 8, 07 x − 136 = 0 std form ( 8, 07 ) − 4 (1)( −136 ) 2 −8, 07   correct subst. into x= quadratic formula 2 (1) x = 8,30 m or x  − 16 ,38 m  length of BC ˆ = 58  BEC OR/OF ˆ EBC = 32  ˆ OR/OF EB̂C  size of BEC BC EC tan BÊC = tan EB̂C = EC BC 8,3 EC = EC = 8,3 tan 32  tan 58  EC = 5,19 m EC = 5,19 m  length of EC DE = 10 ,81 m DE = 10 ,81 m  answer (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 18 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF OR/OF ˆ sin105,61 sin CBA =  correct subst. into 19 15 sine rule ˆ = 49,5 CBA ˆ  BAC ˆ = 24,89 BAC BC 19  correct subst. into = sine formula sin 24,89 sin105, 61  length of BC BC = 8,3 m ˆ = 58  ˆ ˆ OR/OF EB̂C  size of BEC BEC OR/OF EBC = 32  BC EC tan BÊC = tan EB̂C = EC BC 8,3 EC = EC = 8,3 tan 32  tan 58  EC = 5,19 m EC = 5,19 m  length of EC DE = 10 ,81 m DE = 10 ,81 m  answer (7) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 19 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 A 1 2 46° D B 2 3 86° 1 C E 9.1 Â1 = 40  [ext.  of a cyclic quad / buite  van kvh] ✓S ✓R (2) ˆ 1 ˆ ✓S 9.2 B̂ = 80   A1 = 2 B    D̂ = 100  [opp s of cyclic quad / teenoorst. e van kvh] ✓ S/R  Ĉ2 = 40 [sum of s in ∆ / binne e van ∆] ✓ S  Ĉ 2 = Â1 = 40   AD = DC [sides opp = s / sye teenoor gelyke ] ✓ R OR/OF (4) ˆ 1 ˆ B̂ = 80   A1 = 2 B  ✓S   AĈE = Â 2 + B̂ [ext  of ∆ / buite  van ∆] ✓ S/R ✓ S  Ĉ2 = 40  Ĉ 2 = Â1 = 40   AD = DC ✓ R [sides opp = s / sye teenoor gelyke ] (4) OR/OF ˆ 1 ˆ ✓ S B̂ = 80   A1 = 2 B     Ĉ3 = 180  − 46  − 80  [sum of s in ∆ / binne e van ∆] ✓ S/R  Ĉ 3 = 54   Ĉ 2 = 180  − 86  − 54  [s on a str. line / e op 'n reguitlyn]  Ĉ 2 = 40  ✓ S  Ĉ 2 = Â1 = 40   AD = DC [sides opp = s / sye teenoor gelyke ] ✓ R (4) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 20 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 10.1 R h2 h1 A B S T 10.1 Construction: Join SB and TA and draw h1 from B ⊥ AR and ✓ construction h2 from A ⊥ RB Konstruksie: Verbind SB en TA en trek h1 vanaf B ⊥ AR en h2 vanaf A ⊥ RB Proof/Bewys: 1 area ΔRAB 2 RA  h1 1 RA  h1 area ΔRAB 2 RA ✓ = = = area ΔASB 1 AS  h area ΔASB 1 AS AS  h1 2 1 2 RA ✓ AS 1 RB  h2 area RAB RB area ΔRAB 2 RB ✓ = = = area ABT BT area ΔABT 1 BT  h BT 2 2 area ∆RAB = area ∆RAB [common/gemeenskaplik] But area ∆ASB = area ∆ABT [same base & height; AB || ST/ dies. basis & hoogte; AB ||ST] ✓ S ✓ R area ΔRAB area ΔRAB  = area ΔASB area ΔABT RA RB  = AS BT (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 21 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne 10.2 P R S O V T W 10.2.1 PR = RS [line from centre ⊥ to chord/ ✓ S ✓R lyn vanuit midpt. sirkel ⊥ op koord] PO = OW [radii / radiusse] ✓ S  OR = 1 WS [midpt theorem/midpt. stelling] ✓ S✓ R 2  OR : WS = 1 : 2 (5) OR/OF ˆ PSW = 90° [ in semi circle/ in halwe sirkel] ✓ S ˆ PRO = 90° [given] ˆ ˆ = PSW  PRO  RO || SW [corresp s = / ooreenk. e =] ✓ S OR/OF [co-int. s suppl / ko-binne e suppl] PO PR ✓ S ✓ R = [prop theorem; RO || SW/ OW RS lyn || een sy van ∆] PO = OW [radii / radiusse]  PR = RS  OR : WS = 1 : 2 [midpt theorem/ midpt. stelling] ✓ R (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 22 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF ∆PRO and ∆PSW ˆ = 90° PSW [ in semi circle/ in halwe sirkel] ˆ = 90° PRO [given] ˆ ˆ = PSW  PRO P̂ is common P Ô R = P Ŵ S [sum of s in ∆/ som van  e in ] ✓ S ✓ R ∆PRO ||| ∆PSW [] PO RO ✓ S  = [||| ∆s / ||| ∆e] PW SW ✓ S but PW = 2 PO [diameter = 2 radius/middellyn = 2 radius] RO PO ✓ S  = SW 2PO 1 = 2  OR : WS = 1 : 2 (5) OV RS 1 10.2.2 = = [prop theorem; RO || SW/ ✓ S /R VT ST 3 lyn || een sy van ∆] RS 1 = 15 3 RS = 5 units ✓ S PR = RS = 5 units [line from centre ⊥ to chord / ✓ S lyn vanuit midpt. sirkel ⊥ op koord]  PT = 25 units ✓ answer (4) [15] 10.2 P R S O V T W Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 23 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 11 A 3 1 2 D 1 2 1 F 2 2 3 1 4 1 2 C B E G 11.1 D̂1 = EÂG = x [tan-chord theorem/  tussen raaklyn ✓ S ✓ R en koord] Ĉ1 = D̂1 = x [tan-chord theorem/  tussen raaklyn ✓ S ✓ R en koord] Ĉ 4 = Ĉ1 = x [vert opp s = / regoorst. e] ✓ S/R AF̂G = E ÂG = x [tan-chord theorem/  tussen raaklyn ✓ S en koord] (6) OR/OF EA = EC [tans from common pt/ ✓ S/R raaklyne vanuit dies. punt] Ĉ1 = EÂG = x [s opp equal sides/ ✓ S e teenoor gelyke sye] Ĉ 4 = Ĉ1 = x [vert opp s = / regoorst. e] ✓ S/R D̂1 = EÂG = x [tan-chord theorem/  tussen raaklyn ✓ S ✓ R en koord] AF̂G = E ÂG = x [tan-chord theorem  tussen raaklyn ✓ S en koord] (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 24 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne A 3 1 2 D 1 2 1 F 2 2 3 1 4 1 2 C B E G 11.2 ˆ = AFG D ˆ =x  S 1  DC || FG [corresp s = / ooreenk e = ]  S/R AG AF = [prop theorem; DC || FG /  S R AC AD lyn || een sy van ∆]  AG.AD = AC.AF (4) OR/OF In ACD and AGF Â 3 is common  S AF̂G = D̂1 = x [proved in 11.1 / reeds bewys]  S ˆ = AGF C ˆ =x [sum  s/binne e ] 2 ACD ||| AGF [ ]  S/R AC AD = [||| s  sides in proportion /  S AG AF ||| e  sye in dieselfde verhouding] (4)  AG.AD = AC.AF 11.3 In  AGF and ΔABC Gˆ =Bˆ [s in the same seg /  S R 2 e in dies. segment] ˆ =C AFG ˆ =x [proved in 11.1 / reeds bewys]  S 1 ˆ A =Aˆ [sum of s in ∆/binne e van ∆]  S 3 2  AGF |||ΔABC [ ] OR/OF R (4) Copyright reserved/Kopiereg voorbehou

Mathematics P2/Wiskunde V2 25 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne 11.4 GF AF = [ AGF ||| ΔABC ]  S/R BC AC BC.AF  GF =  S AC  ACD |||ΔFGC [ ]  S AC AD  =  S GF FC AD.FG  AC = FC AD.FG  GF = BC.AF   S FC FC GF = BC.AF  AD.FG  S BC.FC.AF  GF2 = AD (6) OR/OF OR/OF  AGF|||ΔABC [ ] GF AF =  S BC AC AF.BC  S GF = AC  ACD |||ΔAGF [ ] AD CD = AF GF AF.CD GF =  S AD AF.BC AF.CD GF×GF = .  S AC AD ΔFCD ||| ΔFAC []  S FC CD = from ||| Δ`s FA AC CD.AF FC =  S AC AF.FC.BC GF2 = AD (6) [20] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou