Mathematics Grade 12 Paper 2 Limpopo Mock 2023 Paper & Memo

Downloaded from Stanmorephysics.com LIMPOPO PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA Tee DEPARTMENT OF EDUCATION NATIONAL SENIOR CERTIFICATE MARKS: 150 TIME: 3 HOURS A This question paper consists of 14 pages and an information sheet. Copyright reserved Please turn over

Downloaded from Stanmorephysics.com INSTRUCTIONS AND INFORMATION Read fhe-following instructions carefully before answering the questions. L This question paper consists of 10 questions. 2. Answer ALL the questions in the SPECIAL ANSWER BOOK. 3. Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in determining your answers. 4. ANSWERS ONLY will not necessarily be awarded full marks. S. You may use an approved scientific calculator (non-programmable and non- graphical), unless stated otherwise. 6. If necessary, round answers off to TWO decimal places, unless stated otherwise. q Diagrams are NOT necessarily drawn to scale. 8. Write legibly and present your work neatly. Copyright reserved Please turn over

Wownloaded from Stanmorephysics.cOM | impopopot/september 2023 NSC QUESTION 1 A mathematics teacher wants to make an unbiased prediction of her Grade 12 learners’ final marks. She uses their SBA mark and notes the final mark. The results are as follows: SBA MA FINAL MARK (%) SBA MARK (%) | FINAL MARK (%) aps 31 B 59 35 43 72 85 69 76 57 63 62 73 25 35 83 85 65 59 75 72 68 75 1.1 Draw the scatter plot for the data on the grid provided in the ANSWER BOOK. (4) 1.2 Calculate the correlation coefficient for the data. (2) 1.3. Is the SBA mark a reliable predictor of the final mark? Provide a reason for your answer. (2) 1.4 Determine the equation of the least squares regression line. (3) 1.5 Predict Toby’s final mark if his SBA mark was 66%. (2) [13] Copyright reserved Please turn over

Downloaded:ifrom Stanmorephysics.com NSC QUESTION 2 The following set of data: 3;4;4;4;6;10; 12; 123 y has 2.1 22 a mean of 7. etermine: 1.1 the value of y .1.2 the median of this set of data points Two additional numbers, 7—n and 7 +77, are added to the data set. 2.2.1 Calculate the mean of these eleven numbers. 2.2.2 Determine the standard deviation if the data points, that are within ONE standard deviation of the mean, lie in the interval 3<.x <1 1. (2) (2) 7]

“Powtiloaded from Stanmorgphysics.com ‘imrepodot/September 2025 QUESTION 3 In the diagram below A, B (—2 ; —7), C (4; 0) and D are the vertices of a kite. E is the midpoint of the diagonal BD and AC | BD at E. The equation of AC is y= -3H+ 2. y+ D A oO £14; 0) 2 B(-2;;=-7) Determine: 3.1 The equation of BD. (4) 3.2 The coordinates of E. @) 3.3. Ifthe ratio CE: EA= 1: 3, determine the coordinates of A. (2) 3.4 Kite PQRS is obtained after the measurements of kite ABCD is d by a scale factor 2, Calculate the area of kite PQRS. (5) [14] Copyright reserved Please turn over

Downloadée'#réh Stanmorephysics.cem QUESTION 4 4.1 In the diagram, the centre of the circle is N (2 ; m) where m <0. The radius of the circle is 17 units. R (—13;5) and S(—13 ; —11) are two points on the circle, Ya R \ mx P 1 4 Oo *N(2; m) s Ne v 4.1.1. (a) Determine the numerical value of m. (4) (b) Determine the equation of the circle in the form (x-a) +(y-b) =7? (1) 4.1.2 Determine the gradients of: (a) NR (2) (b) NS () 4.1.3 The tangents at S and R intersect at P. Calculate the size of B. (6) Ha 4.1.4 Circle N is reflected about the x—axis and the slated 2 units upwards to obtain circle M. Determine the equation le M in the form (x-c) +(y-d) =r a Copyright reserved

Dewrtsaded from Stanmorephysics.com LimpopoDoE/September 2023 4.2 An infinite number of circles, each touching the next, are drawn between C and O. The centres of all the circles lie on the negative x-axis. The radius of the largest circle, centred at A, is 4 units and the radius of each circle thereafter is halved. B is a point e largest circle. rea B _— ee —>. [al 1 2 A 4 @) * v 4.2.1 Show that OC = 16 units. (2) 4.2.2 If BC is a tangent to circle A at B, write down the size of ABC, providing a reason for your answer. (2) 4.2.3 Hence, determine tan c. (4) 4.2.4 Determine the equation of BC. (3) [27] Copyright reserved Please turn over

Mathemati Downloaded from Stanmorephysies.com QUESTION 5 5.1 In the diagram below P (3 ; 4) and R (m ; —12) are two points as indicated. POW =a and ROW=f. Ss f P(3 ; 4) * (o| a ———> x [e) WwW R(m ; - 12) | Answer the following questions without using a calculator. 5.1.1 Write down the value of tana. () 5.1.2 Determine the value of sin(90°+@). (3) 5.1.3. Determine the value of m if it is given that 12+13sin 8 =0. (4) (3) 5.1.4 Determine the value of cos(a +f). 5.2 Simplify the following: 5.2.1 y4'"2°*"°* without using a calculator. (5) tan(180°+x)cosx , taba as ; sin(180° + x) cos. x—cos(540°+ x) cos(90° + x) © 2 single trigonometric expression. 0) I—cos2x-sinx 5.3. Prove that: ———————-= tanx (4) sin 2x—cos.x 5.4 It is given that P and Q are both acute angles, solve for P and Qi 7) NRi- I r sinPsinQ —cosPcosQ = a and sin(P-Q)= [33] Copyright reserved Please turn over

h ics B2 . . 4 202 Downloaded from Stanmorephtysics.com — ‘inpopodot-sepiember 2023 QUESTION 6 In the diagram, the graphs of J (x) =asinx and g(x) =coshx are drawn for x [0° ; 180°]. ya ° L 1 & * 0 90° 180° ™ 6.1 Determine the values of a and b. Q) 6.2 Consider the interval x €[0° ; 180°]: 6.2.1 Calculate the value(s) of x where asinx — cosbx =0. (2) 6.2.2 For which value(s) of x will g(x)./"(x) 20. (2) 6.2.3 Determine the values(s) of y for which p=27/°". (2) [8] Copyright reserved Please turn over

Downloaded’ frei Stanmorephysics.com QUESTION 7 The diagram shows AACE with A=0 and E=a. Points B, D and F lie on AC, CE and AE respectively so that BC =3x , CD=4x, DE=* and AF=y. BD 1 AC and D=90°. ABF=30° and DFE=30°. 7.1 Write BF in terms of @ and y. (3) 72 Write DF in terms of @ and x. (2) 7.3 Hence, prove that BD* =4x" sin? @+4y" sin? @ () _ | 4y’.sin? 0 7.4 Hence, prove that x = Pay rae (3) 19] Copyright reserved

Mathematics P2 Downloaded from Stanmorephysics.com LimpopoDob/September 2023 QUESTION 8 In the diagram, ABC is the tangent to the circle cent ntre circle. EF=BF. ABF=37°. A c Determine, with reasons, the sizes of the following: 81 EB, (2) _ 40s 82 F i (2) 83 (2) 84 6, (2) asf, ®) [10] Please turn over Copyright reserved

Mathematics P2 . 2 LimpopoDoE/September 293 Downloaded ‘from Stanmorephysics.cam : QUESTION 9 9.1 Oiis the centre of the circle. Points K, L, Mand N are on the circle. Use the diagram to prove the theorem that states that the opposite angles of a cyclic quadrilateral are supplementary, ie. K-+M =180° (5)

Dowriléaded from Stanmorephysics.com ern NSC 9 © diag: i i 9.2 Inthe diagram, RS is a tangent to the circle at R. SAB is a line that passes through the circle and RT || SAB, MT = AB. Ss 9.2.1 If ABR=.x, write down THREE other angles in the diagram which are also equal to x. Provide reasons. (6) 9.2.2 If ARB= y, provide a reason why MRT = y? (1) 9.2.3 (a) Write A, in terms of x and y. (1) (b) Write N, in terms of x and y. (1) 9.2.4 Prove that ASAR II! AKNR. (GB) 9.2.5. Prove that SAKR is a cyclic quadrilateral. (2) [19] Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Mathematics P2 14 NSC QUESTION 10 In the diagram, P is a point on side AB of AABC. The circle through P, B and C cuts AC at Q. QP prddaged cuts the circle passing through A, B and C at R. Prove that: 10.1 2 =A,+B, (3) (5) [10] RAND TOTAL: 150

“pewhidaded from Stanmorephysics.com LimpopoDoE/September 2023 INFORMATION SHEET: MATHEMATICS =b+Vb? ~ dae 2a A=P(1+ni) A= P(1=ni) T, =a+(n- tea s _alr"-1 parley -1 I (x) = tim LE+M-f@) Fe) ect h d=¥(x,-%,)? +(y,-y,)? y=me+e y-y =m(x-x;) (x-a)’ +(y-b) =r? In AABC: se ycih 2c sind sinB sinC a* =b +c —2be.cos A areaAABC = 5 absin Cc sin(a + f)=sina.cos#+cosa.sin 8 cos(a + B)= cosa.cos #—sin a.sin 8 cos’ a@—sin’ @ cos2a =41—2sin’?a@ 2cos*a@-1 Copyright reserved r-1 A=P(l-i)" A=P(1+i)" By = 5 Ga+(n-Nd) y rel § 2-2 peret l-r paclstt) "J T M x +X Yy+V2 2 " 2 Pe as m=tanO amas | sin(a — 8) = sin a.cos 8 — cosa.sin B cos(a— £)=cosa.cos B +sin a.sin B sin 2a@ = 2sina.cosa@ Ye-x oa

Downloaded from Stanmorephysics.com LIMPOPO PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA DEPARTMENT OF EDUCATION NATIONAL SENIOR CERTIFICATE MARKS/PUNTE: 150 This memorandum consists of 19 pages/Hierdie nasienriglyne bestaan uit 19 bladsye.

Downloadednfnom Stanmoreph SICS . COMimpopo DoE/September 2023 Memo NS! QUESTION/VRAAG 1 Ll PREDICTION OF FINAL MARKS = FINAL MARK i HIE TW 0 10 20 30 40 50 60 70 80 90 100 v¥vvVall dots correct/alle punte korrek ¥v¥v9-—11 dots correet/ punte korrek ¥V6-8 dots correct/ punte korrek ¥ 1 —6 dots correct/ punte korrek SBA MARK (4) 12 | r=0,94 ¥ ¥ answer/antw (2) 1.3. | The strong correlation between the SBA mark and the final v reason/rede mark implies that the points lie close to the least squares ¥ conclusion/konklu regression line. Hence the prediction is reliable. sie (2) 14 | a=14,49 ¥ value/waarde a 5=0,86 ¥ value/waarde b y =14,49+0,86x Y equation/vgl a) 1.5 | y=14,49+0,86(66) Y subst/vervang y=71,25% ¥ answer/antw Q) OR/OF y=7L18% ¥ ¥ answer/antw (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore ph SICS . COMimpopo DoE/September 2023 Memo NS! QUESTION/VRAAG 2 Copyright reserved/Kopiereg voorbehou 2.1.1 34+44+4464+10+12412+44y _ 4 9 554 v>t¥_4 9 9 55 3 Vz ¥ value of/waarde y (2) 2.1.2 | Median = 6 v6 Q) 2.2.1 padre 44+ 6484104124124 7-n4 74H 11 _ 77 v71 y= 11 z-7 “1 2 OR/OF @) x Oot FontTen v77 11 __77 y= 11 =7 v7 (2) 2.2.2 | ¥- 6, =3 ¥ equation/vg/ 7-0, =3 o,=4 Y answer/antw (2) OR/OF ¥+o,=11 ¥ cquation/vg/ Tto, =11 o,=4 Y answer/antw (2) 7] Please turn over/Blaai assebliefom

Dowslemdedninom Stanmorephysics . coMmimpopo Dot/September 2023 Memo NSC QUESTION/VRAAG 3 ya D A « 63,43) o C4; 0) x B(- 2; - 7) 3.1 I 1 YY, =m(x—-x,) m(AC)=—— ¥ m(AC) =-5 y-(-7) = 2(x-(-2)) m(BD) = 2 ¥ m(BD) =2 yt7=2(x+2) ¥ subt m and point B /verv m en punt B BD: y=2x-3 vanswer/antw (4) OF/OR 1 y=mx+e m(AC) =~ -7 =2(-2)+e m(BD) = 2 bt m and point B —7=-4+¢ c=-3 BD: y=2x-3 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore ph SICS . COMimpopo DoE/September 2023 Memo NS! Copyright reserved/Kopiereg voorbehou 3.2 1 ~3tt2 =2x-3 ¥ equating/vg/ -x+4=4x-6 —S5x=-10 ¥ value of/waarde x BD: y = 2(2)—3 y¥ value of/waarde y (3) 3.3. | A(-4; 4) through translation Y value of/waarde x ¥ value of/waarde y (2) OR/OF Find a point P so that CE = EP ¥P(0; 2) PO; 2) Then CP = PA VA(—4;4) A(-4; 4) (2) 34 | aca (-4-4y +(4-0) =4V5 V length/lengte AC BE= (2-2) +(-7-1) =45 Y length/lengte BE Area of kite = 2x area of AABC _ 1 7 2«( F455 | v method/metode =80 v80 Enlarged by scale factor 2: 80 X 4 = 320 sq unit Y answer/antw (5) OR/OF 2 v AC= (-4-4y +(4-0) =4/5 length//engte AC PR =8V5 ¥ length/lengte PR BE =,/(-2—2)' +(-7-1) =4V5 “.QS=8V5 ¥ length/lengte OS Area of kite = 2xarea of APQR =2x (3 8V5 x85 } thod/metode =320 wer/antw aA (5) [14] Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore ph SICS . COMimpopo DoE/September 2023 Memo NS! QUESTION/VRAAG 4 4.1 R(- 13;5) Ya \ S(- 13; - 11) and *N(2: m) 4.1.1 | (13-2) +(5-m) =177 @) 225+25-10m+m°? = 289 m -10m-39 =0 (m-13)(m+3)=0 m=13 or m=-3 NA OR/OF (-13-2)' +(-ll-m)" =177 225+(-11-m)’ =289 (-ll-my =64 -ll-m=48 m=-19 or m=-3 NA ¥ subst R and N/verv RenN ¥ simplify/vereenv Y std form/std vorm ¥ value/value m (4) ¥ subst S and N/verv SenN plify/vereenv form/std vorm lue/waarde m (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore ph SICS . COMimpopo DoE/September 2023 Memo NS! OR/OF (-13-2)' +(-I1-m)’ = (-13-2)' +(5—m) v¥ equating 2254? +22m+121=2254+m? -10m4 25 ¥ simplify/vereenv 32m m= ¥ value/waarde m (4) OR/OF RSI y-axis“. x =-13 same x — values Vuse line x =—13 Draw line NK | RS v¥ Use NK .. NK bisect RS at (-13 ; -3) line from centre of circle 1 | “midpt of NK to chord “N(2; 3) vy=m=-3 (4) 4.11} (x2) +(y +3) = 289 Y answer/antw (b) (1) 4.1.2 3-5 ¥ subst in gradient (a) m(NR)= 2—(-13) formula/verv in 8 gradiént formule m(NR)=— ¥ gradient of NR 15 (2) 4.12 4 —3-CI) | ™ONS)= 555) m(NS) = & ¥ gradient of NS/ 15 gradiént van NS i) 4.1.3 | NRLPR NS LPS 15 15 vtana def “.m(PR)= e .m(PS) = | ¥ value/waarde a tana = 25 tanp=—-> Vtan PB def 8 8 Yvalue/waarde B a =61,93° B =180°-61,93° B=118,07° P, =118,07°—61,93° = 56,14° ¥ method/metode «. P, =180°—56,14° = 123,86° swer/antw (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Dowslemdedninom Stanmorephysics . CoMmimpopo Dot/September 2023 Memo NSC OF/OR Inclination of NS = 28,07° Inclination of NR = 151,93° Opp 2's suppl ¥ 28,07° V151,93° Vv RNS = 2x 28,07° v56,14° ¥ Opp 2's cyclic quad/teenoorst 4'e kvh Y answer/antw (6) 4.1.4 | Reflection about x — axis: (2; —3)—>(2; 3) Shift 2 units up: (2 ; 3) (2 ; 5) Circle M: (x-2)" +(y-5) = 289 ¥ ¥ equation/vgl (2) 4.2 , ” B « 4.2.1 | Diameter =8+4+24+1 = “OC =16 ¥ sum of diameters/ som middellyne ¥ subst in sum formula/verv in som formule laa? (2) 4.2.2 | 90°, radius 1 tangent Vianswer/antw Ia Cs ason/rede lanay (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore pA SICS . COMimpopo DoE/September 2023 Memo NS! 4.2.3 | AC=16-4=12 vAC AB=4 vAB BC? = AC? — AB? (Pyth) BC? 2_ 4? BC? BC “BC 4 2 tanC =e ¥ tan ratio (4) 4.2.4 | y-y,=m(x-%) y-0= 22 C16) ¥ subst/verv m 4 Y subst point/verv B punt y= 7 x 44/2 ¥ equation/vgl (3) OR/OF a2 vee y 4 ¥ subst/verv m 2 0= we (-16)+c¢ ¥ subst point/verv punt c=4V2 Lye Bey 42 ¥ equation/vg/ (3) [27] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore pr SICS . COMimpopo DoE/September 2023 Memo NS! QUESTION/VRAAG 5 5.1 My P35 4) MA + 0 i R(m; - 12) A 5.1.1 4 tana = 3 ¥ answer/antw () 5.1.2 | sin(90°+@) = cosa ¥ reduction/reduksie 3 vr=5 “5 ¥ answer/antw (3) 5.13 | 12+13smB=0 . 12 sin B= “B ¥ std form/std vorm m? =13? -(-12") (Pyth) ¥ subst into Pyth 2 =25 Y simp]/vereenv m=t5 m=—5 ¥ answer/antw (4) 5.1.4 | cos(a+f) =cosacos #-sinasin B ¥ expansion/witbrei 3)\(-5 4\(-12 _{3)\(—5)_(4)\(/-2 v (2)( 3] (2)( B } subst/vervang 3 48 =—-—+— 13. 65 _ 33 65 swer/antw (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore ph SICS . COMimpopo DoE/September 2023 Memo NS! 5.2.1 | [asint50° 93tan225° 2 8in30° Shana Vj ° = (2 ) Bian sin 30 ¥tan 45° ¥ special values/ spesiale waardes ¥ simpl/vereenv ¥ answer/antw (5) 5.2.2 tan(180° + x) cos.x sin(180° + x) cos x —cos(540° + x) cos(90° + x) -— (tan x)(cos x) Vtan x (sin x)(cos x) —(—cos x)(—sin x) v —sinx sin x ¥ —cos x COS X P _ cos x vy —sinx —sin xcos x —cos xsin x y Sin sinx cos x ~ —2sin xcosx --_! ¥ answer/antw 2cosx (6) 5.3 1—cos2x—-sinx sin2x—cosx _ 1-(1=2sin? x)—sinx ~ 2sinxcosx—cos x _ L-142sin’ x-sinx ~ 2sinxxcosx—cosx __2sin’? x—sinx ~ 2sinxcosx—cosx _ sin x(2sin x—1) 7 cos x(2sin x—1) sinx COS NX =tanx Ycos 2x expansion/ uitbrei vsin 2x expansion/ uitbrei ¥ simpl/vereenv Y factors/faktore (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore ph SICS . COMimpopo DoE/September 2023 Memo NS! 5.4 sin Psin Q cos PeosQ => -.cosPcosQ sin PsinO= Y rearrange terms/ herrangskik terme ¥ cos identity/ Copyright reserved/Kopiereg voorbehou identiteit ¥ 2" quadrant/ kwadrant ¥ equation/vgl v‘1* quadrant/ 7. 2P =150° kwadrant P=75° ° Vv 75° Q=45 v45° (2) [33] QUESTIONVRAAG 6 6.1 a=2 and b=1 ¥ value/waarde a Y value/waarde b (2) 6.2.1 | 2sinx—cosx=0 2sin x =cosx sinx 1 cosx 2 1 v 1 tan x =— tanx = 3 x = 26,57° ¥ answer/antw (2) 6.2.2 | xe [oc : 180°] Vv interval ¥ notation/notasie (2) 6.2.3 | paper € _l 57 y 2° vv answer/antw (2) [8] Please turn over/Blaai assebliefom

Downleadednfinom Stanmorephysics .comimpopo Dot/sept QUESTION/VRAAG 7 A 7 In AABF: BF = AF snA sinB BF y . iar ¥ correct subst in sin@ sin30° . . sine rule/korrekte pr-232 a verv in sinreel 1 ¥ subst special 2 value/ verv BF=2ysin0 spasiale waarde ¥ answer/antw QB) re In AEDF: DF = DE snE sinE DF =- * = ¥ correct subst in sin@ sin 30 sine rule/korrekte DF= ssn a verv in sinreel 2 DF=2xsing i inswer/antw (2) 73 BD? = BF’ + FD? (Pyth) BD? =(2ysin 0) +(2xsina)" BD? = 4y’ sin? 0+4x° sin’ a FZ substiver in Pyth (1)

Downsloadednfénom Stanmore pr SICS . COMimpopo DoE/September 2023 Memo NS! 74 In ABDC: BD? = (4x)? — (3x)? BD? =16x° —9x? BD? =7x? ¥ BD’ =7x° 7x * sin? 6+4x° sin? @ ¥ equating 7. 2 otal = 4 2 oin2 (] on aan sm ¥ simpl/vereenv x sin’ @) =4y" sin’ 6 ve 4y’ sin’? 6 (7—4sin’ a) ~ (3) OF/OR CD? = BD? + BC? 16x? = 4y’sin’@ + 4x°sin?a, + 9x" ¥ subst/verv in 16x* —9x* —4x°sin’a = 4y’sin’0 Pyth Tx’ —4x’sin°a =4y’sin’® 2(4_ Acin2y) = Awrejn2 x (7 4sin a.) 4y°sin“0 ¥ simplivereenv xe 4y’sin’0 7—4sin’a Y ito x ve 4y’sin?0 7—4sin’a (3) 19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore ph SICS . COMimpopo DoE/September 2023 Memo NS! QUESTION/VRAAG 8 A A Cc 8.1 ABF =E, =32° tan-chord theorem YSVYR — (2) 8.2 E, =B, =32° ZS opp equal sides vS/R F=116° sum 2’s ofA vS/R (2) 8.3 D+F=180° opp Z’s of cyclic quad YSYR D=64° - - (2) 8.4 O, =2xD midpt Z=2 X circumf Z YSYR , = 128° : (2) 8.5 E,+0,+B, =180° sum Z’s ofA YSIR E, =B, Z’s opp equal sides VS/R 2B, +128° =180° 2E, =52° B, =26° (2) OR/OF ABF+ B, + B, =90° rad tangent S/R 32°+32°+B, =90° Loy B, =26° A B, = E, = 26° Z’s opp equal sides — YSIR (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Sta nmore, pre ISICS . COMimpopo DoE/September 2023 Memo NS QUESTION/VRAAG 9 9.1 Construction: Join LO and ON ¥ constr/konstr O, =2xK Zatcentre=2 X Z at circumf YSVYR 0, =2xM Zatcentre=2 Z at circumf VYS/R 0, + 0, = 360° revolution YR 2M+2K =360° equating M+K =180° G) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore ph SICS . COMimpopo DoE/September 2023 Memo NS! Copyright reserved/Kopiereg voorbehou 9.2 9.2.1 B=M,=x Z’s in the same segment YSVYR B=R,=x alt Z’s, TRU BS vVSVYR B=R,=x tan-chord theorem “SYR (6) 9.2.2 Equal chords equal circumf 72’s YR () 9.2.3(a) A =x+y ext Z of A vSIR ) a 9.2.3(b) | Nv=x+y ext Zof A YSIR 1) ( 9.2.4 In ASAR and AKNR (i) R,=R, both= x vs (ii) A,=N, both= x+y vs (iii) K, =8 sum Z’s ofA vs SA ASARINAKNR ZZ 2 (3) OR/OF In ASAR and AKNR (i) R,=R, both = x vS (ii) A, =N, both = x+y vS ASARIILAKNR ZZ 2 fire : (3) 9.2.5 | R,=8 ASARIIIAKNR pa s SAKR is acyclic quad ext Z of quadrilateral OR YR converse ext Z of cyclic quad Q) [19] Please turn over/Blaai assebliefom

Downloadednfnom Stanmoreph ‘SCS . COMMimpopo DoE/September 2023 Memo NS QUESTION/VRAAG 10 10.1 P ext Z of cyclic quad opp Z’s of cyclic quad YSYR Y¥SYR ¥ method/metode (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai assebliefom

Downsloadednfénom Stanmore pr SICS . COMimpopo DoE/September 2023 Memo NS! Copyright reserved/Kopiereg voorbehou OR/OF vSVYR a a - VSVYR ARB =180°-(A, +B, ) ¥ method/metode ,)=180°-C (5) 10.2 AABR v identify/ common 2 identifiseer A’s =A, +B, proven vs a i v 180°-(A, + B,) method/metode R, sum Z’s ofA AABR LLZ vR Ill A’s vs OR/OF (5) In AARP and AABR (i) A=A common Z ¥ Identify/ Gi) B =A,+B, proven identifiseer A’s 180°—P, =180°-(A, +B vs oe -( + :) ¥ method/metode B,=R, +R, -. AARP Ill AABR ZLiz AR _ AP Ill Ars YR AB AR v8 AR? = AB.AP (5) [10] Please turn over/s/aai asseblef om