Mathematics Grade 11 P1 National Exemplar November 2019 QP

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Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 2 DBE/November 2018 CAPS/KABV — Grade/Graad 11 - Memorandum NOTE: e Ifacandidate answered a question TWICE, mark only the FIRST attempt. e Ifacandidate crossed out an answer and did not redo it, mark the crossed-out answer. e Consistent accuracy applies to ALL aspects of the marking memorandum. ¢ Assuming values/answers in order to solve a problem is unacceptable. LET WEL: As 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na. As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, sien die deurgehaalde antwoord na. Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. Dit is onaanvaarbaar om waardes/antwoorde te veronderstel om 'n probleem op te los. QUESTION/VRAAG 1 4 | 12 | 13 | 16 | 17 | 18 | 20 | 22 | 22 | 25 1.1 | 4 minutes/ minute Y answer/ antwoord Q) . v ba Mean/ gemiddel: =169 169 169 10 ¥ answer/ antwoord _ (2) 1.3 | Standard deviation/ Standardafwyking = 5,79 ¥ answer/ antwoord Q) 1.4 | (16,9-2x5,79;16,9 +2x 5,79) VY X-26 (5,32 ; 28,48) V¥X+20 .. | member of the team completed the obstacle race outside of 2 ¥ answer/ antwoord standard deviations of the mean./ 1 lid van die span het die hundernisbaan buite twee standardafwykings (3) van die gemiddeld voltooi. 1.5 169+x+5 _J¢ ¥169+x45 20 x =18x 20-174 dividing by 20/ deel deur 20 x= 186 ¥ answer/ antwoord @) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 QUESTION/VRAAG 2 3 CAPS/KABV — Grade/Graad 11 — Memorandum DBE/November 2018 Cumulative frequency graph of the ages of soccer players/ Kumutlatiewe frekwensie -grafiek van sokkerspelers se ouderdomme | 50 EERE Se er 4 | TI] 2 Ea = — = === =a SEER 2 40 SSSSSe Soccee === 2 Ss =t REE EEEEFEEH yy == == | s == == = — & 30 Rot 3 t = aS =o t | > T Ba : =F SSSSenee & f = : | £ 10 ao 4 3 t Arn 4 J | S| | | a 1 9 I oo I == 0 5 10 15 20 25 30 35 40 Ages of soccer players/ Ouderdomme van sokkerspelers 2.1.1 | 50 players/ spelers ¥ answer/ antwoord 1) 2.1.2 | 40-10 =30 players/ spelers v'40 and/ en 10 ¥ answer/ antwoord 2) 2.1.3 Class interval/ Frequency/ Cumulative Klas interval Frekwensie frequency Kumulatiewe Srekwensie 15sx<20 4 4 20<x<25 9 13 255% <30 24 37 30<x<35 10 47 35<x<40 3 50 ¥ two correct values/ twee korrekte waardes ¥ three correct values/ drie korrekte waardes ¥ all correct valucs/ al die waardes korrek (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai am asseblief

Downloaded from hlayiso.com DBE/November 2018 2.1.4 Mathematics/P2/Wiskunde V2 4 CAPS/KABV — Grade/Graad 11 — Memorandum Class interval/ Class midpoint/ Frequency/ Klas- interval Klas- middelpunt Frekwensie 15S<x<20 17,5 4 [ 20<x<25 22,5 9 25<x<30 27,5 24 30<x<35 32,5 10 35s x<40 37,5 3 Frequency polygon of the ages of soccer players/ Frekwensie- veelhoek van sokkerspelers se oudermme Y using midpoints At least 25% of class A have lower marks than any learner in class B./ ten minste 25% van klas A het laer punte as enige leerder in klas B. 25 + / gebruik [ middelpunte 20 Y plotting the & iY points correctly/ < korrekte punte S15 P a geplot S we x 10 Y points joined by — v4 straight line/ punte 2 4 : 4 5 ! a verbind met ‘n B55 4 reguitlyn 2 A : iy Y grounding at/ 0- TT AVL | = LON geanker by 0 5 10 15 20 25 30 35 40 (12,5;0) and/ en Ages of soccer players/ Ouderdomme van sokkerspelers (42,5 ; 0) (4) 2.2 | The claim is not valid. / Die bewering is nie geldig nie Y claim not valid/ Range of class/ Omvang van Mas A= 60 ewering me Range of class/ Omvang van Has B = 40 getalg nie The range of class A is bigger than the range of class B. Therefore the “ comment on the marks of class A are more spread out than the class B./ overall spread/ Die omvang van klas A is groter as die omvang van klas B. Dus is die punte | kommentaar oor in klas A meer verspreid as klas B die algehele verspreiding Y comparison of the lower marks/ Class A performed worse at the bottom end. / vergelyk laer punte Klas A het slegter gevorder aan die onderste groep GB) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 5 DBE/November 2018 CAPS/KABV — Grade/Graad 11 - Memorandum QUESTION/VRAAG 3 y B(6; 5) A(-2; 2) E N (50) x M C(O; -3) D(x) 3.1 B(6;5) C(0;-3) OR/OF mtg, = 222! mm, <2 x, —X, ‘BC oH 5-3) -3-5 ~e.0 ¥ subst into correct 7 0-6 8 grad.form / verv in _ -8 = 6 grad form. = ; ¥ answer/ antwoord “3 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 6 DBE/November 2018 CAPS/KABV - Grade/Graad 11 — Memorandum 3.2 4 4 Man = Mac = 5 (AD || BC) Mp = = 4. +e 3 2-42) +e Y subst of mand 3 point (-2;2) / 14 very. m en punt —=e 3 (-2;2) a4 414 Yass ¥ answer/ antwoord OR/OF G) 4 my = 5 (ADI BC) ¥ my =z y-2=4(¢-(-2)) Y subst of mand 3 point (-2;2) / yoda verv. m en punt —2;2)/ er (-2:2) “y 30 3 Y answer/ antwoord (3) 3.3 _4 xt 14 rhe) 0 =4,,14 Y subst/verv. y=0 3. 3 zl4_4 3. 3 azl4_-7 ¥ answer/ antwoord 4 2 (2) 3.4 2 2 AN= (x. -%) +(9,-y,) ny ; ¥ subst. in distance = () - (-2)] +(2- 0) formula/ very. in : i afstand formule 4 => 2 ¥ answer/ antwoord Q) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 7 CAPS/KABV — Grade/Graad 11 —- Memorandum DBE/November 2018 35 3y -3= 4. + 14 Y equating/ vergelyk 8 3 3 23 _ 23 ¥ simplifcation/ 24 "= 3 vereeny. x=-8 v x- value/ waarde 4 14 =—(—8)4— agar =-6 v y- value/ waarde D(-8;-6) (4) ® 6-(-2) 8 B= 8 Mag = Myc ~ ABDC Y ABIIDC but/maar AD||BC .. ABCD isaparallelogram — [oppsidesare||/teenoorst sye is||] Vteason/ rede OR/OF @B) 2 2 AD= (x. -%) +(9%)-¥) 2 2 = \((-2)-(-8)) +(2- 6) = °° v length of AD/ 7 lengte van AD 2 2 BC= (x. -x,) +(%,-») =, (6-0) +(5-(- 3) y Y length of BC/ = 100 lengte van BC =10 -. AD =BC but/maar AD||BC A _ ¥ reason/ rede «. ABCD isa parallelogram 2 oppsides are =and|i teenoorst syeis =en || 6) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 8 CAPS/KABV — Grade/Graad 11 — Memorandum DBE/November 2018 Mis the midpoint of AC M is die middelpunt van AC (-2)+0 2+(3) Mh see) 2 2 a! 2 is the midpoint of BD is die middelpunt van BD XR RE > 2 2 M -1-4) 2 di Is bisect each oth -. ABCD isa parallelogram iagonals bisect each other hoeklyne halveer mekaar Y midpoint of AC/ middelpunt van AC ¥ midpoint of BD/ middelpunt van AC Vv reason/ rede (3) 3.7 | Mis the midpoint of AC [diagonals bisect] M is die middelpunt van AC [hoelyne hat veer| a C22, 229) 2 2 Y Substitution into the correct formula/ Verv. in korrekte form. ¥x- value / waarde vy- value / waarde @G) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 9 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum QUESTION/VRAAG 4 * G(5;2) < B 04,04 . x F(2;-2) C;y) 4.1 _ 2-(-2) Y subst. into correct Meg = 5-2 gradient form./ vervang 4 in gradiént formule =5 Y answer (2) 42 tye = = @ciF6) Y mg= a yt2 -3 ¥ equating gradients/ 6-2 4 stel gradiénte gelyk ys ¥answer/ antwoord (3) OR/OF Copyright reserved/Kopiereg voorbehou Please tun over/Blaai om asseblief

Mathematics/P2/Wiskunde V2 Downloaded from hlayiso.com DBE/November 2018 10 CAPS/KABV ~ Grade/Graad 11— Memorandum Mp X Mpg =—-1 (FC-LFG) yr2 4 6-2 3 A(y+2)=-12 Yo Myo X Meg =I yt+2=-3 ¥ substitution/ verv. yas vanswer/ antwoord (3) 43 | tanp=4 Y tanp=4 3 3 B=53,13° ¥ B=53,13° 0=104,04°—53,13° [ext Zof A/ buite Zvan A] 0 =50,91° ¥ answer/ antwoord GB) ¥ subst. into distance 44 | FC=\(x,.-x,F +(y.-») su ( : J ez ») formula/ very. in afst. = \(6-2) +52) form Y length of FC / =v16+9 =5 sin? = FC NC sin 50,91° = — NC ca 3 sin 50,91° = 6,44 unit lengte van FC Y sin 50,91° = — NC Y answer! antwoord (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 1 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum QUESTION/VRAAG 5 ty T(x;15) P C 23 a) re oe Oo x 5.1.1 x+y=r? [Pythagoras] (x)? +(15)? $17? v subst in pyth/ verv x =64 in pyth x=-8 (P isin quadrant 2/ is in kwadrant 2) Y answer! antwoord (2) 5.1.2 15 tand-~— ¥ answer! antwoord QQ) 5.1.3 cos(180°-8) =—cos@ ¥Y -cos@ -_ (# 17 8 = ¥ answer/ antwoord (2) 5.1.4 sin °0 = (S) ¥ substitution/ 17 vervanging = 72) v / antwoord 289 answer/ antwoor (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 12 DBE/November 2018 CAPS/KABV — Grade/Graad 11 —- Memorandum 5.1.5 | tan@= = 15 Y tanga —2 -8 -2 | a 15 : | TT 18 Y equating/ stel gelyk a5 ¥ answer/ antwoord 4 (3) OR/OF _15 8 15 15 v yar y=—x _ -8 15 (2) Y substitution of ~_gt P(-2;a)/ 15 vervanging van a P(-2;a) ¥ answer! antwoord @) 5.2 LH — 82 !20°.cos 210° tan 315°.cos27° ~ cos 540°.sin 63° Y sin 60°/ cos 30° vo Oo sin60°,(—cos30°)(-tan45)°.sin63° ¥ oe = ~tan cos180°.sin 63° . ¥ sin 63°/cos27° 3-3.) Y cos180° = 22 Y special angle 3 tatios/ spesiale hoeke = 4 verhoudings ¥ answer/ antwoord (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 13 DBE/November 2018 CAPS/KABV ~ Grade/Graad 11 — Memorandum 53 LHS = 1 cosd cos@ 1+sin0 Y common : 2 _ 1+sin @-cos* 6 denominator/ gemene cos al +sin 6) noemer _1+sin@-(1-sin? 9) - cos O(1+sin9) v1-sin’@ _sin@(1+sin 6) ¥ simplification/ cos al +sin 6) vereeny. sin@ = cos | ¥ factors/ faktore =tan@ | : “Rus y Sin Gi ~ cos OR/OF () LHS= 1 _ cos6 cos@ 1+sind y common : 2 _ l+sin@—cos" @ denominator/ gemene cosO(1+sin 0) noemer _ (1-cos” 9 +sind ; cosO(1+sin8) v1-cos' 0 __ sin? O+sind ¥ simplification/ cos@(1+sin 6) vereeny. _ sin@(1+ sind) v cosd(14sind) @(t-+sind) factors/ faktore _ sind v sind cos@ cosé =tand =RHS ©) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 14 DBE/November 2018 CAPS/KABV — Grade/Graad 11 - Memorandum LUs= 1 __ cos cos@ 1+sin0 _ l+sin@-cos’ 8 cos@(1+sin@) _ sin? 0+cos’ 0 +sin@-cos’ 0 cos al +sin 0) _ sin’ O+sin0 cos@(1+sin A) sin O(1+ sin) ———Ss ¥ common cos@(1+sin 8) : . denominator/ gemene sind = noemer cos@ =tan@ ¥ sin? @+cos’ 0 =RHS ¥ simplification/ vereeny. ¥ factors/ faktore v sind cos? (5) 54 3 sinx= aun 3sinx=2 x 2% sinx cos x v 3sin xcosx=2sin x cos 3sin xcosx—2sinx=0 Y factors/ faktore | . sin x(3cosx—2)=0 sinx=0 : 3 Y both equations/ x= 360°K, kez beide vergelykings or x =180°+360°k, kEZ ¥ both general or solutions/ beide | algemene oplossings | Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 15 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum cosx == 3 x =48,19° + 360°.K, KEZ Y both general or solutions/ beide _ ° ° algemene oplossings x =311,81°+360°A, kKEZ VkeZ (6) OR/ OF 3sinx= 2tanx. sin x 3sin x=2 . sin x=2 x / sinx 3sin xcosx=2sin x cosx 3sin xcosx—2sin x=0 sin x(3cosx—2)=0 Y¥ factors/ faktore sinx=0 | . x=180°k, ke Z | 7 both equations/ 4 beide vergelykings cosx = 3 Y general solution/ x=+48,19°+360°K, ke Z algemene oplossing Y¥ both general solutions/ beide algemene oplossings VkeEZ (6) [28] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 16 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum QUESTION/VRAAG 6 a | tal N 99 Q =F . | Y end points/ eindpunte Y notation/ notasie (2) 6.2 | c=1 Y answer/ antwoord (1) 63 | a=3, b=1 Yan3 Vb=l1 (2) 6.4 | Q(122°;-1,6) Y x- value/ waarde ¥ y- value/ waarde (2) 6.5 K(-45°: 3V2 Y coordinates of/ c >? koGrdinate van K. M(—45°:-1) KM 02 +1 3 2 +2 - tenets off lengte 2 =312 2) 6.6 | f(x)=3cos(@-45°) 3 v —45° (2) {11] Copyright reserved/Kapiereg voorhehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum QUESTION/VRAAG 7 3cm, 67° 92cm R Q 7A QR? =PR? + PQ? —2PR.PQcosP QR? =(3) +(9,2) ~2(3X9,2)cos67° QR = (3) + (9,27 -26)9,2)c0s67° QR =8,49em ¥ using cos rule/ gebruik cos reél ¥ substitution/ vervanging ¥ answer/ antwoord GB) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 18 DBE/November 2018 CAPS/KABV — Grade/Graad 11 - Memorandum A Up 2 () a Cc B 72.1) ACD=0-a ¥ answer/antw. (1) 7.2.2 D. =90° A D, =90° +@ YD, =90' +a sin(90°+a@) _ sin (0 - a) h AD v sin (90° +a) cosa _ sin(9-«) h h AD in (0 -«) . sin (9 -a@ AD= hsin(6-a) v AD cosa ¥ sin (90° + a@)=cosa (4) oF Oo _ ° vo 723 | yy — 17sin (58° —23°) subst/ very. cos 23° v J AD = 10,59 units answer/antw. (2) ts Areaof AADC = 4x ADx ACxsinA Y correct area rule/ ; 2 korrekte area reél -—_— 1 ° = 5 %10,59x17 xsin32 v302 = it? = 47,70 unit Y answer/antw. (3) OR/ OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 19 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum sin58° = AB 17 AB =17sin58° =14,41682.... BD =14,41682....-10,59289... = 3,82393... _ 3,82393... CD 3,82393... Vlength of BD / De sin 23° lengte van BD = 9,78660... sin 23° 1 . “length of CD/ Areaof AADC = 3% CD x AC x sin35° 1 lengte van CD = 5 x 9,78660...x 17 x sin35° = 47,71 unit? Y answer/antw. (3) [28] Copyright reserved/Kopiereg voorbehou Please tum over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 20 CAPS/KABV — Grade/Graad 11 — Memorandum QUESTION/VRAAG 8 DBE/November 2018 8.1 r? =13?—h? (Pythagoras) Y using theorem of r =169—h? pythagoras/ gebruik stelling van pythagoras v= xan i Yr? =169-h° =-arr’h 3 Y¥ substitution/ =30(169—h? Yh vervanging 169h—ah> Y simplification/ = 3° ~~” vereenvoudig (4) 8.2 r= 13-12? (Pythagoras) | v value off waarde =5 van r Total surface area/ buite oppervlakte = nr? +21 s = (5°) +2(5)(13) =90r ¥ subst. / verv. = 282,74 cm? Y answer/ antwoord (3) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 21 DBE/November 2018 CAPS/KABV — Grade/Graad 11 —Memorandum QUESTION/VRAAG 9 91.1 6, = 78° [angle at centre = 2x Z at circumference ] YSVYR [middelpuntshoek = 2x omtrekshoek | (2) 9.1.2 0, =78° [equalchords; equal 2*/ gelyke koorde; gelyke hoeke] YSYR (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 22 CAPS/KABV — Grade/Graad 11 —- Memorandum DBE/November 2018 R 9.2 angle at centre = 2x Z at circumference/ x —8°+180°= 2(2x-40°) middelpuntshoek = 2x omtrekshoek 4x—80° =172°+x 3x = 252° x = 84° OR/OF JoinT and R/ verbind T enR - opp Z's of cyclic quad/ T=180° —(2x-40°) teenoorst.Z° van koordevierhoek R=T=220°-2x la opp.=sides /Z* teeoor gelyke sye ] sumof int Z° of A x —8°+220°—2x +220°-2x =180° som binne Z° van A —3x=- 252° x=84° YSVYR v simplification/ vereenvoudiging ¥ answer/ antwoord (4) Y¥SVYR vs Y answer/ antwoord (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 23 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum 9.3 AB=BC=4x line fromcentre | to chord/ YSYR lynvanmiddelpunt . aankoord OA? = (4x)? +(2x)? [Pythagoras] ¥ Substitut _ ; 5 ubstitution/ OA=Vi6x? +4x vervanging = 20x? =2V5x v OD=O0A=2V5x _ [radii) length of OA / lente van OA BD=2¥5 x -2x v =2x(/5-1 *WS5-1) BD=2V5 x -2x (5) (13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 24 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum QUESTION/VRAAG 10 10.1 Construction: Draw radii OR and OT Y construction/ Konstruksie: teken raduise OR en OT konstruksie Let/ Jaat: =x and/en B = y A A it =2ti i 6, =28 angle at centre = 2 times angle at circumference/ YS VR middelpuntshoek = 2 keer omtrekshoek 6, =2x Similarly/ in die selfde manier: 0, =2y vs 2x+2y = 360° [anlgesarounda pt/hoeke om'n punt] v SIR x+y =180° -.8+B=180° (3) L_ _ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 25 CAPS/KABV — Grade/Graad 11 - Memorandum DBE/November 2018 10.2.1(a) 6=8, =48° ext Z of cyclic quad/ VS YR at buite Z van 'n koodervierhoek Q, =Q, =24° [QS bisects/ halveer PQN] VS (3) 10.2.1(6) | R= Q, = 24° [za in the same segment/ in dieselfde segment | YS YR (2) 10.2.1(c) | Mf, = Q = 48° [corresp/ ooreenkomst Z*, PQ|| SR] YSYR OR/OF (2) 8, = Q, = 24° [alt Z°/ ooreenkomst Z°, PQ|| SR] ¥S/R M, = 48° [ext Zof A/buite Zvan A] Y answer/ antwoord (2) 10.2.2 M, =§, =4g° vs ..STisa tangentto circle MNS.{converse tan—chordtheorem]| yp «ST is 'n raaklyn aan MNS lomgekrd raaklyn-koord st] (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 26 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum QUESTION/VRAAG 11 1d fi =x [tan — chord theorem/ raaklyn-koord st| v SIR 4 Ox Z atcentre= 2x Z at circumference/ vS ' middelpuntshoek = 2 keer omtrekshoek VS P, =90°—x [tan L diameter/raaklyn 1 middellyn] R, =90°- x|Z opp.=sides/ Z° teeoor gelyke sye| VR, =90°—x Rak, Us PR bisects /halveer ORN 6) OR/ OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 27 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum RPT =90° [in the semi circle/ in dieselfide segment] P= x =x [tan—chord theorem/ raaklyn-koord st| R= =90' 0° ~ x[sum int Z* of A/ som binne Z° van A] R, =90' 0° —x [sum int Z° of A/ som binne Z! van A] vs wR =R, Y SR PR bisects/ halveer ORN vs vs OR/ OF r, =x {tan — chord theorem/ raaklyn-koord st] vS Pp, = x[Z opp.=sides/ Z* teeoor gelyke sye | 6, =2x [ext Z of A/buiteZ van A] (5) s 5 180°- . v SIR R, =P, = ——— 2k Ls opp.=sides/ Z* teeoor gelyke sye ] 2 v A Ss R, =90°-x F) ways . s vs R, =90°-x [sum int Z* of A/ som binne Z* van Al wR = R, “Ss PR bisects/ halveer ORN vs G) 11.2 PAT =90° + x | OPP Z's of cyclic quad/ VS VR teenoorst. hoeke van koordevierhoek A or: vS YR S, =90° [Linefrom centre to chord / dyn van mdpt | aanko ROS=90° +x [ext Zof A/buite ZvanA] VSIR -, ROS =PAT OR/ OF (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 28 DBE/November 2018 CAPS/KABV — Grade/Graad 11 — Memorandum RPT =90° [Zin the semi circle/ in dieselfde segment] NPT = 90° + x PAT =NPT = 90° + x [tan — chord theorem / raaklyn-koord s B, =x|2* opp.=sides / 2° teeoor gelyke sye] s, =90°[Line from centre L to chord / lyn van mdpt |.aankoord 0, =90°—x|sum of int 2° of A] ROS= 6, +0, =2x+90°-x =90°+ x v8 “. ROS =PAT VS VR YSIR vs 6) {10} TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou