Mathematics Grade 12 Paper 2 NSC June 2013 Memo

Downloaded from hlayiso.com basic education Department: Basic Education REPUBLIC OF SOUTH AFRICA SENIOR CERTIFICATE EXAMINATION MATHEMATICS P? | HIGHER GRADE | ™: 2013 POSSIBLE ANSWERS _ MARKS: 200 This memorandum consists of 31 pages.

Downloaded from hlayiso.com Mathematics/HG/P2 2 DBE/2013 SCE-Final Memorandum 14 May 2013 QUESTION 1 [26] AC1;4)t B(p;-2) C5 ; 4) v 1.1 | tan 135° =-1 “A 1A tan 135° =- vM VA 4-€2) 1) vM -l-p l+p=6 VCA p=5 OR tan 135° = -1%A YM v y-4 = -1@-C1) VA a. Sub: forB: y=-2 -2=-x+3yce4 fs Jee! b) OR 1M gradient formula 1 correct subs 1M equating the gradient simplification tan 135° =— value of m equation of line substitution into the line simplification hh l4-05-1Y

Downloaded from hlayiso.com Mathematics/HG/P2 3 SCE-Final Memorandum 14 May 2013 DBE/2013 11 Assume B is (5 ; — 2) Assumption Cont. vA —~ 4-¢2)_ 6 _ Gradient of line mB" 1-5 6 “1 WM VA correct gradient YM vA Og inclination of AB is: tan(— 1) = 135° calculating inclination conclusion fd So) G) 12 |AC = BC YM the distance AC2 = BC2 vA LHS sub C5-CbDP + @- 4 = 5-5 +@-C2DY VA | RHS sub 16+ @—-8q+16 =100+q’+4q+4vS Multiplication —8q-4q =104-32v§ Simplification —12q = 72 q=-6YCA (6) | value of g 13 : 7 ett at (xy sv u) = ( 2° 2 oe ae 4+(2) vM correct substitution id 2 2 into mid-point formula M(2;1) “CA VCA correct x and y values vw) _ 6-1 © he YCA correct subst into =e) gradient formula _ gradient CM map=—lva gradient AB =-1 YCA mem * map =—I product of gradients CM Ll AB OR os 1+5 ~4+¢2) vM correct substitution te! 2 7 2 into mid-point formula M(2;1) YCA VYCA correct x and y values AC = (-5-(-1)? + (6 - 47 = 116VA Correct length CM = (5-27 + - 17 =98 VCA Correct length AM’ = (-1-27 + (4-17 = 18 vCA Correct length TER 2 ¥CA —L EDUCATION SIC ] Pythagoras i CML AB on0i3 -06- 24 bene ETORIA 8004 PUBLIOE eect | i XAMINATIONS PRIVATE BAG Kk 40 a we? Llir L 14-05-13

Downloaded from hlayiso.com Mathematics/HG/P2 4 DBE/2013 SCE-Final Memorandum 14 May 2013 1.3 Cont. —14+5 4+(-2) vM correct substitution = 3 a into mid-point formula M(2;1) VCA VCA correct x and y values BC?=116 VA Correct length CM? =98 VCA Correct length 2 BM? = 18VCA Correct length BC? = CM? +BM? VCA Pytha goras “CM 1 AB 7) 1.4 | Ax=xa-xc = xp—xB -1+5=xp-5 YM change in x 9 =xp YCA x value of D Ay=ya-Ye = W-ya v : 446 = yt? M change in y 8 =y VCA D(9;8) y value of D OR Ax=xp—Xc =Xp—XA vM OR Oa oe Gy) change in x : =9 YCA gt Sw x value of D Ay=ya—Yo=Yo-Ya ie ~ a OAR (4) “Mi change in y “Yo y value of D OR M is the point of intersection of the vertices of the parm. Let D be (xp; yp) = *%c+%p xy 2 —~54+x, Sub into midpoint 2= ee vM formula *> = 9 VCA x value of D — Yct+¥d ym 2 i= 29+2> ve Sub into midpoint 2 formula yo = 8 VCA y value of D YVA Answer bint DQ;8) “%A ; EDUCATION (4) 2613 -06- 24 —PUGIC ExaMinarions polo \t3 1

Downloaded from hlayiso. com Mathematics/HG/P2 5 DBE/2013 SCE-Final Memorandum 14 May 2013 1.5 -6-4 5 VCA gradient of AC Mc = TG -5-(-l) 2 5 mM y- 62) = > &-5) va . 2 sub into line formula y= 5 x- 25 _ 2 correct gradient used 2 2 5 29 y any oe simplification OR AC //BD {ACBD is a parm) -8-©2) _5 vea gradient of BD 1D 9-5 2 v) | bY sub into line formula y-(2) = 3 @-5) vA correct gradient used 5 25 =2y- 2-2 4 2 . 2 =35 oS 29 V¥CA simplification 2 2 (4) pore EDUCATION BASIC 2013 -05- 24 PRIVATE BAG x 179 PRETORIA o004 —LUBLIC EXAMINATIONS] ran Satnemencens ow np” Hort 14-05-13

Downloaded from hlayiso.com Mathematics/HG/P2 6 DBE/2013 SCE-Final Memorandum 14 May 2013 QUESTION 2.- [25] 2.1.1 v4 A (6; 3) B(C-8;1) | \ . Cc v 2.1.1 | Let centre of circle be M M= (* + xy data) 2 2 ere Using midpoint formula - (2 #1) = (-1;2)VA VM Coordinates of M. 2 2 vM R= (6—CDY + (3-2 OR RB’ = C8-CY + (1-2) | Use of formula to = 50VCA calc radius Equation of circle: Correct substitution (x41P + (y-2P = 50 YCA into equation of circle 6) formula LEPARTMENT OF BASIC | EDUCATION i 2013 -06- 2 PUMLIC EXAMINATIONS WA 3 14-05 -(B

Downloaded from hlayiso.com Mathematics/HG/P2 Z SCE-Final Memorandum 14 May 2013 DBE/2013 2.1.2 For y-intercepts,x=0 “A (OF1P +(y-2P =50 “A (y-2" = 49 YEA y-2=7 0 ory = 7 YCA y=9 or y =-5 C= (0;-5) CA OR YA Let C be (0 ; &) mac X mpc =~] k-3 _k-1 YA 0-6 0-(-8) R-4k+3-48=0VvCA -'4k — 45 = 0 (k-9)(k+5) = 0 YCA k=-5 or9 C= (0;-5) YEA OR “A Let C be (0 ; &) AC? + BC? = AB? (6-0)? + 3-H? + (8-0) + (1-wY = 200 “A 36+9-6k+h +64+1-2k+K -200=0 VYCA 2h —8k -90=0 BP - 4k -45=0 (k-9) (+5) = 0 =-5 or 9 YCA C= (0;-5) YCA 6) Concept substitution simplification calculation coordinates of C Concept substitution simplification calculation coordinates of C Concept substitution simplification calculation coordinates of C __PUELIG E PRAM NAVIONS

Downloaded from hlayiso.com Mathematics/HG/P2 8 DBE/2013 SCE-Final Memorandum 14 May 2013 2.1.3 Mean = YorYm _ ~5- 2 -7M YCA Correct snbst into Xo —x, O+1 gradient form. i ‘ M value of gradient 7 vCA Gradient of tang. y=oxtec VCA OR (v-y) = m(@&-x) Correct subst into line formula 1 1 y= zs YCA Y-5))= 5(4-9 Equation 1 = >x-5 25 OR Equation of circle: x? + 2x + —-4y-45=0 VCA aii equation of circle. Point of contact of tangent C (0 ; -— 5) Equation of tangent: YM YCA Correct subst into form. x0) + WS) + 1@+0)-2(y-5)-45=0 Correct values used -Sy+x-2y+10-45=0 VCA oe : -ly+x-35=0 simplification y= axes YCA Equation (5) 14-05-13

Downloaded from hlayiso.com Mathematics/HG/P2 9 DBE/2013 SCE-Final Memorandum 14 May 2013 2.2.1 Mgs = Mts k-2_ _ k-(-4) MIA concept of collinear 1-(3)—s1-5 K-2 _kt4 vA Correct substitution 4 -4 vw k-2 = -k-4 — simplification 2k = 2 k =-1VCA value of k OR VMIA Mar = Mrs concept of collinear 4-2 _ k-C) vA Correct substitution 5-(-3) 1-5 id ues 8 4 8k + 32=24 VYCA simplification 8 = -8 k= -1 CA value of k OR Mpr=mgs YM concept of collinear 2-(-4) _ 2-k Si Gn A are (-3)-5--3-1 Correct substitution 6 _2-k 3 4 —16+8k =—24 “CA Simplification 8 = -8 k =-1V%CA value of k OR 2-(4)_ 6 3 Mer = ae tgs gM gradient of RT eq of line: Yeas ~4e +3) Correct substitution 3 1 into equation of line ys=-2x-+ VCA 4 4 Simplification subst (1;): 3 1 v Later Vere a value of k j PERARTMENT OF Basie (4) = ERUCATION i i 2013 -06- 21 : PRIVATE GAG x 149 b \ v3 \y oo ETORIA 6004 4-05-13 eS a Rnetaee ntact rec acca 1?) 4-0 ‘__EMBLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P2 10 SCE-Final Memorandum 14 May 2013 DBE/2013 2.2.2 PO; RP = 2PT “A RE3;2) ad TS 3-4) . YM @-C3Y +Q-2)7 =4e-sP+O-Cyyy VA P+OX+9 + —4y+4 =4(" — 10x +25 + + By + 16) YCA XP +Oxty—4yt 13 =42 —40x4+ 16444 +32y YCA 3x? 46x +37 +36 +151=0 YS (6) correct ratio distance formula correct substitution LHS multiplication RHS multiplication Simplification 14-05-13

Downloaded from hlayiso.com Mathematics/HG/P2 W DBE/2013 SCE-Final Memorandum 14 May 2013 QUESTION 3 [22] 3.1.1 | 2tana +k =0 -tana=—* vA value of tana Pythagoras 27+ (-k? =" VM subst into pyth. wre fP+4 VCA evs YCA calculation of r value of seca seca = OR k vA (oe. value of tana, ‘sec? a=1+ ten’ a’¥M identity . KY vca zu (-£ substitution 2 2 sca =.jtt4 = K+4 von simplification 4 2 (4) 3.2 sin (900°—x) . sec(180°+ x) . sin 58° tan(180°—x) . sin 61°. sin 29°. cos420° v AY A numerator = sin x .(-sec x) . sin 58° VAYAYA (-tan x) . cos29°. sin 29°.cos 60° denominator PER sinx 1. sinsse CA identity x SS CA identity Si 69829°, sin29°4 ; cos x 2 YCA cos60° = 3 _ 4sin29°.cos29 YCA identity cos 29°. sin29° =4 YCA Simplification

Downloaded from hlayiso.com Mathematics/HG/P2 : 2 DBE/2013 ‘ SCE-Final Memorandum 14 May 2013 3.3.1 LHS: 08 *_ + tanx sinx +1 = _cosx sinx ya identity sinx+1 cosx cos x(cosx) + sinx(sinx +1) YCA numerator cos x(sinx +1) YCA denominator YA cos?x +sin?x + sinx ed | =F multiplication cos x(sinx + 1) ee identity cos x(sin x +1) 1 A al = secx = RHS (6) | identity cos x 3.3.2 | undefined when tan x is undefined or cos x = 0 or sinx =—1 vA VA x = 90° OR 270° two angles Q) [4-05-13

Downloaded from hlayiso.com Mathematics/HG/P2 13 DBE/2013 SCE-Final Memorandum 14 May 2013 QUESTION 4 [24] 4.1 | sin2x =cos(x—-30°) [ ; sin 2x = sin (90°-(x—30°) “4 YA co-function S 2x = 120°-x+h.360%A or 2x =180°—(120°—x) 1A solution 1" quad 3x = 120°+ &.360° 2x=60° +x +k360° 1 A solution 2™ quad. x = 40° +4.120° ;keZ x =60°+k.360° wv wv vi x= BO" 40s 160° —- x =60° YCA 4A — correct angles OR sin 2x = cos (x—30°) vA co-function cos (90° — 2x) = cos (x — 30°) ° re? fe) 90° —2x =x-—30°+k.3: or 90°—2x =—(x-—30°)+ k.360° 1A solution 1° quad — 3x =-120°+ £.360° —x=— 60° + 360° | 1 A solution 4” quad. x = 40° +4120° keZ x =60° + k.360° YCA v v x = —80°; 40° Heo" a x = 60° YCA | 44 —comect angles @ 42 Ty id || it |] iy j hs F " ~ ie a oN de ge ix gq 8 o7 | 457 |o is 3 5 = os + f g All points to be visually amplitudev’A Turning point (30°; 1)“A read. shapevA shapevA Shape includes end-points y-interceptVA y-interceptVA and turning points visually x-intercepts“A x-interceptsY”A read. (8) each interval VVWCA 4.3.1 | x@(-80°; 40°) or (60°; 160°) -80° < x < 40° or 60° < x < 160° @) | “A notation 4.3.2 | xe [-90° ; 30°) interval YCA -90° < x < 30° (2)_| “A notation each intervalYVVCA 4.3.3 | xe (—90°;-60°) or (0°; 90°) or (120°; 180°) 90° < x < -60° or 0° < x < 90° or 120° < x < 180° “A notation (4) ML [4-05-13

Downloaded from hlayiso.com Mathematics/HG/P2 14 DBE/2013 SCE-Final Memorandum 14 May 2013 QUESTION 5 [14] 5.1 2 cos?x-3cosxt+1=0 “A quadratic equation factorisation (2 cosx-1)(cosx—1)=0 YCA =tan 15° VA = tan (45°— 30°) “CA __tan45° ~ tan30° 1+ tan 45°,tan30° YCA YCA 3 ee YCA v3 1+1.— 3 OR 3-3 . 3-V3 343 3-3 y¥M = 9-6¥3+3 vou 6 =2-3 OR 1 “CA YCA 2 values for cos x cosx = 3 or cosx = 1 vA YCA : x =60° + &360° or 240° + £.360° OR x = 360°; ke Z| solutions Penalty — 1 omitted on oR oA VA full marks x =60° + 360° OR x= 0°+k360°; ke Z OR 300° + &.360° (6) 5.2.1 _ tanA-—tanB EG) Tea Rey vA recog of iden ) 5.2.2 | cot 255° =cot 75° co-function double angle expansion sub tan 45°; tan 30° Rationalising denom. Simplification DEPARTMENT OF BASIC EBUCATION 2613 -05- 21 PRIVATE BAG X 119 ~ PRE! ORIA 8001 PUBLIC EXAMINATIO s\ob \ (G-0s~ 13

Downloaded from hlayiso.com Mathematics/HG/P2 IS DBE/2013 SCE-Final Memorandum 14 May 2013 5.2.2 | cot 255° =cot 75° Cont. =tan 15° VA co-function = tan (60° - 45°) YCA double angle — tan60°-tan45° vo, Te tan? tana expansion 3-1 a 1+v3 4°" sub tan 60°; tan 45° = 3 -1 7 V3 -lvM Rationalising denom. V3 +1 v3 -1 YCA - OR cot 255° = cot 75° vA =cot(45°+30°) = 1 YCA tan(45°+30°) _1—tan45°.tan30° tan 45°+ tan 30° YCA (3) ¥3 -1) 4341 YCA YCA 43-1 /M 3-1 YCA - 3-2V3+1 ” Simplification Use of inverse double angle expansion sub tan 30°; tan 45° Rationalising denom Simplification 203 -05- 21 PRIVATE BAG X 140 nn PRETORIA DCO DS PUBLIC EXAMINATIONS Atul (4-0S- [2

Downloaded from hlayiso.com Mathematics/HG/P2 16 DBE/2013 SCE-Final Memorandum 14 May 2013 [ QUESTION 6 {22} 6.1.1 1 D D B . Construction: h=AD.L BCD vc Construction peu = sin(180°—B) = sinB ep = -cosB vA e e “AD = c. sinB BD = —c.cosB By Pythagoras b? =CD? + AD? = (a + BD) + 2—BD?, YM VA = @ 4+2.aBD+ BD?+ ¢?-BD? “A =@ +? +2aBD VA = a'+c’—2.ac.cosB v OR Construction: h=AD1 BCD . BD —=-cosB c BD = —c.cosB < By Pythagoras b? =CD?+ ap? “M = (a+ BD)’+(ADY VA @ +2aBD+ (BD?+AD?) YA @+c? -2.accosB VA CEPARTHENT OF EDUCATION BASIC 2013 -05- 21 H PRANYATE ¢ i PRET BAG X 119 Correct ratios Use of Pythagoras Substitution Expansion Identity Construction Correct ratios Use of Pythagoras Substitution Expansion Identity Mu [4-0y~~ f3

Downloaded from hlayiso.com DBE/2013 Mathematics/HG/P2 _ SCE-Final Memorandum 14 May 2013 yt ; If no axes drawn in A(c.cosB;c. binBYM max 5/6 b c B x C(a;0) vA By distance formula . “M Use of formula b? = (cos B—a)* + (c.sin B—0)* vA Substitution = c’cos’B — 2.ac.cosB + a” + c’sin’B VA Expansion = @& + A(cos’B +sin’B)— 2.a.c.cosB YA _| Identity = &+¢’-2.a.c.cosB 2 6) a ly jag (S| Ta o fs le Zz 4 =siB! 6.1.2 GG al <sis he Fe .| o2;3 iba @ Ss/8 Cc I ‘ el ms | PB peg BE Sy Fae A E k & ised 8 ; “* zt B @ | BCE =90°+9 “A Value of BCE BE’ =BC’+ CE’ -2.BC.CEcos BCE VA Peas = p+ p— 2p) p) 205(90° + 4) ee Us = 2p? + 2p’sin 0 Si a ti = js YCA VYCA implification See ak oN, Common factor BE = p,j2(i+sin@) ©) 6.1.2] (3p= pJj2i+sind) VM p#0 Equating ©) | 3 = 2+ 25indvA Squaring : 1 sing 7) 0 =30°VCA (3) | Value of 8 Ne) MUS nd\or\ Y (4-00-13

Downloaded from hlayiso.com Mathematics/HG/P2 18 DBE/2013 SCE-Final Memorandum 14 May 2013 6.2 A h ¥y B Cc D 6.2.1 A =siny VA Correct ratio Cc Correct subject - vea ’ sin y OR xe =cosecy VA Correct ratio AC = hcosecy VCA (2) | Correct subject 6.2.2 a DAC = 180°-2x “A angle DAC In A ADC: AC _ cD M/A subst into sine rule sin x sin (180°-2x) AC rsinx = YCA simplification sin 2x = rsinx 7 r 2 sinx.cos x, CA 2 cos x double angle formula h r Aeron —_— = simplification sin y 2eosx “CA E , = 7siny 2cosx OR - 1 Es TOF Ease i EOUCATION i 208 -05- | ; : j PRIVATE BAG XY ‘a f PRETORIA o00, ’ tb \ St bin 2 A ES3 ¢ ag ~ pe LUPUSLIC EXAM!

Downloaded from hlayiso.com Mathematics/HG/P2 19 DBE/2013 SCE-Final Memorandum 14 May 2013 622 | DAC = 180°-2e v Cont | in & ADC: “ angle DAC DC? = AC? + AD? — 2,AC.AD cos (180° — 2x) a r 2h? = (-cos2x) “M/A subst into cos rule ? sin’? y =2h( 1 +cos 2x) YCA ~ Ginyy? * Giny? Giny) r’ sin? y simplification w= 2(1+2cos* x - 1) CA double angle formula r’ sin? y cap : w= SD YVCA simplification 4cos” x h rsiny 2cosx (5) 6.2.3 — 500. sin27,4°vM Substitution 2cos 65,3° _ = 275326 VA Simplification =275m YCA (3)_| Rounding for paper ware BAG y an DAG X14, TORIA Dany : AA MINATIONS [4-05-12

Downloaded from hlayiso.com Mathematics/HG/P2 20 DBE/2013 SCE-Final Memorandum 14 May 2013 [QUESTION7 fig) __ | ys D TAA! a vs YR M; = 90° (Line from centre to midpoint of chord is L) N, = 90° (given) M, + N, = 180° vg YR ANOM is acyclic quad (opp Z’s supplementary) (4) TA2 Me ite ““) Oy = 180° — A (opp Z’s of cyclic quad suppl) C = 180° A (opp 2’s of cyclic quad suppl) YSIR O=C @) 7.1.3 | In A ABM and A DBM AM = MD (given) vs BM = BM (common) vs M, = M,_ (Line from midpt of chord is 1) A ABM = ADBM (S,A,S) VYS/R B, = B, — (corresp Z’s of = A’s) YS/R (4)

Mathematics/HG/P2 Downloaded from hlayiso.com 2) SCE-Final Memorandum 14 May 2013 DBE/2013 72 BCD = 180° - A = 180° — x - svR 0, =2y VS (Z at centre = 22 at circum) 6; = BCD (opp 2’s of rhombus =) WS/R y= 180° -y YS 3y = 180° y = 60° vS OR C = 180°—y (opp Z’s of cyclic quad) YSR Bi = 90° -y (sumZ’sofA) YS Bi = Bz = 90° — y (proved) Bs = Ba (rhombus) € + OBC = 180° (co-int Z’s)_ vg 180°— y +180° —2y= 180° VS y= 60° VS 6) OR B, = 90°-y (sum 2Z’s A) BCD = 180° -y YS vw diagonals bisect angles and opposite angles are equal: : YSIR 4(90° — y) + 2(180° —y) = 360° (sum 2’s of rhombus) 6y = 360° VS y=60° vg (5) ES eee | DEBARTHENT OF BASIE 7 EDUCATION PRIVATE BAG X 119 PRETORIA GUOT PUBLIC EXARI NATIONS | i 2013 -05- 21 | j ns sy) (4-05-13

Downloaded from hlayiso.com Mathematics/HG/P2 22 DBE/2013 SCE-Final Memorandum 14 May 2013 [QUESTION 8 21) | 8.1 M 2 1 9) R 2 1 A P Construction: Draw POMandQM “YC Proof: P;+ P,=90° “S tant radius YR A A P,=90°-P, A a vs A YR Qi t+ Q, =90° angle in the semi circle M =90°-P, sum Z’s A M = R vs ’s in same segment yp APQ= PRQ OR : DERE DRERF Se j IMENT OF BASIC 7 : ERUCATION | 3 i 2613 -08- 21 j PRIVATE BAG X 110 I ermine PRETORIA 0001 Lm MGIC EXAMINATIONS I Vole 14. -05-[3 Jeo

Downloaded from hlayiso.com DBE/2013 Mathematics/HG/P2 23 SCE-Final Memorandum 14 May 2013 8.1 Cont. Be O 1 2 1 A P : xC Construction: Draw POM and MR YR _ Proof: P,;+ P,=90° YS (tan 1 radius) A a P,= 90°-P ; ws vR = 90°-Ri (2’s in same segment) . A ; YR R,+ R,=90° YS (angle in the semi circle) R, = 90°-Ri APQ= PRQ (¢-0s- 13

Downloaded from hlayiso.com DBE/2013 Mathematics/HG/P2 24 SCE-Final Memorandum 14 May 2013 8.1cont oO 1 R 2 1\\3 A P Construction : Draw radii QO, RO and PO vc A A vS Proof: Pi+ P,=90° tan 1 radius VR P, =90°-P, Boek Ae . Q,+Q,+Ri+R2+P3+P2 =180° Zsof AYSR AQOP and AQOR and APOR are isosceles equal radii VS BoB : 2R,+2R2+2P2 =180° “8 2R, +2R2 = 180° - 2P2 Doe le R,+R2=90°-P2 VS Boa : 7 R=Pi=APQ -. APQ= PRQ OR 1 A P Constr: Join OQ and OPY'C 2P,+ QOP=180° (sum angles in triangle) VS/R P,+ R=90°, But P,+ P2=90° S (tangent L radius) “R a Pi=R-. APQ= PRO R = 596? YS (ang at centre = 2x ang at circum) VR _ Q,=P2 (equal radii) YS/R 1G) if ple? pL (y-0S- /3

Downloaded from hlayiso.com Mathematics/HG/P2 25 DBE/2013 SCE-Final Memorandum 14 May 2013 8.2 8.2.1 A i.) v C, = x ™ (tan-chord theorem) YR §= <x YS/Rtan-chord theorem) * v) T= xV¥S (2’s in same segment/Sin-chord) (5) 822) 7 = 6, “8 (both = x) @) VR BC is a tangent to circle ACT (Converse tan-chord theorem) 2 Q) a DEPARTMENT OF BASIC : EDUCATION 2013 -06- 21 i PHVAT i PRETO! | PUBLIC EXAMINATIONS ig-os- 1% ot \v \7

Downloaded from hlayiso.com DBE/2013 Mathematics/HG/P2 26 SCE-Final Memorandum 14 May 2013 8.2.2 B, = A; +A, o (tan-chord theorem) S/R (b) SRA = A, + A, (tan-chord theorem) “S/R - * YR B, = D,vYS_ (ext Z cyclic quad= inter opp 2) SRA = BD, = CDT. OR § =¢, =x (proved) “S CB //RS_ (corresp Z’s=) VS SRA =B, (correspond. 2’s;//lines) “S/R YS B, = Dd, (ext Z cyclic quad = inter opp 2) yR SRA = D, OR A, = ART (tan-chord theorem) YSIR A, = 6 (tan-chord theorem) “S/R A, = SRT (2’s in same segment) YSIR 2 D, = €, + SRT (@xtZ ofA) VSR SRA = ART + SRT vs i SRT (5) TRENT OF BASIC EDUCATION y-os- 1%

Downloaded from hlayiso.com Mathematics/HG/P2 27 SCE-Final Memorandum 14 May 2013 DBE/2013 sy [ QUESTION 9 9.1 B ar" R Q Construction : Draw BQ and CR. Draw heights k and h YC 1 —PBA YS Proof : Areaof APBC 2 Areaof BRC I pp, 2 | a BR eck Areaof APBC_ 2. ~ Area of ABCQ — leas 762 = PC yg CQ but PB _ PC BR CQ AreaofAPBC _ Area of APBC Areaof ABCR Area of ABCQ Area of ABCR= Area of ABCQ , they have same base and must have same height. They lie between parallel lines vg BC/RQ ieee BUCATION vs OR

Downloaded from hlayiso.com DBE/2013 Mathematics/HG/P2 28 SCE-Final Memorandum 14 May 2013 9.1 cont vC Construction : On PQ mark off point D such that BD // RQ Proof: = . to (given) VS/R vs PB = PD (BD //RQ)YR BR CQ =PC vg CQ vS But this is impossible unless D and C are the same point. But BD/RQVYC BC/RQ ) ri i GEPARTHENT OF BASIC 71 i . j 2013 ~08- 21 i PRI AG X10 i: 4 ORIA 9004 | PUsLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P2 29 DBE/2013 SCE-Final Memorandum 14 May 2013 92 R s 12 r r ar M re) N o Pp ee oe ws (proportionality theoremy R SR _ jr YS 12 2r SR = 4units “S (4) 9.2.21 SN MN smn ///ARMP) YSIR RP MP SNL 1 fr YS SN = ox 12 = 9units vs (3) 9.2.3 A InAMSN, MSN = 90° (Z’sinasemicircle) “S/R SM = 12; SN = 9 MN? = MS? +SN? (pythagoras) VS/R = 12? + 9 = 225 MN = 15 =2rvs = 7,5 units | PAPRRTRENT OF Bisre— EDucation |= } OR - | 2 = MR? + RP? = MP? ; ve i 2013 -86- 24 | 16? +12? = (22r)? (pythagoras) i PRivere i al = (22r YS } PRETORIA G 400 = (22r) Ppp een 20 = 22r Amend XAMINATIONS 15 =7 “8S line (4-0S-1>

Downloaded from hlayiso.com Mathematics/HG/P2 30 DBE/2013 SCE-Final Memorandum 14 May 2013 QUESTION 10 {12] 10 “ T mm N oO R } i B ! Pp } Cc 10.1 RC = NR (Ris midpt of NC) VS BE = PC (Pismidpt of BC) VS v -. BT || PR (converse proportionality theorem) -. PR {| ON OR RC=NR=x (given) AN =NC=2x (medians ofa A are concurrent) YS AO _ 2 (median AABC VSR OP 1 AN _2 NR 1 - AO SR oP PR || ON (sides in proportion) “S/R (4) Pome SBUSATIO rm sflolvr> Pesce i

Downloaded from hlayiso.com Mathematics/HG/P2 3f DBE/2013 SCE-Final Memorandum 14 May 2013 10.2 vs AN = NC (medians ofa A are concurrent) ON =NT (given) “S/R AOCT is a parm (diagonals bisect each other) “S/R YR 4 10.3 | Area of A AON 4x AO x AN x sinPAC YS Area of A APC OR “8 4 4 1 AAON = r AAPR (2 arc) vs 4 | AreaAAON 3A4PC “"AreahAPC AAPC 3 4x APx ACx sin PAC YS AO | AN —— x —— AP AC 2p 1 fx 3p 2 1 3 vs vs (median concurrency point) vs YS @ TOTAL: 200 2613 ~85- 21 Fea . OEPARTHENT OF BASIC 7 i PRETORIA EDUCATION } y | u\v | j 3X 140 i e6n1 LW PUBLIC EXAMINATIONS 1¢-O5-1)