Mathematics Grade 12 Paper 2 KZN Mock 2023 Paper & Memo

Downloaded from Stanmorepnysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE f £000] N.B. This question paper consists of f2%pfages”and 1 information sheet. This paper/hascan;Answer Booklet. Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmore A Sics. com?" 2023 Preparatory Examination INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This questiom/paper consists of 11 questions. 2. Answer ALL the questions. 3. Number the answers correctly according to the numbering system used in this question paper. 4. Clearly show ALL calculations, diagrams, graphs, etc. that you have used in determining your answers. 5. Answers only will NOT necessarily be awarded full marks. 6. | You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. 7. Ifnecessary, round off answers to TWO decimal places, unless stated otherwise. 8. Diagrams are NOT necessarily drawn to scale. 9. An information sheet with formulae is included at the end of the question paper. 10. Write neatly and legibly. Copyright Reserved Please Turn Over

Mathempsiss Vhloaded from Stanmore ki Sics. com?" 2023 Preparatory Examination QUESTION 1 Mr Siphokazi supplements his pension by mowing lawns for customers. He measures the areas (x) (in m?) of 12 of his.customers' lawns and the time (y) in minutes, it takes him to mow these lawns. He works 8 a day. He recorded the data. Area (x) & 120 | 845 | 602 | 1 190 | 530 | 245 | 486 | 350 | 1.005 | 320 | 250 (m”) . Time (y) 50 | 28 | 130} 75 | 120 | 95 | 55 |70 | 48 | 110 | 55 | 60 (minutes) Lt Determine the equation of the least squares regression line. (3) 12 Calculate the value of r, the correlation coefficient for the data. (2) ‘i Dal 1.3 okazi charges a flat call out fee of R150, as well as R50 per half f), estimate the charge for mowing a customer's lawn that has an (For example: 100 minutes would be taken as 2 hours). (3) 1.4 The local high school wants Mr Siphokazi to mow their rugby field which is rectangular, 100 meters long by 70 meters wide. 1.4.1 Use the regression equation found in 1.1 to calculate the time it would take to mow this area. (1) 1.4.2 Is it possible for him to complete this job in a day? Give a reason for your answer. qd) [10] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmore fi Sics. com?" 2023 Preparatory Examination QUESTION 2 The following table gives the frequency distribution of the daily travelling time (in minutes) from home to work for the employees of a certain company. Daily travelling time x (in minutes) Number of employees 0) Midpoint of Interval O0<x<10 20 10<x<20 35 20<x<30 30<x<40 30 10 40 <x <50 5 2.1 Calculate the estimated mean travelling time. 2.2. Write the modal class of the data. 2.3. Anogive was drawn for the given data. Construct a box-whisker plot for the-data in the ANSWER BOOK. 1 Number of employees 00 50 y G3) (2) (3) me (ea 10 20 30 40 Daily travelling time (in minutes) 2.4 State whether the following statements are TRUE or FALSE. x 50 minutes 2.4.1 The distribution of these travelling times is positively skewed. () 2.4.2 The inter-quartile range for the data is 25. 2.4.3 35 employees take less than 20 minutes. Copyright Reserved ()) () {11} Please Turn Over

Mathenpsiss nloaded from Stanmore A Sics. com?" 2023 Preparatory Examination QUESTION 3 Trapezium ABCD is drawn below with AD || BC is drawn. The coordinates of the vertices are A(1 ; 7), Bp ; gq), C6 = -8) and D(-4 ; -3). BC intersects the x — axis at F. DCB =a. AD intersects the y =axis at E. yt adn E Bp sq) eT 0 > x F D(-4; -3) N a C(-2 ; -8) v 3.1 Calculate the gradient of AD. (2) 3.2. Determine the equation of BC in the form y = mx + c. (3) 3.3. Determine the coordinates of F. (2) 3.4 AMCD isa parallelogram, with M on BC. Determine the coordinates of M. (2) 3.5. Show that a = 48,37". (4) 3.6 Calculate the area of ADCF. (4) {17] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmore i Sics. com?" 2023 Preparatory Examination QUESTION 4 4.1 The cquation of a circle is x? + y? — 8x + 6y = 15. 4.1.1 Show-that P(2 ; -9) lies on the circle. (2) 4.1.2 Deterhtine the equation of the tangent to the circle at point P(2 ; -9). (6) 4.1.3 A tangent is drawn from Q(-10 ; 12) to the circle. Calculate the length of (4) the tangent. 4.2 The circle, with centre T, and equation (x — 3)? + (y + 2)? = 25 is given below. B is the y — intercept of the circle. a’ B Cc ‘ [e) > x T v 4.2.1 Determine the coordinates of B. (4) 4.2.2 Write down the coordinates of C, if C is the reflection of B in the line x = 3. (2) 4.2.3 Another circle with centre M and equation (x — 12)? + (vy— 10)? = 100 is given. (a) Calculate the distance, TM, between the centres. (2) (b) Do these circles touch or intersect each other? Justify your/answer. (2) [22] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmore A Sics. com?" 2023 Preparatory Examination QUESTION 5 5.1 If sin 38° = p, determine the valuc of the following, without using a calculator: 5.1.1 cos 268° (3) 5.1.2 cos 4, (3) 5.1.3 sin26 cos 26° (2) 5.2 Evaluate the following trigonometric expression without using a calculator: 2sin 165° cos 195° (5) cos 45° sin 15° — cos 15° sin 45° 5.3. Given: K = ¥3cosx + sinx. 5.3.1 Write K in the form of ¢sin («+6). (3) 5.3.2 Hence, calculate the value of t and 0. (1) 5.3.3. Write down the maximum value of K. (1) 5.4 Prove the identity: 5.4.1 2tan @—sin 26 = tan 8 (6) 2 sin? @ = tan 5.4.2 Hence, determine the values of 8, Qe [l 80°; 360°] which will make the above identity undefined. (2) [26] QUESTION 6 6.1 Sketch the graphs of f(x) = 2sinx and g(x) = cos(x — 30°) forx € [—180°; 180°] on the grid in the ANSWER BOOK. Indicate the intercepts with the axes and also the turning points. (6) 6.2 Use your graphs to answer the following questions: 6.2.1 Write down the period of g. (1) 6.2.2. Netermine the values of x for which f(x) > g(x). (4) 6.2.3. Write down the values of x for which f(x) = 1,5 + g(x). (2) [13] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmore fi Sics. com?" 2023 Preparatory Examination QUESTION 7 In the diagram below, PQ is a vertical mast. R and S are two points in the same horizontal plane as the foot of tl inast, Q. QRS = a, QSR = 6 , SR=8 — 2x and SQ =x. The angle of clevati OF P, the top of the mast from R, is 6. foaar P ee Ss 7A Express PQ in terms QR and a trigonometric ratio of 0. (1) 7.2 Show that: PQ = 7222s"? (4) sina 7.3 If B = 60°, show that the area of AQSR = 2y3x — 2 x?, GB) 74 Determine the value of x for which the area of AQSR will be at a maximum. (3) faa? (yy lana? donot don i Copyright Reserved Please Turn Over

Mathempyss loaded from Stanmore A Sics. com?" 2023 Preparatory Examination QUESTION 8 8.1 In the diagram O is the centre of the circle and M is a point on the circumference of the circle. B subtends AOB at the centre of the circle and M at the circumfere: f the circle. M A B Use the diagram to prove the theorem that states that AOB=2M. (5) 8.2 Froma point A outside the circle, center O, two tangents AD and AV are drawn.AO and VD intersect at M. BOD is a diameter of the circle. BV and VO are drawn. VAD = 40° 8.2.1 Prove that quadrilateral VODA is cyclic. (2) 8.2.2 Calculate the magnitude of 0. (2) 8.2.3. Prove that BV || OA. (5) [14] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmore f Sics. com?" 2023 Preparatory Examination QUESTION 9 9.1 Complete the following statement: A line drawn parallel to one side of a triangle ... (2) 9.2 In the figufeSICL || QR. M and N are points on QR such that KN || PR and LM || PQ. /-PK= 3 units, PL = 4 units, LR = 6 units and MN = 1,8 units. P K L Q M oN R 9.2.1 Calculate the length of KQ. (2) 9.2.2 Prove that QM = NR. (2) [6] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmore } Sics. com?" 2023 Preparatory Examination QUESTION 10 In the figure, two circles intersect at A and B. AB produced to M bisects QAR. Tangents MQ and MR meet the-circles at Q and R respectively. QBR is a straight line. AQ and AR are drawn. Prove: 10.1 AMQA ||| AMBQ. (3) 10..2 MR? = AM.MB (3) 10.3 MQ? MR2 1 (4) [12] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmore f j Sics. com?" 2023 Preparatory Examination QUESTION 11 11 AABC is right angled at C. ED 1 AB with E on CB and D on AB. AC = 4,8 cm-and AB = 8 cm. AD = DB. Cc E A D B 11.1 Calculate BC, correct to 1 decimal digit. (2) 11.2 Complete: ABAC ||| .... (1) 11.3. Hence, or otherwise calculate the area of ADEC. (5) [8] GRAND TOTAL: 150 Copyright Reserved Please Turn Over

“=te®6Whloaded from Stanmorepiiysics.com September 2023 Preparatory Examination. INFORMATION SHEET: MATHEMATICS INLIGTING BLADSY _—b+Vb7 Ade 2a A=P(l+ni) 4 = P(l—ni) A= P(1-i)" A=P(1+i)" T, =a+(n—Nd S, = 5 2a+(n-Nd) T =a"! a(r" -1) -— 7%. | =ar s -* r#l S,,.=——3-l<r<l " r-l l-r pase = pasa") i —_—_—— S(Xth)- f(x) "(x)= lim ——7-— LG) h>0 h xy tx yyty d= (x, 4)? +02 -y,)? eed yy — Vy. y=mxte y-y, =m(x-x) m=22 1 m=tan0 xX -X (s-a) +2) =P In AABC: 4 = = £ a? =b? 407 —2be.cos A area ABC = Lab.sin C sn A snB sinC 2 sin(a + B) = sin a.cos # + cosa.sin 8 cos(a + f)= cos@.cos #—sin a.sin 8 cos’ asin? @ cos2a ={1—-2sin* a 2cos* a@-1 M 2 Rl Il _n(A) n(S) p=ath P(A) Copyright Reserved sin(a - B) = sin a.cos # —cosa.sin B cos(a - B) =cosa@.cos #+sina@.sin 2 sin 2a =2sina@.cosa P(A or B) = P(A) + P(B) - P(A and B) dXO-*o-y~) 65-0’ yo-

Downloaded from Stanmorephysics.com < am OINAL, ; KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA MARKS: 150 This marking guideline consists of 14 pages. Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer f yvsigsigom 2023 Preparatory Examination NB: CA APPLIES TO ALL SUB-QUESTIONS IN THIS MARKING GUIDELINE. QUESTION 1 Ll VA Value of a YA Value of b as ; y = 34,90 + 0,08x ¥CA_ Equation (3) Answer only: Full Marks 1.2 | r=0,88 vAYA Answer (2) 1.3. |y = 34,90 + 0,08x y = 34,90 + 0,08(560) ¥CA_ Substitution y = 79.7 minutes = 1.33 hours YCA_ 79,7 minutes Total cost = R150 + R150 = R300 ¥CA_ Answer (3) 1.4.1 | Area = 100 x 70 = 7000 square meters y = 34,90 + 0,08(7000) y = 594.9 minutes = 9.92 hours ¥CA Calculation (dl) 1.4.2 | No. The time taken will exceed his daily 8 hour working YCA Justification hours. (1) [10] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer } yvsigsigom 2023 Preparatory Examination QUESTION 2 Daily travelling Number of employees Midpoint of Interval fx time x(in minutes) a) (x) : 0<x<10 20 5 100 10<x<20 35 15 §25 20<x<30 30 25 750 30<x< 40 10 35 350 40< x <50 5 45 225 Total 100 1950 y 100 rat J an 8 Va 3 ai E V4 2 50 ge 3 [= =] z HH x 10 20 30 40 50 minutes Daily travelling time (in minutes) 2.1 Estimated Mean = 2 = 19,5 YA 1950 i190 vA 100 Answer only: Full Marks “CA, Answer @) 2.2 | 10<x<20 VAWA Answer (2) 2.3 See Diagram YA — Minimum and Maximum value YA 1 and 3™ Quartiles YA 24 Quartile (3) I ‘Tn BE 4 Five number summary: 0; 12; 18; 26; 50 Accept: +1 deviation on quartiles Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer fh yvsigsigom 2023 Preparatory Examination 2.4.1 | True ¥A___Answer dd) 2.4.2 | False vA Answer da) 2.4.3 | True YA Answer qd) [11] QUESTION 3 hag yt agen E Bip:q) > x F | D(-4;-3) C(-2;-8) 3.1 _7+3 10 3 YA Substitution of points A and D Ma = T4475 7 YCA Answer (2) 3.2. | map = Mgc = 2 .....(AD || BC) YCA_ Gradient of BC y=mxeic -8 =2(-2)+c YCA_ Substitution of point C and gradient Lear —4=c ———) i : y=2x-4 ¥CA_ Answer Ima @) Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer } yvsigsigom 2023 Preparatory Examination 3.3 Let y=0: 0=2x-4 YCA 0=2x-4 x=2 YCA x-—value F(2;0) Q) M@G3 ; 2) YA x-—value (2) YA y-value 3.4 3.5 mp = —34+8 3 YA Gradient of CD P4420 -2 Inclination of CD: tan 0, = 180° — 68,2° = 111,8° YA Inclination of CD Inclination of CF: tan 6, =2 YA Inclination of CF 62 = 63,43" Therefore YA subtraction of the angles a = 111,8° — 63,43° = 48,37" OR DF - /2@+ 2+ 0+ 3)2 YA Distance of DF = V45 YA Distance of DC Dc = 29 CF = 80 VA Distance of CF a 2 2 YA subst. into cosine rule (V45) = (V29) + (¥80)? — 2V29 V80cos a (4) a = 48,37° 3.6 | pc= (3+ 8)? + (—4 + 2) = ¥29 YA Length of CD FC =/(0 + 8)? + (2 + 2)? = V80 YCA Length of FE Area of ADCF = ; (V29)(V80) sin 48,37° ¥CA_ Substitution into area formula = 18 square units. ¥CA_ Answer (4) [17] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer p yvsigsigom 2023 Preparatory Examination QUESTION 4 4.1.1 | x? + y? —8x + 6y = 15 LHS = (2)? + (—9)? — 8(2) + 6(-9) YA Subst. of point = 44 81> 16-54 YA Simplification =15 (2) = RHS 41.2 | oct 4 [OP 4 + 43" =40 Point of contact (x:y) vA writing the equation as (x — 4)? + (y +3)? = 40 (x — 4)? + (vy + 3)? = 40 (6) YCA_ Centre of circle Centre: C(4;-3) P(2;-9 entre: C( ) PC ) ¥YCA_ Gradient of radius _ —34+9 _ 6 _ MRadius ~ ~q 2 27 YCA_ Gradient of tangent Mrangent = “3 Equation of Tangent: y=mx+e YCA_ Substitution -9==5(2) +c 25 38 ¥CA_ Answer Pes Y= Tata 4.1.3 | r?: = 4)2-+4(y + 3)? = 40 YCA Calculation of r? r? = 40 (distance Q to the centre)? = (—10 — 4)? + (12 + 3)? Y¥CA Bilnce calculation = 421 ¥CA JTangent calculation (Length of tangent)? = 421 — 40 = 381 (4) Length of tangent = ¥381 ¥CA_ Answer Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer i yvsigsigom 2023 Preparatory Examination 4.2.1 | (x—3)? +(y +2)? = 25 Let x = 0: (0-3)? + (y +2)? =25 vA Letting x=0 (y +2)? = 16 y+2=44 YA — Simplification y=-6 or jar2 YCA y-—values B(O; 2) y ¥CA_ Answer (4) 42.2|C(6;2) aia YCA x—value (2) 7 YCA_y—value 4.2.3 (a) T(3 ;—2) and M(12; 10) YA Coordinates of M TM? = (12-3) + (10+2) = 225 T™ = 15 units ¥CA_ Answer (2) (b) Radius, center T = 5 units and Radius, center M = 10 units ¥CA_ Sum of radii Sum of radii = 15 units Circles touch . ¥CA_ Justification (2) TM= Sum of radii [22] QUESTION 5 5.1 52° 1 P 38° 1-p? 5.1.1 | cos 218° YA Calculation of 1 — p? = —cos 38° YA Reduction 1—p? rl ¥CA_ Answer (3) Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer } yvsigsigom 2023 Preparatory Examination 5.1.2 | cos 14° = cos(52° — 38°) YA Writing as difference = cos 52’ cos 38° + sin52° sin38" vA Expansion = ¥CA_ Answer 3) cos 24° = sin 76° ¥ sin 76° = sin2(38°) = 2sin 38° cos 38° v =2p Jie double angle v answer 5.1.3 | sin26° cos 26 1. YA Double angle ==sin52 2 = fi-p ¥YCA_ Answer 2 (2) 5.2 2sin165 cos 195" cos 45° sin 15° — cos 15° sin45° _ 2 sin 15°.(—cos 15°) ; ~ cos 45° sin 15° — cos 15° sin 45° vA —cos15 —2sin 30° y ; = ————— A —2sin30 sin(15° — 45 ° ° ( ) YA sin(15 —45) _ —2sin 30° ~ sin(—30°) = T2sin 30° YA —sin 30° —sin 30 =2 ¥CA_ Answer (5) 5.3.1] K =V3cosx+sinx v3 1 } K=2 (Fess +zsinx 2 2 vA 2(Scosx +4sinx) K = 2(sin60 cosx + cos60 sinx) YA sin60° and cos 60° K = 2sin (60° +x) YA 2sin(60°+ x) (3) Copyright Reserved Please Turn Over

Mathempyss loaded from Stanmerephysics;gom 2023 Preparatory Examination ° 5.3.2|}t =2 and @ = 60° YCA t-—value and 6 = 60 qd) 5.3.3 | 2 ¥CA_ Answer qd) S41 LHs = 2 tan 0 —sin 20 nO (] in cosd va sind A Suny _ _\cosé x cos0 cos8 pee cos8 VA 2sin 0 cos 6 _ 2sin0 ~2sin @cos* @ 7 2sin20cos0 YA __ simplification . re _ 2sin0(I-cos’0) 2sin’Ocos0 _ 2sin® . sin’6 sin20 2sinO . cosd fain iysics.co _ sin@ YA __ simplified to sind cos co = tan 0 = LHS (6) 5.4.2 |2sin? 6 = 0 sind = 0 =. 0 = 180° and 360° Y 180° and 360° () 0 = 270° ¥ 270° [26] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer lysics.com 2023 Preparatory Examination QUESTION 6 6.1 2 15 1 ai -150— 120 -90 “60 30 O30 60 90 12 150 “05 4 “1S St 2 Graph of f: LA mark for x — intercepts 1A marks for minimum and maximum points 1A mark for shape Graph of g: 1A mark for end points 1A mark for x — intercepts 1A mark for y— intercept (6) 6.2.1 | 360° YA Answer (1) 6.2.2 [-180° ;-150°) x € [-180° ;—150°) U (30°; 180°] YA: Notation YCA : values (4) (30°; 180°] YA: Notation “CA: values 6.2.3 | f(x) = 1.5 + g(x) fF) —g@)=15 YA f@)—g@) =15 x = 90° orx = 150° YA Answer (2) [13] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer L (sigsgom 2023 Preparatory Examination QUESTION 7 P [1 0 Q a? R x 8-—2x B Ss 71 PQ aR = tand PQ = QRtan@ vA Answer qd) 7.2 | QR_ QS VA _— Sine rule formula sinS sinQRS . YA _ Subs. Sine rule QR x sinB sina VA Making QR a subj. of the i formula vA Subst. of QR x sin B tan 6 PQ = sina : 73 | Area of AQSR = 5 (x)(8 — 2x) sin 60° YA _— Subst. into Area rule “1 ic = 5 (x)(8 - 2x) (F) YA =4(x)(6 - 2x) = 1 VA = V3(x) (2-34) v3 = vax @) Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer Mh (sigsgom 2023 Preparatory Examination 74 For Max Area: x = — 2a YA Formula ___(2v3) 7 v3 X= 2 Y A Substitution into formula | (3) v CA Answer [11] QUESTION 8 8.1 M A B Constr: Join MO and produce to D. vA Construction AOD = OAM+ AMO... (Ext. 0 ZfA) YA S/R BOD = OBM+ BMO... (Ext. Z of A) But OAM = AMO and OBM = BMO ...(Radii =) vA .. AOD + BOD = 2AMO + 2BMO vA AOB = 2(AMO + BMO) vA S AOB = 2M (5) NOTE No construction : No marks 8.2.1 | OVA = ODA = 90 .... (Radius L Tangent) YA S/R VODA is a cyclic quad. (Converse of opposite angles of VA LR (2) quad. Supplementary) 8.2.2 | 0, = 40" ...(Exterior angle of cyclic quad = int. opp. Angle) | ¥S&=“4R (2) Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer } ysigs,com 2023 Preparatory Examination 823) 9 = WO = = 70" ....(sumof Zs of A; radii) Y oSR 180-40" ADM = = 70° ....(sumof Zs of A; Tangents drawn Vv from a common point A) S/R loa? ADM = 0, bu ... (Angles subtended by common chord VS YR AV) BV ||OA pg weeeeee (Converse of Alt Zs or Alt Zs are=) | v R (5) OR In AOVA and AODA OV=OD......... (radii) vA AOVA = AODA.......(SSS) Al= Adve (= As) =20° A2=V).......(Zs in the same segment ) vA SIR =20° V=90° — 20°=70° VA V=10° 0.=180° -[ovA+hr] =180° -[90°+20° | = 70° VA O2=70° wy, t 62.1... (both - 70°) .. BY||OA...... (altZs are = or conv.altZs ) YA R [14] 40.00 Copyright Reserved Please Turn Over

Mathempsiss Vhloaded from Stanmer hysics.com 2023 Preparatory Examination QUESTION 9 91 Divides the other two sides, proportionally. YS divides the other two sides YS ___ proportionally (2) 9.2 P 4 L 6 Q M oN R 1,8 9.21 | eae (Prop. Thm; KL || QR) vy SIR KQ = 4,5 units v Answer (2) 9.2.2 | KL = QM ....(Opposite sides of |" QKLM are v SIR equal) v¥ S/R KL=NR .... (Opposite sides of ||" NKLR are (2) equal) Therefore QM = NR. ox - os R a - “ Prop Thm, KN||PR Q tar Q 45_3 OR 3? MR RL MR_ RL —_—=— Prop Thm, ML||QP QM =oL Prop Thm, ML||QP QM PL -6_3 4 2 QN _MR NR QM QM+1,8 _ NR+1,8 NR QM “QM (QM+1,8)=MR (NR+1,8) / OMELS _ NRH8 “. QM=NR NR QM [16] Copyright Reserved Please Turn Over

Mathenpsiss nloaded from Stanmer livsics.com 2023 Preparatory Examination QUESTION 10 10.1 | InA’s MQA and MBQ M,= My... (Common) v S/R Ay =Qy wees (Tan-Chord Theorem) v JR AQM = By ee (Remaining 4s of A’s) AMQA ||| AMBQ ...... (AAA) v R GB) 10.2 | In A’s MAR and MRB My = Mg oes (Common) s Ag = Ry weeeeee eee (Tan-Chord Thm) S/R ARM = RBM .......... (Remaining Zs of A’s) AMAR ||| AMRB ...... (AAA) v R MA MR —_— = *s simi YS AYR MR MBO (A’s similar) MR? = AM.MB ©) 10.3 |MQ_ MA MA —-= — v — MB MQ (from 10.1) MQ MQ? = MB.MA =.......( 1) Also MR? = AM.MB v = MB. MA r> Now Ma: = MBMA yy MQ _ MB.MA MR2 AM.MB MR2 ~~ AM.MB _ MQ? ” MR2 (4) {12} Copyright Reserved Please Turn Over

Mathempsiss Vhloaded from Stanmer if (sigsgom 2023 Preparatory Examination QUESTION 11 Cc enn 40) lniny 4.8 A D B 8 11.1 | BC? = 8? — 4,8? ........ ( TOP) v S BC =6,4cm v S (2) 11.2 | ABED ys () 11.3 A AC Cc Ba _ AC LBC ABAC ||| ABED vA S BE ED BD 8 48 64 BE ED 4 48x4 YCA_ Value of ED a7 3cm Area of AEDB 1 =5 (4)(3) = 6 cm? YCA_ Area of AEDB Area of AABC 1 = 5 4,8)(64) = 15,6 cm? YCA_ Area of AABC Therefore, Area of ADEC = 15,6-6=9,6cm? (5) ti GRAND TOTAL: 150 Copyright Reserved