Mathematics Grade 11 Paper 2 KZN November 2023 Memo

Downloaded from hlayiso.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE �----------------------- ------------------------� �: MATHE ATICS P2 :: II II II II II II II II II II � � � �-----------------------------�----------------� � [ NAME: CLASS: ] MARKS: 150 TIME: 3 hours For educator use only. Question Ql Q2 Q3 Q4 Q5 Q6 Q7 QB Q9 Ql0 Qll Total Marks obtained Question total 10 12 18 13 25 11 10 6 11 13 21 150 This question paper consists of 24 pages, including pages with additional space and a formula sheet.

Downloaded from hlayiso.com Mathematics P2 . KZN/November 2023 ownloaded from Stanmonephysias.com INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This questio r consists of 11 questions. 2 Answer AL stions. * 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. 5. An approved scientific calculator (non-programmable and non-graphical) may be used, unless stated otherwise. 6. If necessary, answers should be rounded off to TWO decimal places, unless stated otherwise. S Diagrams are NOT necessarily drawn to scale. 8. Write your answers in the spaces provided. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 . ownloaded from Stanmowepleysias.com KZN/November 2023 QUESTION 1 Aphelele plays for his school’s cricket team. The number of runs scored by Aphelele in each of the eight games that he batted in, is shown below. (Aphelele was given out in all of the games.) aca} 8 19 7 15 32 14 12 La Determineithe \verage number of runs per game scored by Aphelele for these eight games. (2) 1:2 Determine the standard deviation for this data set. (1) 1.3 Aphelele wants to be selected for the District trials. The ds to have at least one score above two standard devi S n y ‘eight games. Will he be selected for the District trials? Mot culations. (2) 14 Aphelele hopes to achieve an average of 20 runs per game for his first 13 games. What should his average number of runs per game be for the last five of these games in order for him to reach this goal? (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 4 . KZN/November 2023 ownloaded from Stanmowepleysias.com a3 Brian plays in the same team as Aphelele. After having played a few matches, he has an average batting score of 37 runs per game, and the standard deviation of his scores is 13. In the next two matches that he played, he scored respectively 20 runs and 0 runs. Will the inclusion of these two scores cause the standard deviation of his scores to increase or Gecraniey ty your answer. (2) peal [10] QUESTION 2 The table below shows the height (in cm) of 250 Grade 11 learners. 4 Number of Cumulative Height (cm) learners (f) frequency 145< x <150 26 6 150< x <155 A 29 60 89 74 B 52 215 32 247 3 250 250 yaa Calculate the values of A and B. (2) 2.2 Calculate the estimated mean height of the learners. (3) 23 Write down the modal class. (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 * KZN/November 2023 ownloaded from Stanmowepleysias.com 2.4 On the grid provided below, draw a cumulative frequency graph (ogive) to represent the data on the height of the Gr. 11 learners. (3) OGIVE ey S s S = 2 8 s £ = 1S) 140 145 150 155 160 165 170 175 180 185 Height of Gr. 11 learners in cm 2.5 Use the ogive to estimate the interquartile range of the data. Show all your calculations. (3) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 * KZN/November 2023 ownloaded from Stanmonepheysias.com QUESTION 3 In the diagram below, the vertices of AABC are A, C(4;—4) and E(-6 ;-9). The angle of inclination of AC is a. ACE = B. The equation of AC is 3x+ y—8=0. y] A a. oO X C(4;-4) E(-6;-9) 3.1 Write down the gradient of AC. (2) 3.2 Calculate the size of a. (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 . KZN/November 2023 ownloaded from Stanmowephysias.com 3.3 Calculate the size of /. (5) 3.4 Determine the equation of EF if F is a point on AC produced such that EFA = 90°. (4) 3.5 Calculate the length of EF. (5) [18] Copyright reserved Please turn over

. Downloaded from hlayiso.com Manemtsownloaded from Stanmonepleysias.com KZN/November 2023 QUESTION 4 Triangle PRS has vertices P(—1;3), R(3;—1) and S(a; b), as shown in the sketch below. 41 Calculate the coordinates of T, the midpoint of PR. (2) 4.2 If the perpendicular bisector of PR passes through S, show that a=b. (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 . KZN/November 2023 ownloaded from Stanmowephysias.com 43 If a<0,b<0 and the area of APRS =12 square units, determine the (7) coordinates of S. [13] QUESTION 5 5.1 If sin33° =m, determine the following, in terms of m, without the use of a calculator: 5.1.1 tan 33° (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 0 . KZN/November 2023 ownloaded from Stanmowephysias.com 5.1.2 cos777° (3) 5.1.3 — sin(-237°) (3) 5.2 Given that 4tan 8+5=0 and £ <[0°;180°]. Determine, with the aid of a diagram, and without the use of a calculator, the value of V41cos 7. (4) 5.3 5.3.1 Simplify the following expression to a single trigonometric ratio: sin(180° — #). cos(90° — 2) -1 cos(-—A) a Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 1 . KZN/November 2023 ownloaded from Stanmowephysias.com 5.3.2 Hence, determine for which value(s) of #, where # <[0°;360°], sin(L80° - ).cos(90°= 8)=1 Witt be undefined. cos(—f) (3) 5.4 5.4.1 Prove the following identity: tana.sina + cosa = — cosa (2) 5.4.2 Hence determine the general solution of: tana.sina + cosa = 3. sina (4) [25] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 2: . ownloaded from Stanmonephyysias.com QUESTION 6 KZN/November 2023 In the diagram, the graph of the function f (x) = cos( 3 is drawn for the interval x <[-180°;180°]. + -180° -150° -1209 -909 = -60° = -30® 30° 60° go° 120° = 150° 180° 1 2: 6.1 Write down: 6.1.1 the amplitude of f. (1) 6.1.2 _ the period of f. (1) 6.2 Draw the graph of g(x) =sin(x—30°) for the interval xe [-180° : 180°] on the axes provided above. Clearly indicate all intercepts with the axes and the turning point(s). (3) 6.3 Write down the values of x in the interval xe [-180° ; 180°] , for which f (x).g(x) 20. (2) 6.4 Write down the values of x in the interval x € [oe ‘ 180°], for which g(x) = f (x) +. (2) 6.5 The graph of h is obtained by reflecting the graph of g in the y-axis. Write down the equation of h. (2) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 3 * KZN/November 2023 ownloaded from Stanmonepheysias.com QUESTION 7 WA. AABC has AB = 11cm, BC = 6 cm and AC = 15 cm. A 15cm 11cm Cc 6cm B Calculate the size of B. (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 4 * KZN/November 2023 ownloaded from Stanmowepleysias.com te In the diagram below, P, Q, R and S are points in the same vertical plane. The angle of elevation of R from Q is a and the angle of elevation of Q from S is 0. PS=xunits and QRS= 7. --2 P x s xsin(a@ +0) Prove that: SR = - cos @sin 2 (6) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 5 . KZN/November 2023 ownloaded from Stanmonephysias.com QUESTION 8 The diagram below shows an open rectangular box, made out of sheet metal. Its measurements are as follows: e length = 27 n_, e breadth = 2 mio © height = 10/¢r H 4200] 10 cm 27cm 24cm 8.1 Calculate the total surface area of sheet metal needed to manufacture this open box. (3) 8.2 4,5 liter water is poured into this box. Calculate the depth of the water in the box. Take note: 1 liter = 1000 cm’. (3) [6] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 6 * KZN/November 2023 ownloaded from Stanmonephysias.com Give reasons for ALL statements in QUESTIONS 9, 10 and 11. QUESTION 9 9.1 O is the centre of the circle. A, B, C and D are points on the circumference. AOB is a straight line@SAC and BC are drawn. OD is drawn parallel to BC, and intersects AC in E. The radius Of-the circle is 10 cm, and AC = 12 cm. Calculate-the!length of ED. (6) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 yi * KZN/November 2023 ownloaded from Stanmonephysias.com 9.2 In the diagram below, JG is a tangent to circle FGH at G. FG, GH and FH are drawn. JGF = x-5° and FGH =3x. Calculate the value of x. (5) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 8 * KZN/November 2023 ownloaded from Stanmowepleysias.com QUESTION 10 In the diagram below, RS is a diameter of the circle centred at O. Chord ST is produced to W. Chord SP produced meets the tangent RW at V. R, =50°. 10.1 Calculate the sizes of the following angles. 10.1.1 R, (3) 10.1.2 Ww (3) 10.13 =P (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 9 . KZN/November 2023 ownloaded from Stanmonephysias.com 10.2 Prove that V, = PTS (4) 10.3 Hence, prove that WVPT is a cyclic quadrilateral. (1) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 0 * KZN/November 2023 ownloaded from Stanmowepleysias.com QUESTION 11 11.1 Use the diagram to prove the theorem which states that P+ R =180° (5) Pp Ss Q R Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 4. * KZN/November 2023 ownloaded from Stanmonephysias.com 11.2 In the diagram two circles are intersecting at B and D. O is the centre of the smaller circle. B, C, D and E are points on the circumference of circle O. A is a point on the circumference of the bigger circle. DEA is a straight line. OB, OD, OA, BA, BD and BE are drawn. OA and BE intersect at F. A, =x. 11.2.1 Name, with reasons, THREE other angles equal to x. (4) 11.2.2 Calculate C in terms of x. (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 2 . KZN/November 2023 ownloaded from Stanmowephysias.com 11.2.3 Prove that AB = AE. (5) 11.2.4 Prove that AB is NOT a tangent to circle BCDE. (3) [21] TOTAL MARKS: 150 Additional space Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 3 . KZN/November 2023 ownloaded from Stanmowephysias.com Additional space Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 fy = way KZN/November 2023 Bownloaded from Stanmowepleysias.com INFORMATION SHEET x= —b+vb* —4ac 2a A=P(1+ni) A=P(1-ni) A=P(1-i)" A=P(1+i)" T, =a+(n—2)d s, =F [a+ (n-2d] T, =ar"* Ss = M1). at S, =); -1<r<1 " re 1-r i i £"(x) = lim +h) = £00 hoo h X+X% Wty d= (x, -x,) +(Y2-y,)? Daa 3 +] —' _ = Yo = y=mx+c y-y, =m(x-x,) m=-2—+ m=tand xy — X& (xa)? +(y-by =r? Inaape; —42_=—P___&_ sinA sinB~ sinc a’ =b? +c? — 2bc.cosA area AABC = sabsin Cc sin(a + #) =sin acosf+cosasin 2 sin(@ — 8) =sinacos B—cosasin B cos(a + #) = cosa cos f —sin asin 8 cos(a — 8) = cosacosf+sin asin B cos’ a—sin’? a cos2a@ =41-2sin’ a sin 2a = 2sinacosa 2cos* a-1 _ yx (x, — x)? x a a= n P(A) = ae P(A or B) = P(A) +P(B) — P(A and B) x-xX\y-V ynanbx p-Dl-RKy-9) Yi (x-xP Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com } KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA 6 ~ MASELON 150 This marking guideline consists of 14 pages. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathemawmaploaded from Stanmorephysics.com KZN/November 2023 GRADE 11 Marking Guideline QUESTION 1 1d ee _ 128 vy 128 = 8 -16 v answer (2) 12 6=7.55 Y answer (1) 13 Two standard deviation above the mean: X-25 x -16 ¥ X4+15 -X+25 + —_—_ | —}—+ 0.9 8,45 16 23.55 31,1 X+25 =16+2x7.55=311 v311 Yes, he will make it to the trials, he has one score (32) that is ¥ yes and motivation more than two standard deviations above the mean (2) 14 Total number of runs in 8 games = 128 Total number of runs in 13 games =13x 20 = 260 ¥ 260 Average in last 5 games = anes y 260-128 5 = 26,4 His average must be 26,4 or approximately 26 runs in his last5 | v 26,4 or 26 games. (3) is Increase. y¥ increase Both 0 and 20 are further than one standard deviation from the mean, and will therefore cause an increase in standard deviation. Y both 0 and 20 are further than one standard deviation from the mean (2) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathemawaploaded from Stanmorephysics.com KZN/November 2023 GRADE 11 Marking Guideline QUESTION 2 2.1 | A=29-6=23 ¥ A=23 B=89+74=163 v B=163 7 (2) 40630 = v 250 v answer (3) v answer 2.3 (1) 2.4 vA all points correctly plotted ¥ grounding ¥ shape 20 _ a 2 a, 2 i 3 3 140 145 150 155 Qhico 165 Q 170 175 180 185 (3) 25 Position Q1: 3 (250) = 63"score Position Q3: = (250) =188"score Q, =158 Note: YQ. Q, =167 Quaccept 157 —- 159 vO 3 “IQR =167—158 Qs accept 166 — 168 ‘ IQR accept 7, 8, 9, 10 and 11 answer =9 (3) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathemawaploaded from Stanmorephysics.com KZN/November 2023 GRADE 11 Marking Guideline QUESTION 3 *- A Note: A learner may introduce any point. 6, Qe" - a S 3.1 | y=-3x+8 : v answer “Me =—3 3.2 | tana=m,, =-3 Y¥ tana =Mao a =180° —71,57 =108, 43° v answer (2) _-4-(-9) 3.30) Mec = 4-(-6) Y substitution 1 3 v value of m., Let @be the angle of inclination of EC. tan@ = 5 C7 ; 0 = 2657" 26,57 [vertically opp. 2 s] [Zs ona straight line] ¥ 180° —108,43° = 71,57° 43° =71,57° =71,57° +26,57° [ext. Z of ABCG] Y answer = 98,14" (5) Copyright reserved Please turn over

Mathemawmaploaded from Stanmorephysics.com GRADE 11 Marking Guideline Downloaded from hlayiso.com KZN/November 2023 34 | m,. =-3 Mig, [AC 1 EF] vm, =2 +) Mee = 3 EF 3 Equation of EF y= F X+C Substitute E -9): -9= 38) c ¥ substitution c=-7 v value of c fn te-7 v answer 8 (4) 3.5 | To determine coordinates of F: ~3x+8= i X=7 Y equating the equations of EF 3 and AF a x=15 3 x=4,5 V x-coordinate of F y =-3x +8 =-3(4,5)+8=-5,5 ¥ y-coordinate of F z bi Y substitution in distance EF =,/| 4,5-(-6) | +] -5,5-(-9 il ( ) [ ( )] formula =11,07em Y answer (5) [18] QUESTION 4 ay P(-1;3 4.1 Y substitution in midpt. formula T(1;1) answer (2) Copyright reserved Please turn over

Mathemawmaploaded from Stanmorephysics.com Downloaded from hlayiso.com GRADE 11 Marking Guideline KZN/November 2023 341 Men “3 ml mM + TT bisector = 4.2 Subst. T (1; d m=lin y=mx+c: 1=1+c »<o ) x b Y gradient of PR Y gradient of | bisector Y subst. S(a; b) and m=1 v y=x (4) 43 | PR=\(-1-3)° +(8+1)° = 42 Buta=b, . Area APSR ST = f(a-1)? +(a-2)? ST = /2(a-1)? =12 square units a ; x base x height = 12 5xPRxST =12 5(4V2)( (2-2 2 ) =12 5x W2x V2 x ‘a—1)? =12 (a-1 =3 (a-1) =9 a-1=43 a=4(N/A) ora=-2 “ b=-2 “.S(-2;-2) Y length of PR ¥ length of ST i.t.o. aand b ¥ length of ST i.t.o. a (or b) only v ; x base x height =12 ¥ substitution of PR and ST v values of a and rejection ¥ answer (2) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathemawaploaded from Stanmorephysics.com KZN/November 2023 GRADE 11 Marking Guideline QUESTION 5 5.4 v vi-m tan33° = ¥ answer vi-m? (2) 5.1.2 | cos777° =cos{ 2(360') +57" | ¥ cos| 2(360') +57" | =cos57° Y cos57° =sin33° =m v answer (3) 5.1.3 | sin(—237°) =-sin 237° Y¥ —sin237° =—-(-sin57°) v -sin57° =y1-m? v answer (3) OR OR sin(—237°) = sin123° ¥ sin123° =sin57° v sin57° = V1_-m2 v answer (3) 5.2. 4tanB+5=0 5 P 5 tan B = “a ; Val Y tanB= “| A, -§ ©) x Vv r=V4l Y value of cos 8 v answer (4) Copyright reserved Please turn over

Downloaded from hlayiso.com MathemMawmaploaded from Stanmorephysics.com KZN/November 2023 GRADE 11 Marking Guideline 531 sin(180°— 8). cos(90°— 8)-1 cos(—f) _ Sin B.sinB-1 Y simplification (numerator) ¥ simplification (denominator) Y¥ use of square identity Y answer (4) 5.3.2 | If cos(—f) =0 ¥ cos(—) =0 cos 8 =0 B=90° or f=270° v 90°; v 270° (3) 5.4.1 tana.sina+Ccosa = SIN® sing +cosa y sina cosa cosa sin’? a = +COSa cosa ine 2 eel are Oe Y simplification cosa _o1 cosa (2) 5.4.2 1 8 yl _ 3 cosa sina cosa. sina sina =3cosa@ tana =3 Y tane=3 ref Z = tan*(3) = 71.57° @ =71.57° + k.180, keZ ¥ a@=71.57°+k.180° V¥keZ (4) [25] Copyright reserved Please turn over

Downloaded from hlayiso.com MathemMawaploaded from Stanmorephysics.com KZN/November 2023 GRADE 11 Marking Guideline QUESTION 6 6.1.1 | 1 v answer (1) 6.1.2 | period = (20° oar v answer (1) " fay “] (120°;1) v¥ Shape v x &y - intercepts Y¥ turning points -180 (—60°;1) (3) -15 6.3 —180°<x<-150° or 30°<x<180° v —180° < x <-—150° ¥ 30°<x<180° (2) 6.4 x=120° or x=180° ¥ 120° y¥ 180° (2) 6.5 h(x) = sin(—x — 30°) ¥ ¥ sin(—x —30°) =—sin(x + 30°) OR sin(x+30°) (2) [11] QUESTION 7 71 AC? = AB? + BC? — 2(AB)(BC)cosB Y applying the cosine rule 15? =11° +6? —2(11)(6)cosB Y substitution = 11° +6? -157 cos B = ———___—— 2(11)(6) 17 a =-33 OR —0,5151..... Y value of cosB B =180° —58,99° =121,01 v answer (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathemfawmaploaded from Stanmorephysics.com GRADE 11 Marking Guideline KZN/November 2023 72 In AQRS: Q, =Q5P =0 [alt 2’s =] vs 2. QSR =(a+0) SR Q sin(a +0) = ¥ substitution in sine rule rS, 3 gp — QS: Sina V sR = QS:Sin(a +4) AY sin B In AQPS: PS cos 0 = Os ¥ correct ratio cos0 = x Qs Xx x Sai, ¥ QS = —— Q cos@ Q cos? x _ sin(a+) a ~ cso * ~sinB ¥ substitution of QS sR xeiilo:t) cos @sinB (6) [10] QUESTION 8 8.1 Total surface area = (length x breadth) + 2(length x height) + 2(breadth x height) | v formula for surface area = (27x24) +2(27 x10) + 2(24 x10) Y substitution = 1668 cm? v answer (3) 8.2 Volume = length x breadth x height Y formula for volume 4500cm* = 27cm 24cm height Y substitution height = 4900 24x27 =6,94cm Y answer (3) [6] Copyright reserved Please turn over

Downloaded from hlayiso.com MathemMawmaploaded from Stanmorephysics.com KZN/November 2023 GRADE 11 Marking Guideline QUESTION 9 9.1 € =90° [zZ in a semi-circle] Y SIR E, =C=90° [corr. 2’s, OEI|IBC] Y SIR AE= AG [line from centre | to chord] vSIR AE =2(1 2 AE =6cm In AOAE: OA? = AE? + OE? [Pythagoras] v SIR 10° = 6? + OE’ OE? = 64 OE=8 cm YS (length of OE) ED = OD-OE ED =10-8 ED=2cm v answer —_ (6) 9.2 H=FGJ [tan-chord-theorem] ¥SYR =x-5 H=F [2s opposite = sides] v SIR =x-5 F+FGH +H =180 [sum of 2s of AFGH] vs x-5 +3x+x-5 =180 5x =190° x = 38" ¥ answer (5) [11] QUESTION 10 10.1.1 | R, +R, =90° [radius | to tangent] VSYR R, = 40° v answer — (3) 10.1.2 S, +S, =50° [tan chord theorem] YSYR W+WRS+§ =180° [sum of Zs of AWSR] W +90° +50° = 180° W =40° v answer (3) OR OR T, =90° [Z in semi-circle] ¥SVYR w=7,-R, [exterior 2 of AWTR] =90° -50° = 40 V answer (3) Copyright reserved Please turn over

Downloaded from hlayiso.com MathemMawmaploaded from Stanmorepkysics.com KZN/November 2023 GRADE 11 Marking Guideline 10.1.3 | PR, =R, =40° [Zs in the same segment] VSYR : (2) 10.2 T, =90° [zZ ina semi-circle] v¥ SIR PTS =F, =90°+T, Vv, +S, =90° +S, [ext 2 of A =sum of int opp zs] | v S/R But: T, [4s in the same segment] YSVYR V, =PTS (4) 10.3 | VY =PTS .. WVPT is a cyclic quadrilateral [converse: exterior 2=intoppz] |“ R (1) [13] QUESTION 11 11.1 Pp Note: No construction: ; Q Ss NN Construction: Draw OQ and OS Y construction 0, =2P [Z @ centre = 2xz @ circumf.] Y SIR 0, =2R [2 @ centre = 2x2 @ circumf.] vs 0, +0, =360° [Z’s around a point] Y SIR J. 2P+ 2R = 360° 2(P +R) = 360° vs 2. P+R =180° (5) 11.2.1 | B,=A,=x [Zs in the same segment] “Ss DB, =B, =x [2s opp. = radii] VS YR A, =0, =x [Zs in the same segment] YSIR (4) OR OR B, =A, =x [2s in the same segment] vs D, =B, =x [2s opp. = radii] YS YR A, =A, =x [equal chords; equal 2s] VSIR (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathemawaploaded from Stanmorephysics.com KZN/November 2023 GRADE 11 Marking Guideline 11.2.2 | -. BOD =180° - 2x [sum of Zs of ABOD] YSVYR é= i. BOD [2 at centre=2x Zatcircumference] | VR v answer : (4) 11.23 E,= [ext. 2 of cyclic quadrilateral] YSYR - =E [ext 2 of AFEA] “S = (90° -x)+x=90 B, =180 (ee A.) [sum of Zs of ABFA] vs =90° - AB = AE [sides opp. = Zs] yR (5) OR OR E, =C=90' -x [ext. Z of cyclic quadrilateral] YSYR F, =180 -(é, + A,) [sum of 2s of AEFA] vs =90 In AABF and A AEF: 1. BF=FE [line from centre 1 to chord] vSIR 2. AF=AF [common] 3. F=F [Zs on a straight line] .. AABF = AAEF [s;2;s] /AB = AE [=As] YR (5) 11.2.4 | B, +B, =(90° -x)+x=90 ¥ “showing that .. OBA = B, +B, +B, = 90° +B, > 90 OBA > 90 For AB to be a tangent to circle BCDE, OBA should be equal to 90°. | “For AB tobea tangent, [converse: tangent radius] .. AB is not a tangent to circle BCDE. OR For AB to be a tangent to circle BCDE, DBA should be equal to OBA should be = 90° (3) OR Y AB will be a tangent if €=90°-x [converse: tan-chord-theorem]. DBA =C =90°- x But: DBA =B,+B, = 90°-x+B, =€+B, ¥¥ showing that .. DBA>C DBA>C AB is not a tangent to circle BCDE. (3) [21] TOTAL: 150 Copyright reserved