Mathematics Grade 11 Paper 2 KZN November 2025 Paper & Memo

~ KWAZULU-NATAL PROVINCE | EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL | SENIOR CERTIFICATE | GRADE 11 | Cebbnnbnn ge OO Le Sean eee, iy | " MATHEMATICS P2 i NOVEMBER 2025 : MITT Tee. Ge, peeeemorephvsies.com iY MARKS: 150 TIME: 3 hours This question paper consists of 13 pages and 1 information sheet. Po

DoWHiededed from Stanmorephysics.com November 2025 le 11 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 10. This question paper consists of 11 questions. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. Clearly show numbering if additional space has been used to answer a question. If NOT, cross out all rough work done in the additional space. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. . Answers only will NOT necessarily be awarded full marks. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. If necessary, round off answers correet'to/TWO decimal places, unless stated otherwise. Copyright reserved Please Turn Over

Dewetended from Stanmorephysics.com November 2025 CAPS - Grade 11 QUESTION 1 Learners collected cans during “Operation Clean Up”. The number of cans collected by 10 classes is shown in the table below. ee 250 | 259 | 265 | 276 | 290 | 290 | 294 | 310 | 315 | 330 1.1 Calculate the mean number of cans collected by these 10 classes. 1.2 Write down the standard deviation of the above data. 1.3 How many classes are within one standard deviation of the mean? 1.4 Draw a box-and-whisker diagram of the data on the scaled line provided in the ANSWER BOOK. 1.5 When another 10 classes added their cans to the ori ginal number of cans collected, the new mean is 250. 1.5.1 Calculate how many cans were collected by the 10 classes that were added. 1.5.2 What effect does adding thecans of these 10 classes have on the original standard deviation? — ek “) ee di | : St Aan i QUESTION 2 Learners were asked the time (to the rest minute) that they usually take to get from home to school each morning. The resulis aré shown in the table below: Time in minutes | Number of learners 5<1r<i0 160 10<t<15 150 15<t<20 110 20<t<25 60 25<t<30 45 30<1<35 15 2.1 Write down the modal class of the data. 2.2 Complete the cumulative frequency column in the table provided in the ANSWER BOOK. 2.3. Draw the ogive of this data on the grid provided in the ANSWER BOOK. 2.4 Estimate the percentage of learners who take more than 25 minutes to get to school. 2.5 Estimate the mean time that these learners take to get to school in the morning. Copyright reserved Please Turn Over (2) (1) (2) (2) (2) (1) [10] () (1) (3) (2) (3) [10]

Downleaded from Stanmorephysics.com November 2025 APS - Grade 11 QUESTION 3 In the diagram below, A(—4 ; 12), B(-7;0),E and D(8; —2) are the vertices of a trapezium. E lies on the y axis such that BE is perpendicular to DE. The equation of AB is y=4x+28. The angles of inclination of AB and AD are @ and @ respectively. M is the midpoint of BD. F(3; 2) is a point on AD. A(-4;12) 4Y¥ F(3 ; 2) ‘ $ N D(8; -2) 3.1. Calculate the length of AD. (2) 3.2 Calculate the coordinates of M, the midpoint of BD. (2) 3.3. Calculate the gradient of BE. (3) 3.4 Determine the equation of DE. ; (3) 3.5. Calculate the value of f, the y-coordinate of F. (3) 3.6 Calculate the coordinates of C, if ABCD, in that order, is a parallelogram. (2) 3.7 Calculate the size of BCD, correct to the nearest degree. (3) [20] Copyright reserved Please Turn Over

Dewnlaaded from Stanmorephysi¢s.com CAPS - Grade 11 November 2025 QUESTION 4 In the diagram below, A is a point on the line with equation y+2x=0. B is a point on OA such that OB = 5/5 units. Bisa point on the y axis such that EB is perpendicular to OA. The equation of OD is y=2x and CE // OA. x > Determine the following: ( 5 4.1 The coordinates of B. (4) | 4.2 The area of the trapezium OBEC correct to the nearest square unit. (8) [12] Copyright reserved Please Turn Over

Downleaded, from Stanmorephysics.com NeveRuex 02s CAPS - Grade 11 QUESTION 5 5.1. In the diagram below, P is the point (V2 ; k) such that OP =3¥3 units. TOP = a is an p obtuse angle. ty P (-v2 ‘ k) Wi 2) vy WITHOUT using a calculator, determine the value of the following: 5.1.1 k. (2) 5.1.2 tan(a-180°) (2) 5.1.3 cos f, if iti hat 2+ =180°. (2) 5.2 If sin34°= p, determ rms of p, the value of tan484° WITHOUT using a calculator. A ‘4 (4) 5.3. Simplify the following WITHOUT. the use of a calculator: tan 205°.cos315°.sin (—45°) (5) sin 210°. cos 150°. tan 25° 5.4 Simplify the following to a single trigonometric term: f sin x sin (180°+ x).cos(x-90°) -—__——______—__ 6 (1809+ x).cos(x-90°) - = eos) tan(180—a) o 5.5 Given the identity: ated + E083 = Z cosx I1+sinx cosx 5.5.1 Prove the identity. (4) 5.5.2 For which values-of x in the interval 0° < x <360° will the above identity be undefined? (2) 5.6 Determine the general solution of the equation: tan x.sin x = -s —cosx (6) sinx [33] Copyright reserved Please Turn Over

Dawnaleaded from Stanmorephysics.com November 2025 CAPS - Grade 11 QUESTION 6 Below is a sketch of the graph of f(x)=acosx for x <[-180° ; 90°]. vt 6.1 Write down the (1) 6.2 Write down the pe De) = i (65) i a 6.3 Sketch the graph of g os 2x for x e[-180° ; 90°] on the set of axes provided in the ANSWER BOOK.’ (3) 64 Use the graphs to write down the value of x for which f (x)-1=g(x)+2. (2) 6.5 Determine minimum value of 1- 2[ sin (90° - x)] , (3) , [10] Copyright reserved Please Turn Over

Downleaded..from Stanmorephysics,com November 2025 CAPS - Grade 11 QUESTION 7 In the diagram below, PQ. QR. QPS =a and QPR =0 and PR =k units. P : O 7A Write down the'size of RPS imterms of @ and 0. () 7.2 Show that RS== (3) 73 Calculate the le : t 7 units, 9=58° and @ = 23°. (2) 74 Show that the area of ARSP = 48 square units, correct to the nearest square unit. (2) [8] Copyright reserved Please Turn Over

Dewmleaded from Stanmorephysiss.com QUESTION 8 CAPS - Grade 11 The diagram below represents an open tank with a square base (side dimensions of x cm) and a height of h cm. The tank has a volume of 490 cm’. 8.1 Determine the heigh 8.2 Show that A, the exte hem xcm of the tank is given by: A =x° 48 cm. x 8.3 Calculate the external surface area of the tank if the height is 10 cm. Copyright reserved Please Turn Over November 2025 (2) (2) (4) [8]

Downlaadadifrom Stanmorephysics.cam November 2¢ CAPS - Grade 11 QUESTION 9 91 In the diagram below, A, B and P are points on the circle with centre O. PA, PB, AO and BO are joined. R BOOK to prove the theorem which states that the ord at the centre of the circle is twice the angle subtended by the same are ¢ 1 circumference, that is, prove that AOB = 2ABB a Copyright reserved Please Turn Over

Dowalaaded from Stanmorephysi¢s.com —— CAPS - Grade 11 9:2 In the diagram below, PQRS is a cyclic quadrilateral. M is the centre of the circle. MS bisects PSR such that §, =$, =42° and PQ=SR. P. 9.2.1 Name, givi : (2) 9.2.2 Calculate, giving (a) PQR (2) () B (3) 9.2.3. Prove that SP// QR. (3) (15) Copyright reserved Please Turn Over

Downloadedafram Stanmorephysics.com CAPS - Grade 11 QUESTION 10 November In the diagram, VT and VZ are tangents to the circle with centre O at T and Z respectively. T, S, R, Zand Y are points on the circle such that TR is the diameter. P is the midpoint of chord TZ. T, =x. Prove that: 10.1 TOZYV is a cyclic quadrilateral. 10.2 RZ// SY 10.3 T==x i 2 10.4 Y¥,=90°+T, Copyright reserved Please Turn Over

Dowrileadead from Stanmorephysios.com November 2025 APS - Grade 11 QUESTION 11 In the diagram below, AD is a chord of circle with centre O. OB is drawn perpendicular to AD and cuts AD at E. OE =k and EB =m. OA and AB are drawn. B 11.1 Prove that AB* = (4) 11.2 Write down, giving a reason, the length of DE in terms of k and m. (2) [6] TOTAL: 150 MARKS Copyright reserved Please Turn Over

Bawhistded from Stanmorephysics. com November 2025 INFORMATION SHEET fee? —4ac 2a A= P(i+ni) A= P(1-ni) A=PQU-—i)" A=P(+i)" Tet blt (n—1)a S, =F 2a+in-nd] T, =ar" Ky _ alr" =1) sr#l S.=pSo3-l<r<l 7 mul pata 1 pat 04i)"] i - 1 . f(xt+h)— f(x) 4 = lim ————— FG) h>0 | h d= V(x, -%,) +02- AT% FM 2 2 y=mxt+ec —y, = m(x— xp» 9 m=22— 2 m=tan @ = Xy -% (ea) +(y-bF =r? In AABC: —2-=—2_ = sinA sinB 1 : area AABC = ry ab.sinC sin(a+ 8) =sina.cos B+cosa.sin Z cos(@+ 8) =cosa@.cos 8—sina.sin B cos’ a—sin? a cos2a@ =41—2sin? a 2cos* a1 Copyright reserved sin(@— #)=sina.cos 8—cosa.sin B cos(a—£)=cosa.cos #+sin q@.sin B sin2a = 2sina.cosa Yo, -D? = isl n o P(A or B) = P(A) + P(B) — P(A and B) Ue-zy-H) 2 Sa

EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL TOR CERTIFICATE MARKS: 150 These marking guidelines consist of 18 pages. NOTE:

Mathefowroaded from Stanmoreph ics.com November 2025 G 11 Marking Guideline If a candidate answers a question TWICE, only mark the FIRST attempt. Consistent accuracy applies in ALL aspects of the Marking Guidelines. Stop marking at the second calculation error. e Assuming answers/values in order to solve a problem is NOT acceptable. NOTA: e = As 'n kandidaat 'n vraag TWEE KEER beantwoord, merk slegs die EERSTE poging. e Volgehoue akkuraatheid word in ALLE aspekte van die Nasienriglyne toegepas. Hou op nasien-by die tweede berekeningsfout. e = Aanvaar van antwoorde/waardes om 'n probleem op te los, word NIE toegelaat nie. GEOMETRY A mark for a correct statement (A statement mark is independent of a reason) Ss 'n Punt vir 'n korrekte bewering _ ('n Punt vir 'n bewe inklik van die rede) if the statement is correct) R ('n Punt word slegs vir e oegeken as die bewering korrek is) Award a mark if statement AND reason are both correct S/R Ken 'n punt toe as die bewering EN rede beide korrek is Copyright Reserved Please turn over

Mathefawrloaded from Stanmoreph ics.com November 2025 GRADE 11 Marking Guideline QUESTION 1 1. Re L879 AY 2879 10 = 287,9 ANSWER ONLY: Full marks CAY answer ial 2) eps AY o =24,45 d) 13 | Interval: (263,45 ; 312,35) CAV (§-0; ¥+0) .. 6 classes Ca anes (2) 1.4 : . AY whiskers ending at 250 & 330 AY box with Q,=265 Q,=290 & Q, =310 (2) 1.5.1 | Let the number of cans add 7 2879+k _ 459 CAV 2879 +k _ 459 20 20 2879 +k =5000 k=2121 .. 2121 cans were added. CAY answer (2) 1.5.2 | The standard deviation will increase. AY increase d) [10] Copyright Reserved Please turn over

Matheewrloaded from Stanmorephysics.com November 2025 G 11 Marking Guideline QUESTION 2 2.1 | 5<t<10 AY answer Q) 22 Time Number of | Cumulative . (in minutes) learners frequency §<t<10 160 160 10<t<15 150 310 15<1<20 110 420 AY all cumulative — frequency values 20<t<25 60 480 correct 25<t<30 45 525 30<1<35 15 540 Q) = 23 550 ES 500 eeenee! eee AY grounding at (5;0) aso Hee = Seecee CAY plotting 400 FERRER aeenae cumulative z oo ; frequency at z 350 et : upper limits & 300 =o Pa t t cot g =: t = AY smooth increasing = 200 = 150 : 5 ; ARERR RP RPEEP EE 0 15 25 30 35 40 (3) f ites 2.4 | No. of learners = 540-48 cr on i AY 60 % of learners = ——x100% 540 =11,11% CAY answer (2) — _(7,5x160) + (12,5150) +(17,5x110) + (22,560) +(27,5x45)+(32,5x15) | AY frequency 2.5 | x= 340 x midpoints _ 8075 AY 8075 _ a : ANSWER ONLY: Full marks. All ACCURACY marks x =14,95 minutes Answer must EXACTLY be 14,95 minutes. CA aa, [10} Copyright Reserved Please turn over

Mathefowroaded from Stanmoreph ics.com November 2025 GRADE 11 Marking Guideline QUESTION 3 3.1. | AD= f =4-8] +[12- (2p A substitution in correct = 340 or 285 units or 18,44 units CAY answer (2) 3.2 w= : a) 2 2 1 1 AY x-coordinate of = M| =; -1 2 2 AY y-coordinate of —1 (2) 33. | my=m,, (BE// AD) AY My, =m = 12-@) | <1 If mg, = Mgy is not stated: || AV substitution —A-8 Award 2 marks here __ 414 di 19 OG CAY answer a 6 1 3.4. | my, = 2(BE LDE) CAY my, =-— 7 Mg Y-yY= m(x-xX) y+2=—(x-8) <—| “Tp m,, =——— is not stated: CAY substitution of D(8 ; —2) & 6 48 BE mpe into equation of a line yt2 “s a Award 2 marks here 6 62 a CAY answer (3) OR OR 6 1 Mpp = —(DE 1 DE) CAY m,, =-— 7. Mop y=mxt+e 2= 5) semi WC _ to isnot stated: CAY substitution of D(8 ; -2) & 62 Mop moe into equation of a line c= a Award 2 marks here 6 62 » ie x a CAY answer @) 3.5 My = Mp Av condition f-12_ _t42 AY substitution 3-(4) 3-8 7t+14=—St+60 12t=46 t=2 or 3> or 3,83 CAV answer 6 6 (3) Copyright Reserved Please turn over

MatheRowrloaded from Stanmoreph ics.com November 2025 GRADE 11 Marking Guideline OR My = t AY condition t=(2)_ 7 AY substitution 3-8 6 t+2= ¢s(-2) 6 = si or 32 or 3,83 CAY answer 6 6 (@) 3.5 | OR Equation of AD: y-l2= —[x-(-4)] or 12= — a+ c AY substitution of A 4 ; 12) & 6 6 map into equation of a line i= 7,28 * 6 mec 2M ar ar) t= -203)+2 AY substitution of F(3 ; t) t= a or re CAY answer 6 (3) OR Equation of AD: y-(2)= 1-8) or —2= _7¢) +é AY substitution of D(8 ; -2) & 6 6 map into equation of a line 7 28 22 yt+2=-—x+— or c=— 6 3 3 ea gg 6 3 t= Lise AY substitution of F(3 ; t) 6 8 = 23 or 3> or 3,83 CAY answer 6 6 (3) 3.6 | C(5; -14) [by translation] AAY V answer OR (2) C(xsy) m(3 s -1) A(-4; 12) xo eee 1 2 2 2 x,-4=1 and yg4i2=-2 w= 5 andy. =-14 .C(5; -14) AAY V answer (2) Copyright Reserved Please turn over

Mathefowroaded from Stanmorephysics.com G 11 Marking Guideline November 2025 OB =55 (x-0)' +(-2x-0)' =5 Vx? +4x? =5V5 5x? =12 x=-5 .. B(-5; 10) 3.7 tand =m, LO 7 CAY tan 9 = mg, 6 9 =130,60° CAY @ tana@=m,, =4 a@=75,96° AY a BAD =130,60°-—75,96° [ext. Z of A] =54,64° CAY BAD .. BCD = BAD =55° [opp. Zs of a parm.] CAY answer (5) [20] QUESTION 4 4.1 | Equation of AO: y =-2x AY substitution of O(0 ; 0) & B(x ; —2x) into distance formula AY equating to 5/5 CAY x value CAY y value (4) Copyright Reserved Please turn over

Mathefowroaded from Stanmorephysics.com G 11 Marking Guideline November 2025 42 | my=s2 gg -5 (EB LOA) Equation BE: y—10 =F (x45) 2 Mpc = -2 [EC //OB] 1 nyesrxt— en Equation of EC: y = 2x42 at C: ~2x4 2 2x Oj} + pe = [-s-0} (10-28) -98 Area OBEC =; (BO+ EC)(BE) AY My -5 CAY e(0 " >} CAY equation of EC CAY (2 F 3) 8 4 CAY length of EC CAY length of BE CAY substitution into area of a trapezium = 51 square units CAY answer mt (8) tanOd=m,, = —2 2. = 180" = tan"(2) 8 =116,57° AY @=116,57° .. BOE = 116,57—90° = 26,57° CAY size of BOE Mop = A 5 Copyright Reserved Please turn over

Mathefowroaded from Stanmoreph ics.com November 2025 GRADE 11 Marking Guideline <. Equation BE: y-10=5 (x +5) wystx+2 ay 2 2 J e(0 } 3) CAY coordinates of E 7 BO = 2 unit and OB= 55 units Area ABOE = L x (2) sf (sv5) x sin 26,57° CAY substitution into area 2 2 formula = 31,26 square units CAY Area ABOE In AEOC: CEO = 26,57’ — [alt Zs=; BO // EC] seat Myce =-2 [EC // OBJ Equation of EC: y=—2x# at E:2x= ~9%4 25 2 25 pom ae 8 25 =o| = * ( 8 =25 v4 (25. 25 ae 8° 4 CAY coordinates of C 2 2 8 4 2 8 Area ABOC = Jigh:2l) | 2eNB0) aa coe 22 8 “dl = 19,53 square units Aaa CAV AEOC OR J either of the OR 1 (25 5 Area AEOC = —x| — |x| = two for the 2 ( 2 ( 7 b area of =19,53 square units AEOC CAV Area AEOC :. Area OBEC = 31,26 +19,53 =51 square units CAY answer (8) [12] Copyright Reserved Please turn over

Mathefowroaded from Stanmorephysics. .com Marking Guideline November 2025 QUESTION 5 5.1.1 (AR) $k? = (33) AY substitution 227 a k=+t5 oe ons | CAY answer (2) 5.1.2 | tan(a—180’ ) = tana AY tand = CAY answer v2 2) 5.1.3 | cos B = cos(180°-a) =-cosa@ AY -—cosa (-¥2) - 3y3 = Re AY answer 3v3 (2) 5.2 sin 34° = p AY third side = 1p? tan 484° = AY tan124° CAY reduction CAY answer (4) OR OR tan 484° = tan 124° AY tan124° _ sin 124° y sine cos 124° cos 124° _ sin(90°+34°) ~ cos(90°+34°) _ cos 34° CAV cos 34° —sin34° —sin 34° 2 = _i-r? CAY answer P (4) Copyright Reserved Please turn over

Matheowrloaded from Stanmoreph ics.com November 2025 GRADE 11 Marking Guideline 5.3 | tan205°,cos315°.sin(—45°) (tan 25°)(cos45°)(—sin45°) | AV reductions in numerator sin210°.cos150°.tan25° — (—sin30°)(—cos30°)(tan 25°) | AY reductions in denominator v2 / v2 CAY substitution of special Z (2 2 values for 45° 7 CAY substitution of special Z values for 30° CAV answer (5) 5.4 : AY sin(180°+x)=—sinx AY cos(x-90°)=sinx AY sin(90°—x)=cosx AY tan(180°-x)=—tanx AY tanx=22* cosx CAY answer ; (6) 5.5.1) LHS= l+sinx coe cosx 1+sinx _ (1+sin x)’ +(cosx) cos x(1+sinx) _ 1+2sinx+sin* x+cos’ x AY numerator cos x(1+sin x) Av LCD __1+2sinx+] AY identity: sin? x+cos* x=1 cos x(1+sin x) _ _ 2+2sinx cos x(1+sin.x) 2(1+sin x) AY factorization ~ cosx(1+sin x) _ 2 cos x 4 = RHS (4) 5.5.2 | Undefined: cosx=0 or sinx=-l x =90° or x =270° or x=270° “x € {90° ; 270°} YA 90° YA 270° (2) Copyright Reserved Please turn over

Matheoswreloaded from Stanmorephysics.com G 11 Marking Guideline November 2025 5.6 _ sinx cosx AY tanx AY sin* x =3cosx—sin x.cos* x CAY factors on LHS CAY equation x=71,57°+k.180; keEZ CAY answer AVY +k.180;keEZ (6) OR OR 3 tan x.sin x =——. sin x sinx sinx | ogy a AY tanx = SX. cosx 1 sinx cosx omy | 2 sin’x+cos’x_ 3 w sin? x + cos’ x cos x . AY E cos x 1 ahimorephysics.com _ CAY sin? x+cos*x=1 cos x sinx _ cos x tanx=3 CAY equation x= 71,57°+k.180; keZ CAY answer AV +k.180; k eZ (6) [33] QUESTION 6 6.1 | a=2 AY answer dQ) 6.2 | Period =180° AY answer (1) 6.3 ——— Graph of g: Pm AY x-intercepts and y-intercept AY turning points \.4, | AY shape 0 nt: Ineeeeeaemenneee innmeenieaninen: ~ (3) Copyright Reserved Please turn over

Mathefowroaded from Stanmoreph ics.com GRADE 11 Marking Guideline November 2025 64 | f(x)=l=g(x)+2 f(x)=g(x)+3 AY f(x) =g(x)+3 “Aa OF AY answer (2) 6.5 | 1=2[sin(90°-x) ]=1-2cosx=—2cosx+1 AY 1—2cosx Minimum value of —2cos x is—2 AY min. of —2cosx Minimum value of 1—2cosx =—2+1 =] AY answer OR (3) 1-2[sin(90°—x) ]=1-2cosx seiiniinenates Maximum value of 2 cos x is AM mana ease . Minimum value of 1-2 co: AY answer Q) [10] QUESTION 7 7.1 | RPS=(0-a) AY answer a) 7.2 §, =(90°+a) [exterior angle of A] RS ik sin(@—a) sin(90°+a) ps — &sin(O-a@) In ARPS: AY §, =(90°+a) AY substitution into the sine rule =32° Area ARSP =>(RS)(PR)sin R =5*10,59%17 xsin32° = 48 square units cos a AY sin(90°+a) =cosa G3) 17.sin (58°—23° 7.3 | RS= ba lle AY substitution cos 23° =10,59 units CAY answer (2) 74 R=90°-—58° [sum of Zs ina A] . A” R=32° AY substitution into the area rule (2) [8] Copyright Reserved Please turn over

Matheoswreloaded from Stanmorephysics.com November 2025 G 11 Marking Guideline 8.1 °h 490 490 xe q 8.2 | A=x?+4xh 490 =x +4x *) (2 > 1960, =x 4 xi +——em - 33 | 4% 49 AY substitution of h = 10 2: x x =49 x=7 AY x=7 A=( 1y pan CAY substitution into external vs surface area = 329 cm* CAY answer (4) [8] Copyright Reserved Please turn over

Mathefowroaded from Stanmoreph ics.com GRADE 11 Marking Guideline QUESTION 9 November 2025 91 [ext. Z of A] Similarly: 0, = 2P, 2.0, +0, = 2B, +2P, [4s opp. = sides] =2(R+B,) ie. AOB = 2APB If there is no construction: : AOB =360°—(POA +POB) =360°-(180°—2P, +180°—28, ) = 2P, + 2P, [Zs opp. equal sides] [sum of 7s of AAPO] OR Construction: Draw PO Proof: A=?, .. POA =180°—2P, B= B, [Zs opp. equal sides] .. POB =180°-2P, [sum of 7s of ABPO] [Zs around a point] = 2(P, +P,) =2APB If there is no construction: : Y construction ¥ S/R ¥ SIR vS vs (5) Y construction ¥ SIR v¥ S/R vs vs (5) Copyright Reserved Please turn over

MatheRowrloaded from Stanmoreph ics.com November 2025 GRADE 11 Marking Guideline 9.2.1 R,=§, =42° [Zs opp. = sides] YSYR 9.2.2 (a) R =96° [ opp. Zs of a cyclic quad] YSYR Dar (2) 9.2.2 (b)-| SMR = 96° [sum of angles of A] VSR LJP = 48° [42 at centre = 2x2 at circumference] YSYR (3) 9.2.3 R, =P =48° [equal chords; equal Zs] YSYR “SP // QR [alt Zs=] YR (3) OR SRQ=96° [opp. 2s of a cyclic quad supplementary] YSYR PSR + SRQ =84°+ 96° = = 180° “SP // QR supplementary] YR (3) [15] QESTION 10 10.1 | OTV=90° {tan 1 radius] YSYR OZV =90° [tan radius] v SIR OTV +OZV =180° -. TOZV is a cyclic quadrilateral [converse opp. 2s of a cyclic quad. suppl.] | YR OR (4) TV= ZV [two tans from same point] YvSIR “.TZV =T, =x [Zs opp. = sides] Y SIR OZV = 90° [tan 1 radius] YSIR 2. Z, =90°—x but P, =90° [line from centre to midpoint] s 0, = {sum of angles of A] -.0,-t, -. TOZV is acyclic quadrilateral [converse Zs in the same segment] YR (4) 10.2 RZT = 90" [Z in semi-circle] YSYR P =90° [line from centre to midpoint of chord] YSYR “RZ //SY [corresponding Zs =] YR OR (5) RIT = 90° [angle in semi-circle] YSYR P,=90" [line from centre to midpoint of chord] YSYR “RZ // SY [co-interior Zs supplementary] YR (5) Copyright Reserved Please turn over

Matheowrloaded from Stanmorephysics.com November 2025 G 11 Marking Guideline 10.3 R =x [tan chord theorem] Y¥SYR 0, =R, =x [alt Zs=;RZ // SY] v SIR Butt -16 [Z at centre = 2xZ at Y SIR uy 2°! circumference] ~ 1 ——A=5% (4) OR R,=x YSYR 6, =R, =x [tan chord theorem] VSIR and T =§ [alt 2s =; RZ// SY] but T, +8, =O, [2s:qpp.~ sides] v S/R a a » [ext. Z of A T+T,=x mg [ext 2 0f A] 29, =x ~ | == 4 5 x (4) 10.4 | §,=T, =5 [Zs opp. = sides] Vv SIR 7, = § - 1 ™ [Zs in the same segment] YSVYR 2 Y, =P,+Z, [ext. Z of A] Y SIR =90°+—x vS =90°+ Th (5) OR OR _ [proved in 10.3] = L radi a a tan 1 radius = 90°—x l ] en but % =t+7 [4s in the same segment] 1 | 2 Y= A agi 90°=5 2) Re — YSIR a 130°-¥, ’ [4s on a straight line] - M4 vs =90°+—x is (5) =904+T, Copyright Reserved Please turn over

Matheewrloaded from Stanmoreph ics.com November 2025 GRADE 11 Marking Guideline OR OR TSR =90° [Z in a semi-circle] YSYR R, =180°-90°-—x [sum of Zs ofa A] YSIR A 1 R,=90°-—x 2; Y, =R,+ R, z 1 [ext. Zof a cyclic quad] YSIR =90°-—x+x 2 =90°+—x VS =90°+T, (3) [15] QUESTION 11 11.1 OA=OB=k+m [radii] ¥ SR In AAOE: AE* = OA? — [Pythagoras] ¥ SR =(k+m) = =k +2k = 2km+m? vS In AAEB: AB? = AE’ + EB” [Pythagoras] = 2km +m? +m? vs 7 =2km+2m* o OR In AAEB: AB? = EB*+ AE? [Pythagoras] v¥ S/R In AOEA: AE? = OA? -—OE? [Pythagoras] .. AB? = EB’ + OA® — OE? and OA = OB f [radii] v S/R .. AB? = EB’+ OB’ —- OE? =EB’+ OB’ =(OB-EB)" vs = EB’+ OB* — OB’ + 20B.EB- EB* = 20B.EB vs =2(k+m).m =2km+2m* (4) 11.2 | DE=AE [line from centre L YR hord] vs = V2km+2m* is Q) [6] TOTAL: 150 MARKS Copyright Reserved 2 igase tut UveL