# Mathematics Grade 11 Paper 2 KZN November 2025 Paper & Memo

> Grade 11 Mathematics Paper 2 question paper and memo from the KZN November Exam 2025, 32 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/kzn-maths-grade-11-november-2025-p2-and-memo/a2aac931-a969-41c8-bde2-85d23035d042
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2025
- Exam period: KZN November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 32

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## About this document

Mathematics · Grade 11 · KZN November Exam · 2025. Question paper and memorandum, 32 pages. Read online or download the PDF.

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## Document text

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### Page 1

~ 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

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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.

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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.

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(2)
(1)
(2)

(2)

(2)

(1)
[10]

()
(1)
(3)
(2)

(3)
[10]

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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]

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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]

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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]

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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]

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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]

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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.

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November 2025

(2)

(2)

(4)
[8]

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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

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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)

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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,

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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

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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

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EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
TOR CERTIFICATE

MARKS: 150

These marking guidelines consist of 18 pages.

NOTE:

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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

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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]

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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}

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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)

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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)

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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)

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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

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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]

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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)

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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)

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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)
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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]

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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]

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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)

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### Page 30

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)

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### Page 31

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,

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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

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2 igase tut UveL

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