# Mathematics Grade 11 Paper 2 KZN June 2024 Memo

> Grade 11 Mathematics Paper 2 memorandum from the KZN June Exam 2024, 24 pages. Read online or get the free PDF download, question paper linked when available.

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- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2024
- Exam period: KZN June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 24

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## About this document

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                                      NATIONAL
                                 SENIOR CERTIFICATE

                                     GRADE 11

                                 MATHEMATICS P2

                                  COMMON TEST

                                    JUNE 2024

      MARKS:           100

      TIME:            2 hours

        This question paper consists of 9 pages and 3 DIAGRAM SHEETS.

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Mathematics P2                        2                                           Common Test June 2024
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 INSTRUCTIONS AND INFORMATION

 Read the following instructions carefully before answering the questions.

 1.        This question paper consists of 7 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. which 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.        If necessary, round off answers correct to TWO decimal places, unless stated otherwise.

 8.        Diagrams are NOT necessarily drawn to scale.

 9.        THREE DIAGRAM SHEETS for QUESTION 1, QUESTION 2, QUESTION 5.1,
           QUESTION 5.2, QUESTION 6.1, QUESTION 6.2 and QUESTION 7 are attached at
           the end of this question paper. Detach the DIAGRAM SHEETS and hand them in
           together with your ANSWER BOOK.

 10.       Write neatly and legibly.

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QUESTION 1

 In the diagram below, D(−10 ; 6) , E, F and G(1 ; 9) are the vertices of a quadrilateral.
 F is a point on the x-axis. The diagonals of the quadrilateral bisect each other at K. The equation
 of diagonal EG is 3x − y + 6 = 0 and β is the angle of inclination of EG.

                                                         y

                                                K

                                                β
                                                       O                     F              x

                                         E

 1.1        Calculate the size of β.                                                                      (3)

 1.2        Calculate the coordinates of F if the equation of DF is x + 3y = 8.                           (2)

 1.3        Calculate the coordinates of E.                                                               (4)

 1.4        Prove that DEFG is a rhombus.                                                                 (3)
                                                                                                          [12]

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QUESTION 2

 In the diagram below, A(−2 ; −1), B(−3 ; 4) and C (1 ; 5) are the vertices of a triangle. BC is
 produced to D(5 ; y), AC cuts the y-axis at F. E is a point on the x-axis such that CE is
 perpendicular to the x-axis.

                                      y

                                  F

                                      O       E

                                                                                                   x

 2.1       Calculate the length of AC. Leave answer in simplified surd form.                               (2)

 2.2       Calculate the gradient of BC.                                                                   (2)

 2.3       Calculate the value of y if B, C and D are collinear.                                           (3)

 2.4       If H is a point such that AH ⊥ BC, determine the equation of AH.                                (3)

 2.5       Calculate the coordinates of G if CBAG, in that order, is a parallelogram.                      (3)

 2.6       Calculate the area of CEOF.                                                                     (4)
                                                                                                           [17]

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QUESTION 3

 If cos 23 = 1 − g 2 , express each of the following in terms of g:

 3.1        sin 23                                                                                            (3)

 3.2        tan113                                                                                            (2)

 3.3        sin ( −653 )                                                                                      (2)
                                                                                                               [7]

QUESTION 4

 4.1       Simplify the following without using a calculator:

           4.1.1       cos 390
                                   − tan ( 360 − x ) .cos (180 + x ) .cos ( x − 90 )                        (8)
                      cos ( −30 )

                      sin120.tan 330
           4.1.2
                          cos 240                                                                             (4)

                                                   cos x.tan 2 x
   4.2     Prove the following identity:                         + cos x = 1                                   (5)
                                                       1
                                                           +1
                                                     cos x

                                            cos 2 x.tan x
 4.3        Given the expression:                         .
                                               sin 2 x
           Determine the value(s) of x, in the interval x  0 ;180 , for which the expression
           will be undefined.                                                                                  (3)

 4.4       Solve for α and  if:

                                 2
            cos ( +  ) = −         for     ( +  )   0 ;180 ; and
                                2
                               1
            cos ( − 2 ) =      for       ( − 2 )  0 ;180
                               2
                                                                                                               (4)
                                                                                                              [24]

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GIVE REASONS FOR YOUR STATEMENTS IN QUESTION 5, QUESTION 6 AND
QUESTION 7.

QUESTION 5

 5.1       In the diagram, O is the centre of the circle. AB and AC are chords, and BO and CO
           have been drawn.

                                                            A

                                                        O

                                B                                            C

           Use the diagram on the DIAGRAM SHEET to prove the theorem which states that
           the angle subtended by an arc or chord at the centre of the circle is equal to twice the
           angle subtended by the same arc or chord at the circumference, that is, prove that
           obtuse BOCˆ = 2BACˆ .                                                                          (5)

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 5.2       In the diagram below, DF is a diameter of the circle with centre O. Chord EG
           intersects DF at H such that GH = HE. Chords EF, GF and DE and radius OE are
           drawn. EGF ˆ = 55 .

                                                              E
                                                        1 2
                                                               3

                                 D
                                                3
                                            H       2
                                                1
                                                          11 2
                                                           O

                                                                              2
                                 G                                          11
                                                                                   F

            5.2.1 Write down the size of D̂ . Provide a reason.                                             (2)

           5.2.2     Write down the size of Ô2 . Provide a reason.                                         (2)

           5.2.3     Calculate, giving reasons, the size of F̂2                                             (2)

           5.2.4     Prove, with reasons, that DE is a tangent to the circle passing through E, F and
                     H at E.                                                                                (3)
                                                                                                           [14]

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QUESTION 6

 6.1       In the diagram below, CD is a diameter and PC is a tangent to the circle at C.
           Chord DK is produced to P. PT intersects KC at Q. CDP ˆ = 40 , DPTˆ = 25 and
              ˆ
           TQC = 65 .
                                                                                 P
                                                         K
                                                                             1
                            D                            1 2

                                                         Q
                                               2
                                               T 1

                                                               1 2
                                                                 C

           Calculate, with reasons, the size of the following angles:

           6.1.1      Ĉ2                                                                             (2)

           6.1.2      K̂1                                                                             (2)

           6.1.3      P̂1                                                                             (2)

           6.1.4     Prove, with reasons, that TC = QC .                                             (2)

 6.2       In the diagram below, O is the centre of circle PTR. N is a point on chord PR such
           that ON ⊥ PR . RS and PS are tangents to the circle at R and P respectively.
                                              ˆ = 62 .
            RS = 15 units , TS = 9 units and RPS

                                                     P

                                                     N                   9
                                           O                                     S
                                                                 T

                                                                     15

                                                     R

           6.2.1                            ˆ .
                     Calculate the size of NOR                                                        (5)
           6.2.2     Calculate the length of the radius of the circle.                                (4)
                                                                                                     [17]

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QUESTION 7

 In the diagram, O is the centre of the circle and PVR is a tangent to the circle at V.
 T is a point on chord SU such that ST = TU. US extended meets PV at R. OV, OR, OS and OT
 are drawn.

                                                               R
                                             V

         P
                                                               S

                                                 O             T

                                                               U

 7.1         Prove that VOTR is a cyclic quadrilateral.                                          (3)

             Prove that OR = OV + TR − ( RU − TR ) .
                          2    2    2                     2
 7.2                                                                                             (6)

                                                                                                 [9]

                                            TOTAL: 100 MARKS

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           Mathematics P2                      10                Common Test June 2024
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            NAME & SURNAME:

           DIAGRAM SHEET 1

           QUESTION 1
                                                         y

                                                 K

                                                     β
                                                         O   F         x

                                         E

TEAR OFF
           QUESTION 2

                                                 y

                                             F

                                                 O       E

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            NAME & SURNAME:

           DIAGRAM SHEET 2

           QUESTION 5.1

                                                         A

                                                     O

                                   B                           C

TEAR OFF

           QUESTION 5.2

                                                         E
                                                   1 2
                                                         3

                                   D
                                           3
                                       H       2
                                           1
                                                    11 2
                                                     O

                                                               2
                                   G                         11
                                                                   F

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            NAME & SURNAME:

           DIAGRAM SHEET 3

           QUESTION 6.1                                                       P
                                                   K
                                                                          1
                                 D             1 2

                                                   Q
                                           2
                                        T 1

                                                       1 2
                                                           C

           QUESTION 6.2

                                               P

TEAR OFF                                       N                  9
                                       O                                      S
                                                             T

                                                                 15

                                               R
           QUESTION 7
                                                                      R
                                               V

                         P
                                                                      S

                                                       O              T

                                                                      U

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                                                                  FINAL

                          MATHEMATICS P2
                            COMMON TEST
                                 JUNE 2024
                      MARKING GUIDELINES

                            NATIONAL
                       SENIOR CERTIFICATE

                                 GRADE 11

MARKS:   100

                  These marking guidelines consist of 12 pages.

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Mathematics P2                      2         com                          Common Test June 2024
                              GRADE 11
                            Marking Guideline

   QUESTION 1
    1.1 Equation of EG:
            3x – y  6  0
            y  3x  6                                                A y  3x  6
             mEG  3
             tan   mEG  3                                         CA tan   mEG  3
            β  71,57                                                CA answer
                                                                                                    (3)
     1.2    x  30  8                                              A substitute y = 0
            x 8
             F 8 ; 0                Answer only: FULL MARKS        A answer in coordinate form
                                                                                                 (2)
     1.3    midptDF  midptGE                [diagonals bisect]
               10  8 6  0 
            K         ;      
               2          2 
            K  1 ; 3                                               CACA co-ordinates of K
                      x 1         y 9
            1           and 3 
                        2            2
             x  3          and y  3
                 E  3 ;  3                                       CA x-value CAy-value
                                        OR                                                          (4)
            x  3y  8
            x  3  3x  6   8
              x  9 x  18  8
                      10 x  10
                         x  1                                       CA xK
             y  3  1  6
              y 3                                                    CA yK
             K  1 ; 3
            x 1           y9
                   1 and      3
              2             2
                x  3 and    y  3             CA only applies if
                                                  both values are
                      E  3 ;  3                 negative         CA x-value CA y-value
                                                                                                    (4)
     1.4    3y  x  8
                  1     8
             y   x
                  3     3                                                       1
                         1
                                                                      A mDF  
            mDF  mGE    3                                                   3
                         3                                            A mDF  mGE  1
                      1
             DF  GE                                                  A conclusion                (3)

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Mathematics P2                      3         com                                       Common Test June 2024
                              GRADE 11
                            Marking Guideline
             DEFG is a rhombus (diagonals bisect at right
            angle)
                                                  OR

            DEFG is a parallelogram                   (diagonals bisect each other)
            DG           10  1   6  9 
                                   2              2

                 = 130                                                                A DG  130

            DE   10   3   6   3 
                                        2                   2

                 130                                                                 CA DE  130

             DG  DE
             DEFG is a rhombus                       (parm. with adjacent sides =)   A conclusion
                                                                                                                (3)
                                                                                                               [12]

   QUESTION 2

            AC          1  (2)    5  (1) 
                                    2                  2
     2.1                                                                              A substitution in correct
                                                                                       formula
                 9  36
                                                                Penalise if answer
                3 5                                            is not simplified     CA answer
                                                                                                                (2)
                       54                                                            A substitution in correct
     2.2    mBC 
                     1   3                                                         formula
                     1
                                                                                     CA answer
                     4                                                                                          (2)
     2.3     mCD  mBC
             y 5 1                                                                   CA mCD
                                                                                     CA equating gradients
             5 1 4
                                                                                      CA answer
                y6
                                                                                                                (3)
                                                  OR

               mBD  mBC
              y4      1                                                              CA mBD
                     
            5   3 4                                                               CA equating gradients
               y4 2
                   
                8    8
                  y6                                                                 CA answer
                                                                                                                (3)

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Mathematics P2                      4         com                         Common Test June 2024
                              GRADE 11
                            Marking Guideline
     2.4

                H

                             (0 ; 3)         I(1 ; 3)

                                                   G(2 ; 0)

            mAH  4              (AH  BC)                           CA mAH  4
            y   1  4  x   2                               CA substitution of
            y  4x  9                                               A  2 ;  1
                                            OR                        CA equation of AH
            mAH  4                                                                              (3)
            1  4(2)  c                                           CA mAH  4
                                                                      CA substitution of
             c  9
                                                                      A  2 ;  1
             y  4 x  9
                                                                      CA equation of AH
                                                                                                  (3)
     2.5    By using transformation:
            B  C :  3 ; 4   1 ; 5 
                         x ; y    x  4 ; y  1                   A rule
                                 G  2  4 ;  1  1
                                    G  2 ; 0                       CA x-value CAy-value
                                        OR                                                        (3)
            B  A :  3 ; 4    2 ;  1
                        x ; y    x  1 ; y  5                   A rule
                                  G 1  1 ; 5  5 
                                  G  2 ; 0                         CA x-value CAy-value
                                    OR                                                            (3)
                              2  1 1  5 
            Midpoint of AC =        ;       
                              2        2 
                              1    
                           =   ; 2                                 A Midpoint of AC
                              2    
            Midpoint of AC = Midpoint of BG
            x 3    1          y4
                      and        2
              2     2           2
                x2     and       y0
                        G  2 ; 0                                   CA x-value CA y-value
                                            Answer only: FULL MARKS                         (3)
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Mathematics P2                      5         com                   Common Test June 2024
                              GRADE 11
                            Marking Guideline
     2.6    F  0 ; 3                                        A coordinates of F & E or
                                                              yF and xE
            E 1 ; 0 
            Area CEOF  Area OEIF + Area FIC
                                                              CA area of OEIF
                                           1      
                                = (3 1)    2 1          CA area of FIC
                                           2      
                                 4square units               CA answer
                                             OR                                               (4)

            F  0 ; 3                                        A coordinates of F & E or
                                                              yF and xE
            E 1 ; 0 
            CEOF is a trapezium………………...[OF∥EC]
                           1
            Area CEOF =  OF +EC   OE                       A formula for area of
                           2                                  trapezium
                          1
                        =  3  5  1                        CA substitution into area of
                          2                                    trapezium
                         4square units                       CA answer
                                                                                             (4)
                                                                                            [17]

   QUESTION 3

                            1 g2 x
     3.1    cos 23             
                             1     r

                                        
                                             2
            y 2  1          1 g 2                            A using x2  y 2  r 2
                     2

            y2  11 g 2
            y   g2
            y g                                                  CA y = g
            sin 23  g                                           CA answer
                                                                                              (3)
                                                         OR

            sin 2 23  cos 2 23  1                             A square identity
            sin 23  1  cos 23
               2                     2

                                                
                                                     2
                          1       1 g2                         CA substitution

                         = 1  1  g 2 
                         = g2
             sin 23  g                                         CA answer
                                                                                              (3)

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Mathematics P2                      6         com                                Common Test June 2024
                              GRADE 11
                            Marking Guideline
     3.2    tan113  tan 180  67 
                          tan 67                                            A  tan 67
                              1 g2
                                                                             CA answer
                               g                                                                            (2)
     3.3    sin  653   sin 67                                             A sin  653   sin 67
                             1 g 2                                           CA answer
                                                                                                            (2)
                                                                                                            [7]

  QUESTION 4

             cos390
    4.1.1                tan  360  x  .cos 180  x  .cos  x  90    A cos390  cos30
            cos  30
                                                                               A cos(30)  cos30
                cos 30                                                        A tan(360  x)   tan x
                          tan x   cos x  sin x 
                cos 30                                                        A cos(180  x)   cos x
                  sin x                                                      A cos( x  90)  sin x
             1          cos x  sin x                                               sin x 
                  cos x                                                      A  tan x         
                                                                                             cos x 
             1  sin 2 x                                                      CA simplification
             cos 2 x                                                          CA answer
                                                                                                            (8)
            sin120.tan 330 sin 60.   tan 30                             A sin 60 and  tan 30
    4.1.2                   
                cos 240           cos 60                                    A  cos 60
                                3 1
                                                                              A substitution of special
                               2     3
                                                                              angles
                                 1
                                 2
                            1                                                 CA answer
                                                                                                            (4)

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### Page 19

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Mathematics P2                      7         com      Common Test June 2024
                              GRADE 11
                            Marking Guideline
                  cos x.tan 2 x
     4.2    LHS                    cos x
                       1
                             1
                     cos x
                           sin 2 x 
                  cos x                            A quotient identity
                           cos 2 x 
                                      cos x
                      1  cos x                      A denominator as a single
                        cos x                         fraction
                   sin x  cos x
                        2
                                        cos x
                   cos x  1  cos x
                    sin 2 x
                             cos x
                  1  cos x
                  1  cos 2 x                        A square identity
                               cos x
                   1  cos x

                
                  1  cos x 1  cos x   cos x   A factorisation
                          1  cos x
                 1  cos x  cos x                  A simplification
                1
                                                                                  (5)
                 RHS
                                          OR
                             2
                  cos x.tan x
            LHS                   cos x
                       1
                            1
                     cos x
                          sin 2 x                    A quotient identity
                  cos x .
                         cos 2 x  cos x
                     1  cos x                       A denominator as a single
                        cos x                         fraction
                      2
                  sin x       cos x
                                     cos x
                  cos x 1  cos x
                   sin 2 x
                             cos x
                  1  cos x
                  sin 2 x  cos x 1  cos x        A writing as a single
                
                           1  cos x                 fraction
                  sin x  cos x  cos 2 x
                      2
                                                    A simplification
                          1  cos x
                  1  cos x
                
                  1  cos x                          A square identity
                1
                 RHS
                                                                                  (5)
     4.3    x  0                                   A x  0
            x  90                                  A x  90
            x  180                                 A x  180
                                                                                  (3)

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### Page 20

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Mathematics P2                      8         com                          Common Test June 2024
                              GRADE 11
                            Marking Guideline
                                   2
     4.4   cos      
                                  2
                                              2
                   180  cos 1         
                                              2 
                   180  45
                   135..............................(1)          A      135

                              1
           cos   2   
                              2

                 2   cos 1  
                                        1
                                       2
                 2   60............................(2)           A   2   60

           (1)  (2) : 3  75
                          25                                         A   25
                        =110                                         A   110
                                                                                                    (4)
                                                                                                   [24]

                                             GEOMETRY  MEETKUNDE
                       A mark for a correct statement
                       (A statement mark is independent of a reason)
              S
                       'n Punt vir 'n korrekte bewering
                       ('n Punt vir 'n bewering is onafhanklik van die rede)
                       A mark for the correct reason
                       (A reason mark may only be awarded if the statement is correct)
              R
                       'n Punt vir 'n korrekte rede
                       ('n Punt word slegs vir die rede toegeken 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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Mathematics P2                      9         com                          Common Test June 2024
                              GRADE 11
                            Marking Guideline
QUESTION 5

    5.1
                                                          A
                                                         xy

                                                         O
                                           x        2x 2y        y
                                      B                               C
                                                    D

            Construction:
            Draw AO extended to D                                                 A construction
            Proof:
              ˆ = ABO
            BAO       ˆ x                [∠’s opp. = sides]                      A S/R
              ˆ  2x
            BOD                           [ext. ∠ of ∆]                           A S/R
            Similarly: ˆ
                       COD  2 y           [ext. ∠ of ∆]
              ˆ  2x  2 y                                                        A S
            BOC
                     2 x  y
                                                                                  A 2 (x + y)
                       ˆ
                     2A                                                                            (5)

    5.2.1         ˆ  55
             D̂  G                       [∠’s in the same segment]               A S A R
                                                                                                    (2)
    5.2.2   Ô 2  2(55)  110          [ at centre = 2 × at circumference]   A S  A R
                                                                                                    (2)
                                   OR
                                                                                        OR
            ˆ  DEO
            D    ˆ  55                  [∠’s opp = sides]
                                                                                  A S/R
             Ô2  110                  [ext ∠ of ∆]                            A S/R
                                                                                                    (2)
                                   OR
            ˆ  DEO
                 ˆ  55                                                                OR
            D                             [∠’s opp = sides]
                                                                                  A S/R
                                          [sum of s in ∆]
             Ô1  70
                                                                                   A S/R
           Ô2  180  70  110       [sum of ∠’s on a st line]                                 (2)
                       ˆ  180
    5.2.3 ˆF2 + Eˆ 3 + O                  [sum of s in ∆]                        A S/R
                         2

            Fˆ2 = Eˆ 3                    [s opp equal sides]
             F̂2  35                                                           A S/R
                           OR                                                                       (2)
             ˆ  90
            DEF                           [ in semi-circle]                      A S/R
            ˆ + DEF
            D    ˆ + Fˆ  180            [sum of s in ∆]
                       2

             F̂2  35                                                           A S/R
                                                                                                    (2)

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Mathematics P2                     10         com                        Common Test June 2024
                              GRADE 11
                            Marking Guideline
    5.2.4    ˆ = ˆ
             H   H 2  90       [line from centre to midpt of chord]    A S/R
               3

              Ê1  35         [sum of s in ∆ OR ext  of ]          A S/R
              Eˆ  Fˆ
                1    2

             ⸫ DE is a tangent   [converse tan chord theorem]            A R
                                                                                         (3)
                                                                                        [14]

QUESTION 6

     6.1.1    Ĉ 2  40                [tan-chord theorem]             A S A R
                                                                                        (2)
     6.1.2    K̂1  90                 [∠ in a semi-circle]            A S A R
                                                                                        (2)
     6.1.3    P̂1  65  40           [exterior ∠ of a ∆]             A S/R
                                                                        A answer
                  25                                                                  (2)
     6.1.4    Tˆ1  D
                    ˆ + DPT
                         ˆ              [exterior ∠ of a ∆]
                =40+25
                =65                                                    A S/R
               TC = QC                 [ sides opp. =∠’s]              A R
                                                                                        (2)
                                        OR
              Ĉ1  50                 [radius ⏊ to tangent]            A S/R
              T̂1  65                 [sum of ∠’s in a ∆]
                       ˆ  65
               T̂1  TQC
               TC  QC                 [sides opp. = ∠’s]              A R
                                                                                        (2)
     6.2.1     SP = SR                  [tans from same point]          A S/R
               ˆ = SPR
              SRP    ˆ  62            [∠’s opp. = sides]              A S/R
                ˆ  90
              ORS                       [tan ⏊ rad]                     A S/R
                  ˆ  28
               ORN                                                     A S
                ˆ  90  28
              NOR                       [sum of ∠’s of ∆]
                ˆ  62
              NOR                                                       A answer
                                                                                        (5)

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Mathematics P2                     11         com                  Common Test June 2024
                              GRADE 11
                            Marking Guideline
     6.2.2 In ∆ORS: let OT  r
            OT  OR  r                    radii
           OS = OT  9  r  9
           OS2  OR 2 + RS2                 Pythagoras
            r  9   r 2  152
                    2
                                                                   A LHS A RHS
            r  18r  81  r  225
              2            2
                                                                  A simplification
            18r  144
               r  8 units
             radius  8 units                                    A answer
                                                                                   (4)
                                                                          OR
                                            OR
              ˆ  NPR
             NRS   ˆ  62                  ∠’s opp = sides       A S/R
                               NR
             In NRS: cos 62 
                               15
                         NR  15.cos 62                         A NR
              ˆ  90  62  28
             NRO                            radius ⏊ to tangent
                                   NR
             In ONR: cos 28 
                                   OR
                   15.cos 62                                     A OR
             OR 
                     cos 28
             OR  7.98 units                                     A Answer
                                                                                  [17]

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Mathematics P2                     12         com                                     Common Test June 2024
                              GRADE 11
                            Marking Guideline
QUESTION 7

    7.1     ˆ  90
           PVO                                      [rad ⏊ tan]                     A S/R
            ˆ  90
           OTR                                      [line from centre to midpoint   A S/R
                                                    of a chord]
          ∴VOTR is a cyclic quadrilateral           [conv. ext. ∠of a cyclic
                                                    quad.]                             A R (Any)
                                                    [conv. opp. ∠’s suppl]                                (3)
                           OR                                    OR
            ˆ  90
           OVR                                      [rad ⏊ tan]
            ˆ  90
           OTR                                      [line from centre to midpoint   A S/R
              ˆ    ˆ  180                         of a chord]
           OVR+OTR                                                                 A S/R
           V OTR is a cyclic quadrilateral         [conv. opp. ∠’s of cyclic
                                                                                    A R
                                                    quad.]
                                                                                                          (3)
    7.2    In ROT:
           OR 2  TR 2 + OT 2                       [Pythagoras]                    A S
           In SOT:
           OT 2  OS2  ST 2                        [Pythagoras]                    A S
           OR  TR  OS  ST
                2      2        2   2
                                                                                    A substitution for
          but OS = OV                               [radii]                         OT2
                                                    [given]
          and ST = TU
           OR 2  TR 2  OV 2  TU 2
          also: TU = RU  TR                                                        A A substitution
                                                                                    for OS and ST
           OR 2  TR 2  OV 2   RU  TR 
                                               2

                                                                                     A TU = RU  TR
                                                                                                   (6)
                                                                                                   [9]

                                                   TOTAL: 100

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