GEOMETRY

S

A mark for a correct statement
(A statement mark is independent of a reason.)

R

A mark for a correct reason
(A reason mark may only be awarded if the statement is correct.)

S/R

Award a mark if the statement AND reason are both correct.

110

112

156

164

167

169

171

176

192

228

278

360

1.1.1

Mean  =  2283
                12
= 190,25
Mean profit = R190 250,00  or 190,25  thousand rands

✓ sum/som
✓ answer
✓ answer in thousands of rands (3)

1.1.2

Median =  169 +171 =170 thousand rands
                        2
= R170 000

✓ answer (1)

1.2

✓ whiskers
✓  quartiles (2)

1.3

IQR = Q₃ − Q₁
= 210 - 160 thousand rands
= R50 000

✓ answer (1)

1.4

Skewed to the right or positively skewed.

✓ answer (1)

1.5.1

σ = 67,04118759 thousand rands
= R67 041,19

✓ answer (1)

1.5.2

x −σ = 123,21 thousand rands
For 2 months the profit was less than one standard deviation below the mean.

✓ lower limit
✓ answer (2)

[11]

CHIRPS PER MINUTE

AIR TEMPERATURE IN °C

32

8

40

10

52

12

76

15

92

17

112

20

128

25

180

28

184

30

200

35

2.1

SCATTER PLOT

• 3 marks: All points correct
• 2 marks:6 - 9  points correct
• 1 mark: 3 - 5 points correct (3)

2.2

The points lie almost in a straight line. This suggests a very strong positive

relationship between the number of chirps per minute and the temperature

of the air.

OR

r = 0,99 so there is a very strong positive relationship between the number

of chirps per minute and the temperature of the air.

✓ justify with straight line (1)
✓ link with r = 0,99 (1)

2.3

a = 3,97
b= 0,15
y = 3,97 + 0,15x

✓a = 3,97
✓ b = 0,15
✓ equation (3)

2.4

Air temperature ≈  15,67°C    (calculator)

OR

y ≈ 3,97 + 0,15(80)
≈ 15,97°C

OR

Air temperature ≈  16°C         (graph: Accept between 15°C and 17°C)

✓✓ answer (2)
✓ substitution
✓  answer (2)
✓✓ answer (2)

[9]

3.1

m_AC = 1−(−4)  OR  − 4−1
          7 −(−3)         −3−7
=  5 = 1 
   10     2

✓ substitution
✓answer (2)

3.2.1

y = ¹/₂x + c      y - y₁ = ¹/₂(x - x₁)
1 = ¹/₂(7) + c    y - 1 = ¹/₂(x - 7)
c = -⁵/₂   OR     y - 1 = ¹/₂x - ⁷/₂
y = ¹/₂x - 21/2    y = ¹/₂x - 2¹/₂

OR

y = ¹/₂x + c      y - y1 = ¹/₂(x - x1)
-4 = ¹/₂(-3) + c    y - (-4) = ¹/₂(x - (-3))
c = -⁵/₂   OR     y - 1 = ¹/₂x - ⁷/₂
y = ¹/₂x - 21/2    y = ¹/₂x - 2¹/₂

✓ substitution M and A(7 ; 1)
✓ equation (2)
✓ substitution M and C(-3 ; -4) 
✓ equation                         (2)

3.2.2

Equation of AC: y = ¹/₂x − 2¹/₂    OR  y =  ¹/₂x  − 2¹/₂
y = ¹/₂ (3)− 2¹/₂                                      −1= ¹/₂ x− 2¹/₂
y= −1                                                      x  = 3
∴M lies on AC

OR

m_CM =  14 + 1  = 1  
              −3−3     2
∴ m_CM = ACm and C a common point
∴ M lies on AC

✓ x coordinate  ✓y coordinate
✓ substitution of x
✓ conclusion (4)

✓ x coordinate
✓ y coordinate
✓ gradient of CM
✓  reasoning & conclusion (4)

3.3

m_BD = 9−(−11) OR   (−11) −9
             −2−8             8−(−2)
= −2
m_BD  ×  m_AC  =  ¹/₂× −2
= −1
∴BD ⊥ AC

✓ correct substitution
✓ m_BD
✓ product of gradients = -1 (3)

3.4.1

tanθ =m_BD  = −2
∴θ =116,57°

✓ tanθ = m_BD
✓ answer (2)

3.4.2

tan β = m_BC
m_BC =  9−(−4)  OR   − 4−9  
         − 2−(−3)          −3−(−2)
= 13
β = 85,6°
∴CˆD = 116,57°−85,60°         [ext ∠ of ∆]
= 30,97°

OR

BD = √500  ; BC = √170 & CD = √170
CD² = BD² + BC² − 2BD.BC.cosCBD
170 = 500+170− 2√500.√170.cosCBD
cosCBD  =  √500  = 0,85749....
                  2√170
CBD = 30,96°

✓ m_BC  =13
✓ value of β
✓ answer (3)
✓  subst into cos rule 
✓ value of cosCBD
✓ answer (3)

3.4.3

✓  correct substitution into distance formula
✓  answer (2)

3.4.4

Area of ∆ABC= ¹/₂base × ⊥height
= ¹/₂ (√125)(√125)
= 62,5 square units
Area of ABCD = 2 × 62,5
= 125 square units

✓ correct substitution into distance formula
✓ BM
✓ substitution into area formula
✓  62,5
✓ 2 ×  ∆ABC (5)

[23]

4.1

M( 0 + 4; 0 + (-6))
         2         2
∴M(2 ; - 3)

✓ 2 
✓ - 3 (2)

4.2.1

x² +y²= 4² +(−6)²
= 52
∴x² +y² = 52

✓  substitution
✓ equation (2)

4.2.2

(x− 2)² +(y +3)² = (√52 )²  = 13
                                 2 
x² − 4x + 4 + y² + 6y + 9 −13= 0
x² + y² − 4x + 6y = 0

✓  substitution of M
✓  substitution of radius =   √⁵² / ₂
✓ answer (3)

4.2.3

m_OP  = -6  = − 3 
              4        2
m_RS  × m_OP        = −1    [radius ⊥  tangent ]
∴ m_RS  = ²/₃
∴ y = ²/₃x

✓ m_OP
✓ m_RS
✓ equation (3)

4.3

x² + y² = 52 and y = ²/₃x
x² +(²/₃ x)² = 52
x² + ⁴/₉x² = 52
1⁴/₉ x² = 52
x² = 36
x = 6
∴ R(6 ; 4)  and  N(-6 ; 4)
∴ NR = 12 units

✓ substitution
✓ simplification
✓  value of x
✓  length of NR (4)

4.4

Let T(x ; 0) be the other x intercept of the small circle
Then OT is the common chord
∴(x−2)² +(0+3)² =13
(x − 2)² =13−9 = 4                    x² −4x+4+9 =13
x − 2 = ±2                                 x −4x = 0
OR
x = 2± 2                                    x(x−4)= 0
x = 4 or 0                                  x = 0    or x = 4
∴length of common chord = OT = 4 units

✓ y = 0
✓ x-values
✓ answer (3) [17]

5.1.1

Given :sinM = ^15/₁₇
MN² =17² −15²   
= 64
MN = 8
∴tanM= ¹⁵/₈

OR

✓  sketch or Pyth
✓ MN = 8
✓ answer (3)

5.1.2

sin M = NP
             MP

NP = 15a
51     17a
∴NP = 45

✓ equating trig ratios
✓ answer (2)

5.2

cos(x−360°).sin(90°+x +cos²(−x) −1
= cosx . cosx +cos² x −1
= cos² x + cos²x −1
= 2 cos² x −1
= cos 2x

✓ cos x  ✓ cos x
✓  cos²x
✓  identity(4)

5.3.1

sin(2x + 40°)cos (x + 30°) − cos(2x + 40°) sin(x + 30°)
= sin[(2x + 40°) − (x + 30°)]
= sin(x +10°)

✓ reduction
✓ answer (2)

5.3.2

sin(2x + 40°) cos(x + 30°) − cos(2x + 40°) sin(x + 30°) = cos(2x − 20°)
∴cos(2x − 20°) = sin(x +10°)
cos(2x − 20°) = cos[90° − (x + 10°)]
2x − 20° = 80° − x+k.360° or 2x − 20° = 360° − (80° − x) + k.360°
3x =100°+ k.360° or 2x − 20° = 280° + x + k.360°
x = 33,33°+k.120° or x = 300° + k.360° ;k ∈ Z

OR

∴cos(2x − 20°) = sin(x +10°)
sin[90° − (2x − 20°)] = sin(x +10°)
110° − 2x = x + 10°+k.360° or110° − 2x = 180° − (x +10°) + k.360°
3x =100°−k.360° or 110° − 2x = 170° − x + k.360°
x =33,33°− k.120° or x = −60° − k.360° ;k ∈ Z

✓ equating
✓ co ratio
✓ 80° - x ✓280° + x
✓ simplification
✓ x = 33,33° + k.120°
✓ x = 300° + k.360° ;  k∈Z (7) 
✓ equating
✓ co ratio
✓ x + 10°  ✓170° - x
✓  simplification
✓ x = 33,33° − k.120°
✓ x = −60°−k.360° ; k∈Z (7)

[18]

6.1

Period = 720°

✓ answer (1)

6.2

y∈[−2 ; 2]
OR
−2≤ y ≤ 2

✓✓ answer (2)
✓✓ answer (2)

6.3

f (−120°) − g(−120°)
=−3sin(− ^120º/₂) −2 cos(−120°−60°)
= 4+3 √3  or 4,60 (4,5980...)
        2

✓  x = −120°
✓ substitution
✓ answer (3)

6.4.1

x-intercepts of g at -90° + 60° = -30° and 90° + 60° = 150°
∴x∈(−30° ; 150°)

OR

x-intercepts of g at -90° + 60° = -30° and 90° + 60° = 150°
− 30° < x < 150°

✓ value
✓ value
✓ answer (3)
✓ value
✓ value
✓ answer (3)

6.4.2

x∈[−180° ; −120°) ∪ (−30° ; 60°) ∪(150° ;180°]
OR
−180° ≤ x < −120°  or − 30° < x < 60°  or 150° < x ≤ 180°

✓ [−180°  ; −120°)
✓  (−30° ; 60°)
✓  (150° ; 180°]
✓ notation for inclusive in the first/last interval (4)
✓ −180° ≤ x < −120°
✓ −30° < x < 60°
✓ 150° < x ≤180°1 mark: each interval
✓ notation for inclusive in the first/last interval (4)

[13]

7.1

In PMQ : tanθ =   x   
                          QM
∴Q M =  x  
           tan θ

OR
   x   =    MQ
sinθ      sin P
MQ = xsin P
           sinθ
= x cosθ
      sinθ
=   x    
   tanθ

✓  trig ratio
✓ answer (2)
✓ sine rule
✓  answer (2)

7.2

In PMR : tanθ =   x   OR     PMQ ≡PMR [AAS/HHS]
                          MR
∴ MR =    x   = QM
            tanθ
QMR =180°− 2β
sin β = sinQMR
MR          12x
sin β ×tanθ = sin(180°−2β)
               x             12x
tanθ = sin 2β  ×   x      
            12x          sin β
tanθ = 2 sin β cosβ  ×  ×    
                 12x            sin β
tanθ = cosβ
              6

OR

In PMR : tanθ =   x    OR  PMQ = PMR[AAS/HHS]
                          MR
MR² = QM² + QR² - 2QM.QR cos β
MR² =(  x  )² + (12x)² - 2(  x  )(12x)(cos β)
           tanθ                       tanθ
  x²    =   x²  + 144x²  24(  x²  )(cos β)
tan²θ   tan²θ                    tanθ
24(  x²  )(cos β) = 144x(cos β)
     tanθ
cos β = 6 tanθ
tanθ =  cos β
                6

✓  MR = QM
✓ correct substitution into the sine rule in ∆QMR
✓ reduction
✓ double angle (4)
✓ correct substitution into the cosine rule in ∆QMR
✓ substitution
✓ MR = QM
✓ simplification (4)

7.3

 x  =  cos β [both equal tan θ ]
QM        6
x = 60cos40
          6
x = 7,66
The height of the lighthouse is 8 metres

✓ equating
✓ subst. QM = 60 and β = 40°
✓ answer (3)

[9]

8.1.1

G = x        [∠ centre = 2× circumference ]
H₁ = x        [alt ∠s; KH || GJ]
GĴH= x [tan chord theorem]

✓S  ✓R
✓S
✓S ✓R  (5)

8.1.2

Ĵ + ˆH₃ =180°−2x             [sum of ∠s in ∆ ]
∴Ĵ₁ = H₃ = 90° - x [∠s opp equal sides ]
∴x +H₂  = 90° OR         [tan ⊥ radius ]

OR

H₂ = 90° - x
∴H₂ =H₃

✓S
✓S  ✓R  (3)

8.2.1

N₂ = y                            [∠s in the same seg ]

✓S  ✓R  (2)

8.2.2(a)

2y + y +87° = 180° [opp ∠s of cyclic quad]
3y = 93°
y = 31°

✓S  ✓R
✓S (3)

8.2.2(b)

TPL = 62°                           [ext. ∠ of cyclic quad]

✓S ✓R  (2)

[15]

9.1

Constr:  Join KZ and LY and draw h₁ from K ⊥ XL and h₂ from L ⊥ XK
Proof :

area ∆XKL   =  ½XK×h₁
area ∆LYK       ½KY×h₁
area ∆XKL = ½XL × h₂ = XL
area ∆KLZ     ½LZ ×  h₂   LZ    
area ∆XKL = area ∆XKL       [common ]
But area ∆LYK = area ∆ KLZ [same base & height ; LK || YZ]
∴area ∆XKL  =  area ΔXKL
   area ∆LYK     area ΔKLZ
∴XK = XL
  KY    LZ

✓ constr 

✓ area ∆XKL
    area ∆LYK
✓ ½XK×h₁
     ½KY×h₁
✓ S ✓R
✓ S (5)

9.2.1

RH

^RF =            [line || one side of ∆ OR  prop theorem; FH || ST]

FS   HT

[Lyn || een sy van ∆ OF eweredigh. st; FH|| ST]

2x −10        4

=

9          x − 2

(2x −10)(x − 2) = 4×9

2

2x −14x −16 = 0

2

x − 7x −8 = 0

(x −8)(x +1) = 0

∴x =8 (x ≠ -1)

OR/OF

RF     RH

=              [line || one side of ∆ OR  prop theorem; FH || ST]

RS   RT

[Lyn || een sy van ∆ OF eweredigh. st; FH|| ST]

2x −10         4

=

2x −1       x + 2

(2x −10)(x + 2) = 4(2x −1)

2

2x −14x −16 = 0

2

x − 7x −8 = 0

(x −8)(x +1) = 0

∴x =8 (x ≠ -1)

✓S/R
✓ substitution
✓ standard form
✓ factors
✓answer with rejection (5)
✓S/R 
✓ substitution
✓ standard form
✓ factors
✓answer with rejection (5)

9.2.2

1

RF×RH sinˆ

area ∆RFH     2

=

area ∆RST      1

RS×RT sinˆ

2

1                     ˆ

×6×4×sin

= 2

¹×15×10×sin ˆ

2

24       4

=         =

150     25

✓ numerator
✓ denominator
✓ substitution
✓answer (4)

[14]

10.1.1

C₁ = 90°             [∠ in semi circle ]
D₁ = 90°           [line from centre to midpt of chord ]
∴ C₁ = D₁
∴FC || OD        [corresp ∠s =]

OR

FO = OE         [radii]
CD = DE         [given ]
∴FC || OD        [midpoint theorem ]

✓ S ✓ R
✓ S ✓ R
✓ R (5)

✓ S ✓ R
✓ S ✓ R
✓✓ R (5)

10.1.2

DÔE = F                       [corresp ∠s =; FC || OD]
BÂE = F                         [∠s in the same seg]
∴ DOE  = BAE

✓ S ✓ R
✓ S ✓ R (4)

10.1.3

In ∆ABE and ∆FCE:
E is common
BÂE = F                            [proved in 10.1.2]
∴ABE = C₁                        [sum of ∠s in ∆]
∴∆ABE ||| ∆FCE                [∠∠∠]
AB =   AE                           [||| ∆s]
FC      FE
AB × FE = AE × FC
But FE = 2 OF                   [d = 2r]
And FC =2 OD                 [midpoint theorem]
AB × 2OF = AE × 2OD
∴AB × OF = AE × OD

✓ S
✓ S
✓ R
✓ S
✓ S
✓ S/R
✓ S (7)

OR
In ∆ODE and ∆ABE

1. E  is common
2. DOE = EAB (proved in 10.1.2)
3. D₁ = ABE           (∠ sum ∆)

∆ODE ||| ∆ABE (∠∠∠)
EO   =   OD  =   ED    (||| ∆s)
EA         AB        EB
∴AB.EO = OD.EA
but  OE = FO              (radii)
∴AB×OF = OD×EA

✓ S
✓ S
✓ R
✓ S
✓ S
✓ S  ✓ R (7)

10.2

AT  =   AC   =  3      [line || one side of ∆ OR  prop theorem; FC || OD]
TO      CD       1
But CD = DE
AE = 5  ∴ AE = 5 CE
CE    2              2
BE   =   AE    [||| ∆s]
CE        FE
BE   = ⁵/₂ CE  
CE          FE
BE×FE =  ⁵/₂CE²
∴5CE² = 2BE.FE

✓ S ✓ R
✓ S
✓ S
✓substitute AE = 5 CE  (5)
                            2

[21]

3.1

No ca if x₂ −x₁
            y₂ − y₁

3.3

MD² +AM²
= [(3−8)² +(−1+11)²]+ [(3−7)² +(−1−1)²]
=125+20
= 145
AD²
= (7−8)² +(1+11)²
= 145
MD² +AM² = AD²

✓ AM² + MD²
✓ AD²
✓ MD² + AM² = AD²   (3)