GEOMETRY 

S

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

R

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

S/R

Award a mark if statement AND reason are both correct

Temperature (in °C)

14

24

26

18

20

28

22

17

12

19

Number of hot drinks sold

410

258

192

324

328

156

280

384

230

280

1.1

As the temperature increases the number of hot drinks sold decreases.
OR
As the temperature decreases the number of hot drinks sold increases. 

answer

(1)

1.2

a = 489,47
b = - 10,37
yˆ = 489,47 -10,37x

• value of a
• value of b
• equation

(3)

1.3

y = 489,47 -10,37x
= 489,47 -10,37(17)
= 313,18
Number of hot drinks sold = 314
Number of litres of milk    = 314
                                             8
= 39,25
= 40 boxes of 1𝑙

substitution

314  (accept 313)

answer as N₀   (3)

1.4

The outlier is the point (12; 230).

(12; 230)  (1)

[8]

x - σ = 82,7
x + σ = 94,1
2x = 176,8
x = 88,4
σ = 88,4 - 82,7   or   σ = 94,1 - 88,4
σ = 5,7                      σ = 5,7

OR

x = 82,7 + 94,1
             2
x = 88,4
σ = 88,4 - 82,7   or σ = 94,1 - 88,4
σ = 5,7                      σ = 5,7 

x = 88,4

answer  (3)

x = 88,4

answer (3)

y = x +c
-2 = 8 + c
c = -10
y = x - 10
k = 4 - 10
k = -6

or

tan θ = m_AB
1 = k - (-2)
       4 - 8
k + 2 = 1 
 -4
k = -4 - 2
k = -6

equation of AB
substitute A in equation  (2)

Substitute A & B into gradient formula
equate to 1    (2)

m_EA = -½

substitution of (4 ; -6)

equation  (3)

tan β = -½

value of β

value of θ   (3)

F (0;2)

substitution

answer   (3)

0 = ½x + 2
x = 4
∴ P(4;0)
∴PA II y-axis 
Area Δ ABF = area ΔABP + area ΔAPF
Area Δ ABF = ½(6)(4) + ½ (6)(4)
Area ΔABF = 24units²

OR

y = x + c
-2 = 8 + c
c = -10
∴ T(0;-10)
Area Δ ABF = area ΔFBT - area ΔAFT
Area Δ ABF = ½(8)(12) + ½ (12)(4)
Area ΔABF = 24units²

OR

mAF = - 6 - 2    = -2   ∴y = -2x + 2
             4 - 0
-2 = -2x + 2
x = 2                   ∴T(2;-2)
Area Δ ABF = area ΔFTB + area ΔTBA
Area Δ ABF = ½(6)(4) + ½ (6)(4)
Area ΔABF = 24units²

OR

A(4;-6)
B(8;-2)

= 24units²

P(4;0)

area ΔABP

area ΔAPF

answer  (4)

C(0;10)

area ΔABT

area ΔAFT

answer  (4)

T(2;-2)

area ΔFTB

area ΔTBA

answer  (4)

AB = √32 = 4√2

area formula

substitution into area formula

anser (4)

RA II Y-axis
CPB = 26,57º
RPB = 90º + 26,57º
RPB = 116,57º
PB II AG
∴ PAG = RPB = 116,57º   [corresp ∠s; PB II AG]

OR

OFP = 153,43º - 90º  [ext ∠ of Δ]
OFP = 63,43º
FEA = 180º - 63,43º [co-interior ∠s; FB II EA]
=116,57º
PAG = 116,57º   [corresp ∠s; FE II PA]

PA II y-axis
PCA = 45º   [vert opp ∠s = ]
PAC = 45º   [∠s of Δ]

PA II BG
BAG = θ = 71,57º   [alt ∠s; PA II BG]
PAG = 45º + 71,57º
PAG = 116,57º

CPB = 26,57º

RPB = 90º + CPB

RPB

answer of PAG  (4)

OFP = 63,43º

FEA = 180º - 63,43º

=116,57º

answer of PAG  (4)

APC = 90º    OR   AP = PC

PAC = 45º

BAG = θ = 71,57º

answer of PAG  (4)

ER = 6 units
EM = 3 units

EM = MP = 3 units [radii]

SM = 5 units
SP² = 5² - 3²  [pythagoras]
SP = 4 units
∴ P(-4;0)
∴M(-4;3)

X_E

Y_E  (2)

X_E

Y_E(2)

x-value

y-value

substitution

lenght of SK

conclusion  (5)

5.1.1

tan α = - 2 = 2
             -1

answer

(1)

5.1.2

OT =  √5
answer         (2)

5.1.3

expansion

substitution of sin α

special angle ratios

answer  (4)

expansion

substitution of sin α

special angle ratios

answer  (4)

5.2

2sin(-20°).sin160° - cos 40°
= 2(-sin 20°).sin 20° - cos 40°
= -2sin ² 20° - (1- 2sin ² 20°)
= -1

 - sin 20°

sin 20°

 1- 2sin² 20°

answer  (4)

5.3.1

3cos x.sin x + tan x.cos² (180° - x)
= 3cos x.sin x + tan x.(-cos x)²
= 3cos x.sin x + sin x .cos² x
                         cos x
= 4cos x.sin x
= 2sin 2x

reduction

identity

simplification

single ratio  (4)

5.3.2

y∈[-2 ; 2]

critical values
notation       (2)

5.4

cos 3x = 4 cos ² x - 3
 cos x
LHS = cos 3x = cos(2x + x)
           cos x        cos x
= cos 2x cos x -sin 2x sin x
                   cos x
= (2 cos ² x -1)cos x - 2sin x cos x sin x
          cos x                        cos x
= 2 cos ² x -1- 2 sin ² x
= 2 cos ² x -1- 2(1- cos ² x)
= 2 cos ² x -1- 2 + 2 cos ² x
= 4 cos ² x - 3
= RHS

compound identity

 2cos² x -1

2sin x cos x

1- cos ² x

expansion   (5)

5.5

3^{2 tan x} - 3^{tan x+1} = 54
3^{2tan x} - 3.3^{tan x} - 54 = 0
(3^{tan x} - 9)(3^{tan x} + 6) = 0
3^{tan x} = 3²     or        3^{tan x} = -6
tan x = 2     no solution
∴ x = 63,43° + k.180°; k ∈ Z

OR

∴ x = 63,43° + k.360°; k ∈ Z or x = 243,43° + k.360°; k ∈ Z

standard from

factors

both equations

tan x = 2

x = 63,43° + k.180°;

k ∈ Z  (5)

OR

x = 63,43° + k.360°;

k ∈ Z

& 243,43° + k.360°;

k ∈ Z   (5)

[27]

6.1.1

x ∈[- 30° ;90°]

endpoints

notation (2)

6.1.2

x = –180° or –60°

 –180°

 –60° (2)

6.2

f (x)= - cos(x + 90°)+ 1
= sin x + 1

 cos x + 90°

 answer (2)

[6]

7.1

sin θ = AD
            5
AD = 5 sin θ
sin 2θ = AD 
               x
AD = x sin 2θ
= x.2 sinθ cosθ
x.2 sinθ cosθ = 5sinθ
x =       5sinθ     
     2 sinθ cosθ
=      5     
   2 cos θ

trig ratio

trig ratio

2sinθ cosθ

equating AD

x as subject         (5)

7.2

use area rule correctly

substitution

answer   (3)

[8]

8.1

Construction: Draw diameter LT and draw TP
SLK = 90° - TLˆK  [radius ⊥ tangent]
TPL =  90°              [∠ in semi-circle]
∴ KPL = P  = 90° - TPK
= 90° – TLK [∠s same segment]
∴ SLK = Pˆ

Constr

S R

S /R

S

R

(6)

8.1

Construction: Draw diameter LT and draw KT 
SLK = 90° - TLK      [radius ⊥ tangent]
LKT =  90°                [∠ in half circle]
∴ Pˆ =  KTL  [∠s same segment]
= 90°- TLK
∴ SLK = P

construction
S /R
S
R
S
S / R  (6)

8.2.1(a)

S₁ = 25°

[tan chord theorem]

S R

(2)

8.2.1(b)

O₁  = 50°

[∠ at centre = 2x∠ at circumference ]

S R

(2)

8.2.1(c)

R₂  = W₃  + W₄  = 65°
P = 60°
R₁ =55°

[∠ s opp = radii ]
[∠ s of equilateral ∆ ]
[opp ∠ of cyclic quad ]

S  R
S / R
S  R

(5)

8.2.2

W₁  = S₂   = 60°
Pˆ = 60°
∴ Wˆ 1  = Pˆ =60°
SP || TW

[tan chord theorem]
[∠ s of equilateral ∆ ]
[alt ∠ s = ]

S / R
S
R

(3)

8.3.1

OG = x + 6
∴ HM = 2x + 6

S
S  (2)

8.3.2

OM ⊥ JL      [line from centre to midp of chord]
OJ² = JM² + OM²          [Pythagoras]
(x + 6)²  =12² + x ²
x ² + 12x + 36 = 144 + x ²
x = 9
r =15 units

OR

OM ⊥ JL      [line from centre to midp of chord]
HJ² = HM² + JM²         [Pythagoras]
(12√5)² =(2x + 6)² +12²
720 = 4x ² + 24x + 36 + 144
0 = 4x ² + 24x - 540
0 = x ² + 6x - 135
0 = (x - 9)(x + 15)
x = 9
r =15 units

S R
subst into Pyth
value of x
length of radius  (5)

S R
subst into Pyth
value of x
radius  (5)

[25]

9.1

TQ = VS               [Prop Th , TV || QS ]
QP    SP
VS = WR             [Prop Th , RS || VW ]
SP     RP
∴ TQ = WR
   QP     RP

S  R

S/R  (3)

9.2

Q₁  = R₁                     [∠s in the same segment  ]
R₁  = W          [ corres ∠s, RS || VW, RS || VW ]
∴ Q₁ = W
Q₁ = T           [ corres ∠s ,TV || QS , TV || QS]
∴ T  = W
∴TPVW is a cyclic quad [ converse ∠s in the same segment ]

S R

S/R

S

R  (5)

[8]

10.1.1

D₁  = x                 [tan chord theorem ]
C₂  = D₁  = x    [Tans from common pt ]
E₁  = 180° - 2x     [sum of int ∠s D]

OR

D₁  = x                 [tan chord theorem ]
C₂  = x                 [tan chord theorem ]
E₁  = 180° - 2x    [sum of int ∠s D ]

S R

S R

R

S R

SR

R

(5)

(5)

10.1.2

In Δ ECD and ΔCBD
C₂  = B  = x         [tan chord theorem ]
D₂  = B  = x        [∠s opp equal sides ]
∴ D₁  = D₂   = x
∴ Δ ECD  ||| ΔCBD       [∠, ∠, ∠ ]

OR

In Δ ECD and ΔCBD
C₂  = B  = x         [tan chord theorem ]
D₂  = B = x        [∠s opp equal sides ]
∴ D₁  = D₂   = x
E₁  = C₃       [3rd ∠ of ∆ ]
∴ Δ ECD ||| ΔCBD

S / R

S

R

S / R

S

R

(3)

(3)

10.2.1

EC = CD = ED             [ Δ ECD ||| ΔCBD]
BC    BD    CD
CD = ED
BD    CD
CD² = ED.BD
ED = CE
∴ CD² = CE.BD

S

CD² = ED.BD

ED = CE  (3)

10.2.2

C₂  = D₂  = x                 [proven ]
BD || CE                      [alt ∠s = ]
FE = FC                  [line || one side of Δ ]
DE    CB
CF² = CB²
EF²    DE²
CF² = DE.BD  [CB = CD]
EF²       DE²
CF² = BD
EF²    DE

S R

S R

squaring

subst

CD² = ED.BD     (6)

[17]