LC Higher Level Solutions Sample Paper 4
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Sample Paper 4: Paper 2
Question 9 (75 marks)
Question 9 (a)
(i) A rhombus is a simple quadilateral whose sides are equal.
(ii) The angles in a quadrilateral add up to 360o. If each angle is equal, each angle will be 90o
and hence a square is the shape formed.
A
Question 9 (b)
30
30
(i)Perimeter P = 2a + 2b
O
D 100
100 B
q
(ii) Area of ∆ADC = 12 ( p )( 12 q ) = 14 pq
o
o
Area of ∆ABC = 12 ( p )( 12 q ) = 14 pq
p
Area of ABCD = 14 pq + 14 pq = 12 pq
Area of ∆ADC = 12 ab sin θ
40
40
Area of ∆ABC = 12 ab sin θ
Area of ABCD = ab sin θ
C
Question 9 (c)
(i)
AC = 302 + 402 − 2(30)(40) cos 100o
2
∴ AC = 54 cm
(ii)
sin( ∠CAD ) sin 100o
=
40
54
40 sin 100o
sin( ∠CAD ) =
54
 40 sin 100o 
o
∠CAD = sin −1 
 = 46.8
54


(iii) sin( ∠CAD ) =
∴ sin 46.8o =
1
2
BD
30
BD
60
BD = 60 sin 46.8o = 43.7 m
(iv) Area (30
=
=
)(40) sin 100o 1182 cm 2
LC Higher Level Solutions Sample Paper 4
Question 9 (d)
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A
C
Draw a line segment [AC] of length 10 cm.
Using a compass draw an arc of radius 10 cm with C as centre.
Using a set square place a rule perpendicular to AC and move the ruler into position such that it
measures a chord of distance of 10 cm. Draw a line along the ruler in this position.
A
D
Draw the kite ABCD.
B
C
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LC Higher Level Solutions Sample Paper 4
Question 9 (e) (i)
A
x
D
B
d
2
d
E
d
2
d h
d
C
Consider the right-angled triangle BEC:
d 
d = h + 
2
d2
d 2 = h2 +
4
d 2 3d 2
∴ h2 = d 2 −
=
4
4
Consider the right-angled triangle BEA:
2
2
2
3d 2
3
=
h=
d
4
2
x = d −h = d −
 2− 3 

3
3
d = d 1 −

 = d 
2
2 
 2 

 2− 3 
d

2 

tan( ∠ABD ) =
= (2 − 3 )
d
2
−1
∴ ∠ABD = tan (2 − 3 ) = 15o
LC Higher Level Solutions Sample Paper 4
Question 9 (e) (ii)
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A
75o 75o
15o
D
o
B
d
2
60
E
d
d
30o
θ
sin θ =
d
2
d
=
1
⇒ θ = sin −1 ( 12 ) = 30o
2
30o
C
∠ADC = 75o, ∠ABC = 75o, ∠BAD = 150o, ∠BCD = 60o
Question 9 (f)
tan 15o = tan(45o − 30o )
tan 45o − tan 30o
1 + tan 45o tan 30o
1
1−
3
=
1
1+
3
1 


1 −
3
3 −1
3

×
=
=
1 

3
3 +1

1 +
3

=
=
( 3 − 1) ( 3 − 1) 3 − 2 3 + 1
×
=
3 −1
( 3 + 1) ( 3 − 1)
4−2 3
2
= 2− 3
=
60o
d
2