LC Ordinary Level Solutions Sample Paper 2
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Sample Paper 2: Paper 2
Question 9 (75 marks)
A
15 m
B
12 m
E
25 m
28 m
C
21 m
D
Question 9 (a) (i)
No, a regular pentagon has five equal sides, whereas a pentagon is a 5-sided figure.
Question 9 (a) (ii)
Perimeter = 15 + 25 + 21 + 28 + 12 = 101 m
Question 9 (b) (i)
The pedometer needs to know the length of Joan’s stride.
Distance walked = Number of clicks × length of stride
Question 9 (b) (ii)
She will find the sum of the areas of triangles AED, ADC and ABC.
Question 9 (c) (i)
Name: Hypotenuse
Question 9 (c) (ii)
Area = 12 × Base × Height = 12 (15)(25) = 187.5 m 2
A
15 m
B
Question 9 (c) (iii)
AC = 152 + 252
2
∴ AC = 152 + 252 = 29.2 m
25 m
Question 9 (c) (iv)
tan( ∠BAC ) =
25
⇒ ∠BAC = tan −1 ( 53 ) = 59o
15
∠BCA = 180o − 90o − 59o = 31o
C
(© Educate.ie)
LC Ordinary Level Solutions Sample Paper 2
Question 9 (d) (i)
A
Area = × Base × Height = (12)(28) = 168 m
1
2
1
2
2
12 m
Question 9 (d) (ii)
AD = 12 + 28
2
2
E
2
∴ AC = 122 + 282 = 30.5 m
Question 9 (d) (iii)
28 m
No, for a triangle, base times height does not depend on
the choice of base. =
( A 12=
bh 12 hb)
Question 9 (e) (i)
Type of triangle: Scalene
D
Question 9 (e) (ii)
A
You will use the Cosine rule. This rule is used
when you are given 3 sides and no angle.
θ
29.2 m
30.5 m
C
21 m
D
Question 9 (e) (iii)
Question 9 (e) (iv)
a 2 = b 2 + c 2 − 2bc cos A
Area = 12 ab sin C
212 = 30.52 + 29.22 − 2(30.5)(29.2) cos=
θ
Area
2(30.5)(29.2) cos θ = 30.52 + 29.22 − 212
=
(30.5)(29.2) sin 41.1o 292.7 m 2
1
2
1781.2 cos θ = 1341.89
 1341.89 
o
θ = cos −1 
 = 41.1
 1781.2 
Question 9 (f) (i)
Area ∆ABC + Area ∆DEA + Area ∆ADC
= 187.5 + 168 + 292.7 = 648.2 m 2 ≈ 648 m 2
Question 9 (f) (ii)
% Error =
648 − 610
= 6%
610