GEOMETRY AND MEASUREMENT BA101/CHAPTER4
1 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
CHAPTER 4: GEOMETRY MEASUREMENT
(Angles & properties of angles in circle)
4.0 INTRODUCTION
Trigonometry was developed by Greek astronomers and has since evolved from its use by
surveyors, navigators, and engineers to present applications involving ocean tides, the rise and fall
of food supplies in certain ecologies, brain waves patterns, and many other phenomenons.
Geometri adalah kajian secara matematik mengenai kedudukan titik, garisan, dan permukaan
sesuatu rajah.
4.1 ANGLES
If you draw two lines with a common vertex, they form an angle. The angle is the rotational limit of a
line from one position to another position. Lets look at Figure 6.
Rotational axis
Figure 4.1: An angles direction of rotation
INPUT
A+
A -
direction of rotation (+)
direction of rotation (-)
4.1 And 4.2
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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Angles are measured in degrees and radians. For example, 35 degrees is written as 35 . 2
radians is written as 2 rad. A straight line is 180 or rad. A full rotation of a circle is 360 or 2
rad. Value of is 3.142 or 7
22
4.2 TYPES OF ANGLES
less than 90
Acute angle ( 0 90 )
Straight line
180 or 2
1 rotation of a circle ( = 180 )
Obtuse angle
more than 900 but less than180
0
( 90 180 )
To change measurement of degrees to radian
360 = 2 rad 180 = rad
Right angle
=900 900 or rotation of a circle
( = 90 )
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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Reflex angle
more than 180 but less than 360
( 180 360 )
complete rotation 360 or 1 complete rotation.
angle
Figure 4.2: Different Types of Angles
4.3 ADDING ANGLES
a b
a + b = 1800
A complete rotation of a circle = 360
4 right angles = 360
a +b +c =360
Figure 4.3: Adding Angles
angle 0 No rotation
a b c
2 right angles = 90 + 90 = 180
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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4.4 TOTAL ANGLE
Pada sudut tegak, PQ berserenjang dengan RS.
Jarak tegak dari titik P ke garis lurus RS ialah
panjang garis serenjang PQ.
Adjacent angles
a + b = 180
4.5 ANGLES IN STRAIGHT LINES
Parallel lines never meet and the distance between them is always the same
Alternate angles
a = c
b= d Adjacent angles
a + d = 180
b + c = 180 Opposite angles
a = b
c = d
Figure 4.5: Angles in Straight Lines
P
Q R
a b
Figure 4.4: Total Angle
distance
b
d c
a
a
c d
b
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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4.6 PROPERTIES OF ANGLES IN A CIRCLE When two straight lines are drawn from two points on the circumference of a circle, the angle
formed at the center of the circle is always double the angle formed at the circumference.
Figure 4.6: Properties of Angles in a Circle 4.7 ANGLES IN A QUADRILATERAL PLACED INSIDE A CIRCLE
2x
x
O
x
2x O
x 2x
x
O
O
2x
Xx
w
x
z
y
P
Q
R
S
x + y = 180
w + z = 180
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Figure 4.7: Angles In A Quadrilateral Placed Inside A Circle 4.8 ANGLES INSIDE A CIRCLE Based on line CF CAF = CBF CDF = CEF A If line DE goes through center O A= B =C = 90
K
N
J
M
L
external
angle JML = internal
angle JKN
A B
C
D E
F
O
A B
C
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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If length of arc JL = length of arc PR, then angle JKL = angle PQR K Q Figure 4.8: Angles Inside A Circle 4.9 TANGENT AND NORMAL OF A CIRCLE In Figure 6.9, O is the center of the circle. The line PQ only touches the circle and it is called a
Tangent. The line OS is a Normal to the Tangent PQ because it is perpendicular to the Tangent.
Figure 4.9: Tangent and Normal
D E
O
P Q S
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4.10 BISECTORS D
A C
C A
W Y if WX = YZ, then AO = OB, if AO = OB, then WX = YZ. X Z
O
B
O
F
B D
O A B
if OD is perpendicular to AB,
then AB = BC.
If AB and CD are bisectors,
then AE = EB, CF =FD,
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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A C if AB = CD, then minor arc AB = minor arc CD.
B D
Figure 4.10: Bisectors
Example 4.1:
State these angles as a fraction of a circle
a) 45 b) 120 c) 200
Solution
a) 45 =
360
45 =
8
1 of a circle b) 120 =
360
120 =
3
1 of a circle
c) 200 =
360
200=
9
5 of a circle
O
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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Example 4.2:
What is the angle, in degrees, as a result of rotating
a) 3
2 of a circle b) 2
15
2 of a circle
c) 8
7 of a circle
Solution
a) 3
2 of a circle =
3
2 360 = 240
b) 215
2 of a circle =
15
32 360 = 768
c) 8
7 of a circle =
8
7 360 = 315
Example 4.3:
Given that AB and CD are straight lines, find the angles of x and y.
25
61 Solution
x + 25 + 61 = 180
x = 180 - 86 = 94
x = 94
x + y = 180
94 + y = 180
y = 180 - 94 = 86
y = 86
B
D
x
y
C
A
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Example 4.4:
Given that PQ is a straight line, calculate the value of x.
Solution
2x + 4x + 3x = 180
9x = 180
x = 9
180 = 40
x = 40
Example 4.5:
Determine the value of x.
Solution
3x + 79 + 37 + 88 = 360
3x = 360 - 204 =156
x =3
156 = 52
4x
3x 2x
Q P
88
37
79
3x
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Example 4.6:
Determine the value of x, given that PQ and RS are parallels.
Solution
x = PRS = 63
Example 4.7:
If the line JK is parallel to LM, and AB is a straight line, determine the values of x and y.
Solution
y + 25 = 73 x + 73 = 180
y = 73 - 25 = 48 x = 180 - 73 = 107
x
63
Q
S
P
R
A
K J
x y
L M
25
B
73
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Example 4.8:
Given that AB is parallel to CD, what is the value of x?
A B
45 O x
25 C D Solution A B
45 O a M b
25 C D Note: You should draw a line OM that passes O that is parallel to AB and CD. Then;
a = 45 dan b = 25
Therefore, x = a + b
= 45 + 25
= 70
Example 4.9: What is the value of angle a? P R
O
a
120
S
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
14 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
B
Solution
The reflex angle POR = 2 PSR = 2 x 120 = 240
Therefore angle a = 360 - 240 = 120
Example 4.10: Given OA = OB = 5 cm. OJ = 13 cm. Calculate bisectors of LM.
L K M Solution AJ2 = 132 52 = 144 JK = 2 x 12 = 24 cm
AJ = 144 = 12 cm LM = JK = 24 cm.
O
A
J
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
15 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION..!
4.1 State the following angles as fractions of a circle
a) 48 c) 210
b) 160 d) 320
4.2 Convert each of the statements below into angles in degrees
a) 6
1 rotation of a circle
b) 3
2 rotation of a circle
c) 115
2 rotations of a circle
d) 38
3 rotations of a circle
4.3 Determine all the angles below
a) b)
ACTIVITY 4a
a
105
x z
35 46
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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4.4 Determine all the angles below a) b) 4.5 Determine the value of x in each situation a)
a
47
b
x
140
2x 4x
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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b) 4.6 Find the value of x a) b) 4.7 Find the value of x a)
4x
56
155
x
2x 3x
77 33
11x
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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b)
4.8 Find the value of x a) b) 4.9 Determine the value of a a)
3x
2x
35
36 36
x
133
x
y
50
a
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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b) 4.10 Calculate the value of m a) b) 4.11 Given that O is the center of the circle, determine the values of x and y a)
85
a
56
26
m
105
120
m
O
y
28 x
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b) 4.12 Find the value of x a)
O
x
54
O
y x
215
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b)
75 68
x
27
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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4.1 a) 152 b)
94 c)
12
7 d)
9
8
4.2 a) 60 b) 240 c) 408 d) 1215 4.3 a) a = 99 b) x = 75, z = 75 4.4 a) a = 43 , b = 90 b) x = 50 4.5 a) x = 60 b) x = 31 4.6 a) x = 115 b) x = 18 4.7 a) x = 4 b) x = 25 4.8 a) x = 108 b) x = 47, y = 47 4.9 a) a = 50 b) a = 95 4.10 a) m = 82 b) m = 135 4.11 a) x = 28, y = 62 b) x = 72.5, y = 55 4.12 a) x = 27 b) x = 112
FEEDBACK
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
23 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
PYTHAGORASS THEOREM 4.11 PYTHAGORASS THEOREM Let us consider a right angle triangle now. A right angle triangle always has a side with a 900 angle. Look at the triangle in Figure 2.10. The angle between side a and side b is a right angle. The side a is called the opposite side, side b the adjacent side, and side c the hypothenus.
Figure 6.11: A Right Angle Triangle
The Pythagorass Theorem state that, for any given right angle triangle,
c2 = a2 + b2
Example 4.11:
Calculate the length of side AC.
Solution: Using Pythagorass Theorem, AC2 = AB2 + BC2 = 32 + 42 = 25
AC = 25 = 5 cm.
b
a
c
A
B
C
3cm
4 cm
4.3
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4.13 Solve the right angle triangles given:
x a) 10 b) w 5 13 8 15 d) z 7 c) 8 y 25 4.14 Determine the values of a and b 24 cm
b cm a cm 8 cm 6 cm
ACTIVITY 4b
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
25 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
4.15 Determine z and y in the cases given below
11 y 4
b) z
a)
y 5
5 13
z
4.16 Dalam rajah berikut, Z ialah titik tengah XY. Cari panjang XW.
V
15 cm 9 cm
X Y
8 cm
4.17 A bus traveled 70 km towards North and then moved 240 km towards East. How far is the
bus from the starting point?
4.18 The diagonal of a rectangle is 100 cm and its breadth is 60 cm. What is its length?
4.19 Muthu and Ali are crossing a river that is 24 meter wide. Muthu chooses the direction AB,
whereas Ali chooses the direction AC. What is the difference in distance traveled by Muthu
and Ali ?
B 7 m C
6 3
A
W
Z
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4.20 A motorcyclist is moving 15 km towards East and then 8 km towards South. How far will
he from the starting point then?
4.21 The diagonals of a rhombus are 12 cm dan 16 cm respectively. Determine the lengths of all
sides.
4.22 A boy is flying his kite using a string that is 149 m long. If the kite is flying at a vertical height
of 140 m, what is the horizontal distance the kite is from him?
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
27 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
4.13 a) w = 6 unit b) x = 12 unit c) y = 17 unit d) z = 24 unit 4.14 a = 10, b = 26 4.15 a) y = 20 unit, z = 12 unit b) y = 4 unit, z = 10 unit 4.16 XZ = 6 unit, XW = 10 unit 4.17 250 km 4.18 80 cm 4.19 1 m 4.20 17 km 4.21 10 cm 4.22 51 m
FEEDBACK
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
28 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
4.1 Determine all the angles in the diagrams given below
a) b) 4.2 Determine the value of x
a)
b)
2x
3x 4x
2x 55
3x
Congratulations to you for making it so far. You are very close to
mastering this unit. Attempt all questions in this section and check your
solutions with the answers provided in SOLUTIONS TO SELF
ASSESSMENT given after this.
SELF ASSESSMENT
b
a
55
x
y
165
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4.3 Determine the value of x
a) b) 4.4 Determine the value of x:
a) b)
4.5 Determine the value of x
a) b) 4.6 Determine the values of a and b:
a)
b)
77
a
b
a
51
x
y 127
133
x
6x 4x
5x 50
20
60
9x
4x
44 54
x
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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4.7 Determine the value of k:
a)
b) 4.8 Given that O is the center of the circle, find x and y .
a) b)
110
130
k
O
25
sm y 7
28
x
110
x
6 cm
y
63
15
k
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
31 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
4.9 Determine the value of x :
a)
b) 4.10 Determine the values of x and y:
a) b )
O
x
65
83 55
x
30
x
y
15
12
20
y
x
9
8
17
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
32 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
4.11 Given that AB = BC = CD = DE = 2 cm. Calculate the length of
a) AC.
b) AE.
4.12 The diagonals of a rhombus are 10 cm dan 24 cm respectively. Determine the lengths of all
sides.
4.13 A yatch is sailing 39 km towards East. It then sails 9 km towards South and finally 27 km
towards West. How far is it from the starting point?m
E
A
D
C
B
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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1. a) a = 35 , b = 55 b) x = 75, y = 15
2. a) x = 20 b) x = 25
3. a) x = 137 b) x = 9
4. a) x = 6 b) x = 11.54
5. a) x = 82 b) x = 53, y = 53
6. a) b = 51 b) a = 103
7. a) k = 78 b) k = 130
8. a) x = 24 cm, y = 62 b) x = 6 cm, y = 20
9. a) x = 32.5 b) x = 125
10. a) x = 15 units, y = 12 units b) x = 16 units, y = 9 units
11. a) AC = 2.83 cm b) AE = 4 cm
12. 13 cm
13. 15 km
SOLUTIONS TO SELF
ASSESSMENT 1
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
34 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
MEASUREMENT
(Length of arc and sector) 3.4
4.4 INTRODUCTION
In our everyday life there are many things made of geometrical patent such as sphere, triangle,
cone and others. Examples for spheres are ball, bicycle tyres, and clock. There basic angles found
in triangles and spheres.
A O B = (symbol)
, can be measured by 1. Degrees
2. Radians
A
B o
A
B
O
Can you help Azmi in finding
the length of the arc for sector
from A to B as shown in the
diagram ?..
4.4
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4.4.1 RADIAN
4. 4.1.1 Relation between radians and degrees:
1 rotation = 360 = 2 radians
2
1 rotation = 180 = radians
a) Conversion from radians to degrees. b)Conversion from degrees to radians.
1 radian =
180 1 = radian
180
Example 4.1
Convert the measurements in degrees to radians.
a) 45 b) 150 c) 65 d) 54 20'
Solution :
a) 45 = 45 x 180
rad
= 0.7855 rad
b) 150 = 150 x 180
rad
A
B
r
o 1 rad
Conversion
formula for
degrees to
radians is
1 = 180
rad
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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= 2.618 rad
c) 65 = 65 x 180
rad
= 1.135 rad
d) 54 20' = 5460
20 x
180
rad
= 54.3 x 180
rad
= 54.3 x 180
142.3rad
= 0.9478 rad
Example 4.2
Convert the measurements in radians to degrees.
a) 3
2rad b)
3
10rad c)
7
rad d) 0.8962 rad
Value of is
3.142
5460
20 =
'
'
60
2054
= 3.054
= 54.3
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Solution:
a) 3
2rad =
3
2x
180
= 3
360
= 120
b) 3
10rad =
3
10 x
180
= 1800
3
= 600
c) 7
rad =
7
x
180
= 180
7
= 25.71
c) 0.8962 rad = 0.8962 x
180
= 51.34
Conversion
formula for
radian to degree
is
1 rad =
180
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
38 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
Example 4.3
Convert each of the following angles from degrees to radians in .
a) 135 = 135 x 180
= rad4
3
b) 270 = 270 x 180
= 2
3rad
4.4.2 LENGTH OF AN ARC OF A CIRCLE
s
Length of an arc , s = r
Where is,
s = length of arc
r = radius
= angle in radians
Example 4.4
Find the length of arc for the figure below which the radius is 7 cm and angle is 1.2 rad.
Solution:
o
r
s
1.2 rad
7 cm
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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Solution:
Angle, , s = r
= r
s
= 4
8
= 0.5 rad
.
Length of arc, s = r
s = (7) (1.2)
s = 8.4 cm
There for, the length of arc for this figure is 8.4 cm.
Example 4.5
The figure below is a wire in sector form of OPQ, with its origin O. Its radius is 0.6m, and angle
is 0.4 rad. Find the length of arc PQ?
.
Example 4.6
The figure below is a rope with radius 8 cm and the length of an arc 4 cm and find the angle for its
figure.
4 cm
8 cm
O
s
0.6 cm
P
Q
O0.4rad
Solution:
The length of an arc PQ , s = r
= (0.6) (0.4)
s = 0.24 m
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Example 4.7
Find the length of an arc, s.
Solution:
11536 = 115.6 3.142
180
= 2.02 rad
There for, the length of arc,
s = r
s = 6 x 2.02
= 12.12 cm
1
= 60 minute
115
36 = 115
+
60
36
= 115.6
11536
6 cm
s
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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Example 4.8
The length of an arc is given, s is 3 cm and angle is 1 rad. Find the radius of this sector.
Solution :
s = r
3 cm = r (1 rad)
1
3 = r
r = 3 cm.
4.4.4 AREA OF SECTOR IN THE CIRCLE
Where , r = radius of circle
= angle in radian
1 rad
r
sO
Area of minor sector
r O
Area of major sector
Area of a Sector , A = r2
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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Area of a sector ,A=2
1r 2
Example 4.9
Find the area of a sector for angle is 210 :
Solution :
Area of a sector , A = 2
1 r2
Given angle 210o = x 210o
180o
= 3.666 rad
Area of a sector = )666.3()6(2
1 2
= 65.99 cm 2
Example 4.10
Rizal cut a piece of paper in a sector form with radius 10 cm menggunting suatu sektor daripada
sekeping kertas berbentuk bulatan berjejari 10 cm untuk dijadikan corong kecil. Jika luas sektor
minor ialah 78.6 cm2, cari sudut sektor tersebut.
Solution :
Area of minor sector = 78.6 cm2
Sector radius = 10 cm
A = 2
1 r2
78.6 cm2 = 2
1 (10)2
78.6 cm2= 50
50
6.78 =
= 1.572 rad
6 cm
2100
O
Angle must be in radians. So, angle in
degrees 210o must be
converting first to
radians 3.666 rad.
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4.15 AREA OF A SEGMENT
Example 4.11
The figure shown a triangle ABC, with AB = 30cm, 35BAC and 90ABC . BD is a length
of an arc in a circle with origin at A. Find
a) Perimeter of a sector ABD
b) The area of shaded region
Solution:
a) Perimeter of a sector ABD
The length of arc BD, s = r
= 30 ( )35180
BD = 18.33 cm
Perimeter of a sector ABD = AB + AD + BD
= 30 + 30 + 18.33
= 78.33 cm
Length of an arc , s = r
Area of a sector , A = 22
1r
Area of a triangle = sin2
1 2r
Area of a segment = )sin(2
1 2 r atau
Area of a segment = Area of a sector AOB Area of A triangle AOB
r
O
B
S
A
A
35 perlu ditukarkan kepada
radian terlebih dahulu dengan
rumus ( )180
B
C
35
D
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
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c) The area of shaded region= Are of ABC - Area of a sector ABD
= )35180
()30(2
1)21)(30(
2
1 2
= 315 274.93
= 40.07 cm 2
Example 4.12
Liza has a piece of cake with shape like a quarter of a circle. A sector with radius 10 cm, which a
radius OA and OB with the length of arc AB. Given the angle of a sector is 60 . Find:
a) The length of arc AB
b)Area of a sector OAB
c) Area of a triangle AOB
d) The area of shaded region
Solution:
a) The length of arc AB
s = r
s = 10 (60
180
)
= 10(1.05 rad)
s = 10.47 cm
b) Area of a sector OAB = 22
1r
= 05.1102
1 2
= 52.36 cm 2
b) The area of shaded region
Tan 35 = 30
BC
BC = 30 (tan 35 )
= 30 (0.700)
= 21.0 cm
O A
B
60o
10 cm
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
45 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
c) Area of AOB = sin2
1 2r
= 60sin)10(2
1 2
= 9.01002
1
= 45 cm 2
d) The area of shaded region = Area of a sector AOB area of AOB
= 52.36 cm 2 - 45 cm 2
= 7.36 cm 2
Area of a triangle is, A= sin2
1 2r
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
46 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
9.1 Convert each following angles to degrees.
a) 3 radian b) 3
2radian c) 3.15 radian
9.2.1 Convert each following angles to degrees. Give the answer for 1 decimal place. (Using
= 3.142).
a) 0.75 radian b) 0.8 radian c) radian5
4
9.3 Convert each following angles to radians in .
a) 105 52 ' b) 6.3 c) 158.5
9.4 Convert each following angles to radians. Berikan jawapan kamu betul
kepada 3 angka bererti. (Using = 3.142).
a) 52 b) 30 '15 c) 25
9.5
Rajah di sebelah menunjukkan seutas dawai berbentuk
sektor bulatan OBC yang berpusat pada O. Panjang
dawai itu , jejari ialah 0.25 m. Di beri panjang lengkok
bagi BC ialah 0.1 m. Carikan :
a) Sudut dalam radian.
b) Luas sektor bagi OBC.
9.6 Sebuah bandul berayun melalui sudut 1.2 radian dari
Kedudukan C ke kedudukan D. Jika panjang lengkok
CD yang dibentukkan oleh bandul itu ialah 17 cm.
Carikan panjang jejari tali tersebut.
O
B
C
O
D C
ACTIVITY 9
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
47 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
9.7 Figure shows the bisection of ball. Find the radius of ball and given the answer with three decimal points.
9.8 From the figure, find the area of segment.
1.8 rad
28 cm
O
A B
70
A B
O
r
cm03
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
48 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
9.1 a.) 3
180
3radian = 540
b)
120180
3
2
3
2
radian
c) 3.15
567180
15.3
radian
9.2 a) 0.75 radian = 0.75
180
= 43
b) 0.8
180
8.0radian =144
c)
4671.15
720180
5
4
5
4
radian
9.3 a) 105 '52 = 105.9 .588.0180
rad
b) 6.3 035.0180
3.6
rad
c) 158.5 rad
881.0180
5.158
9.4 a) 52 = 52 rad907.0180
b) 30 rad528.018060
153015 '
c) 25180
25
= 0.436 rad
9.5 a) Sudut dalam radian , s = r 0.1 m = 0.25 m
= 0.4 rad
b) Luas sektor OBC, A = 22
1r
FEEDBACK
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
49 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
= )4.0()25.0(2
1 2
= 0.0125 cm 2
9.6 Panjang jejari tali tersebut ialah 14.2 cm
9.7 Sudut dalam radian = 4.48 rad Jejari bola = 6.25 cm
9.8 i. Sudut AOB = 70180
=1.222 rad
ii. Luas sektor AOB = 22
1r
= )222.1()15(2
1 2
= 137.475 cm 2
iii. Luas segitiga AOB = sin2
1 2r
= )70(sin)15(2
1 2
= 105.716 cm 2
iv. Luas segmen berlorek = Luas sektor AOB Luas segitiga AOB
= 137.475 cm 22 716.105 cm
= 31.8 cm 2
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
50 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
Tahniah ! Anda telah menghampiri kejayaan. Sebelum anda berpuas hati dengan pencapaian anda, sila cuba semua soalan dalam bahagian ini dan semak jawapannya pada maklum balas yang telah disediakan. Sekiranya terdapat sebarang kemusykilan, sila dapatkan khidmat nasihat pensyarah anda. Selamat mencuba dan semoga berjaya !!!.
9.1 Tukarkan setiap yang berikut kepada darjah.
a) 2 radian b) 9
21 radian c) 3.3 radian
9.2 Tukarkan setiap sudut yang berikut kepada radian dalam sebutan .
a) 540 b) 400 c) 41 '15
9.3 Tukarkan setiap sudut yang berikut kepada radian tanpa melibatkan . Berikan jawapan anda betul kepada 3 angka bererti.
a) 36 b) 133.5 c)
3
195
9.4 Sebuah taman bunga yang berbentuk bulatan dengan jejari 49 m dikelilingi oleh pagar. Hitungkan panjang dawai yang digunakan untuk memagar taman bunga tersebut.
9.5
Rajah di sebelah menunjukkan sebuah bulatan berpusat O.
Jika jejari bulatan itu ialah 9 m dan lengkok PQ ialah 4 m.
Kirakan sudut bagi y?.
O
y
P Q
SELF
ASSESSMENT 9
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
51 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
9.6
9.7
9.8
72
28m
Bunga
Ros
Bunga
Kekwa
Sebuah taman berbentuk bulatan dengan diameter
56m dibahagikan kepada 2 sektor bulatan yang di
tanam dengan pokok-pokok bunga kekwa dan ros.
Hitungkan luas kawasan yang di tanam dengan
bunga ros jika kawasan itu mencangkum sudut 72
pada pusat taman itu. (Ambil nilai = 3.142)
100
O
A B
C Dapatkan luas segmen ABC yang berjejari 10 cm untuk
rajah di sebelah.
Rajah di sebelah menunjukkan 2 sektor bulatan berpusat
O. Di beri bahawa perimeter ABCD ialah 23 cm. Di
mana panjang AD ialah 4 cm dan panjang OA diwakili
dengan h cm. Carikan:
a) nilai h b) luas rantau yang berlorek
0.75
rad
h
cm
B
C
A D
O
100
O
A B
C
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
52 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
9.1 a) 360
b) 220
c) 1893
9.2 a) 3 rad
b) 2.2 rad
c) 0.23 rad
9.3 a) 0.628 rad b) 2.33 rad
c) 1.66 rad
9.4 406 m
9.5 Y=80
9.6 Area=492.7 m2.
9.7 area of segment ABC= 38 cm2
9.8 area of shaded reagent = 30 cm2.
FEEDBACK
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
53 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
CHAPTER 4:GEOMETRY MEASUREMENT
( Perimeter and Area) Perimeter
The perimeter is the distance or length around the outside of a shape.
It is calculated by adding the lengths of all the sides together, ie in diagram (left):
Perimeter
= 2 + 6+ 4+ 6
= 18 cm
EXAMPLE 1
What is perimeter?
As perimeter is a length or distance it is measured in units of length, eg mm, cm, etc.
Area
The area is the surface space contained within the edges of a 2-D shape (see coloured area of shape left).
EXAMPLE 2
The shape, left, has been divided into squares. Each square is 1 cm2 (1 cm x 1 cm).
The area of the rectangle is calculated using the formula:
Area = l x b, where: l = length, b = breadth Here, l = 5 cm and b = 2 cm. So:
Area = 5 x 2
= 10 cm2
2cm
m
5cm
4.5
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
54 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
Area is measured as the number of squares of a particular unit, eg mm2, cm2, m2 etc.
a)
a) Parallelograms
Perimeter = 2(a+b) Area = b x h
b) Triangles
Perimeter = a+b+c Area = x b x h
d) Trapeziums
Perimeter = a+b+c+d Area = (b+d)h
d) Right angled Triangles
Perimeter = b+d+l
Area = x l x b
b
l
a
b
h
b
b
a
b
c
h
a
h
d
d
a c
b
l
d
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
55 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
EXAMPLE 3
Calculate the area and perimeter of the diagram shown.
EXAMPLE 4
10cm
12c
m
6cm
8cm
Devide the
diagram
into part P
and Q
P
p
Q
q
12c
m
4cm
4c
m
8cm
6cm
Area of rectangle P = 12cm x 6cm
= 72cm
Area of Square Q = 4cm x 4cm
= 16cm
Area of diagrams = 72cm + 16cm
= 88cm
Perimeter =
4cm+4cm+4cm+8cm+6cm+12cm+6cm
= 44cm
10cm
4cm
m
6cm
12cm
8c
m
Area of big rectangle
=12cm x 10cm
=120cm
Area of small reactangle R
= 8cm x 4cm
= 32cm
Area of diagram = 120 cm-32 cm=88
cm
=120cm - 32cm
= 88cm
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
56 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
i) Calculate the area of the triangle ABC.
ii) Find the trapezium shown
iii) Given that area of triangle ABC is 30cm, find its height, h.
6cm
B D C
A
Solution
Area of ABC = x Base x Height
= x 10 x 6
=30cm
16cm
20cm
12cm
Solution
Area of trapezium = (a+b) h
= (16 + 20)
12
= 216cm
h
Solution
Area of triangle = x Base
x Height
30 = x 10
x h
30 = 5h
h = 6cm
A
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
57 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
EXAMPLE 5
i) In the diagram shown, ABCD is a parallelogram and CDEF is a rectangle.
Find the total area of the whole diagram. ii)_The diagram shows a trapezium PQRS.
Given that the area of the trapezium is 100cm, find the value of x
10cm C B
A B
C D
E F
Solution
Total area of the whole diagram
= Area of parallelogram ABCD
+ Area of rectangle CDEF
= ( 5x3 ) + ( 5x2 )
= 25cm
P
p Q
R S
xcm
10cm
8cm
Solution
Area = ( a+b) h
100 = (x+8) 10
100 = (x+8) 5
20 = x+8
x = 12cm
5cm
3cm
2cm
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
58 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
Activity
1. Calculate the area of the trapezium PQRS shown above
2. Find the perimeter of the shape
3. What is the area of the triangle below having a base of length 5.2 and a height
of 4.2?
15cm
7cm
3cm P Q
R S
8cm
6c
m
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
59 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
Answer 1. 33cm 2. 66cm 3. 50cm 4.4.6 Volume A measure of the amount of space enclosed by a three-dimensional geometric figure. The volume equals the number of cubic units contained inside the figure Table 6
SHAPES FORMULA
Cuboid Volume = Length X Width X Height V = lwh Surface = 2lw + 2lh + 2wh
Prisms Volume = Base X Height v=bh Surface = 2b + Ph (b is the area of the base P is the perimeter of the base)
Cylinder Volume = r2 x height V = r2 h Surface = 2 radius x height S = 2 rh + 2 r2
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
60 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
Pyramid V = 1/3 bh b is the area of the base Surface Area: Add the area of the base to the sum of the areas of all of the triangular faces. The areas of the triangular faces will have different formulas for different shaped bases.
Cones Volume = 1/3 r2 x height V= 1/3 r2h Surface = r2 + rs S = r2 + rs = r2 + r
Sphere Volume = 4/3 r3
V = 4/3 r3
Surface = 4 r2
S = 4 r2
EXAMPLE 6 a) Calculate the surface area of a solid cylinder with diameter 14 cm and height 10 cm. (Use
= 22/7)
EXAMPLE 7 a) Calculate the surface area and volume of a spere of radius 2.8 cm.
10 cm 14
cm
Solution
S=2rh + 2r2
= 2 x22/7 x 7 x 10 + 2 x 22/7 x 72
= 748 cm2
Solution
Surface area of the spere = 4r2
S = 4 x 22/7 x 2.82
= 98.56 cm2
V = 4/3 r3
= 4/3 x 22/7 x 2.8
= 91.99 cm
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
61 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
r = 2.8 cm Activity
1. Find the volumes of the following right cylinders.
a) b)
2. Find the height of the following solids.
a) Volume of cylinder = 528mm
Area of cross section = 44mm
b) Volume of cylinder = 1 540cm
9mm
40mm
10.5c
m
r=5cm
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
62 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
Radius of cross section = 14cm
3. A cylindrical solid with base radius 5cm and height is melted down to form 12 identical
cones with base radii 4cm. calculate the height of each cone.
4.
4cm
In the right cylindrical container shown, the base radius is
8cm. if the height of the water level is 20cm, finf the
volume of the water in the container ( Use = 3.142)
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
63 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
MEASUREMENT (Area & Volume of Similar Shapes)
General Objective : Learn and understand the area and volume of similar shapes
Specific Objectives : At the end of the unit, students should be able to :-
State relations between corresponding sides, area, and
volumes of similar shapes.
Calculate area of similar shapes.
Calculate volume of similar shapes.
4.6
OBJECTIVE
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
64 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
4.6.1 SIMILAR SHAPES
There are 3 elements explaining about similar shapes which are:
1. Similar shapes are solids that have the same and similar shape if the ratio of its
linear measurements is equal.
For example, lets look at two different sizes of ice cream cones in figure 11.1.
These two ice cream cones can be categorize as similar because it has the same
shape although different in size.
Cone 1 Cone 2
Figure 11.1
h1
r1
r2
h2
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
65 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
Based on the figure above, those 2 ice cream cones have a ratio of :
2
1
2
1
r
r
h
h
where;
h1 = height of cone 1
h2 = height of cone 2
r1 = radius of cone 1
r2 = radius of cone 2
2. Area of similar shapes is proportional to the square of its linear measurements.
(a) If the two spheres of radius r1 and r2 and each has a surface area of
As1 and As2, so:
2
2
2
1
2
1
)(
)(
r
r
As
As
(b) If two cones of radius r1 and r2 and height h1 and h2 each has a
surface area As1 and As2, so:
2
2
2
1
2
2
2
1
2
1
)(
)(
)(
)(
h
h
r
r
As
As
3. Volume for similar shapes is directly proportional to the cube of corresponding
linear measurement.
(a) 3
2
3
1
2
1
)(
)(
r
r
V
V
(b) 3
2
3
1
2
1
)(
)(
r
r
V
V =
3
2
3
1
)(
)(
h
h
4. As the weight (W) is directly proportional with the volume (V), so:
2
1
2
1
V
V
W
W
Example 11.1
Ooo.. now I
understand
what similar
shapes means
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
66 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
Mak Limah owns two spherical balls. Mak Limah has difficulty in calculating the surface
area of the ball to be stored in the appropriate box. Calculate the surface area of the ball,
with the diameter is 152.4 mm. Also calculate the surface area of the ball which has a
diameter of 76.2 mm?
Solution :
Ball 1 Ball 2
The ball is spherical in shape
The area of a sphere formula is 4 2r
Say the surface area of ball 1 with diameter 152.4 mm = As1
area of ball 1 with diameter 76.2 mm = As2
Therefore, the area, As1 = 42r = 4 (3.142)(
2
4.152) 2
= 72.96 2310 mm
2
2
2
1
2
1
r
r
As
As
2
1
2
2
1
2
r
r
As
As
Therefore,
As2 = 121
2
2 Asr
r
= 32
2
1096.72)2.76(
)1.38(
As2 = 18.24 2310 mm
Example 11.2:
152.4 mm 76.2mm
mm
The value of the radius of each ball is
obtained by dividing the diameter of
each ball with 2.
Example : 2
2.76 mm = 38.1 mm
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
67 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
There are two types of cylinders to which the cylinder A and B each have different height
and length of radii. Calculate the height of the cylinder B if the height and radius of the
cylinder A is given.
Solution :
Cylinder A Cylinder B
height of cylinder A, h1 = 381 mm
base diameter of cylinder A, = 254 mm
base diameter of cylinder B, = 127 mm.
To find the base radii of the cylinder :
CYLINDER
A
CYLINDER
B
Base diameter
= 254 mm
Base radii = 2
254mm
= 127 mm
Base diameter
= 127 mm
Base radii = 2
127mm
= 63.5 mm
Based on the ratio of the cylinder;
mm254
mm127
mm381 h2
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
68 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
2
1
2
1
r
r
h
h
Therefore;
mm
mm
h
mm
5.63
127381
2
127 (h2) = 24193.5 mm
h2 = 127
5.24193 mm
h2 = 190.5 mm
Therefore, the height of the cylinder B can be obtained which is as high as 190.5 mm.
Example 11.3:
Aziz would like to calculate the volume of liquid contained in his cone jar for a science
projects. The volume of a cone jar with 342.9 mm high is 1.8 Calculate the volume of a
cone with a height similar with a height of 182.9 mm.
Solution:
3
2
3
1
2
1
h
h
V
V
Formula for volume
Value for height and volume of
cone is substitute in the formula
If we memorize the
formula for similar shapes,
then we can answer the
questions easily
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
69 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
3
36
)9.182(
)9.342(
V
101.8
2
V2= 6
3
3
108.1)9.342(
)9.182(
V2= 273.23310 mm
With that, volume for the similar cone is 273.2 x 103 mm
3
1. Two solid cylinders of cylinder A and B are similar but have different size and
height. Calculate the height of the cylinder A.
Cylinder A Cylinder B
h1 190.8 mm
mm352
mm176
ACTIVITY FOR
SIMILAR SHAPE
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
70 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
2. Calculate the surface area of a sphere with a diameter of 150.2 mm. What is the
surface area for a sphere of diameter 75.1 mm?
3. There is a cone with a height of 352.9 mm and the volume is 1.9 x 106 mm3.
Calculate the height of a similar cone in which a volume of 283.3 x 103 mm
3.
Volume 1.9 x 106 mm
3 Volume 283.3 x 10
3 mm
3
4. In the figure ABCD is similar to PQRS since (i) the sides corresponding to the
two diagrams are proportional, and (ii) the two diagrams is the same angle.
Calculate the area of figure ABCD?
352.9 mm h2
150.2 mm 75.1 mm
B
C D
A
P Q
R S
3 cm 6 cm
5 cm
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
71 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
1. Height of cylinder A, h1 = 381.6 mm.
2. surface area of sphere = 17.72 x 103 mm3.
3. Height of cone, h2 = 187.13 mm.
4. 7.5cm2 .
You are about to achieve success. I hope you try all the questions in this self-assessment
and check your answers on the feedback provided. If there are problems that cannot be
completed, please talk to your friends or lecturers. Good luck ....
1. 2 pieces of a spherical globe has a diameter of 18cm and 30 cm. If the ratio of the
volume of the globe is represented by the ratio x: y. Find the value of the ratio of
the fraction of the simplest form.
2. 2 similar bottles has a height of 9 cm and 14 cm. If the volume of the first bottle is
1458 cm. find the volume of the second bottle?
ANSWER
18 cm 30 cm
SELF-ASSESSMENT
GEOMETRY AND MEASUREMENT BA101/CHAPTER4
72 Prepared By : Siti Aminah Abdul Halim and Mohd Hafis Bin Yunus
3. A container, the height is 14 cm, has a volume of 1260 cm. A container similar to
a height 10 cm. Calculate the volume of the container?
4. There are two similar tables of tables A and B. When two sides of the table
corresponding to the A and B is 1.8 m and 0.4 m. Find the ratio of the upper part
of both the table?
1. 125
27
2. 5488 cm 3
3. 459.3 cm 3
4. 81 : 4
10 cm 14 cm
Volume = 1260 cm2
ANSWERS