ba101 engineering mathematic chapter 4 geometry and measurement

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



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 )



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



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



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



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



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



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,



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



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



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



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



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



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



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



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



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



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



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



b) 4.12 Find the value of x a)

O

x

54

O

y x

215



b)

75 68

x

27



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



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



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



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



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?



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



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



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



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



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



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



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



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



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



= 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



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



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



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



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



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



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.



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



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



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



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



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



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



= )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



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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



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



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



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



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



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



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



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



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



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



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



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



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)



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



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



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



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



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



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



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



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



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



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

ba101 engineering mathematic chapter 4 geometry and measurement

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