Mathematics Grade 9 Weeks 1-5 - Term 3 - education.gov.gy

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MINISTRY OF EDUCATION

SECONDARY ENGAGEMENT PROGRAMME

MATHEMATICS

TERM 3

GRADE 9 WEEK ONE LESSON ONE

TOPIC: Number Theory – Number Bases

.

For example:

1. 100011 is a base 2 number ( notice only the digits 0 and 1 are used and the base is written at the

bottom).

2. 102210 , 2001 , 1002 , 11 , are base 3 numbers (again take notice).

3. 100011 , 3021 , 2223 , are all base 4 numbers.

4. 101 , 104011 , 14031 and so on are base 5 numbers





When representing quantities, various number bases are used. For example, base

2, base 3, base 4, base 5, …, base 8, base 9 and base 10, etc.

Each base has a specific number of digits; i.e. base 2 has two digits (0, 1), base

3 has three digits (0, 1, 2) base 4 has four digits (0, 1, 2, 3) and so on.

Normally we use the base 10 number system. Guess how many digits this

system has? Yes! Ten digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)

When writing numbers these digits are used.

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9. 2983, 9, 784, 3100, 29133, are all base 10 numbers. Notice that it is not compulsory that we write the

base at the bottom for base 10 numbers.

𝑁𝐵. 𝐹𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑏𝑎𝑠𝑒 𝑒𝑥𝑐𝑒𝑝𝑡 𝑏𝑎𝑠𝑒 10 𝑤𝑒 𝑚𝑢𝑠𝑡 𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟

Practice writing 5 numbers for each base from base 2 to base 10.

Place values

Each digit in a number has a specific value. For example 77379.701 is a base 10 number; the four 7s have different

values.

7 7 3 7 9 . 7 0 1

(7 tenths) the place value is tenths

70 (7 tens) the place value is ten (10)

7000 (7 thousands) the place value is thousands (1000)

70000 (70 thousands) the place value is tens of thousands (10 000)

Notice that the place values are written in terms of powers of tens

10 10 10 10 10 10

Decim

al point

10 10 10 H

undreds

thousands

Tens

thousands

Thousands

Hundreds

Tens

Ones

One tenth

One

hundredth

One

thousandth

100 000 10

000

1 000 100 10 1 1

10

1

100

1

1000

7 7 3 7 9 7 0 1

All bases carry the same pattern where the place value is written in terms of powers of the base.

Base 2

Activity 1

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In the number 100011.110 , all the digits have a specific place value written in terms of power of 2.

2 2 2 2 2 2 2

Decim

al point

2 2 2

Sixty

fours

Thirty

twos

Sixteen’

s

Eights

Fours

Tw

os

One

One

half

One

quarter

One

eights

64 32 16 8 4 2 1 1

2

1

4

1

8

1 0 0 0 1 1 1 1 0

Base 5

In the number 4301 , all the digits have a specific place value written in terms of power of 5.

Example:

What is the place value of the 3?

Yes 25. And the place value of 4 is 125.

Note that the place value for each base is written as base 10.

5 5 5 5 5

.

5 5

625 125 25 5 1 1

5

1

25

4 3 0 1

2 is the place value of the 1 Notice that

the power keep increasing by

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1.

State the place value of each of the 7 in the base 10 numbers below.

a) 537 b) 3756 c) 79374 d) 387.379 e) 94.7007

2. State which has more value in quantity.

𝑎)11111001 𝑏)324 𝑐)57 𝑑)633.443

3. Write down the place value of each 2 in the given numbers:

𝑎)121 𝑏)2210 𝑐)324 𝑑)23324 𝑒)23425 𝑓) 32 472

4. What is the difference between the values of the sevens in 738 735?

5. What is the difference between the values of the 6 in 263 163?

6. Find the value of the digit mentioned in each of the following numbers

(i) 324 ( find the value of 2)

(ii) 9712 ( find the value of 7)

Activity 2

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WEEK ONE

LESSON TWO

TOPIC: Number Theory – Multiplication in Base 2

Multiplication of number bases except base 10 is done similarly to that of multiplying in base 10

Base 2

Example 1: Calculate 111 × 10

Solution 111

× 10

1110

+ 000

1110


Solution 11011

× 11

110110 2 = 10 , 2 = 10 , 2 = 10 , 3 = 11 , 2 = 10

11011

1010001

Calculate :

1. 11 × 10 8. 11101 × 101

2. 10011 × 11 9. 111100 × 111

3. 11111 × 110 10. 111000 × 1011

4. 100111 × 111 11. 110011 × 101

5. 1111 × 11 12. 110011 × 101

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WEEK ONE

LESSON THREE


Base 5


Solution 321 8 = 13 , 12 + 1 gives 13 = 23

× 4

2 334


Solution 321 8 = 13 , 12 + 1 gives 13 = 23

× 42

2 3340 6 = 11

+ 1142 8 = 13 , 5 = 10 , 5=10

30032

Calculate:

1. 21 × 3 16. 4232 × 32

2. 132 × 14 17. 2243 × 43

3. 432 × 32 18. 443334 × 423

4. 3223 × 23

5. 4443 × 321

6. 3421 × 141

7. 4332 × 44

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WEEK ONE

LESSON FOUR


Base 8


Solution 361 12 = 14

× 2

742


Solution 3216 42 = 52 , 12 = 1 4 , 13 = 15 22 = 26

× 7

26542


Solution 327 28 = 34 , 11 = 13 , 13 = 15

× 46

15340 42 = 5 2 , 17 = 2 1 , 20 = 24

+ 2412 8 = 13 , 5 = 10 , 5=10

17752

Calculate:

1. 21 × 3 7. 432 × 44

2. 732 × 24 8. 42732 × 65

3. 634 × 7 9. 6743 × 43

4. 673 × 76 10. 4434 × 436

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WEEK TWO

LESSON ONE

TOPIC: MEASUREMENTS – Length of an arc of a circle

What is an Arc of a Circle? An arc of a circle is any portion of the circumference of a circle. To recall, the circumference of a circle is the perimeter or distance around a circle. Therefore, we can say that the circumference of a circle is the full arc of the circle itself. The arc of a circle is a fraction of the circumference of that circle.

The length of the arc is 𝑳 = 𝟐𝝅𝒓 × 𝜽

𝟑𝟔𝟎 𝒐𝒓 𝝅𝒅 ×

𝜽

𝟑𝟔𝟎

π = pi = 3.14

θ = the angle (in degrees) subtended by an arc at the center of the circle.

360 = the angle of one complete rotation.

Example:

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If the circumference of the following circle is 54 cm, what is the length of the arc ABC?

Solution:

Circumference = 2πr = 54cm

L = 2πr × °

L = 54cm × °

°

= 18cm

Example 1

If the radius of a circle is 5 cm and the measure of angle subtended by the arc at the centre is 110˚, what is the length of the arc?

Solution:

L = 2πr × °

L = 2 × × 5cm × °

°

= 9.6cm

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Exercises

Make a sketch of each circle and then calculate the length.

1. Radius = 1.6cm, Angle = 297°, Arc Length =

2. Radius = 7.3cm, Angle = 234°, Arc Length =

3. Radius = 9.2m, Angle = 143°, Arc Length =

Example 2:

The length of an arc is 35 m. If the radius of the circle is 14 m, find the angle subtended by the arc.

Solution

The length of an arc = 2πr(θ/360)

35 m = 2 x 3.14 x 14 x (θ/360)

35 = 87.92θ/360

Multiply both sides by 360 to remove the fraction.

12600 = 87.92θ

Divide both sides by 87.92

θ = 143.3 degrees.

Example 2

Find the radius of an arc that is 156 cm in length and subtends an angle of 150 degrees to the circle’s center.

Solution

The length of an arc = 2πr (θ/360)

156 cm = 2 x 3.14 x r x 150/360

156 = 2.6167 r

r = 59.62 cm.

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Exercise:

1. The length of an arc is 26m. Find the angle subtended by the arc if the radius of the

circle is 12m.

2. Find the radius of a circle if it subtends an angle of 129° at the centre when the arc is

129cm in length.

1. A semi-circle has diameter 20cm. Taking π = 3.14, calculate the perimeter of the

semi-circle

2.

A circle has area 81π. Give the length of a 90∘ arc of the circle

3. If a sector covers of the area of a given circle, what is the measure of that sector's

central angle?

4. A semi- circle has a diameter of 12.5cm. Taking π = 3.14, calculate the perimeter of

the semi-circle

5. If a sector covers of the area of a circle, what is the measure of that sector’s central

angle?

6. Liza orders a slice of pizza. Its central angle is 68° . The distance from the vertex of

the pizza to the edge of its crust is 18cm. Find the approximate length of the

crust(arc) of the pizza

Challeng

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WEEK TWO

LESSON TWO

TOPIC: MEASUREMENTS – Area of sector of a circle

Area of sector of a circle

𝑨 = 𝝅𝒓𝟐 × 𝜽

𝟑𝟔𝟎° =

𝝅𝒓𝟐 × 𝜽

𝟑𝟔𝟎°

Example:

If the radius of a circle is 5 cm and the measure of the arc is 110˚, what is the area of the minor sector?

Solution:

𝑨 = 𝝅𝒓𝟐 × 𝜽

𝟑𝟔𝟎° 𝑨 =

𝟐𝟐

𝟕𝟓𝒄𝒎𝟐 ×

𝟏𝟏𝟎°

𝟑𝟔𝟎° = 48 𝒄𝒎𝟐

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Find the area of the sector of the circle for each given:

1. Radius = 1.6m, Angle = 297°, Area of sector =

2. Radius = 7.3cm, Angle = 234°, Area of sector =

3. Radius = 9.2mm, Angle = 143°, Area of sector =

4. Radius = 6m, Angle = 155°, Area of sector =





9. Diameter = 4.3m, Angle = 20°, Area of sector =

Diameter= 7.8m, Angle = 229°, Area of sector =

Word Problems:

1. A slice of pizza has a radius of 9 cm. It forms an angle of 30° at the centre. Find

the area of the slice of pizza.

2. A pendulum swings from A to B through an angle of 120°. The length of the string

is 8cm. Determine:

(i) the area of the minor sector defined from each swing from A to B

(ii) the area of the major sector defined by each swing from A to B.

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WEEK TWO

LESSON FOUR

TOPIC: MEASUREMENTS – Area of triangle

𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 = 𝟏

𝟐 𝒂𝒃 𝑺𝒊𝒏. 𝑪

This formula is used when two sides and an angle between those two sides is given.

Example 1.

9cm 1050 10cm

𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ = × 𝑎𝑏 × 𝑠𝑖𝑛𝐶

= × 10𝑐𝑚 × 9𝑐𝑚 × 𝑠𝑖𝑛105°

= 43.47cm2

Find the area of each of the following triangles

1. 1270

11cm 13cm

2.

12.3cm 8.2cm

1120

7.5cm

3. 340

13cm

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WEEK TWO

LESSON FOUR

TOPIC: MEASUREMENT – Area of segment of a circle

To find the area of the segment of a circle we must first find the area of the sector, then find the area of the triangle then we subtract the area of triangle from area of sector.

Example 1.

Find the area of the minor segment of the above circle given the radius is 10cm

Solution: 𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ = × 𝑎𝑏 × 𝑠𝑖𝑛𝐶

= × 10𝑐𝑚 × 10𝑐𝑚 × 𝑠𝑖𝑛120° = 43.3cm2

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𝑨𝒓𝒆𝒂 𝒐𝒇 𝒔𝒆𝒄𝒕𝒐𝒓 = 𝝅𝒓𝟐 × 𝜽

𝟑𝟔𝟎°

= × 10𝑐𝑚 × °

°

= 104.76cm2

Area of the segment = area of sector – area of triangle

= 104.76cm2 - 43.3cm2

= 61.46cm2

Find the area of the minor segment for each circle given below:

1.

620

7cm

2.

870

9.3cm

3.

5m

450

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WEEK THREE

LESSON ONE

TOPIC: Transformation Geometry – Translation

In Mathematics, the term “transformation” refers to some kind of change in relation to geometry. It is usually a

movement of a plane figure from one position to another.

Types of transformation: translation, reflection, rotation, enlargement, shear, stretch.

For weeks 2 and 3 we will look at Translation, Reflection and Rotation.

Translation in Mathematics means a “glide” or “shift” of an object from one point to another. For example, the

sliding of a book across a table, the pushing of a desk across the floor, etc. A translation is a movement, along a

straight line, in a fix direction without any turning.

It can be described informally as a glide or a slide. When an object undergoes a translation, all points on the

object undergoes a translation, all points on the object move the same distance and the same direction.

Ie.

That is 𝑇 =

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Determining the image under translation, using the column vector, 𝑇 = ,by calculation

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Example 5

EXERCISE

1. The vertices of triangle ABC are (−3, 2), (−1, 1) and (−3, 5) respectively. Determine the vertices of triangle

AʹBʹCʹ under translation 𝑇 =5

−3.

2. The vertices of quadrilateral ABCD are (−5, 1), (−1, 0), (−3, 4)and (−4, 6) respectively. Determine the

vertices of triangle AʹBʹCʹDʹ under translation 𝑇 =−3 3

.

3. The vertices of quadrilateral PQRS are (1, 1), (4, 1), (5, 3)and (2, 3) respectively. Determine the vertices of

triangle PʹQʹRʹSʹ under translation 𝑇 =−3−5

.

4. Under the translation T = 𝑥𝑦 , the point X(3, -4) is mapped onto Xʹ (-5, 2). Determine the translation,

T = 𝑥𝑦 .

5. Rectangle WXYZ is mapped onto rectangle WʹXʹYʹZʹ with vertices Wʹ(2, 1), Xʹ(6, 1), Yʹ(6, -2), Zʹ(2, -2) under

the translation 𝑇 =−45

. Evaluate the vertices of WXYZ.

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WEEK THREE

LESSON TWO

TOPIC: Transformation Geometry – Reflection

A reflection is a transformation in which a figure is flipped, or reflected, over a line of reflection.

Reflection in the x- axis and y- axis

Reflection in the line 𝒙 = 𝟎 (𝒚 − 𝒂𝒙𝒊𝒔)

Example 1

Triangle ABC with vertices A(−3, 2), B(1, 4) and C (−1, 1) is reflected in the 𝑦 − 𝑎𝑥𝑖𝑠. Determine the

coordinates of AʹBʹCʹ, the image of ABC.

Solution

Since 𝐴 = → 𝐴 for reflection in the y- axis

𝐴 𝐵 𝐶−3 1 −1 2 4 1

→𝐴′ 𝐵′ 𝐶′3 −1 12 4 1

𝑨 = (𝟑, 𝟐) 𝑩 = (−𝟏, 𝟒) 𝑪 = (𝟏 , 𝟏)

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Reflection in the 𝒚 = 𝟎(𝒙 − 𝒂𝒙𝒊𝒔)

EXERCISE

1. Given triangle ABC with coordinates A(-4, 1), B(11, -12) and C(-7, -9), find the image of ABC after a

reflection over the 𝑦 − 𝑎𝑥𝑖𝑠 and the 𝑥 − 𝑎𝑥𝑖𝑠

2. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the quadrilateral

in the 𝑥 − 𝑎𝑥𝑖𝑠. Write down the co-ordinates of A'B'C'D'.

3. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN after a reflection

in the line 𝑥 = 0.

4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2) and reflect

this triangle in the line𝑦 = 0 to ∆A'B'C'. Find the co-ordinates of the vertices of the triangle A'B'C'.

5. (a) Given triangle UCJ with coordinates U(-12, 7), C(4, 2), and J(-3, 9), find the image of UCJ after a

reflection over the 𝑦 − 𝑎𝑥𝑖𝑠.

(b) Reflect triangle 𝑈′𝐶′𝐽′ in the 𝑥 − 𝑎𝑥𝑖𝑠.

WEEK THREE

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LESSON THREE

TOPIC: Transformation Geometry – Reflection (continued)

Reflection in the line 𝒚 = 𝒙 and 𝒚 = −𝒙

Reflection in the line 𝒚 = 𝒙

Example

Triangle ABC with vertices A(−4, 2), B(2, 3) and C (−2, 1) is reflected in the line y= 𝑥. Determine the

coordinates of AʹBʹCʹ, the image of ABC.

Solution

Given 𝐴 = → 𝐴 for reflection in the line y = x

𝐴 𝐵 𝐶−4 2 −2 2 3 1

→ 𝐴′ 𝐵′ 𝐶′ 2 3 1−4 2 −2

𝑨 = (𝟐, −𝟒) 𝑩 (𝟑, 𝟐) 𝑪′(𝟏, −𝟐)

Reflection in the line 𝒚 = −𝒙

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EXERCISE

1. Given triangle ABC with coordinates A(-4, 1), B(11, -12) and C(-7, -9), find the image of ABC after a

reflection in the line 𝑦 = −𝑥.

2. Draw a quadrilateral PQRS where P (1, 1), Q (1, 4), R (3, 3) and S (3, 2). Now reflect the quadrilateral in

the line 𝑦 = 𝑥. Write down the co-ordinates of A'B'C'D'.

3. Given triangle MNO with coordinates M(4, 5), N(-1, -7), and O(-7, 8), find the image of JBN after a

reflection in the line 𝑦 = −𝑥.

4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1),

C (-1, 2) and reflect this triangle in the line 𝑦 = 𝑥 to ∆A'B'C'. Find the co-ordinates of the vertices of the

triangle A'B'C'.

7. Given triangle UCJ with coordinates U(-12, 7), C(4, 2), and J(-3, 9), find the image of UCJ after a

reflection over the in the line 𝑦 = −𝑥.

WEEK THREE

25

LESSON FOUR


Reflection in the origin

EXERCISE 1. State the co-ordinates of the following points under reflection in origin:

(i) (−2, −4) (ii) (−2, 7) (iii) (0, 0)

2. The point P (𝑎, 𝑏) is mapped onto P’ (−17, 10) on reflection in the origin. Find the values of 𝑎 𝑎𝑛𝑑 𝑏.

3. Write down the coordinates of the image of the point (3, −11) when reflected in the origin.

4. The point P (1, 8) is reflected in the origin to get the image P’. Write down the coordinates of P’.

5. Plot the points A (1, 0), B(3,2), C(5, 1) and D(4, -1). Reflect these points in the origin.

WEEK FOUR

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LESSON ONE


Reflection in a line

Reflection in the line 𝒙 = 𝒂

When reflection is done in the line𝑥 = 𝑎, the axis of symmetry (mirror line) is a vertical line passing through

the 𝑥 – axis at 𝑎. In the example below the red vertical line passes through the 𝑥 − 𝑎𝑥𝑖𝑠 at -2.

EXERCISE

1. Given triangle ONA with coordinates O(-4, 1), N(11, -12) and A(-7, -9), find the image of ONA after a

reflection in the line 𝑥 = −1.

2. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the quadrilateral

in the line 𝑥 = 2. Write down the co-ordinates of A'B'C'D'.

3. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN after a

reflection in the line 𝑥 = −2.

4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2) and

reflect this triangle in the line 𝑥 = 1.5 to ∆A'B'C'. Find the co-ordinates of the vertices of the triangle

A'B'C'.


reflection over the in the line 𝑥 = −2.5.

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WEEK FOUR

LESSON TWO


Reflection in the line 𝒚 = 𝒃 The lines 𝑦 = 𝑏 is parallel to the 𝑥 − 𝑎𝑥𝑖𝑠. (runs horizontal and passes through the 𝑦 − 𝑎𝑥𝑖𝑠)

EXERCISE 1. Given triangle ONA with coordinates O(-4, 1), N(11, -12) and A(-7, -9), find the image of ONA after a

reflection in the line 𝑦 = −1. 2. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the quadrilateral

in the line 𝑦 = 2. Write down the co-ordinates of A'B'C'D'. 3. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN after a

reflection in the line 𝑦 = −2. 4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2) and

reflect this triangle in the line 𝑦 = 1.5 to ∆A'B'C'. Find the co-ordinates of the vertices of the triangle A'B'C'.


reflection over the in the line 𝑦 = −2.5. WEEK FOUR

28

LESSON THREE

TOPIC: Transformation Geometry – Rotation

A rotation is a transformation in which the object is rotated about a fixed point called the centre of rotation.

The direction of rotation can be clockwise or anticlockwise.

The amount of rotation made is called the angle of rotation.

For any rotation, we need to specify the centre, the angle and the direction of rotation.

Anti-clockwise rotation of 90⁰ about the origin

1. A parallelogram ABCD is defined by the points 𝐴(3, 3), 𝐵(7, 3), 𝐶(5, 1) 𝑎𝑛𝑑 𝐷(1, 1). Calculate the

points defining the image when the parallelogram ABCD undergoes an anti-clockwise rotation about the

origin of 90°.

2. ABC is a triangle with vertices (3, 2), (5, 4) and (4, 7) respectively. Find the image of triangle ABC

under an anti-clockwise rotation of 900 about the origin.

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WEEK FOUR

LESSON FOUR

TOPIC: Transformation Geometry – Rotation


Example

A parallelogram is defined by the points 𝐴(2, 1), 𝐵(3, 3), 𝐶(6, 3) 𝑎𝑛𝑑 𝐷(5, 1). Calculate the points defining the

image when the parallelogram ABCD undergoes an anti-clockwise rotation about the origin of 180°.

Solution

Since the rotation is 1800 about the origin in an anti-clockwise direction, we change the sign of the coordinates

of the object therefore:𝑨 (−𝟐, −𝟏) 𝑩 (−𝟑, −𝟑) 𝑪 (−𝟔, −𝟑) 𝑫′(−𝟓, −𝟏)

30


Exercise

1. A kite KLMN is defined by the points 𝐾(−3, 2), 𝐿(−5, 0), 𝑀(−3, −4) 𝑎𝑛𝑑 𝑁(1, 0). Calculate the points

defining the image when the trapezium PQRS undergoes an anti-clockwise rotation about the origin of

180°.

2. KLM is a triangle with coordinates (-3, -5), (-4, -3) and (-5, -6) respectively. Determine the image of

triangle KLM under an anti-clockwise rotation of 1800 about the origin.

3. A trapezium PQRS is defined by the points 𝑃(−5, 3), 𝑄(−3, 5), 𝑅(−1, 5) 𝑎𝑛𝑑 𝑆(−1, 3). Calculate the

points defining the image when the trapezium PQRS undergoes an anti-clockwise rotation about the

origin of 270°.

4. Determine the image of quadrilateral WXYZ with coordinates W (2, 1), X (4, 1),

Y (3.5, 4) and Z (5, 3) under an anti-clockwise rotation of 2700 about the origin.

WEEK FIVE

31

LESSON ONE

TOPIC: TRIGONOMETRY – Bearing

Reading of bearings from diagrams and sketching of diagrams

A bearing is an angle, measured clockwise from the northern direction.

Note: Always express your answers as three-figure bearings (so 30° would be 030°).

Examples

In this diagram the bearing of In this diagram the bearing of B from A is 050 P from Q is 2550 Example 1.

Given that the bearing of Y from X is 1000, make a sketch with this information.

To sketch this diagram:

We first identify our starting point. (Y from X) so our starting point is X.

We then draw our North line and from this north line we measured 1000 in a clockwise direction.

Exercise

32

1. State the bearing of A from B

N

A

47°

B

2. Draw a rough sketch to illustrate each of the following bearings. Mark the angle in your sketch.

(a) From a point P, the bearing of a point Q is 30°.

(b) From a point A, the bearing of a point B is 140°.

(c) The bearing of a point K from a point L is 250°.

(d) The bearing of a place M from a place N is 330°.

3. The bearing of C from A is 190°, and C is 5km from A. The bearing of D from A is 270°, and D is 7km from

A.

Draw a rough sketch to illustrate the information above.

4. George needs to take the route as follows. 6km on a bearing of 080° from A to B, 5km on a bearing of 160°

from B to C. Draw a rough sketch to illustrate George’s route.

5. Starting at point A, Freya makes a journey as follows. 8km on a bearing of 135° to B, 4km on a bearing

of 180° to C, 6km on a bearing of 315° to D. Draw a suitable diagram to represent her journey.

WEEK FIVE

33

LESSON TWO


Calculating bearings

Example 1

The bearing of P from Q is 1200. Find the bearing of Q from P.

Firstly, make a sketch of the diagram

Then look for alternate angles which will help to determine the bearing

From the second drawing, we see that the bearing of Q from P is 2700 + 0300 = 3000

Example 2

The bearing of a boat B from a harbour H is 2500. Calculate the bearing of the harbour H from the boat B

34

Exercise

1. The bearing of a point P from a point Q is 70°. Calculate the bearing of Q from P. 2. The bearing of a point P from a point Q is 70°. Calculate the bearing of Q from P. 3. The bearing of a point A from a point B is 145°. Calculate the bearing of B from A. 4. The bearing of a ship S from a yacht Y is 220°. Calculate the bearing of the yacht from the ship. 5. The bearing of an airport A from a plane P is 310°. Calculate the bearing of P from A. 6. The bearing of a ship S from a harbor H is 339°. Calculate the bearing of H from S. 7. An aircraft flies 245 km at a bearing of 070°. How far north and east of its starting point is it?

North ………………………… East …………………………………

8. A boat sails 32 km at a bearing of 284°. It then turns to a bearing of 200° and sails another 46 k. To the

nearest 100 m, how far west of its starting place is it?

35

WEEK FIVE

LESSON THREE


Application of bearings

Example

36

Exercise

1. The bearing of an airport A from a plane P is 310°. Calculate the bearing of P from A.

2. The bearing of a ship S from a harbor H is 339°. Calculate the bearing of H from S.

3. A boat leaves a dock at point A and travels for a distance of 15km to point B on a bearing of 1350.

The boat then changes course and travels for a distance of 8km to a point C on a bearing of 0600.

Illustrate the above information in a clearly labeled diagram. The diagram should show the:

a) north direction

b) bearings 1350 and 0600

c) distance 8km and 15km

4. Sketch a diagram to represent the information given below.

Show carefully all measurements and any north-south lines that may be required.

A, B, and C are three buoys. B is 125m due east of A and the bearing of C from B is 1900 and

CB = 75km.

5. A ship leaves Port R, sails to Port S and then to Port T. The bearing of S from R is 1120 while the

bearing of T from S is 0330. The distance RT is 75km and the distance RS is 56km. draw a diagram

showing the journey of the ship from R to S to T.

6. From a point P, the bearing of a tree, T, is 600. From a second point Q, which is 200m due east of P, the

bearing of the tree is 3000. Use a scale of 1cm to represent 20m to make a scale diagram and determine

the distance of the tree from P.

37

WEEK FIVE

LESSON FOUR

TOPIC: TRIGONOMETRY – Trig. Ratios

Application to Trigonometric Ratios

From previous worksheet, we learnt:

Acronym for remembering ratios

𝑺𝑶𝑯 𝑪𝑯𝑨 𝑻𝑶𝑨

𝑡𝑎𝑛 𝜃 =

𝑐𝑜𝑠 𝜃 =

𝑠𝑖𝑛 𝜃 = 𝑜𝑝𝑝

ℎ𝑦𝑝

OR Sine: The value of the sine of an acute angle in a right angle triangle is always

S – Some

O – old Sine

H – horses

C – Can

A – always Cosine

H – hear

T – Their

O – owner’s Tangent

A – approach

Sine: The value of the sine of an acute angle in a right angle triangle is always less than 1, since the opposite side length is always less than the hypotenuse.

The value of the sine ratio increases as the angle goes from 0˚ to 90°

Cosine: The value of the cosine of an acute angle in a right angle is always less than , since the adjacent side is always less than the hypotenuse

The value of the cosine ratio decreases as the angle goes from 0° to 90°

38

1.

2.

3.

39

4.

1

MINISTRY OF EDUCATION

SECONDARY ENGAGEMENT PROGRAMME

MATHEMATICS

TERM 3

GRADE 9 WEEK SIX

LESSON ONE

TOPIC: ALGEBRA – Simplifying Algebraic Fractions

Addition and Subtraction of algebraic fraction (when denominators are whole numbers)

Procedure:

1. Determine the LCM

2. Apply procedure as when adding numeric fractions

3. Reduce result if necessary

Example: Simplify each of the following algebraic expression.

a) + c) +

=( ) ( )

=( ) ( )

= =15𝑥 + 4𝑦

6

b) − d) + −

=( ) ( )

=( ) ( ) – ( )

= = –

=

Exercises:

Simplify each algebraic fractions below:

2

1. + 6. −

2. + 7. + − −

3. − 8. + −

4. − 9. + −

5. ( )

– ( )

10. −

3

WEEK SIX

LESSON TWO


Addition and Subtraction of algebraic fraction ( when denominators are algebraic expressions)

Example: Simplify each of the following algebraic expression.

a) + c) +

=( ) ( )

=( ) ( )

= =15𝑥2 + 4𝑦2

6𝑥𝑦

b) − d) + −

=( ) ( )

=( ) ( ) – ( )

= = –

Exercises:

Simplify each algebraic fraction below:

1. + 7. −

2. + 8. + − −

3. − 9. + −

4. − 10. + −

5. +

6. −

4

WEEK SIX

LESSON THREE


Multiplication of algebraic factions

Procedure:

1. Multiply numerators together then denominators together

2. Cancel factors that are common

Example:

Simplify each of the following algebraic expressions.

a) ×

= × × × × × ×

× × × × ×

=

b) × ×

= × × × × × × × ×

× × × × × ×

=

Exercises: Simplify each algebraic fraction below:

1. × 6. ×

2. × 7. × × ×

3. × 8. × ×

4. ×

5. ×

5

WEEK SIX

LESSON FOUR


Division of algebraic factions

Procedure:

1. Change division sign to multiplication sign and upturn the proceeding fraction

2. Multiply numerators together then denominators together

3. Cancel factors that are common

Example:

Simplify each of the following algebraic expressions.

c) ÷

= ×

= × × × × × ×

× × × × ×

=

d) ÷ ÷

= × ×

= × × × × × × × × ×

× × × × × × ×

=

Exercises:

Simplify each algebraic fraction below:

1. ÷ 3. ÷ 5. ÷

2. ÷ 4. ÷ ÷ ÷ 6. ÷

6

WEEK SEVEN

LESSON ONE

TOPIC: ALGEBRA – Solution of linear equations

Equations containing fractions

Procedure: When the equation contains fractions, we first find the LCM of the denominators, then multiply

each term by the LCM, simplify and solve.

Example:

1. − = 3. + = 1

× 35 − × 35 = × 35 × 10 + × 10 = 1 × 10

7𝑥 − 15 = 7 2𝑥 + 15 = 10

7𝑥 = 7 +15 2𝑥 = 10 − 15

7𝑥 = 22 2𝑥 = −5

𝒙 =𝟐𝟐

𝟕 𝑥 =

2. − =

× 9 − × 9 = × 9 1

4𝑥 − 3 = 4

4𝑥 = 4 + 3

4𝑥 = 7

𝒙 =𝟕

𝟒 Solve each equation below :

1. + = 3. − = 5. + = −

2. − = 4. + = 4 6. + −

7

WEEK SEVEN

LESSON TWO

TOPIC: ALGEBRA – Solution of linear equations

Equations containing indices.

Note: 𝒂𝒎 = 𝒂𝒏

Then 𝒎 = 𝒏

Make the bases the same first, and then remove the base leaving the indices to solve for the unknown.

Example:

1. 64 = 16 2. 625 = 5

(4 ) = 4 (5 ) = 5

4 = 4 5 ( ) = 5

3𝑥 = 2 4𝑥 − 8 = 1 + 𝑥

𝑥 = 4𝑥 − 𝑥 = 1 + 8

3𝑥 = 9

𝑥 =

𝑥 = 3

Solve each of the following equations.

1. 2 = 128 2. 3 = 243

3. 5 = 625

4. 7 = 2401

5. 2 = 1024

6. If 64 = 4 , evaluate p.

7. If 27 = 3 , calculate the value of q.

8. Given that 243 = 3 , evaluate the magnitude of p.

9. Solve 729 = 3 ( ) for s.

8

WEEK SEVEN

LESSON THREE

TOPIC: ALGEBRA – Solution of quadratic equations

By method of factorization

General form of a quadratic equation is 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎

Factorize as per normal then solve for values of x

Example 1. Solve 𝑥 + 5𝑥 + 6 = 0

𝑥 + 5𝑥 + 6 = 0

𝑥 + 3𝑥 + 2𝑥 + 6 = 0

(𝑥 + 3𝑥) + (2𝑥 + 6) = 0

𝑥(𝑥 + 𝑥) + 2(𝑥 + 3) = 0

(𝑥 + 3)(𝑥 + 2) = 0 this means one of the pair of brackets is equal to zero.

So 𝑥 + 3 = 0 𝑜𝑟 𝑥 + 2 = 0

𝑥 = −3 𝑜𝑟 𝑥 = −2

Solution: 𝒙 = −𝟑 𝒐𝒓 𝒙 = −𝟐

Exercises.

Solve the following equations

1. 𝑦 + 6𝑦 + 5 = 0

2. 𝑎 + 14𝑎 + 13 = 0

3. 7𝑥 + 15𝑥 + 2 = 0

4. 12 + 8𝑦 + 𝑦 = 0

5. 4𝑦 + 10𝑦 − 6 = 0

6. 8𝑥 + 13𝑥 − 6 = 0

7. 𝑥 + 11𝑥 − 60 = 0

8. 𝑥 + 𝑥 − 6 = 0

9

WEEK SEVEN

LESSON FOUR

TOPIC: ALGEBRA – Solution of quadratic equations

By method of quadratic formula.

𝒃± 𝒃𝟐 𝟒𝒂𝒄

𝟐𝒂

Where a is the coefficient of x2, b is the coefficient of x and c is the constant.

Using the same example from lesson 4.

Solve 𝑥 + 5𝑥 + 6 = 0 , where a = 1, b = 5, c = 6

𝒙 =𝒃± 𝒃𝟐 𝟒𝒂𝒄

𝟐𝒂

=𝟓± 𝟓𝟐 𝟒(𝟏)(𝟔)

𝟐(𝟏)

=𝟓±√𝟐𝟓 𝟐𝟒

𝟐

=𝟓±√𝟏

𝟐

=𝟓±𝟏

𝟐

=𝟓 𝟏

𝟐 𝒐𝒓

𝟓 𝟏

𝟐

=𝟒

𝟐 𝒐𝒓

𝟔

𝟐

= −𝟐 𝒐𝒓 − 𝟑

Solution: 𝒙 = −𝟐 𝒐𝒓 𝒙 = −𝟑

Solve each of the following equation using the quadratic formula.

1. 𝑦 + 6𝑦 + 5 = 0 4. 12 + 8𝑦 + 𝑦 = 0 7. 𝑥 + 11𝑥 − 60 = 0

2. 𝑎 + 14𝑎 + 13 = 0 5. 4𝑦 + 10𝑦 − 6 = 0 8. 𝑥 + 𝑥 − 6 = 0

3. 7𝑥 + 15𝑥 + 2 = 0 6. 8𝑥 + 13𝑥 − 6 = 0

10

WEEK EIGHT

LESSON ONE

TOPIC: MATICES – Revision

Addition, subtraction and scalar multiplication of matrices.

Remember:

1. We can only add or subtract matrices if they have the same order.

2. Each term in the matrix must be multiply by the scalar quantity.

Exercises

1. Given A = −12 1215 1110 9

and B = 2 13

−14 75 11

, find A + B.

2. Given X =

13 14

4−13

13 7

−10 14

23197

15

and Y =

1 13 4−1

9 8 2

−15

91313

, compute X + Y.

3. When A = 3 2 54 7 6

11 11 8 and B =

−2 −3 −24 5 33 2 1

then evaluate A – B

4. If A = 2 −1 4−2 3 5

and B = 0 −1 23 0 1

, compute 𝐵 − 𝐴

5. Given that A = 2 −3−11 4

, B = 0 128 −6

and C = 𝑎 9−3 𝑏

, find:

(a) 2A

(b) 6B

(c) A + 2B

(d) 2A – 3B

(e) B + A

11

WEEK EIGHT

LESSON TWO

TOPIC: MATICES –Matrix multiplication

Multiplication of Matrices

For multiplication of matrices the number of columns in the first matrix must be equal to the number of rows in the second matrix.

Ie. 𝑨 × 𝑩 = 𝑪

𝒎 × 𝒏 𝒏 × 𝒑 = 𝒎 × 𝒑

The table below demonstrates this principle

Order of A Order of B Order of C

(result of A x B)

𝟑 × 𝟐 𝟐 × 𝟑 𝟑 × 𝟑

𝟐 × 𝟑 𝟑 × 𝟐 𝟐 × 𝟐

𝟒 × 𝟐 𝟐 × 𝟑 𝟒 × 𝟑

2 x 2 2 x 2 2 x 2

Note: if two matrices cannot be multiplied, they are said to be non-conformable to multiplication.

When we are finding the product of two matrices, we first check if they are compatible for multiplication. The

number of columns in the first matrix must be the same as the number of rows in the second matrix.

2 3 10 4 5

X 9 87 62 4

2 X 3 3 X 2

No. of columns in matrix 1 = No. of rows in matrix 2, hence we can proceed to multiply these matrices

The order of the product will consist of the no. of rows of the first matrix and the no. of columns of the second

matrix.

2 3 10 4 5

X 9 87 62 4

2 X 3 3 X 2 Order of multiplication (the order of the product)

12

= 2 X 2

When we multiply two matrices, we use the row of the first with the column of the second.

Example 1.

Given A= 2 3 10 4 5

and B= 9 87 62 4

= 2 3 10 4 5

× 9 87 62 4

We know that our product should be a 2 x 2 matrices. Hence,

𝑅1𝐶1 𝑅1𝐶2𝑅2𝐶1 𝑅2𝐶2

= 2 3 10 4 5

X 9 87 62 4

= (2 × 9) + (3 × 7) + (1 × 2) 𝑅1𝐶2

𝑅2𝐶1 𝑅2𝐶2

(2 × 9) + (3 × 7) + (1 × 2) (2 × 8) + (3 × 6) + (1 × 4)

(0 × 9) + (4 × 7) + (5 × 2) (0 × 8) + (4 × 6) + (5 × 4)

= 18 + 21 + 2 16 + 18 + 40 + 28 + 10 0 + 24 + 20

= 41 3838 44

EXERCISES:

Given the following matrices

P =[2 −6], Q = 54

, R = 3 55 4

, S = 9

1110

and T = 4 2 −10 −3 −1

.

Find the product of the following:

1. PQ 4. ST

2. QP 5. SP

3. TS 6. QS

13

WEEK EIGHT

LESSON THREE

TOPIC: MATICES (continued) - Determinant of a 2 x 2 matrix

Determinant of a 2 x 2 matrix

Given 𝐴 = 𝑎 𝑏𝑐 𝑑

, then the determinant of the matrix A is,

|𝑨| = 𝒂 𝒃𝒄 𝒅

= 𝒂𝒅 − 𝒃𝒄

Example:

1. Given 𝐵 = 3 21 4

, 2. Given 𝐶 = −2 15 −3

,

evaluate the determinant of B evaluate the determinant of B

|𝑩| = 𝟑 𝟐𝟏 𝟒

= 𝟑 × 𝟒 − 𝟐 × 𝟏 |𝑩| = −𝟐 𝟏𝟓 −𝟑

= (−𝟐 × −𝟑) − (𝟏 × 𝟓)

= 𝟏𝟐 − 𝟐 = 𝟔 − 𝟓

= 10 = 1

Exercise 1: Evaluate the determinant of each of the following matrix.

1. 𝐶 = 2 15 3

2. 𝐷 = −1 21 −6

3. 𝐸 = 2 2

−3 −7 4. 𝐹 =

−2 15 −3

Adjoint of a 2 x 2 matrix

Given 𝐴 = 𝑎 𝑏𝑐 𝑑

, then the determinant of the matrix A is,

𝐴 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 = 𝑑 −𝑏

−𝑐 𝑎

1. Given 𝐵 = 3 21 4

, 2. Given 𝐶 = −2 15 −3

,

evaluate the determinant of B evaluate the determinant of B

𝐵 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 = 4 −2

−1 3 𝐶 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 =

−3 −1−5 −2

Exercise 2: Evaluate the adjoint of each of the following matrix.

1. 𝐵 = 2 15 3

2. 𝐶 = −1 21 −6

3. 𝐷 = 2 2

−3 −7 4. 𝐸 =

−2 15 −3

14

WEEK EIGHT

LESSON FOUR

TOPIC: MATICES (continued)

Inverse of a non-singular 2 x 2 matrix

To determine the inverse of a non-singular 2 x 2 matrix we must first find the determinant of that matrix then the adjoint of that matrix.

Ie. Given 𝐴 = 𝑎 𝑏𝑐 𝑑

, then the inverse of the matrix A is

𝐴 = 𝑎 𝑏𝑐 𝑑

=| |

× 𝐴 𝑎𝑑𝑗𝑜𝑖𝑛𝑡

𝐴 = 𝑑 −𝑏

−𝑐 𝑎

Example

Given 𝐴 = 4 −51 −2

, then the inverse of the matrix A is

𝐴 = 3 −51 −2

= ( × ) –( × )

−2 5−1 3

= −2 5−1 3

= − −2 5−1 3

= − × −2 − × 5

− × −1 − × 3

= −

−1

Evaluate the adjoint of each of the following matrix. Then find the inverse.

1. 𝐶 = 2 15 3

2. 𝐷 = −1 21 −6

3. 𝐸 = 2 2

−3 −7 4. 𝐹 =

−2 15 −3

15

WEEK NINE

LESSON ONE

TOPIC: STATISTICS - Introduction to Statistics

Statistics is a branch of Mathematics that deals with data collection, summary, analysis and interpretation. In our daily lives, we are bombarded with problems to be investigated on a range of topics such as health, sports, student performance, and opinions on issues, population trends, income growth and many other areas. The setting up of scientific experiments to tests hypotheses also rely on knowledge of statistics.

Statistics equips us with tools to carry out an investigation in a systematic way and to arrive at a solution based on informed judgments.

Population and Sample

A population is the entire group that you want to draw conclusions about. A sample is the specific group that

you will collect data from. The size of the sample is always less than the total size of the population.

Types of Variables

The type of data we collect will depend on the characteristic we are investigating. We may be interested in the mass of a new born baby, test scores, type of transport or the opinion of persons on a social issue. Each characteristic is associated with a variable whose value may change from one object to the other in a population.

16

Variables can be classified as:

Qualitative (non-numerical) - Qualitative variables are usually categorical in nature. There are two

types:

a) Nominal

b) Ordinal

Quantitative (numerical) –In quantitative data, the value of the characteristic is a number. Numerical

data can be of two types:

a) Discrete, if it can only take particular values within an interval. e.g. The number of skittles in a pack,

shoe sizes, etc. There are no values between any two consecutive values.

b) Continuous, if it has an infinite number of possible values between any two points on the

measurement scale. e.g. distance, weight, etc. The exact value of the measure is not possible to

determine. The degree of accuracy depends on the measuring device.

17

Exercise

2.

18

WEEK NINE

LESSON TWO

TOPIC: STATISTICS - Frequency Distributions

Data that is not processed or organized is called raw data. It is difficult to draw conclusions from raw data by

simply looking at the values (observations). A common method of organizing raw data is to construct a

frequency distribution which can be ungrouped or grouped. This is usually the first and simplest step in

analyzing data.

Ungrouped Frequency Distributions

A frequency table is used to summarize a set of data. It records the frequency or number of times each

observation occurs. The frequency table usually has two to three columns. The first column describes the data

(scores - height, marks, age, shoe size, etc.) in ascending order. The second column is optional; it records the

frequency of each observation in tally form. The last column records the numerical frequency of each

observation.

Example

The test scores of 20 students in Grade 10 are as follows: 23, 26, 24, 28, 29, 29, 23, 30, 24, 24, 25, 28, 25, 27, 23, 25, 29, 25, 26, 26 Use the information given to construct a frequency table Solution

The least score is 23

The highest score is 30

Score Tally Frequency 23 ||| 3 24 ||| 3 25 |||| 4 26 ||| 3 27 | 1 28 || 2 29 ||| 3 30 | 1

Total 20

19

Grouped Frequency Distributions Sometimes the data under consideration has such a large range of values that it is most useful to collect these

values into groups (or classes).

Example

The gown sizes (cm) of 40 women are listed below:

28 36 37 35 29 36 35 29 42 30

31 30 38 28 33 30 32 38 29 36

32 33 31 41 29 36 37 27 30 28

35 40 35 33 40 42 39 32 30 34

Use the information to construct a grouped frequency table

Solution

The data on gown sizes has 15 possible values and if represented in an ungrouped distribution, there will be 15 rows. This will not permit meaningful results, therefore it is better to group the data so as to reduce the number of rows.

We can select five classes each containing 3 values. These are: 28 – 30, 31 – 33, 34 – 36, 37 – 39 and 40 – 42.

Now, use these groups to construct the frequency table

Gown Size Tally Frequency 28 – 30 |||||||| ||| 13 31 – 33 |||| ||| 8 34 – 36 |||| |||| 9 37 – 39 |||| 5 40 – 42 |||| 5

TOTAL 40 Class Interval

A class interval is defined as a grouping of statistical data. From the frequency table above:

i. The first class interval is 28 – 30 ii. The second class interval is 31 – 33, and so on

Class Limits

The class limits are the end values of a class interval. The class interval 28 – 30 has two limits.

i. The lower class limit (LCL) which is 28 ii. The upper class limit (UCL) which is 30

20

Class Boundaries

It was mentioned earlier that for continuous data, the true value of the measure is not really known due to lack of precision in measuring instruments.

Let us assume that a class interval 150 – 154 represents heights rounded to the nearest cm. This class interval can actually contain values from 149.5 and up to but not including 154.5.

Another class interval 155 – 159 will have heights from 154.5 and up to but not including 159.5.

These end points are the class boundaries. In the first class interval 150 – 154, the lower class boundary is 149.5 and the upper class boundary is 154.5

Class Size (Width)

The width of a class interval (Class Size) = the upper class boundary – the lower class boundary

Alternatively, the class interval 28 – 30 has data values 28, 29 and 30 falling in this category. Therefore, the class size is 3

Class Mid-point (Class Mark)

Class mid-point =

Or class mid-point =

21

Exercise

(b) What is the class size?

(c) What is the upper class limit for the third class interval?

2. Construct ungrouped frequency tables for each of the following.

(a) 4, 3, 6, 5, 2, 4, 3, 3, 6, 4, 2, 3, 2, 2, 3, 3, 4, 5, 6, 4, 2, 3, 4

(b) 6, 7, 5, 4, 5, 6, 6, 8, 7, 9, 6, 5, 6, 7, 7, 8, 9, 4, 6, 7, 6, 5

3. The marks obtained by 30 students in an examination are given below.

20 6 23 19 9 14 15 3 1 12

10 20 13 3 17 10 11 6 21 9

6 10 9 4 5 1 5 11 7 24

Represent the above data on a grouped frequency table starting with the class interval

22

WEEK NINE

LESSON THREE

TOPIC: STATISTICS - Statistical Diagrams

Displaying data

When data is collected and organized into frequency tables, it is easy to summarise, interpret and make

decisions. However, displaying the data using charts is an even more effective means of organizing it. There are

various types of charts (bar chart, pie chart, line graph, pictogram, and frequency diagram)and these can be

drawn by hand or using computer programmes, such as EXCEL.

In displaying data, we must consider whether the data is discrete or continuous. This will influence our choice

of the chart. In addition to the type of data, the purpose of the display is also an important consideration when

making choices about the type of chart to select.

Types of graphs and charts

The type of graph you use depends on the type of data you want to represent.

Examples

Discrete data is best represented using bar charts.

Temperature graphs would usually be line graphs because the data is continuous.

When you are graphing percentages of a distribution a pie chart would be suitable.

The other thing you need to remember is that all graphs need ‘SALT’. Make sure you remember the following information:

S = Scale A = Axes L = Labels T = Title

Pie Chart

Pie Charts are circular diagrams. They are useful when we wish to make simple comparisons or show how parts

compare with a whole. It is usually used for representing categorical data when the number of categories is

relatively small.

When data is represented on pie charts, the sizes of the sectors are proportional to the categories sampled.

23

𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐒𝐞𝐜𝐭𝐨𝐫𝐬 =𝐟𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝐲 𝐨𝐟 𝐚𝐜𝐭𝐢𝐯𝐢𝐭𝐲

𝐭𝐨𝐭𝐚𝐥 𝐟𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝐲 × 𝟑𝟔𝟎°

Example 1

The following table shows the numbers of hours spent by a child on different events on a working day.

Represent the information on a pie chart

Activity No. of Hours

(frequency)

School 6

Sleep 8

Playing 2

Study 4

T. V. 1

Others 3

TOTAL 24

The sector angles for various observations can be calculated as:

Activity No. of Hours

(frequency) Measure of sector angle

School 6 (6/24 × 360)° = 90°

Sleep 8 (8/24 × 360)° = 120°

Playing 2 (2/24 × 360)° = 30°

Study 4 (4/24 × 360)° = 60°

T. V. 1 (1/24 × 360)° = 15°

Others 3 (3/24 × 360)° = 45°

The pie chart below shows the numbers of hours spent by a child on different events on a working day.

24

Example 2

A pie chart shows that in a survey of 60 people, those who chose mango as their favourite fruit were represented

by an angle of 48⁰. Calculate how many people chose mango.

Solution

The mango sector is represented as of the circle

So the number of people choosing mango is of 60

=48

360× 60

= 8 people

School

SleepPlay

Study

TV

Others

School Sleep Play Study TV Others

25

Exercise 1. The favourite flavours of ice-cream for the children in a locality are given in percentage as follow. Draw

a pie chart to represent the given information

2. Priya lists down her monthly expenditure as follows:

Expenditure Amount ($)

Rent 4000

Food 5400

Clothing 2800

Savings 400

(a) How much is Priya’s total expense per month?

(b) Draw a pie chart for her monthly expenses.

3. The favourite sports of students in a Grade Ten class are seen in the table below:

Football Hockey Cricket Basketball Badminton

10 5 5 10 10

(a) Determine the number of students in the Grade Ten class

(b) Use the information given to construct a pie chart

4. In a pie chart showing favourite ice cream, the angle for coconut is 54⁰. If 40 people took part in the survey, how many choose coconut?

5. At a career guidance seminar, a survey was done to find out the type of careers that Form 5 students were likely to choose.

Flavours % of Students Prefer the Flavours

Vanilla 25 %

Strawberry 15 %

Chocolate 10 %

Pumpkin Spice 30 %

Mango Zap 20 %

26

6. Students in a group were asked to name their favourite sport. Their responses are shown on the pie chart below.

(a) Calculate the value of 𝑥

(b) What percentage of the students chose cricket?

(c) Given that 40 students chose tennis, calculate the TOTAL number of students in the group.

27

WEEK NINE

LESSON FOUR


Line Graphs

Line graphs are useful means of displaying statistical data when we wish to examine trends or growth over a

period of time. A line graph is drawn by plotting points and connecting them with straight lines. Usually, if time

is one of the quantities, it is placed on the horizontal axis.

Example

Exercises:

1. A boy’s height was measured over several years and the data were recorded below

Age (years) 9 11 13 15 17 19 21

Height (cm) 130 135 150 160 167.5 175 175

28

(a) Draw a line graph to represent the information given in the table above

(b) Use the line graph to answer the following questions

i. During which period did the boy’s height increase the least?

ii. During which period did the boy’s height increase the most?

iii. State the periods during which the boy’s height increased by the same amount

Locate and state the period when the boy’s height was constant

2. A boy’s height was measured over several years and the data were recorded below.

Height (cm) 130 135 150 160 167.5 175 175

Age (years) 9 11 13 15 17 19 21

(a) Draw a line graph to represent the information in the table above

(b) Use the line graph to answer the following questions

(i) During which period did the boy’s height increase the least?

(ii) During which period did the boy’s height increase the most?

(iii) State the periods during which the boy’s height increased by the same amount.

(iv) Locate and state the period when the boy’s height was constant.

Bar Graphs (Bar Charts)

Bar graphs are used for qualitative data or discrete quantitative data, but not for continuous data. To draw bar

graphs, the following must be taken into consideration:

The spaces between the bars are equal The widths of the bars are the same The height of each bar gives the frequency The bar graph is best drawn on graph paper The bars can be drawn horizontally or vertically

Example

A company uses many reams of paper in a year. The table below gives the data for last year:

Quarter Reams of paper used Jan – Mar 270 April – Jun 330 Jul – Sept 190 Oct – Dec 450

Use the information above to draw a bar graph.

29

First we need to choose a scale. When choosing a scale, we choose a round number for each square on

the y-axis.

The biggest number is 450, so we need a vertical scale that shows at least 450.

A scale of one square to 50 reams will take 9 squares

We can then draw the bar chart as shown below

Exercise

1. The marks obtained by 50 students in a quiz out of five questions are displayed in the table below. Draw

a bar graph to represent the data

Marks 0 1 2 3 4 5

Frequency 1 4 7 10 18 10

2. What are the advantages of a bar chart?

3. The table below shows the money spent, in millions of dollars, on education in a Caribbean country

during the Covid 19 Pandemic. Construct a bar graph to display the data

Year Expenditure (in $m)

1980 2.0

1981 3.5

1982 1.5

1983 5.0

1984 4.0

30

4. The table below shows the number of bananas, to the nearest tonne, produced annually on a farm over a

period of 6 years.

(a) Draw a bar chart to represent the data given in the table above using a scale of 1cm to represent 1 year

on the x- axis and 1cm to represent 25 tonnes on the y-axis.

(b) (i) During which year was there the greatest production of bananas?

(ii) How is this information shown on the bar chart?

(c) (i) Between which two consecutive years was there the greatest change in the

production of bananas?

(ii) How is this information shown on the bar chart?

(d) Give ONE reason why the bar chart is unsuitable for predicting the number of bananas produced in

2016.

Year 2010 2011 2012 2013 2014 2015

Production

(tonne)

150

275

100

40

125

210

31

WEEK TEN

LESSON ONE


Histograms

Histograms are used to display continuous data and are useful in showing the shape of a distribution. It resembles a bar graph except that there are no ‘gaps’ between the bars. This is because the histogram represents data in which the variables are continuous and as such, the variables can take any value in the interval on the horizontal axis. So, where one bar ends the other starts.

Example 1

The heights (in cm) of 40 students are shown below. Use the information to draw a histogram Height (cm) Frequency 150 – 154 6 155 – 159 5 160 – 164 7 165 – 169 9 170 – 174 8 175 – 179 5 TOTAL 40

Solution: Insert another column to show the class boundaries

Class boundaries Height (cm) Frequency 149.5 ≤ 𝑥 < 154.5 150 – 154 6 154.5 ≤ 𝑥 < 159.5 155 – 159 5 159.5 ≤ 𝑥 < 164.5 160 – 164 7 164.5 ≤ 𝑥 < 169.5 165 – 169 9 169.5 ≤ 𝑥 < 174.5 170 – 174 8 174.5 ≤ 𝑥 < 179.5 175 – 179 5

TOTAL 40

32

Example 2 The table below shows the ages of some people on a bus. Use the information to construct a histogram.

Age (years) Frequency 10 – 19 4 20 – 29 6 30 – 39 7 40 – 49 5

The data are continuous, as you are classed at 19

right up to the day before your 20th birthday. So

the 10 – 19 bar occupies the space from 10 right

up to 20 on the chart.

Exercise

1. The table below shows the masses (kg) of members in a sports club. Construct a histogram using this

data.

Masses 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99

Frequency 6 8 12 14 7 3

2. The following data represents the lengths of leaves on a tree (to the nearest cm).

9 16 13 7 8 4 18 10 17 18 9 12 5

9 9 16 1 8 17 1 10 5 9 11 15 6

14 9 1 12 5 16 4 16 8 15 14 17

(a) Draw a grouped frequency distribution for this data using appropriate class intervals

33

(b) Hence, construct a histogram

(c) How many leaves had the same length of 16 cm?

(d) How many leaves were found to be the shortest?

Frequency Polygons

A frequency polygon is used to represent quantitative data, usually continuous data. If the mid-points of the bars

of a histogram are joined by straight lines, a frequency polygon is formed.Therefore, in a frequency polygon,

we plot the mid-points of the class intervals against the frequencies.

Like the histogram, the frequency polygon is useful in displaying the shape of a distribution.

Example The scores of students in a test are given in the table below. Use it to draw a frequency polygon Frequency polygon showing the scores of students

For this data:

The points are (10.5,2), (30.5, 4), (50.5, 7), (70.5, 5) and (90.5, 3)

The next inteval on either side would be -19 to 0 and 101 to 120. So the end points are

(-9.5, 0) and (110.5, 0)

34

Exercise

1. The frequency distribution below shows the masses, in kg, of 50 adults prior to the start of a fitness

programme.

Using a scale of 2cm to represent 5 units on the x-axis and 1cm to represent 1 unit on the y-axis, draw a

frequency polygon to represent the information in the table.

2.

35

WEEK TEN

LESSON TWO

TOPIC: STATISTICS - Measures of Central Tendency

Mean The mean of a set of scores is the sum of scores divided by the number of scores in the set. The mean is another name for what is commonly called the arithmetical average. When data is presented in the raw form, we can calculate the exact value of the mean by simply adding up the scores and dividing the total by the number of scores. Example 1

Example 2 On five Chemistry tests, John received the following scores: 72, 86, 92, 63 and 77. (i) What is his mean score on the five tests? (ii) What test score must John earn on his sixth test so that his mean score for all six tests will be 80?

36

Mean from ungrouped frequency distributions

Example

The scores obtained from tossing a die a total of 40 times are recorded in the table below.

To obtain the mean score, we need to calculate the sum of all 40 scores. Since the frequency tells us how many

times each score occurs, we can obtain the total for each of the individual scores by multiplying each score by

its frequency. Rearranging the table in vertical form, we insert a third column to record these totals under the

heading, 𝑓 × 𝑥.

37

Exercise 1. Calculate the mean of the following: a) 8, 3, 52, 48, 7, 78. b) 55, 22, 33, 59, 2, 4. c) 18, 98, 9, 88, 59, 98. 2. The mean of three numbers is 7. Two of the numbers are 6 and 12.

What is the third number?

3. The mean of four numbers is 5. Three of the numbers are 3, 5, and 8. What is the third number? 4. Three numbers have a mean of 23. Two of the numbers have a mean of 12 and two of the numbers have a mean of 30 What are the three numbers? 5. The table below shows the number of tickets bought per person for a calypso show.

No. of tickets bought per person for a calypso show (x)

Frequency (f) 𝑓 × 𝑥

1 2 3 4 5 6

12 35 44 18 8 3

Calculate the mean number of tickets bought.

Mean from grouped frequency distributions

When the data is grouped, it is not possible to reconstruct the raw data. Hence, we cannot obtain an exact value

for the mean. However, it is possible to obtain a very good estimate of the mean using the following method.

Consider the following data in which the masses of peas were recorded.

38

We know that 3 peas had masses that were between 3-7 grams. This means that there masses could have been 3,

4, 5, 6, or 7 grams. To obtain the total mass of the 3 peas in this interval, we assume that their average mass is 5

grams and multiply 3 by 5 to obtain a total mass of 15 grams.

We make an assumption that on the average, the set of values in an interval will approximate to the midpoint of

the interval. In this case, 5 is the midpoint of the interval 3-7. We now treat the midpoint as the variable score,

x, and proceed to use the same formula as was done for ungrouped data. The computations are shown in the

next table.

Exercise

1. The frequency distribution of the marks awarded to 100 candidates in an examination is as follows:

Marks Number of candidates 1- 10

11- 20 21- 30 31- 40 41- 50

13 23 36 20 8

Estimate the mean mark to the nearest whole number.

2. The table below shows the distribution of masses of 100 adults measured to the nearest kilogram.

Mass (kg) Frequency 50- 59 60- 69 70- 79 80- 89 90- 99

100- 109

5 9

28 33 17 8

Calculate the mean mass.

39

3. The distribution of marks obtained by 120 candidates in an examination are shown in the table below.

Marks Frequency

1- 10

11- 20

21- 30

31- 40

41- 50

51- 60

61- 70

71- 80

81- 90

91- 100

2

7

10

22

25

29

12

8

3

2

Calculate the mean mark.

4. The frequency distribution of the lengths of 100 steel rods measured in millimeter is given in the table below.

Length (mm)

Frequency

200- 204

205- 209

210- 214

215- 219

220- 224

225- 229

230- 234

235- 239

4

9

10

17

25

21

9

5

Calculate the mean length

40

WEEK TEN

LESSON THREE


Median

The median of a set of scores is the middle value when the scores are arranged in either ascending or

descending order of magnitude. If there are two middle values, then the median is the mean of the two middle

two values.

Example 1

Example 2

41

If we have a large number of scores on a frequency table, then it will be difficult and time consuming to list and

rearrange them all to find the middle. We can use a simple formula to calculate the rank of the median. For a set

of n scores, the middle score is the 𝑡ℎ 𝑠𝑐𝑜𝑟𝑒.

When n is odd, this works out to be a single score.

For example, when 𝑛 = 41, the median is the = 21𝑠𝑡 score.

When n is even, this works out to be the mean of two scores.

For 𝑛 = 40, the formula gives ( )𝑡ℎ 𝑠𝑐𝑜𝑟𝑒 = 20.5th score.

This means the median is halfway between the 20th and 21st score. The median is obtained by calculating the

mean of the 20th and 21st scores.

Median from ungrouped frequency distributions

Using the data on the scores obtained from tossing a die a total of 40 times; let us now obtain the median.

On the frequency table we insert another column (cumulative frequency). The cumulative frequency is

calculated by adding each frequency to the sum of its predecessors.

Score Frequency Cumulative frequency

1 8 8

2 5 8 + 5 = 13

3 9 13 + 9 = 22

4 6 22 + 6 = 28

5 7 28 + 7 = 35

6 5 35 + 5 = 40

Total 40

42

We then find the rank of the median by applying the formula the th term.

Median position is = 20.5𝑡ℎ 𝑡𝑒𝑟𝑚, therefore the median lie on the 20th and 21st position.

If we were to list the scores we will get 8 ones, followed by 5 twos, then 9 threes and so on.

By looking at our cumulative frequency column we noticed that the score‘3’ will lie on the 14th to the 22nd position.

Hence the median score is 3 since it lies on the 20th and 21st rank and the mean of 2 threes is 3.

Exercise

1. Find the median of the following scores:

a) 2, 1, 1, 5, 6

b) 7, 7, 0, 14, 0, 7, 14

c) 3, 12, 18, 12, 7, 12, 6, 4

2. The marks obtained by 40 students in a test are shown in the table below.

Marks 1 2 3 4 5 6 7 8

Frequency 3 5 6 9 5 2 6 4

Find the median of the marks shown in the frequency distribution given above.

2. In a shooting contest in which 50 people participated, the following frequency table was obtained.

Score 1 2 3 4 5 6 7 8

Frequency 3 1 4 10 15 9 3 5

Find the median score for the information above.

4. The table below shows the number of children per family in the families of pupils in a class.

No. of children

per family

1 2 3 4 5 6 7

Frequency 2 3 9 5 6 4 1

Calculate the median number of children per family.

43

Median from grouped data

To determine the median from a grouped frequency table, we use a cumulative frequency curve (Ogive).Before

drawing the cumulative frequency curve, we need to calculate the cumulative frequencies. The cumulative

frequency of a particular score, x is the number of scores less than or equal to x. In calculating the cumulative

frequencies, we use the upper-class boundaries as the highest possible score in an interval

Example

Consider the data on scores of students in a test out of 50 shown in the table below. In the table below, the

number of scores that is less than 10.5 is 0, the number less than 15.5 is 0+2 = 2, 20.5 is 2+5 = 7 and so on. So,

the cumulative frequency of 20.5 is 7. By adding the frequencies, in turn, we obtain their cumulative

frequencies.

To draw a cumulative frequency curve, we use the following steps:

Determine the upper-class boundaries for each interval.

Calculate the cumulative frequencies.

Plot the cumulative frequency against the upper-class boundary of each interval.

Join the points with a smooth curve.

The cumulative frequency curve for the data is shown below.

The points plotted are (10.5, 0), (15.5, 2), (20.5, 7), (25.5, 17)…

44

The estimated median is determined by drawing a horizontal line at 25 (half of 50) and a vertical line from the

point where the line meets the curve. The estimated median is where the vertical line cuts the horizontal axis.

In the above example, an estimated median score of 28.3indicates that one half the number of students in the

class scored below 28.3on the test. Of course, it also means that half of them scored more than 28.3 marks.

Exercise: The distribution of marks obtained by 120 candidates in an examination are shown in the table

below.

Marks Frequency

1- 10

11- 20

21- 30

31- 40

41- 50

51- 60

61- 70

71- 80

81- 90

91- 100

2

7

10

22

25

29

12

8

3

2

45

a) Draw up a cumulative frequency table

b) Hence draw the cumulative frequency curve on a graph paper, using 1cm to represent 10 units on both axes.

c) From the curve estimate the median

2. The frequency distribution of the lengths of 100 steel rods measured in millimeter is given in the table below.

Length (mm) Frequency

200- 204

205- 209

210- 214

215- 219

220- 224

225- 229

230- 234

235- 239

4

9

10

17

25

21

9

5

a) Draw a cumulative frequency curve to represent the above data.

b) Hence estimate the median length of the steel rods.

3. A survey was conducted in a school. The masses of 100 students are shown in the table below.

Mass (Kg) Frequency

50- 59

60- 69

70- 79

80- 89

90- 99

100- 109

14

31

35

13

5

2

Draw a cumulative frequency curve to represent the information above and use it to determine an estimate of the median mass.

4. The marks of 100 students who took an art examination is shown in the frequency table below.

Marks 0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99

Frequency 5 7 18 26 14 10 8 6 4 2

Represent the information on a cumulative frequency curve and determine an estimate of the median mark.

46

WEEK TEN

LESSON FOUR


Mode

The mode is the value (number) or category that occurs the most or is the more frequent occurrence in a

distribution.

In a set of observations, there can be no mode, one mode, or multiple modes. A distribution with two modes is

called bimodal. A distribution with three modes is called tri-modal.

Examples

1. Determine the mode of the set of scores: 8 10 12 15 16 16 17 18 19 19 19 19 19 19

Clearly, the score 19 occurs the most frequently. Hence, 19 is the mode.

2. Determine the mode of the set of scores: 17 7 3 12 3 3 5 5 12 10 5 24

Since the scores 3 and 5 both occur ‘the most frequently’, there are two modes. These are 3 and 5

3. Using the frequency table below, state the modal score

The score 3 has the highest frequency of 9, indicating that it occurred most often. The mode is therefore 3.

4. Using the grouped frequency table below, determine:

a) The modal class

b) The modal mass

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Solution a) The class interval 13-17 has the highest frequency of 12. The modal class is therefore 13-17.

The steps below will allow us to find an estimate of the actual mode:

Construct a histogram andidentify the tallest bar. This represents the modal class.

Join the tips of this bar to those of the neighbouring bars on either side, with the one on the left joined to

that on the right and vice-versa. The lines used to join these tips cross each other at some point in this

bar.

Drop a perpendicular line from the tip of the point where these lines meet to the base of the bar

(horizontal axis). The point where it meets the base is the mode

Read off the value at the base using the estimation method

From the above histogram the estimated mode for the mass of the peas is 16 grams.

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Exercise

1. Find the mode of the following scores:

3 7 4 5 6 5

2. The heights of 10 girls stated in centimeters are:

153 156 154 161 148 155 154 152 149 154

Determine their modal height.

3. The table below shows the number of children per family in the families of pupils in a class.

No. of children per family

1 2 3 4 5 6 7

Frequency 2 3 9 5 6 4 1

Determine the Mode.

4. The frequency distribution of the masses of 100 students are shown in the table below.

Mass Frequency

50- 59

60- 69

70- 79

80- 89

90- 99

100- 109

14

31

35

13

5

2

(a) State the modal class

(b) Determine the modal mass

5. The distribution of marks for 100 students in a test are shown in the table below.

Marks 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99

Frequency 5 7 18 26 14 10 8 6 4

(a) State the modal class

(b) Construct a histogram to represent the information on the table above

(c) Hence determine a

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