3.1 Circles II

Posted on January 1, 2018 by user

3.1 Circles II

(A) Properties of Circles Involving Symmetry, Chords and Arcs

(B) Identify the Properties of Chords

(1) A radius that is perpendicular to a chord divides the chord into two equal parts.

(C) Properties of Angles in Circles

(1)

(2)

(3)

(4)

(5)

2.2.2 Polygons II, PT3 Practice

Posted on December 22, 2017 by user

Question 6:
In diagram below, PQRSTU is a regular hexagon QUV is a straight line.

Find the value of x.

Solution:

\begin{array}{l} Each interior angle \\ = \frac{(6 - 2) \times 180^{o}}{6} \\ = 120^{o} \\ ∠ P U Q = \frac{180^{o} - 120^{o}}{2} = 30^{o} (△ P U Q is a isosceles triangle) \\ x^{o} = 180^{o} - 30^{o} \\ = 150^{o} \\ x = 150 \end{array}

Question 7:
In diagram below, PQRSTU is a hexagon. UPE, PQF, QRG, RSH and UTJ are straight lines.

Find the value of x.

Solution:

\begin{array}{l} Sum of all the exterior angles of any polygon = 360^{o} \\ 25^{o} + 3 x^{o} + 2 x^{o} + 40^{o} + 75^{o} + 55^{o} = 360^{o} \\ 5 x^{o} + 195^{o} = 360^{o} \\ 5 x^{o} = 360^{o} - 195^{o} \\ = 165^{o} \\ x^{o} = 165^{o} \div 5 \\ = 33^{o} \\ x = 33 \end{array}

Question 8:
In Diagram below, A, B, C, D and E are vertices of a 9 sided regular polygon.

Find the value of x.

Solution:

\begin{array}{l} Exterior angle = \frac{360^{o}}{9} = 40^{o} \\ Interior angle = 180^{o} - 40^{o} = 140^{o} \\ Sum of interior angles of pentagon A B C D E \\ = (5 - 2) \times 180^{o} \\ = 540^{o} \\ x^{o} + 140^{o} + 140^{o} + 140^{o} + x^{o} = 540^{o} \\ 2 x^{o} = 540^{o} - 420^{o} \\ = 120^{o} \\ x^{o} = 60^{o} \\ x = 60 \end{array}

2.2.1 Polygons II, PT3 Practice

Posted on December 17, 2017 by user

2.2.1 Polygons II, PT3 Practice

Question 1:

Diagram below shows a pentagon PQRST. TPU and RSV are straight lines.

Find the value of x.

Solution:

\begin{array}{l} Sum of interior angles of a pentagon \\ = (5 - 2) \times 180^{o} \\ = 3 \times 180^{o} \\ = 540^{o} \\ ∠ T S R = 180^{o} - 70^{o} \\ = 110^{o} \\ ∠ T P Q = 180^{o} - 85^{o} \\ = 95^{o} \\ x^{o} = 540^{o} - (110^{o} + 105^{o} + 115^{o} + 95^{o}) \\ = 540^{o} - 425^{o} \\ = 115^{o} \\ x = 115 \end{array}

Question 2:

In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines.

Find the value of x + y.

Solution:

\begin{array}{l} ∠ Q P U = 180^{o} - 160^{o} = 20^{o} \\ Reflex ∠ P U T = 360^{o} - 80^{o} = 280^{o} \\ ∠ U T S = 180^{o} - 120^{o} = 60^{o} \\ ∠ T S R = 180^{o} - 35^{o} = 145^{o} \\ Sum of interior angles of a hexagon \\ = (6 - 2) \times 180^{o} \\ = 720^{o} \\ x^{o} + y^{o} + 145^{o} + 60^{o} + 280^{o} + 20^{o} = 720^{o} \\ x^{o} + y^{o} = 720^{o} - 505^{o} \\ = 215^{o} \\ x + y = 215 \end{array}

Question 3:

Diagram below shows a regular hexagon PQRSTU. PUV is a straight line.

Find the value of x + y.

Solution:

\begin{array}{l} Size of each interior angle of a regular hexagon \\ = \frac{(6 - 2) \times 180^{o}}{6} \\ = 120^{o} \\ x^{o} = \frac{180^{o} - 120^{o}}{2} = 30^{o} \\ y^{o} = 180^{o} - 120^{o} \\ = 60^{o} \\ x^{o} + y^{o} = 30^{o} + 60^{o} \\ = 90^{o} \\ x + y = 90 \end{array}

Question 4:

In the diagram below, KLMNP is a regular pentagon. LKS and MNQ are straight lines.

Find the value of x + y.

Solution:

\begin{array}{l} Size of each interior angle of a regular pentagon \\ = \frac{(5 - 2) \times 180^{o}}{5} \\ = 108^{o} \\ ∠ P K S = ∠ P N Q = 180^{o} - 108^{o} = 72^{o} \\ Reflex angle ∠ K P N = 360^{o} - 108^{o} = 252^{o} \end{array}

\begin{array}{l} Sum of interior angles of a hexagon \\ = (6 - 2) \times 180^{o} \\ = 720^{o} \\ ∴ x^{o} + y^{o} + 72^{o} + 252^{o} + 72^{o} + 100^{o} = 720^{o} \\ x^{o} + y^{o} = 720^{o} - 496^{o} \\ = 224^{o} \\ x + y = 224 \end{array}

Question 5:

In Diagram below, PQR is an isosceles triangle and PRU is a straight line.

Find the value of x + y.

Solution:

\begin{array}{l} x^{o} = 180^{o} - 20^{o} - 20^{o} = 140^{o} \\ ∠ P R S = 180^{o} - 110^{o} = 70^{o} \\ y^{o} + 85^{o} + 75^{o} + 70^{o} = 360^{o} \\ y^{o} + 230^{o} = 360^{o} \\ y^{o} = 130^{o} \\ x^{o} + y^{o} = 140^{o} + 130^{o} \\ = 270^{o} \\ x + y = 270 \end{array}

2.1 Polygons II

Posted on December 15, 2017 by user

2.1 Polygons II

2.1.1 Regular Polygons

1. A regular polygon is a polygon where

(a) all its sides are of equal length, and

(b) all its interior angles are of equal size.

2. The number of axis of symmetry of a regular polygon is equal to its number of sides.

Example:

2.1.2 Exterior and Interior Angles of Polygons

1. The exterior and interior angles at a vertex of a polygon is supplementary.

2. The sum of interior angles of an n-sided polygon is $(n - 2) \times 180^{o}$

3. Each interior angle of a regular n-sided polygon is $\frac{(n - 2) \times 180^{o}}{n}$

4. The sum of all exterior angles of a polygon is 360^o.

5. Each exterior angle of a regular n-sided polygon is

\frac{360^{o}}{n}

1.2.1 Angles and Lines II, PT3 Practice

Posted on December 3, 2017 by user

1.2.1 Angles and Lines II, PT3 Practice

Question 1:

In Diagram below, PQRS, ABQC and KRL are straight lines.

Find the value of x.

Solution:

\begin{array}{l} ∠ K R B + ∠ A B R = 180^{o} \leftarrow Sum of interior angles \\ ∠ K R B + 105^{o} = 180^{o} \\ ∠ K R B = 180^{o} - 105^{o} \\ ∠ K R B = 75^{o} \\ ∠ B R Q = 40^{o} \leftarrow Corresponding angle \\ x^{o} = ∠ S R L = ∠ K R Q \leftarrow \begin{array}{l} Vertically \\ opposite angle \end{array} \\ x^{o} = 75^{o} + 40^{o} \\ = 115^{o} \\ x = 115 \end{array}

Question 2:

(a) In Diagram below, PQR and SQT are straight lines.

Find the value of x.

(b) In Diagram below, PQRS is a straight line. Find the value of x.

Solution:

(a)

\begin{array}{l} ∠ P Q T = ∠ R Q S \\ x^{o} + 90^{o} = 155^{o} \\ x^{o} = 155^{o} - 90^{o} \\ = 65^{o} \\ x = 65 \end{array}

(b)

\begin{array}{l} ∠ Q R T = 57^{o} \\ x^{o} + 2 x^{o} = 180^{o} - 57^{o} \\ 3 x^{o} = 123^{o} \\ x^{o} = \frac{123^{o}}{3} \\ x^{o} = 41^{o} \\ x = 41 \end{array}

Question 3:

In Diagram below, PQR is a straight line. Find the value of x.

Solution:

\begin{array}{l} x^{o} = ∠ Q R T + ∠ T R S \\ ∠ Q R T = 40^{o} \\ ∠ T R S = 180^{o} - 135^{o} \\ = 45^{o} \\ Hence, x^{o} = 40^{o} + 45^{o} \\ x^{o} = 85^{o} \\ x = 85 \end{array}

Question 4:

In Diagram below, find the value of x.

Solution:

40^o + 80^o + x^o+ x^o = 180^o

2x^o = 180^o – 120^o

2x^o = 60^o

x^o= 30^o

x = 30

Question 5:

In Diagram below, PQR is an isosceles triangle and QRS is a straight line.

Find the values of x and y.

Solution:

\begin{array}{l} ∠ P R Q = 180^{o} - 110^{o} = 70^{o} \\ ∠ P R Q = ∠ P Q R = 70^{o} \\ y^{o} = 180^{o} - 70^{o} - 70^{o} \\ = 40^{o} \\ y = 40 \\ x^{o} = 110^{o} - 40^{o} \\ = 70^{o} \\ x = 70 \end{array}

1.1 Angles and Lines II

Posted on December 1, 2017 by user

1.1 Angles and Lines II

Identifying Parallel, Transversals, Corresponding Angles, Alternate Angles and Interior Angles

(A) Parallel lines

Parallel lines are lines with the same direction. They remain the same distance apart and never meet.

(B) Transversal lines

A transversal is a straight line that intersects two or more straight lines.

(C) Alternate angles

(D) Corresponding angles

(E) Interior angles

PT3 Smart TIP

Alternate angles are easily identified by tracing out the pattern “Z” as shown.

Corresponding angles are easily identified by the pattern “F” as shown.

Interior angles are easily identified by the pattern “C” as shown.

Example 1:

In Diagram below,PQ is parallel to RS. Determine the value of y.

Solution:

Construct a line parallel to PQ and passing through W.

a = 40^o and b = 50^o← (Alternate angles)

y = a + b

= 40^o+ 50^o

= 90^o

Example 2:

In Diagram below, PSQ and STU are straight lines. Find the value of x.

Solution:

\begin{array}{l} ∠ W S Q = 180^{o} - 150^{o} \\ = 30^{o} \leftarrow Supplementary angle \\ ∠ X T U = ∠ W S Q + ∠ x \leftarrow Corresponding angle \\ 75^{o} = 30^{o} + ∠ x \\ ∠ x = 75^{o} - 30^{o} \\ ∠ x = 45^{o} \\ x = 45 \end{array}

13.2.3 Statistics (I), PT3 Focus Practice

Posted on November 23, 2017 by user

Question 7:
Diagram below is an incomplete pictograph showing the sales of books for a duration of five months.

(a) The sales in May is ¼ of the total sales in January and February.
Complete the pictograph in the Diagram.
(b) Find the total number of books sold before April.

Solution:
(a)

\begin{array}{l} Sales in May \\ = \frac{1}{4} \times 8 \\ = 2 \end{array}

(b)
Total number of books sold before April
= (5 + 3 + 6) × 15
= 14 × 15
= 210

Question 8:
Diagram below shows an incomplete line graph of the number of eggs sold in four weeks. The number of eggs sold on week 1 is 2000 and 4000 on week 4.

(a) Complete the line graph in the Diagram.
(b) Complete the pie chart in the second Diagram to represent sales from Week 1 to Week 4.

Solution:
(a)

(b)

\begin{array}{l} First week \\ = \frac{2000}{2000 + 1500 + 2500 + 4000} \times 360^{o} \\ = \frac{2000}{10000} \times 360^{o} \\ = 72^{o} \\ Fourth week \\ = \frac{4000}{10000} \times 360^{o} \\ = 144^{o} \end{array}

13.2.2 Statistics (I), PT3 Focus Practice

Posted on November 20, 2017 by user

13.2.2 Statistics (I), PT3 Focus Practice

Question 4:

Table below shows the profit made from the sales of coconut at a stall over five days.

Day	Day 1	Day 2	Day 3	Day 4	Day 5
Profit (RM)	280	200	200	360	320

On diagram in the answer space, draw a line graph to represent all the information in the Table. Use the scale 2 cm to RM80 on the vertical axis.

Answer:

Solution:

Question 5:

Table below shows the number of students who read four types of books.

Types of books	Fiction	Language	Technology	Religion
Number of students	18	16	8	14

On diagram in the answer space, draw a bar chart to represent all the information in the Table.

Answer:

Solution:

Question 6:
Diagram below is an incomplete bar chart which shows the number of cars sold in a shop for duration of five months.

(a) The number of cars sold for the five months is 150 units. The number of cars sold in February is equal to the number of cars sold in April.
Complete the bar chart in the Diagram.
(b) Find the percentage of cars sold in January.

Solution:
(a)
(b)

\begin{array}{l} Percentage of cars sold \\ = \frac{45}{150} \times 100 % \\ = 30 % \end{array}

13.2.1 Statistics (I), PT3 Focus Practice

Posted on November 17, 2017 by user

13.2.1 Statistics (I), PT3 Focus Practice

Question 1:

Diagram below is a pictograph showing the number of pizzas sold in three months.

The price of one pizza is RM18. The total sales for those four months are RM2250.

Calculate the number of pizzas sold in April.

Solution:

Total number of pizzas sold in four months

\begin{array}{l} = \frac{R M 2250}{R M 18} \\ = 125 \end{array}

Total number of pizzas sold in first three months

= 9 × 10

= 90

Number of pizzas sold in April

= 125 – 90

= 35

Question 2:

Diagram below is a bar chart showing the number of new houses sold over four days.

The total number of new houses sold on Wednesday and Thursday is 56.

Calculate the total number of new houses sold on Monday and Tuesday.

Solution:

Wednesday: 12 units

Thursday: 16 units

Number of houses represented by 1 unit

\begin{array}{l} = \frac{56}{12 + 16} \\ = 2 \end{array}

Monday: 14 units

Tuesday: 18 units

Total number of new houses sold on Monday and Tuesday

= (14 + 8) × 2

= 64

Question 3:

Table below shows the number of companies that gave contributions to an orphan fund over a period of five days.

Days	Monday	Tuesday	Wednesday	Thursday	Friday
Number of companies	8	14	m	17	11

It is given that 20% of the total contribution was made on Tuesday.

Calculate the value of m.

Solution:

Let C = Total contribution

\begin{array}{l} \frac{20}{100} \times C = 14 \\ C = 14 \times \frac{100}{20} \\ = 70 \\ m = 70 - (8 + 14 + 17 + 11) \\ = 70 - 50 \\ = 20 \end{array}

12.2.1 Solid Geometry (II), PT3 Focus Practice

Posted on November 3, 2017 by user

12.2.1 Solid Geometry (II), PT3 Focus Practice

Question 1:

Diagram below shows closed right cylinder.

Calculate the total surface area, in cm², of the cylinder.

(π = \frac{22}{7})

Solution:

Total surface area

= 2(πr²) + 2πrh

\begin{array}{l} = (2 \times \frac{22}{7} \times 7^{2}) + (2 \times \frac{22}{7} \times 7 \times 20) \\ = 308 + 880 \\ = 1188 c m^{2} \end{array}

Question 2:

Diagram below shows a right prism with right-angled triangle ABC as its uniform cross section.

Calculate the total surface area, in cm², of the prism.

Solution:

\begin{array}{l} A B = \sqrt{5^{2} - 3^{2}} \\ = \sqrt{25 - 9} \\ = \sqrt{16} \\ = 4 c m \end{array}

Total surface area

= 2 (½× 3 × 4) + (3 × 10) + (4 × 10) + (5 × 10)

= 12 + 30 + 40 + 50

= 132 cm²

Question 3:

Diagram below shows a right pyramid with a square base.

Calculate the total surface area, in cm², of the right pyramid.

Solution:

h²= 10² – 6²

= 100 – 36

= 64

h = √64

= 8cm

Total surface area of the right pyramid

= (12 × 12) + 4 × (½× 12 × 8)

= 144 + 192

= 336 cm²

Question 4:

The diagram shows a cone. The radius of its base is 3.5 cm and its slant height is 6 cm. Find the area of its curved surface.
( π= 22 7 )

Solution:

Area of curved surface

= π × radius of base × slant height

= πrs

\begin{array}{l} = \frac{22}{7} \times 3.5 \times 6 \\ = 66 c m^{2} \end{array}