Focus On Maths Grade 7

186 CHAPTER 5 Linear Equations and Inequalities in One Variable 5.5 Solving Inequalities in One Variable A Solving inequalities by addition and subtraction When a number is added to or subtracted from both sides of an inequality, the condition of the inequality is unchanged as shown below. (a) 7 . 3 (b) –4 , –1 7 + 5 ? 3 + 5 –4 – 3 ? –1 – 3 12 . 8 –7 , –4 Solving a linear inequality in x means to find the values of x that satisfy the inequality. Add 5 to each side. Subtract 3 from each side. The inequality is unchanged. LHS . RHS The inequality is unchanged. LHS , RHS EXAMPLE 28 Determine whether each of the following relationships is a linear inequality in one variable. (a) x – 4 > 3 (b) m , n + 4 (c) x 2 , x – 7 Solution: (a) Yes (b) No (c) No EXAMPLE 29 Solve the following inequalities. (a) x + 6 . 10 (b) y – 4 , 9 (c) y + 12 > 7 Solution: (a) x + 6 . 10 x + 6 + (–6) . 10 + (–6) Add –6 to each side. x . 4 (b) y – 4 , 9 y – 4 + 4 , 9 + 4 Add 4 to each side. y , 13 (c) y + 12 > 7 y + 12 + (–12) > 7 + (–12) Subtract 12 from each side. y > –5 There is 1 variable with power of one. There are 2 variables. The power of x is 2. ©Praxis Publishing_Focus On Maths

187 Linear Equations and Inequalities in One Variable CHAPTER 5 B Solving inequalities by multiplication or division When both sides of an inequality are multiplied or divided by a positive number, the inequality is unchanged as shown below. (a) 8 . 5 (b) –6 , –4 –6 2 ? –4 2 –3 , –2 Divide both sides by 2. The inequality sign is unchanged. LHS , RHS 8 × 2 ? 5 × 2 16 . 10 When both sides of an inequality are multiplied or divided by a negative number, the inequality sign will be reversed. For example, (a) 12 . 8 (b) –12 , –8 –12 –2 ? –8 –2 6 . 4 Divide both sides by –2. Reverse the inequality sign. LHS . RHS 12 × (–2) ? 8 × (–2) –24 , –16 Multiply both sides by 2. The inequality sign is unchanged. LHS . RHS . changes to , and , changes to ., > changes to < and < changes to >. Multiply both sides by –2. Reverse the inequality sign. LHS , RHS EXAMPLE 30 Solve the following inequalities. (a) x 2 . 4 (b) 3y < –15 (c) – y 3 < –6 (d) –6y . 24 Solution: (a) x 2 . 4 (b) 3y < –15 x 2 × 2 . 4 × 2 3y × 1 3 < –15 × 1 3 x . 8 y < –5 (c) – y 3 < –6 (d) –6y . 24 – y 3 × (–3) > –6 × (–3) –6y × 1– 1 6 2 , 24 × 1– 1 6 2 y < 18 y , –4 5 1 Multiply both sides by –  1 6 . The inequality sign is reversed. 1 –24 , –16 –30 –20 –10 0 10 20 30 ©Praxis Publishing_Focus On Maths

188 CHAPTER 5 Linear Equations and Inequalities in One Variable C Solving linear Inequalities using a combination of operations When solving linear inequalities involving more than one operation, we apply the procedures used in solving linear equation. When multiplying or dividing by a negative number, the inequality sign must be reversed. EXAMPLE 31 Solve the following inequalities. (a) 3x – 4 . 11 (b) – 5p 3 < 20 (c) 3(2y + 5) , 4y – 3 Solution: (a) 3x – 4 . 11 (b) – 5p 3 , 20 – 5p 3 × 3 , 20 × 3 –5p , 60 –5p 5 > 60 –5 p > –12 3x – 4 + 4 . 11 + 4 3x . 15 3x 3 . 15 3 x . 5 (c) 3(2y + 5) , 4y – 3 6y + 15 , 4y – 3 6y + 15 – 15 , 4y – 3 – 15 6y – 4y , 4y – 4y – 18 2y , –18 2y 2 , –18 2 y , –9 EXAMPLE 32 Solve each of the following inequalities. (a) 1 + 3x < x – 7 (b) 5(4 + x) . –10 (c) 1 – 2x 3 < 7 (d) 3 2 x – 1 > 2x + 1 Solution: (a) 1 + 3x < x – 7 1 + 3x – 1 < x – 7 – 1 3x < x – 8 3x – x < x – 8 – x 2x < –8 2x 2 < –8 2 x < –4 –1 –x ÷2 ©Praxis Publishing_Focus On Maths

189 Linear Equations and Inequalities in One Variable CHAPTER 5 (b) 5(4 + x) . –10 5(4 + x) 5 . –10 5 4 + x . –2 4 + x – 4 . –2 – 4 x . –6 (c) 1 – 2x 3 < 7 1 – 2x 3 × 3 < 7 × 3 1 – 2x < 21 1 – 2x – 1 < 21 – 1 –2x < 20 –2x –2 > –20 –2 x > –10 (d) 3 2 x – 1 > 2x + 1 3 2 x – 1 + 1 > 2x + 1 + 1 3 2 x > 2x + 2 3 2 x – 2x > 2x + 2 – 2x – 1 2 x > 2 x < –4 ÷5 –4 ×3 –1 ÷(–2), the inequality symbol is reversed. +1 –2x ×(–2), the inequality symbol is reversed. D Constructing inequalities from given information To construct inequalities from given information, we follow the steps below. 1 Construct the basic inequality in the form of x . a, x , a, x > a or x < a where a is a number. 2 Based on the information given, ‘+’, ‘–’, ‘×’ or ‘÷’ both sides of the inequality by the same number to obtain a new inequality. 3 Simplify the new inequality. EXAMPLE 33 (a) Few months ago, Raju bought two fish as pets. One of them has a mass of x kg and was heavier than the other fish that has a mass of 45 g. Now the mass of each fish is five times as heavy as compared to a few months ago. Construct a relationship between the present masses of the two fish. ©Praxis Publishing_Focus On Maths

190 CHAPTER 5 Linear Equations and Inequalities in One Variable (b) Ann bought x pens for $4.50 each. She paid $30 and received a balance that is more than $3. Construct an inequality based on the given information. Solution: (a) x . 45 1000 Change g to kg. x × 5 . 45 1000 × 5 Mass of the fish is 5 times the original mass. Therefore, 5x . 0.225 (b) Balance . 3 30 – 4.5x . 3 EXAMPLE 34 A machine produced 1750 marbles on a certain day. Among the marbles, 18 are substandard. All the substandard marbles are removed and the remaining marbles are packed into x containers. Each container can accommodate 40 marbles. Calculate the minimum number of containers needed. Solution: Stage 1: Understand the problem List the facts and the question. Facts: A machine produce 1750 marbles. Among marbles produced, 18 are rejected. The remaining marbles packed into x containers. Each container can fill with 40 marbles. Question: Find the minimum number of containers needed. Stage 2: Think of a plan • x is number of containers needed to fill remaining marbles. • 18 marbles are rejected from 1750 marbles. • Minimum number of containers needed, so 40 × x > 1750 – 18 Total cost of x pens = 4.5x ©Praxis Publishing_Focus On Maths

191 Linear Equations and Inequalities in One Variable CHAPTER 5 Stage 3: Carry out the plan 40x > 1750 – 18 40x > 1732 x > 1732 40 x > 43.3 The possible number of containers used is 44, 45, 46, … Therefore, the minimum number of containers needed is 44. Stage 4: Look back Work backwards to check. 40(44) = 1760 44 containers are enough for the remaining 1732 marbles. Practice 5.5 Basic Intermediate Advanced A Solve the following inequalities. (a) x + 4 . 9 (b) x + 3 , 11 (c) x + 2 > –6 (d) x + 4.5 < 8 (e) 3 , x + 7 (f) – 8 > x + 3 (g) – 3.4 < x + 6 (h) x + 5.6 , –4.4 (i) 3 2 3 . x + 1 1 3 B Solve the following inequalities. (a) x – 2 , 3 (b) x – 5 . –8 (c) x – 8 < 13 (d) x – 3.2 > –1.6 (e) 9 , x – 6 (f) – 5.4 > x – 3 (g) x – 3 2 5 . 4 1 5 (h) – 3.6 < x – 5.4 ( i) 5 5 8 . x – 1 3 4 C Solve the following inequalities. (a) 3x . 21 (b) 4x , – 32 (c) 6x < 9 (d) – 26 > 4x (e) 33 < 3x ( f) 1 2 y . 5 (g) y 3 , – 9 (h) 2 3 y > – 4 9 (i ) 32 5 > 8 15 y D Solve each of the following inequalities. (a) –2x , 16 (b) –4x > 20 (c) –7x < – 35 (d) –6x . 27 (e) – x 3 < 8 ( f) – y 7 , – 3 2 (g) –24 > – 4 5 y (h) 5 4 > – 15 8 y (i ) – 3 2 < – 27 4 y E Solve each of the following inequalities. (a) 3 2 x – 4 , –7 (b) 5 – 2 3 x > 3 (c) 4x – 5 , 2x + 9 (d) 3(2x – 5) 4 < 6 (e) 5(x – 2) < 2(x + 4)( f) 2(3 – 2y) 3 . –8 F Emma and Su Lin have $y and $9 in their purses respectively. Emma has less money than Su Lin. Each of them donates $5 to a charity. Construct a relationship between the amount of money left in Emma’s and Su Lin’s purses. G Mrs Teng intends to cook 200 g of vegetables and m kg of meat for each of her family members. The mass of the meat is greater than the mass of the vegetables. Construct a relationship between the masses of meat and vegetables that Mrs Teng has to cook for her family of six members. ©Praxis Publishing_Focus On Maths

192 CHAPTER 5 Linear Equations and Inequalities in One Variable Summary Summary Summary Integers Linear Equations Linear algebraic expressions A combination of two or more linear algebraic terms by addition or subtraction. For example, 6x + 5 and 7x – 8y. Linear equations in one variable An equation that has only one variable which is raised to the power of 1. For example, 5x – 12 = 23 and 7y = 9. Solutions of linear equations in one variable A numerical value is the solution of a given linear equation in one unknown if the value satisfies the equation. For example, 5 is the solution of x – 4 = 1. Linear algebraic terms A term with only one variable which is raised to the power of 1. For example, 3x and –7p. Inequality Inequality is the relationship between two unequal quantities. Linear inequality in one variable A linear inequality in one unknown is an inequality with an variable raised to the power of one. Solving linear inequalities in one variable A linear inequality can be solved by finding an equivalent inequality in its simplest form. Operations involving linear inequalities • The inequality symbol remains unchanged (a) when any number is added to or subtracted from both sides of the inequality. (b) when both sides of the inequality is multiplied or divided by a positive number. • The inequality symbol is reversed when both sides of the inequality are multiplied or divided by a negative number. Representation on number line A linear inequality can be represented on a number line. For example, (a) x a a (b) x a a (c) x a a (d) x a a Integers Linear Equations and Inequalities in One Variable ©Praxis Publishing_Focus On Maths

193 Linear Equations and Inequalities in One Variable CHAPTER 5 5 Section A 1. Which of the following is a linear expression? A 4pq – 6q B 3x – 2y C 2y = x + 3 D 1 – 2m2 2. When 4 is subtracted from 5 times a number x, the result is 30. Which of the following represents the statement? A 4 – 5x = 30 B 5 – 4x = 30 C 5x – 4 = 30 D 4 + 5x = 30 3. Katina’s age is x years and Carol is 4 years younger than Katina. The sum of their ages is 46 years. The linear equation that represents the above is A 2x – 4 = 46 B 2x – 8 = 46 C x – 4 = 46 D 4 – 2x = 46 4. Given that 4p – 3 = 21 + 2p, then p = A 10 B 11 C 12 D 13 5. Solve the equation 3m 5 = m – 2 2 . A –10 B –5 C 5 D 10 6. Madam Lam bought a packet of sugar which cost $4 and x kg of flour which cost $3 per kg. She paid $22 altogether. Find the value of x. A 4 B 5 C 6 D 8 7. P Q R The diagram shows three solids P, Q and R on a balance. The masses of P and R are 1500 g and 1.2 kg respectively. Find the mass, in g, of solid Q such that the balance is in equilibrium. A 300 B 350 C 400 D 450 8. Solve the following linear inequality. 2(5 – g) > 12 A g < –1 B g > –1 C g < 1 D g > 1 9. Given that 6x – 5 . 11 and x is a whole number. Find the smallest value of x. A 1 B 2 C 3 D 4 10. If 2m > 5, what is the possible value of m? A 0 B 1 C 2 D 3 11. Find the solution for the inequality 8 + x > 13. A x > 5 B x < 5 C x > 21 D x < 21 ©Praxis Publishing_Focus On Maths

194 CHAPTER 5 Linear Equations and Inequalities in One Variable 12. The number line that represents the solution for – 4 , 2 – x, 5 is A –6 3 B –6 3 C –3 6 D –3 6 Section B 1. (a) Given that h = 3h 7 – 8, find the value of h. (b) Given that 2x – 1 5 = x + 3 3 , find the value of x. (c) Solve the equation 5q 24 = – 3 8 . 2. (a) Given that m = 5m 7 – 4, find the value of m. (b) Given that k 3 – 3 = 2 – k, find the value of k. (c) Given that 3m + 2 m – 4 = 5, find the value of m. 3. Solve each of the following equations. (a) k 12 – 2 3 = 0 (b) 2x 5 – x = 6 (a) 2x + 24 = 5x (b) 1 2 (16 – 3y) = y + 7 4. (a) The sum of three quarter of a number and 12 is 54. Find the number. (b) $480 is shared among three persons, Kate, Mike and Fay. Mike receives $120 and Kate’s share is three times as much as Fay’s share. How much is Kate’s share? 5. Given that 24 – 5x < –10, (a) find the least value of x, (b) find the least integer value of x. 6. Given that – 8 9 x < 2 2 15, (a) solve the inequality, (b) state the first three possible solutions if x is an integer. 7. Given that 6 – 2 5 x < –2x + 2 3 5 , (a) find the greatest value of x, (b) find the greatest integer value of x. (c) Solve the linear inequality 12 – 1 3 p , –2p + 3 2 3 . Hence state the greatest value of p if p is an integer. 8. The price of an apple is $y which is greater than $1.20. (a) Construct an inequality for the price of 15 apples. (b) Find the lowest possible price for 15 apples, to the nearest $. 9. (a) Write an equation to represent each of the following statements. (i) The sum of p and 6 is 12. (ii) The difference between q and –6 is 25. (iii) The product of 5 and –w is 35. (b) The table below shows the price of 2 sets of furniture at a shop. Set Price P : A round table and 4 chairs. $400 Q : A round table and 6 chairs. $500 Given that the model of the table and the chair in set P and set Q are the same. How much Mr Kamal needs to pay if he wants to buy a chair only? ©Praxis Publishing_Focus On Maths

195 Linear Equations and Inequalities in One Variable CHAPTER 5 10. (a) (i) Write two true relationships between –1.24 and –1 1 2 by using the symbol ‘.’ and ‘,’. (ii) Mr Jamil’s monthly salary consists of basic salary of $3500 and food allowance. The food allowance depends on the number of working days and the maximum amount of the food allowance is $500. Mr Jamil spent $b this month. Construct an inequality based on the above situation. (b) The length of a rectangle is 80 cm and the perimeter of the rectangle is smaller than 500 cm. To find the width of the rectangle, Chong wrote a linear inequality and represented the inequality on a number line as shown in the diagram below. 60 70 80 90 100 110 120 Is Chong’s answer correct? Give your justification. 11. (a) Represent each of the following inequalities on a number line. (i) x , 3 (ii) x . –6 (iii) x > –5 (iv) x < 8 (b) List all the integers x that satisfy the inequalities. 3x + 5 > –1 and 1 – 2x . –13 (c) A bus accommodates a few adults, 10 children and 12 senior citizens. The bus can accommodate 60 passengers. Find the possible number of adults in the bus? (d) A box contains (x + 3) blue marbles and red marbles are twice the number of blue marbles. If the total number of blue marbles and red marbles are 24, find the value of x. 12. (a) When a number is multiply with 3 and divide with 5, the result is 18. Find the number. (b) John buys 70 stamps with price of $39.00. Parts of the stamps are 50 cents stamp and the rest are 60 cents stamps. How many 50 cents stamp did he bought? (c) (x – 3) cm (2x + 1) cm The perimeter for the rectangle is 38 cm. Find x. 13. (a) Ganeswaran’s age is 7 years older than Hanif. Lee Yun’s age is 3 years younger than Hanif. If the sum of their age is 58 years, what is Ganeswaran’s age? (b) The price for 3 Mathematics books and 4 Science books are $96. If the price for one Mathematics book and Science book are $x and $15, find the value of x. ©Praxis Publishing_Focus On Maths

Applications of this chapter The concept of lines and angles are used in our daily life. Architects and engineers use angles to design houses, machines, buildings, roads, and bridges. Various types of angles are used in yoga positions, games fields, paintings and so on. We can see various lines in classrooms on the floor, doors, windows, and zebra crossing on the road. Identify the parallel line segments in the photo above. How can you check if they are parallel? 6LINES AND ANGLES 196 ©Praxis Publishing_Focus On Maths

Maths History Euclid of Alexandria was the most prominent mathematician best known for his expertise in geometry. His work named Elements is one of the most influential work in the world of mathematics. In the Elements, Euclid deduced the theorems (now known as Euclidean geometry) from a small set of axioms. He is also known for his works about spherical geometry, number theory and mathematical rigour. Relationship between Lines Measures of Angles Constructing Angles Dividing Line Lines Angles Lines and Angles • Point • Line • Plane • Angle • Cone • Curve • Curved surface • Right angle • Acute angle • Obtuse angle • Reflex angle • Complementary angle • Supplementary angle • Opposite angle Concept Map Learning Outcomes Learning Outcomes Key Terms • Understand the concept of parallel and perpendicular lines. • Recognise, compare and draw different types of angles. • Understand the concept of angles. • State the properties of the vertically opposite angles, complementary angles and supplementary angles. • Recognise angles associated with transversals. • Solve problems involving angles associated with transversals. • Understand and use the properties of angles associated with intersecting lines to solve problems. • Construct and explain steps of basic geometric constructions such as line segments, perpendicular lines, angles and angle bisectors. Relationship between Angles Supplementary and Complementary Angles Opposite Angle 197 ©Praxis Publishing_Focus On Maths

198 CHAPTER 6 Lines and Angles Flashback 1. Name this tool. What angles can you find on each patchwork? An art college hosts a festival of patchworks each August. Look at these patchworks. (a) What shapes do you see? (b) Which shapes have sides that are perpendicular? How do you know? 1 170 180 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 10 90 0 Critical Thinking 2. What are the sizes of the angles in a (a) rectangle? (b) square? 3. Give two examples that have parallel lines. ©Praxis Publishing_Focus On Maths

199 Lines and Angles CHAPTER 6 6.1 Relationship between Lines A Points, lines and planes Points, lines and planes are undefined items that starting the topic of geometry. The most fundamental object in geometry is a point. A point is usually represented by a dot and frequently named by a capital letter. However, a line is a connection between two or more points. A plane is a flat, two dimensional surface that extends infinitely far. It can be visualised as a large sheet of paper. A plane also can be represented in drawings of four sided figure. For example, Plane P Plane Q Objective : To investigate the features of points, lines, rays and line segments. Instruction : Do this activity in groups of four. 1. Open the file Points, Lines, Rays and Segments using GeoGebra. 2. Use the following tools to create each of geometric objects. 3. Observe and discuss with your group members what you see from each object. Tool Object Observations from the object Point Point Segment Line Segment Ray Ray Line Line 4. What are the differences between lines, segments and rays? 5. Points and lines lie on a flat surface. What is this flat surface? 6. Discuss with your group members and present your findings in class. 1 Scan or click the above QR code to download this activity file. GeoGebra ©Praxis Publishing_Focus On Maths

200 CHAPTER 6 Lines and Angles The objects that we have explored in Activity 1 can be summarised in the following table. Description of objects Notation Point • has position • has no size Point A Point B Line • has an infinite number of points • has no width • can be straight or curved Line I Curved line L Line segment • is a part of a straight line between and including two end points • has length Line segment AB A B Ray • is a part of a line with one end point • has one end indicated by an arrowhead to show that a ray can be extended in that direction indefinitely Ray AB or ray L A B L Ray MN N M Point of intersection • is the point where two lines meet A is a point of intersection. A Plane • is a flat surface • has no thickness Points A, B and C lie on the same plane. A B C A line refers to a straight line while a curve refers to a curved line. When a surface is not flat, it is called a curved surface. B Parallel line, transversal and perpendicular line (I) Parallel line Parallel lines are lines that will never meet (intersect) no matter how long the lines are extended. The symbol ‘//’ is used to represent ‘is parallel to’. ©Praxis Publishing_Focus On Maths

201 Lines and Angles CHAPTER 6 For example, A C B D Z Y W X AB // CD WX // ZY and WZ // XY Single arrowhead or double arrowhead can be used to show that the lines are parallel. A set square and a ruler can be used to determine whether two lines are parallel. For example, in the diagram, PQ is parallel to RS because the edge of the set square coincides with PQ and RS as it slides along the ruler. (II) Transversal A transversal is a line which intersects two or more straight lines. For example, B A (III) Perpendicular line Perpendicular lines are two lines that are at right angles to each other. The symbol ‘ ’ is used to represent ‘is perpendicular to’. A protractor or a set square can be used to determine whether a line is perpendicular to another line. For example, A D C B P Q R AB CD PQ QR P Q R S Parallel lines are lines with the same direction. They remain the same distance apart and never meet. A perpendicular line is a straight line that meets or intersects with another straight line at a right angle, 90°. AB is a transversal EXAMPLE 1 For the diagram given, determine whether (a) JK is parallel to LM, (b) PQ is perpendicular to LM. K P Q M J L ©Praxis Publishing_Focus On Maths

202 CHAPTER 6 Lines and Angles Solution: (a) J L K M (b) M L P Q 170 180 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 10 9 0 0 JK is not parallel to LM. PQ is perpendicular to LM. EXAMPLE 2 In the diagram, AOE and BOF are straight lines. Determine the lines that are perpendicular to each other. Solution: OD intersects BOF at 90°. Therefore, OD and BOF are perpendicular lines. A B C D E F G O Practice 6.1 Basic Intermediate Advanced 1 Using a set square and a ruler, determine whether each of the following pairs of straight lines are parallel. (a) (b) B For each of the following diagrams, use a protractor or a set square to determine the lines which are perpendicular to each other. (a) P Q R S U W T V (b) P L J K M QR C Based on the diagram, determine the lines that are perpendicular to each other. P Q R S • The distance between two parallel lines is always the same. • The distance between two parallel lines is the perpendicular distance from any point on one of the lines to the other line. ©Praxis Publishing_Focus On Maths

203 Lines and Angles CHAPTER 6 6.2 Dividing Line Segment into Multiple Parts A Constructing line segments (I) Constructing line segments with given length A straight edge is a construction instrument with straight edges. Geometrical constructions are precise drawings of geometrical figures which satisfy certain conditions, using only a straight edge and a pair of compasses. For example, when an architect draws a house plan, he is performing geometrical constructions. A line segment is part of a straight line between two points with a finite length. For example, A B 3 cm AB is a line segment of length 3 cm. The length of a line segment can be estimated by using another object of a known length. For example, if the length of a key is 3 cm, then the length of line AB can be estimated to be approximately 11 cm. The length of a line segment can be measured more accurately by using a ruler. Recall that a rhombus has all equal sides and equal opposite angles. Each diagonal divides the rhombus into 2 congruent isosceles triangles. How do you know the triangles are isoceles? How do you know the triangles are congruent? Discuss with your friends and investigate ways to cut line segments into 2 equal parts. INTERACTIVE ZONE 3 cm 3 cm 3 cm 3 cm ≈ 2 cm A B EXAMPLE 3 Construct a line segment MN of length 6 cm. Solution: Step 1 Draw a straight line using a ruler. Mark the point M on the line as shown. M Constructing line segments using a pair of compasses and a ruler can reduce the error when marking the measurements on the line segments. ©Praxis Publishing_Focus On Maths

204 CHAPTER 6 Lines and Angles Step 2 Open the compasses and measure on a ruler so that the opening is exactly 6 cm wide. Step 3 Place the point of the compasses at M. Draw a small arc which intersects the straight line. Mark the intersection point as N. A line segment MN of length 6 cm is constructed. M N Therefore, MN is a line segment with a length of 6 cm. M 6 cm N fi ff 0cm1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Do not erase the construction lines when doing a construction. (II) Constructing line segments with different ratios A given line segment can be divided internally in a given ratio. For example, P 12 cm Q A line segment of length 12 cm can be divided in the ratio 2 : 1. PQ is the line segment of length 12 cm and R divides the line in the ratio 2 : 1. To divide the line in the ratio 2 : 1, let RQ = x and PR = 2x, where PR + RQ = 2x + x 12 = 3x x = 4 So, PR = 8 cm and RQ = 4 cm. Mark point R 8 cm away from point P. P 8 cm R 4 cm Q EXAMPLE 4 Construct a line segment AB in the ratio 3 : 2 if the given length of AB is 20 cm. ©Praxis Publishing_Focus On Maths

205 Lines and Angles CHAPTER 6 Solution: Step 1 Draw a straight line of 20 cm using a ruler. Mark points A and B on the line as shown. A 20 cm B Step 2 Let C be the point that divides line AB in the ratio 3 : 2. To divide line in the ratio 3 : 2, let AC = 3x and CB = 2x, where AC + CB = 3x + 2x 20 = 5x x = 4 cm Thus, AC = 3(4) = 12 cm and CB = 2(4) = 8 cm. Open the compasses and measure on a ruler so that the opening is exactly 12 cm wide. cm1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Step 3 Place the point of the compasses at point A. Draw a small arc which intersects the straight line. Mark the intersection point as C. A line segment AC of length 12 cm is constructed. A C B Therefore, line CB is 8 cm. A 12 cm 8 cm C B ©Praxis Publishing_Focus On Maths

206 CHAPTER 6 Lines and Angles B Constructing perpendicular lines When a line makes an angle of 90° with another line, it is known as a perpendicular to the line. For example, C D A B C D A B CD is the perpendicular of AB or AB is the perpendicular of CD. When a line divides another line into two equal parts, it is known as a bisector of the line. For example, C B A D CD is a bisector of AB. When a line bisects another line at 90°, it is known as the perpendicular bisector of the line. For example, C D A B CD is the perpendicular bisector of AB. (a) Identify perpendicular line segments in these photos. (b) How can you check if they are perpendicular? 2 Describe 3 different methods you can use to draw a line perpendicular to a given line segment. Which method is the most accurate? Explain your choice. INTERACTIVE ZONE ©Praxis Publishing_Focus On Maths

207 Lines and Angles CHAPTER 6 EXAMPLE 5 Construct the perpendicular bisector of the line MN. Solution: Step 1 Adjust the compasses so that the width of the opening is slightly more than half the length of MN. Place the point of the compasses at M and draw two small arcs, one above the line MN and the other below the line MN. Step 2 Without altering the width of the opening, place the point of the compasses at point N and draw another two arcs intersecting the arcs marked in Step 1. Step 3 Join the two intersections to form a straight line. The line is the perpendicular bisector of MN. M N M N M N M N A perpendicular bisector is also known as a mediator. The width of the opening of the compasses in Step 1 must be more than half the length of MN. Otherwise the arcs will not intersect in Step 2. In the construction of the perpendicular bisector of MN, we are actually constructing a rhombus with MN as one of the diagonals. M N EXAMPLE 6 Construct the perpendicular to the line MN passing through point P. Solution: Step 1 Place the point of the compasses at P. Draw two arcs that intersect the line MN at two points which are equidistant from P. Label the two intersections as A and B respectively. Step 2 Place the point of the compasses at A and adjust the compasses so that the width of the opening is slightly greater than the length of AP, then draw a small arc as shown. M N P M N A P B M N A P B ©Praxis Publishing_Focus On Maths

208 CHAPTER 6 Lines and Angles Step 3 Without altering the width of the opening, place the point of the compasses at B and draw another arc intersecting the arc marked in Step 2. Step 4 Join the intersection to P. The line is perpendicular to MN. M N A P B M N A P B EXAMPLE 7 Construct a line passing through A which is parallel to the line PQ. P Q A C Constructing parallel lines Parallel lines are lines that never meet even when extended infinitely. (a) Identify parallel line segments in these photos. (b) How can you check if they are parallel? 3 Describe 3 different methods you can use to draw a line segment parallel to a given line segment. Which method is the most accurate? Explain your choice. INTERACTIVE ZONE ©Praxis Publishing_Focus On Maths

209 Lines and Angles CHAPTER 6 We can also use a set square and a ruler to construct parallel lines. P Q A Step 1 Adjust the set square so that one edge of the set square is aligned with the line PQ. Step 2 Align the ruler on the other edge of the set square. Step 3 Without moving the ruler, slide the set square along the ruler. Step 4 Draw the line passing through A. Solution: Step 1 Place the point of the compasses at P and adjust compasses so that the width of the opening is equal to the distance of A to P. Then, place the point of the compasses at Q and draw an arc as shown. Step 2 Adjust the compasses so that the width of the opening is equal to the exact length of PQ. Then, place the point of the compasses at A and draw a small arc that intersects the other arc as shown. Step 3 Join point A to the intersection. The line is parallel to the line PQ. P Q A P Q A P Q A Practice 6.2 Basic Intermediate Advanced 1 Estimate the length of each of the following line segments. Explain the methods used to make the estimation. (a) A B (b) P Q B Measure the length of each of the following line segments. (a) K L (b) M N C Determine whether line segments PQ and RS in each of the following are congruent. (a) (b) 3.5 cm 3.5 cm P Q R S P Q 4.3 cm 4.2 cm S R (c) P Q S R 32 cm 32 cm D Construct each of the following line segments. (a) PQ = 6.8 cm (b) AB = 8.4 cm (c) R 5.9 cm S ©Praxis Publishing_Focus On Maths

210 CHAPTER 6 Lines and Angles E Construct each of the following line segments. (a) AB = 5 cm (b) 5.8 cm (c) 6.9 cm (d) 7.4 cm F Name the following diagrams that has parallel lines. (a) (b) (c) (d) G Copy and construct the line that passes through M and perpendicular to PQ. (a) P Q M (b) P Q M H For each of the following, copy and construct the line that passes through M and parallel to PQ. (a) P Q M (b) P Q M I U T S R Q P Determine one pair of parallel lines in the diagram. 6.3 Angles A Denoting and labelling angles An angle is a measurement of a rotation. An angle is formed when two straight lines meet at a point. • The point is called the vertex. • The two straight lines are called arms. There are three ways of denoting and labelling an angle. For example, Diagram Three capital letters One capital letter One small letter x P Q R /PQR or PQ ^ R /RQP or RQ ^ P /Q or ^ Q x The symbols ‘/’ and ‘^’ are used to denote angles. When labelling an angle with three capital letters, the letter representing the vertex of the angle must be written in the middle. Angle Arm Vertex Arm ©Praxis Publishing_Focus On Maths

211 Lines and Angles CHAPTER 6 B Measuring angles Angles are measured in units called degrees (°). A protractor is used to measure an angle. (a) The inner scale measures angles in an anticlockwise direction from 0° to 180°. (b) The outer scale measures angles in a clockwise direction from 0° to 180°. Steps to measure a given angle: (a) /PQR (angle less than 180°) 1 Place the centre of the protractor on vertex Q. 2 Align the arm PQ with the baseline of the protractor. 3 Start measuring from 0°, that is the outer scale. 4 Read from the outer scale to obtain the value of /PQR. Note: Make sure the arms PQ and QR are extended. Therefore, /PQR = 115°. (b) /JKL (angle more than 180°) 1 Place the centre of the protractor on vertex K. 2 Align the arm JK with the baseline of the protractor. 3 Start measuring from 0°, that is the inner scale. 4 Read the value from the inner scale, that is 120º. Therefore, /JKL = 360° – 120° = 240° 170 180 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 10 9 0 0 Outer scale Baseline Centre Inner scale K L J 120° 240° 170 180 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 10 9 0 P 0 R Q fiff ffl ffi 170 180 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 120 70 130 60 50 140 40 150 30 160 20 170 10 180 0 10 9 0 0 L K J ffl ff fi ffi Maths LINK Science It takes about 365 days for the Earth to make one complete revolution around the Sun. The number of degrees in a complete turn is 360º. So, the Earth travels about 1º around the Sun each day. or /JKL = 180° + 60° = 240° K L J 60° 180° ©Praxis Publishing_Focus On Maths

212 CHAPTER 6 Lines and Angles C Drawing angles A protractor can also be used to draw angles. Steps to draw an angle of a given size: (I) /ABC = 84° 1 Draw a horizontal line and mark two points, A and B on it. 2 Use B as a vertex, place the protractor so that its centre is on the vertex B and align the arm AB with the baseline of the protractor. 3 Find 84° on the outer scale and mark a point C. 4 Remove the protractor. Join marked point C to vertex B with a straight line. Label /ABC as 84°. (II) /STU = 210° 1 T S 4 210° U T S 30° 180° U T S 150° 210° U T S A B A B C 84° 210° can be either: (a) 210° = 180° + 30° Therefore, find 30° on the inner scale and mark point U. or (b) 210° = 360° – 150° Therefore, find 150° on the outer scale and mark point U. A B C 170 180 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 10 9 0 0 fi ff 170 180 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 120 70 130 60 50 140 40 150 30 160 20 170 10 180 0 10 9 0 0 fi ff T S ©Praxis Publishing_Focus On Maths

213 Lines and Angles CHAPTER 6 D Type of angles (I)  Acute, right, obtuse and reflex angles Angles are classified according to their sizes. 54° 125° 230° Acute angle • More than 0° but less than 90° Right angle • Exactly 90° Obtuse angle • More than 90° but less than 180° Reflex angle • More than 180° but less than 360° (II) Angles on straight lines Sum of angles on a straight line = 180° In the diagrams below, PQR and ABC are straight lines. P Q R p q A B C a b c p + q = 180° a + b + c = 180° Name 4 objects in your classroom that have: (a) an angle greater than 100º (b) an angle less than 60º Use a protactor to check your answers. team work EXAMPLE 8 In the diagram, PQR is a straight line. Find the value of x. P Q 34º x 78º R Solution: x + 78° + 34° = 180° x + 112° = 180° x = 180° – 112° = 68° The sum of angles on a straight line is 180°. ©Praxis Publishing_Focus On Maths

214 CHAPTER 6 Lines and Angles (III) One whole turn of a rotation O P Q P O R O P In the diagram, arm OP rotates at O to OQ, then to OR, in an anticlockwise direction until it is back to its original position, OP. Arm OP has gone through one whole turn of a rotation. One whole turn of a rotation is 360°. Therefore, x + y + z = 360°. Angles x, y and z are angles at point O. Sum of angles at a point = 360° EXAMPLE 9 Find the value of y in the diagram. Solution: y + 2y + 30° + 123° + 90° = 360° 3y + 243° = 360° 3y = 117° y = 39° 2y y 30° 123° The sum of angles at a point is 360°. Practice 6.3 Basic Intermediate Advanced 1 In the diagram, PRS is a straight line. Copy the diagram and mark separately for each of the following angles. P Q R ST (a) /Q (b) /QRS (c) Q ^ RT (d) P ^ RT B For each of the following diagrams, name the marked angles using three capital letters. (a) x y A D B C (b) a b P O Q R S (c) A B C D E r p q s (d) K J G E H F x y w z P R Q O x y z ©Praxis Publishing_Focus On Maths

215 Lines and Angles CHAPTER 6 C Measure each of the following angles using a protractor. (a) (b) (c) p (d) p D Draw and label each of the following angles using a protractor. (a) 50° (b) 88° (c) 112° (d) 156° (e) /P = 270° (f) /A = 192° (g) /KLM = 307° (h) /EFG = 328° E Classify each of the marked angles in the diagram below as acute, right, obtuse or reflex angle. s r q p u t F In each of the following diagrams, POQ is a straight line. Find the value of x. (a) P O Q x 42° (b) 38° 126° x P O Q (c) 78° x P O Q (d) 2x 6x 28° P QO G Find the value of y in each of the following diagrams. (a) 52° y (b) 64° 150° y (c) (d) 42° 4y y y y 58° 136° 72° 6.4 Relationship Between Angles A Vertically opposite angles Two lines intersect if they meet (cross) each other at a point. The point is called the point of intersection. For example, P S R Q T A C B D Point of intersection When two straight lines intersect, the opposite angles are equal. These angles are called vertically opposite angles. For example, a b a b a and b are vertically opposite angles. Therefore, a = b. ©Praxis Publishing_Focus On Maths

216 CHAPTER 6 Lines and Angles EXAMPLE 10 In the diagram, POQ and ROS are straight lines. Find the values of x and y. Solution: x + 40° = 120° x = 120° – 40° = 80° y + 120° = 180° y = 180° – 120° = 60° P S O R Q y x 120° 40° Vertically opposite angles The sum of the angles on a straight line is 180°. B Complementary and supplementary angles Objective : To investigate the properties of complementary angles, supplementary angle and conjugate angles. Instruction : Do this activity in groups of four. 1. Open the file Complementary Angles, Supplementary Angle and Conjugate Angles using GeoGebra. 2. Tick the checkbox ‘Exploration 1’ and calculate the sum of the two angles displayed. Observe and record the results. 3. Click and drag the black points on the screen displayed and repeat Step 2. Discuss with your group members and state a conclusion. 4. Observe and state another pair of corresponding angles. 5. Repeat Steps 2 and 3 for ‘Exploration 2’ and ‘Exploration 3’. 6. Present your findings in class. 2 Scan or click the above QR code to download this activity file. GeoGebra ©Praxis Publishing_Focus On Maths

217 Lines and Angles CHAPTER 6 From the findings in Activity 2, it is found that (i) in Exploration 1, the sum of two angles is always 90º. These angles knowns as complementary angles. (ii) in Exploration 2, the sum of two angles is always 180º. These angles knowns as supplementary angles. (iii) in Exploration 3, the sum of two angles is always 360º. These angles known as conjugate angles. If two angles add up to 90°, these two angles are called complementary angles. For example, 25° 65° P Q R S 30° P S Q Q 60° S R /PQS is the complement of /RQS and vice versa. (/PQS + /RQS = 90°) If two angles add up to 180°, these two angles are called supplementary angles. For example, 45° 135° B C D A 155° A B C 25° C A D /ACB is the supplement of /ACD and vice versa. (/ACB + /ACD = 180°) EXAMPLE 11 P Q R T S In the diagram, PQR is a straight line. Find the angle which is (a) complementary to /PQT, (b) supplementary to /PQT. Solution: (a) /SQT is complementary to /PQT. (b) /RQT is supplementary to /PQT. ©Praxis Publishing_Focus On Maths

218 CHAPTER 6 Lines and Angles The sum of two adjacent angles on a straight line is 180°. For example, p and q are adjacent angles on a straight line ABC. Therefore, p + q = 180°. p q A B C EXAMPLE 12 x P Q R 126° In the diagram, PQR is a straight line. Find the value of x. Solution: 126° + x = 180° x = 180° – 126° = 54° D Angles associated with transversals and parallel lines (I) Transversals A transversal is a line which crosses two or more lines at different points. For example, AB is a transversal. B A C Adjacent angles Adjacent angles are any two angles which are next to each other. For example, A O B C D (a) Since /AOC is next to /COD, /AOC and /COD are adjacent angles. (b) Since /COD is also next to /BOD, /COD and /BOD are also adjacent angles. ©Praxis Publishing_Focus On Maths

219 Lines and Angles CHAPTER 6 EXAMPLE 13 Find the values of x and y in the diagram. Solution: x = 110° x y 100° 110° x 110° The pair of angles which lie between two parallel lines and are at opposite sides of a transversal are known as alternate angles. They are equal in size. (II) Corresponding, alternate and interior angles A transversal forms 8 angles with two parallel lines. For example, a d b c p s q r The pair of angles lying at a similar positions on two parallel lines with respect to a transversal are known as corresponding angles. They are equal in size. For example, b a a b a and b are corresponding angles. Therefore, a = b. Corresponding angles can be identified by looking for the pattern 'F'. 100° + y = 180° y = 180° – 100° = 80° y 100° 100° ©Praxis Publishing_Focus On Maths

220 CHAPTER 6 Lines and Angles The pair of angles which lie between two parallel lines and are at the same side of a transversal are known as interior angles. They add up to 180° (supplementary). For example, u v u v u and v are interior angles. Therefore, u + v = 180°. EXAMPLE 15 Find the value of x in the diagram. Solution: x + 85° = 180° x = 180° – 85° = 95° x 85° x 85° Solution: x = 85° y = 140° x 85° y 140° EXAMPLE 14 Find the values of x and y in the diagram. Alternate angles can be identified by looking for the pattern 'Z'. x y 140° 85° For example, p q p q p and q are alternate angles. Therefore, p = q. Interior angles can be identified by looking for the pattern 'C'. ©Praxis Publishing_Focus On Maths

221 Lines and Angles CHAPTER 6 Objective : To investigate the properties of angles formed in two parallel lines and a transversal. Instruction : Do this activity in groups of four. 1. Open the file Geometric Angles in Parallel Lines using GeoGebra. 2. Drag the points on the screen displayed. Identify the parallel lines and the transversal. 3. Tick the checkboxes ‘Show corresponding angles 1’ and ‘Show corresponding angles 2’. 4. Observe and explore the properties of corresponding angles. 5. Now click the ‘Show size of angles’ button and drag the points on the screen displayed if necessary. 6. Observe and state another pair of corresponding angles. 7. Discuss among your group members and explain the properties of corresponding angles. 8. Tick the checkboxes ‘Show alternate angles’ and ‘Show interior angles’ and continue exploring alternate and interior angles. 9. State another pair of alternate angles another pair of interior angles. 10. Discuss with your group members and explain the properties of alternate angles and interior angles. 11. Present your findings in class. 3 From the findings in Activity 3, it is found that when a transversal line intersects with parallel lines, (i) the corresponding angles are equal, (ii) the alternate angles are equal, and (iii) the sum of the interior angles is 180º. (III) Determining parallel lines Two lines are parallel if • corresponding angles are equal, • alternate angles are equal, • the sum of interior angles is 180°. For example, If a = c, b = d or b + c = 180°, PQR and STU are parallel. a b c d P S Q T R U V W Scan or click the above QR code to download this activity file. GeoGebra ©Praxis Publishing_Focus On Maths

222 CHAPTER 6 Lines and Angles a b c d b + c = d EXAMPLE 16 Determine whether ABC and DEF are parallel in each of the following diagrams. (a) G B 115º 105º E A D H F C (b) G B 75º 105º E A D H C F Solution: (a) /GBC and /BEF are corresponding angles. Since /GBC (115°) and /BEF (105°) are not equal, ABC and DEF are not parallel. (b) /CBE and /FEB are interior angles. /CBE + /FEB = 75° + 105° = 180° Since the sum of /CBE and /FEB is 180°, ABC and DEF are parallel. (IV) Problem -solving EXAMPLE 17 P R Q T x S 130° In the diagram, PQR is a straight line and PQ = QT. Find the value of x. ©Praxis Publishing_Focus On Maths

223 Lines and Angles CHAPTER 6 EXAMPLE 18 A B C E x D 100° 150° Find the value of x in the diagram. Solution: ∠BCD + 100° = 180° ∠BCD = 180° – 100° = 80° Therefore, x + 80° + 150° = 360° x = 360° – 80° – 150° = 130° Interior angles Solution: ∠SRQ + ∠QPT = 180° 130° + ∠QPT = 180° ∠QPT = 180° – 130° = 50° Since PQ = QT, ∠PTQ = ∠QPT = 50° Therefore, x + 50° + 50° = 180° x = 180° – 50° – 50° = 80° Interior angles Practice 6.4 Basic Intermediate Advanced 1 Identify the transversal in each of the following diagrams. (a) T Q S P R U (b) G H C A D B ©Praxis Publishing_Focus On Maths

224 CHAPTER 6 Lines and Angles E For each of the following diagrams, determine whether PQ and RS are parallel. (a) 70° 110° P Q R S (b) 120° 62° P R Q S (c) 73° 107° P R Q S (d) 96° 85° P R Q S F Find the value of x in each of the following diagrams. (a) 108° x 30° (b) 40° 65° 50° x (c) 73° 32° x (d) 65° 80° x 7 x P S T V U N M RQ 100° 130° In the diagram, PQRS, TUV and MQUN are straight lines. Find the value of x. B Find the values of x and y in each of the following diagrams. (a) x y 85° (b) x 100° y 75° (c) x y 80° 70° (d) x y 110° 60° C Find the value of x in each of the following diagrams. (a) 120° x (b) 72° x (c) 65° 100° x (d) 50° 105° x D Find the values of x and y in each of the following diagrams. (a) y x 65° 115° (b) x y 95° 125° (c) 98° 80° x y (d) 85° 75° y x ©Praxis Publishing_Focus On Maths

225 Lines and Angles CHAPTER 6 8 x AF CDE B 70° In the diagram, EDCB is a straight line. Find the value of x. 9 2x 4y 26° P R O Q S T ∠QOT is a right angle. Find the the value of y – x.  120° 130° 38° x y y y y y y Find the value of x + y in the diagram. 6.5 Constructing Special Angles A Constructing 60° and 90° angles (I) Constructing angles and angle bisectors EXAMPLE 19 Construct /KMN = 60°. Solution: Step 1 Draw the line MN. Step B Place the point of the compasses at M and draw a long arc that intersects the line as shown. Mark the intersection as A. Step C Without altering the width of the opening, place the point of the compasses at A and draw a small arc that intersects the long arc marked in Step 2. Label the intersection as K. Step D Join K to M. /KMN is 60°. In the construction of /KMN = 60°, we are actually constructing an equilateral triangle. Each interior angle of an equilateral triangle is 60°. M A K 60° 60° 60° M N M A N A M N K A M N K 60° In the diagram, POR is a straight line and ©Praxis Publishing_Focus On Maths

226 CHAPTER 6 Lines and Angles B Constructing 90° angles EXAMPLE 20 Construct the perpendicular to the line MN passing through Q. Solution: Step 1 Place the point of the compasses at Q. Draw two arcs that intersect the line MN at two points which are equidistant from Q. Label the two intersections as A and B respectively. Step B Place the point of the compasses at A. Adjust the compasses so that the width of the opening is more than half the length of AB, then draw a small arc below the line MN as shown. Step C Without altering the width of the opening, place the point of the compasses at B and draw another arc, intersecting the arc marked in Step 2. Step D Join the intersection to Q. The line is perpendicular to the line MN. M N Q M N A B Q A B M N Q A B M N Q A B M N Q ©Praxis Publishing_Focus On Maths

227 Lines and Angles CHAPTER 6 Solution: Step 1 Draw the line QR. Step B Place the point of the compasses at Q and draw a long arc that intersects the line as shown. Mark the intersection as A. Step C Without altering the width of the opening, place the point of the compasses at A and draw a small arc that intersects the long arc marked in Step 2. Label the intersection as B. Step D Without altering the width of the opening, place the point of the compasses at B and draw another arc that intersects the long arc as shown. Mark the intersection as P. Step E Join P to Q. /PQR is 120°. Q R Q R A Q R A B Q R A B P Q R A B P 120° In the construction of /PQR = 120°, we are actually constructing two side-to-side adjoining equilateral triangles. /PQR is the sum of two adjacent angles of 60°. Q A BP 60° 60° D Constructing angle bisectors The bisector of an angle is the line that divides the angle into two equal angles. For example, A P O B x x If OP is the bisector of /AOB, then /AOP = /POB. C Constructing 120° angles EXAMPLE 21 Construct /PQR = 120°. ©Praxis Publishing_Focus On Maths

228 CHAPTER 6 Lines and Angles EXAMPLE 22 Construct the bisector of /POQ. Solution: Step 1 Place the point of the compasses at the vertex of the angle, that is, point O. Adjust the opening of the compasses to about two-thirds the length of the shorter arm and draw an arc crossing both arms of /POQ. Mark the intersections as A and B respectively. Step B Place the point of the compasses at A and draw a small arc. Without altering the width of the opening, repeat this for point B so that both arcs intersect. Mark the intersection as M. Step C Join O to M. The line OM is the bisector of /POQ. O P Q O P A B Q A B O P Q M A B O P Q M Practice 6.5 Basic Intermediate Advanced 1 Copy each of the following lines and construct (i) /PQR = 60°, (ii) /SQR = 120°. (a) (b) Q R B Construct the following angles using only a pair of compasses and a ruler. (a) /AOB = 90° (b) /POQ = 30° Q R C Copy and construct the line that passes through M and perpendicular to PQ. (a) (b) P Q M D Copy and construct the bisector of each of the following angles. (a) (b) P Q M ©Praxis Publishing_Focus On Maths

229 Lines and Angles CHAPTER 6 E Copy the following diagram and construct the perpendicular bisector of the lines MN and NP. Mark the intersection of the two perpendicular bisectors as A. Measure /MAP. M N P F Copy and construct the perpendicular bisector of each of the following lines. (a) K M (b) P Q G Copy the following diagram and construct the line that passes through A and perpendicular to BC. Construct another line that passes through B and perpendicular to AB. Mark the intersection of the two perpendiculars as P. Measure /APB. B A C 6.6 Similarity A Identifying if given shapes are similar Two geometrical shapes are similar if (a) the corresponding angles are equal and (b) the corresponding sides are proportional. For example, (i) ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, ∠D = ∠S (ii) BC QR = 2 3 , CD RS = 2.4 3.6 = 2 3 , DA SP = 1.4 2.1 = 2 3 , AB PQ = 1.6 2.4 = 2 3 Therefore, quadrilaterals ABCD and PQRS are similar. If the corresponding angles of two triangles are equal, then the corresponding sides are proportional. For ∆ ABC and ∆ EFG, ∠A = ∠E, ∠B = ∠F and ∠C = ∠G. C G A B E F Therefore, AB EF = BC FG = AC EG. 2.4 cm 2 cm 3.6 cm 3 cm 2.4 cm 2.1 cm 1.6 cm 1.4 cm C S P Q R D A B ©Praxis Publishing_Focus On Maths

230 CHAPTER 6 Lines and Angles Use dot paper. (a) Draw two different (i) equilateral triangles, (ii) squares and (iii) regular hexagons. (b) Are all regular polygons of the same type similar? Justify your answer. (c) Are all circles similar? Justify your answer. Critical Thinking EXAMPLE 23 Determine if rectangles ABCD and EFGH as shown are similar. A D B C 1cm 1.6 cm H G E 3.2 cm F 2 cm Solution: ∠A = ∠E, ∠B = ∠F, ∠C = ∠G, ∠D = ∠H AB EF = 1.6 3.2 = 1 2 , BC FG = 1 2 , CD GH = 1.6 3.2 = 1 2 , AD EH = 1 2 Therefore, rectangles ABCD and EFGH are similar. EXAMPLE 24 The diagram shows two similar triangles, ABC and DEC. Find the value of (a) x, (b) y. Solution: (a) BC EC = AB DE ABC and DEC are similar. (b) CD CA = DE AB x 6 = 6.4 4.8 y 12 = 4.8 6.4 x = 6 × 6.4 4.8 y = 12 × 4.8 6.4 = 8 cm = 9 cm B Calculating of the lengths of unknown sides of two similar shapes B A E D C x y 12 cm 6 cm 4.8 cm 6.4 cm The symbol for "is similar to" is ; . When ∆ABC and ∆DEC are similar, we can write it as ∆ABC ; ∆DEC. We can then obtain the ratio of the corresponding sides easily, that is, AB DE = BC EC = AC DC . ©Praxis Publishing_Focus On Maths

231 Lines and Angles CHAPTER 6 Practice 6.6 Basic Intermediate Advanced 1 Each pair of the triangles are similar. Write down all pairs of matching sides. (a) 45° 45 70° 70° 65° 65° A B L M K C ° (b) 30° 110° 40° 30° 110° 40° E F G R Q P (c) 33° 110° 37° 110° °33 °37 M N O U S T B In the diagrams below, each pair of the polygons are similar. Find the values of x, y and z. Give your answer in 3 significant figures where necessary. (a) 14 cm 13.5 cm x y 24 cm 9 cm (b) x 8 cm 9 cm y cm 24 cm 42 cm (c) x y z 8 m 4 m 7 m 3.8 m 10 m C For each pair of the similar triangles below, find the value of x. Give your answer in 3 significant figures where necessary. (a) A B C D E x 6 m 4 m 6 m (b) Q P R T S x 7 cm 3 cm 6 cm (c) x 70° 20° 14 mm 4 mm 2 mm 10 mm D In the given diagrams, find the values of x and y. (a) A C E B D x y 7 cm 5 cm 12 cm 8 cm (b) x y F E D A B C G 6 cm 3 cm 4 cm 10 cm 5 cm 9 cm ©Praxis Publishing_Focus On Maths

232 CHAPTER 6 Lines and Angles Angles • Measurement of a rotation. • Measured in degrees (°) with a protractor. • Sum of angles on a straight line is 180°. p q r p + q + r = 180° • Sum of angles at a point is 360°. y x z x + y + z = 360° Perpendicular lines Transversal A transversal is a line which crosses a set of parallel lines. Transversal Intersecting lines Point of intersection Right angle • 1 4 turn of a rotation, 90° Reflex angle Between 180° and 360° Parallel lines Acute angle Less than 90° Obtuse angle Between 90° and 180° Vertically opposite angles b d ca a = c and b = d Supplementary angles x y x + y = 180° Complementary angles p q p + q = 90° Adjacent angles a b • a and b are two adjacent angles. • Sum of adjacent angles on a straight line is 180°. a + b = 180° Corresponding angles x y x = y Alternate angles x y x = y Interior angles x y x + y = 180° Integers Lines and Angles Integers Lines Summary Summary Summary Similar shapes • Two shapes are similar if (a) their corresponding angles are equal and (b) their corresponding sides are proportional For example, ABCD and STUV are similar if ∠A = ∠S, ∠B = ∠T, ∠C = ∠U, ∠D = ∠V and ST AB = TU BC = UV CD = SV AD • If the corresponding angles of two triangles are equal, then the corresponding sides are proportional. Hence, two triangles are similar if the corresponding angles are equal or the corresponding sides are proportional. T U VS C B A D ©Praxis Publishing_Focus On Maths

233 Lines and Angles CHAPTER 6 1. 2x 3x 3x 100° P Q R In the diagram, PQR is a straight line. The value of x is A 10° C 60° B 40° D 80° 2. 25° 60° J K x O N L M In the diagram, KON is a straight line and ∠JOL is a right angle. Find the value of x. A 50° C 65° B 55° D 85° 3. x x x x x y y y P Q R In the diagram, PQR is a straight line. Calculate the value of x + y. A 72° C 108° B 96° D 144° 4. 150° 58° 34° x y In the diagram, the value of x + y is A 62° C 118° B 92° D 174° 5. 138° 48° S Q T P R Based on the diagram, which of the following statements is not true? A ∠PQS and ∠PQT are complementary angles. B ∠PQS and 138° are supplementary angles. C PQR is a straight line. D ∠RQT = 138° 6. The diagram shows the construction of angle A 30° C 90° B 60° D 120° 7. Which of the following angles cannot be constructed using only a ruler and a pair of compasses? A 15° B 22.5° C 110° D 135° 8. A B I II III IV The diagram shows the construction of 90°. Which of the following is the correct sequence of the construction steps? A I, II, III, IV B II, I, III, IV C II, III, I, IV D III, I, II, IV 6 Section A ©Praxis Publishing_Focus On Maths

234 CHAPTER 6 Lines and Angles 9. A D O B C 40° x 26° In the diagram, OB is the bisector of ∠AOD. The value of x is A 10° C 16° B 14° D 20° 10. y The diagram shows the bisection of an angle. Which of the following angles could be the angle y? A 30° C 60° B 45° D 75° Section B 1. Using a protractor, measure the following angles. (a) ∠PQR P R Q (b) ∠XYZ X Y Z 2. (a) S T P Q R In the diagram, name the obtuse angle. (b) The diagram shows a straight line PQ. Using a protractor and a ruler, draw ∠PQR = 134°. P Q 3. (a) In the diagram, x is equal to 1 5 of a whole turn. Find the value of x. x (b) Find the value of y in the diagram. y 230° 120° 4. (a) The diagram shows line KL drawn on a grid. On the same diagram, draw a straight line which is parallel to line KL. Label the line as RS. K L (b) x 42° P Q R In the diagram, PQ is perpendicular to QR. Find the value of x. ©Praxis Publishing_Focus On Maths

235 Lines and Angles CHAPTER 6 5. (a) A D B C E In the diagram, state all the intersecting lines. (b) x J L K 50° In the diagram, JKL is a straight line. Find the value of x. 6. (a) 72° 5° x P R Q T S In the diagram, PQR and SQT are straight lines. Find the value of x. (b) 3y 4x 120° N J M K L In the diagram, JNL and KNM are straight lines. Find the values of x and y. 7. (a) If x is the complement of y and y is the supplement of 125°, find the values of x and y. (b) k is the supplement of n. If n and 80° are vertically opposite angles, find the values of k and n. 8. (a) If p and q are two adjacent angles on a straight line and p is the complement of 25°, find the values of p and q. (b) x P V W R S U T Q 70° 30° In the diagram, PQR, VQW and STU are straight lines. Find the value of x. 9. (a) x P R S U T Q 60° 40° In the diagram, PQR and QTU are straight lines. Find the value of x. (b) x 65° Find the value of x in the diagram. 10. x 45° 130° 30° In the diagram, calculate the value of x. ©Praxis Publishing_Focus On Maths