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Get Ahead Mathematics Bilingual Teaching Guide for Book 7 Parveen Arif Ali Contents Unit I: Sets..................................................................................01 Unit II: Operations on Directed Numbers...................................07 Unit III: Rational Numbers..........................................................19 Unit IV: Square Roots.................................................................37 Unit V: Percentage.....................................................................43 Unit VI: Ratio and Proportion.....................................................47 Unit VII: Algebra.........................................................................57 Unit VIII: Geometry.....................................................................79 Unit IX: Information Handling.....................................................81 iii Introduction Get Ahead Mathematics is a series of eight books from levels one to eight. The accompanying Teacher’s Guides contain guidelines for the teachers. Answer keys to Books 6, 7 and 8 are included in the textbooks and should be referred to by the teachers. The teacher should devise means and ways of reaching out to the students so that they have a thorough knowledge of the subject without getting bored. The teachers must use their discretion in teaching a topic in such a way as they find appropriate, depending on the intelligence level as well as the academic standard of the class. Encourage the students to relate examples to real things. Don’t rush. Allow time to respond to questions and discuss particular concepts. Come well prepared to the class. Read the introduction to the topic to be taught in the pupils’ book. Prepare charts if necessary. Practice diagrams to be drawn on the blackboard. Collect material relevant to the topic. Prepare short questions, homework, tests and assignments. Before starting the lesson make a quick survey of the previous knowledge of the students, by asking them questions pertaining to the topic. Explain the concepts with worked examples on the board. The students should be encouraged to work independently, with useful suggestions from the teacher. Exercises at the end of each lesson should be divided between class work and homework.The lesson should conclude with a review of the concept that has been developed or with the work that has been discussed or accomplished. Blackboard work is an important aspect of teaching mathematics. However, too much time should not be spent on it as the students lose interest. Charts can also be used to explain some concepts, as visual material helps students make mental pictures which are learnt quickly and can be recalled instantly. Most of the work will be done in the exercise books. These should be carefully and neatly set out so that the processes can easily be seen. Careful setting out leads to better understanding. The above guidelines for teachers will enable them to teach effectively and develop in the students an interest in the subject. These suggestions can only supplement and support the professional judgement of the teacher. In no way can they serve as a substitute for it. It is hoped that your interest in the subject together with the features of the book will provide students with more zest to learn mathematics and excel in the subject. iv v Unit I: Sets Pages 1-5 Discuss the use of collective nouns such as a pack of cards, a team of players, etc in our day-to-day lives. Words such as pack, flock, team, group, etc are used to denote a collection of objects. Explain that the word ‘set’ is used in mathematics to describe a ‘collection of objects’. ‘A set is a well-defined collection of objects’. The phrase ‘well-defined’ means that a set must have some specific property so that we can easily identify whether an object belongs to the given set or not. A book, for example, does not belong to a tea set, and a ball does not belong to a set of playing cards. So we can say that a tea set and a set of playing cards are well-defined sets. Write a set of vowels of the English alphabet on the blackboard. Explain that it is a well-defined set because we know that specific letters are being referred to: i.e. a, e, i, o, u. Now consider the example of ‘a set of interesting books in the library’. It is not a set because the term interesting is not clearly defined by any fixed standards. A book may be interesting to one person but boring to another. Thus, this is not well defined and hence it is not a set. Explain that in a well-defined set, the objects are not repeated. Notation of a set We generally use capital letters A,B,C etc. to denote sets. 1 2 Elements of a set The objects of a set are called ‘members’ or ‘elements’ of a set. All the elements of a set are enclosed in brackets { } and separated by commas e.g. a set of vowels of the English alphabet; A = {a, e, i, o, u}. The elements of a set are said to ‘belong to’ the set. Write the symbol for ‘belongs to’ on the board. Write a set of even numbers on the board. B = {0, 2, 4, 6, 8}. Ask: Does 1 belong to this set? Explain that 1 is not an even number so it does not belong to the set of even numbers. Write the symbol for ‘does not belong to’. Null or empty set Ask the pupils if they can list the elements of the following set: ‘The set of cats with two tails’. Explain that the set does not have elements which can be listed. Such a set is known as an ‘empty set’ or a ‘null set’. An empty set is denoted by { } or the Greek letter ‘phi ’. Write the set, A = {0}, on the board. Ask: Is this an empty set? Explain that it is not an empty set because it contains the element ‘0’. Methods of writing sets Explain that sets are described or written in two ways. These are: 1. Tabular form In this form, all elements of the set are listed, e.g. A = {a, e, i, o, u} (a set of vowels). 2. Descriptive form In this form, the elements of the set are not listed, but a specific property satisfied by the elements of the set is given, 3 4 e.g. A = {the vowels of the English alphabet}. Write some standard sets on the board and explain their properties. Types of sets Sets can be differentiated into three types depending on the number of elements they have. 1. Finite sets A set that has a limited number of elements is called a ‘finite set ’, e.g. A = {1, 2, 3, 4}. Set A has 4 elements. 2. Infinite sets A set that has an unlimited number of elements is called an ‘infinite set’, e.g. A = {0, 2, 4, 6, 8,…}. This set is an infinite set because it contains endless even numbers. 3. Empty sets It is a set that has no members, as explained earlier. Union of two sets A union of two sets is a set which contains the elements of both sets. A and B are written as one set in which none of the elements are repeated e.g. A = {1, 2, 3}, B = {2, 3, 4}. The union set of A and B is {1, 2, 3, 4}. Intersection of sets The intersecton of two sets A and B is a set which contains the common elements of both the sets e.g. A = {1, 2, 3}, B = {2, 3, 4}. The intersection of A and B is {2, 3}. 5 6 Commutative property of union and intersection of sets For any two sets A and B, the union of A and B is equal to the union of B and A e.g. A = {1, 2}, B = {1, 3}. A union B = {1, 2, 3} and B union A = {1, 2, 3}. Both the sets are equal. For any two sets A and B, the intersection of A and B is equal to the intersection of B and A e.g. A = {1, 2, 3}, B = {2, 3}. A intersection B = {2, 3} and B intersection A = {2, 3}. Both the sets are equal. Difference of two sets Difference of sets is like subtraction. The elements of set B are subtracted from A and the elements left in A is the solution set e.g. A = {1, 2, 3, 4} and B = {2, 3, 4, 5}. The difference between A and B is A – B = {1}. The difference between B and A is B – A = {5}. Unit II: Operations on Directed Numbers Pages 6-12 The number line When we count or measure, we use real numbers. These numbers can be pictured as points on a line, called a ‘number line’. To construct a number line Choose a starting point on a line and label it ‘0’ (zero). This point is called the ‘origin’. The origin separates the line into two horizontal sides, the ‘positive side’ and the ‘negative side’. If the line is horizontal, the side to the right of the origin is taken to be the positive side and the side to the left as the negative side. Mark off equal units of distance on both sides of the origin. On the positive side, pair the end points with positive integers +1, +2, +3, +4,… and so on. 7 8 On the negative side, pair the end points with negative integers –1, –2, –3, –4,… and so on. The positive integers, the negative integers and zero make up the set of ‘integers’. The positive integers and zero are often called ‘whole numbers’. Any number that is either a positive number, a negative number or zero is called a ‘real number ’. When we see the pairing of points on a number line, the paired points are at the same distance from the origin but on the opposite sides of it. Each number in a pair such as +7 and –7 is called the ‘opposite of ’ the other number. Order of integers On a horizontal number line, the numbers increase from left to right and decrease from right to left. The ‘symbols of inequality’ are used to show the order of pairs of real numbers. < means ‘less than’ > means ‘greater than’ By studying a number line we can see that –5 is less than –2 and 0 is greater than –5. –5 < –2 and 0 > –5 +5 > –5 and 3 > 0 Absolute value of numbers In any pair of non-zero opposites, such as –5 and +5, one number is negative and the other is positive. The positive number of any pair of non-zero real numbers is called the ‘absolute value’ of each number in the pair. 9 –5 < –2 0 > –5 +5 > –5 3 > 0 10 The absolute value of a number ‘a’ is denoted by |a| e.g. |–5| = 5 and |+5| = 5. Notice that the absolute value of a real number ‘a’ is ‘a’ if ‘a’ is non-negative and also if ‘a’ is negative. The absolute value of 0 is defined as 0 itself. |0| = 0. Working with real numbers The rules used in adding and multiplying real numbers are based on several properties. These are: 1. Every pair of real numbers has one and only one sum that is a real number. 2. Every pair of real numbers has one and only one product that is a real number. 3. When you add two real numbers, you get the same sum no matter what order you use in adding them. 4. When you multiply two real numbers, you get the same product no matter what order you use in multiplying them. Addition of integers We already know how to add two positive numbers. We can use a horizontal number line to help us find the the sum of any two real numbers. e.g. To find the sum of –2 and –5, draw a number line. Starting at the origin move the pencil along the number line 2 units to the left. Then from that position move the pencil 5 units to the left. Moves to the left represent negative numbers. Together, the two moves amount to a move of 7 units to the left from the origin. Thus: –2 + (–5 ) = –7. The expression –(2 + 5) represents the opposite of the sum of 2 and 5. 11 12 Since 2+5=7 –(2 + 5) = –7 The expression –2 + (–5) represents the sum of the opposite of 2 and the opposite of 5. Using the number line we know that –2 + (–5) = –7. It follows that –(2 + 5) = –2 + (–5). Using the property of the opposite of a sum and the familiar addition facts for positive numbers, we can compute sums of any real numbers without thinking of a number line. e.g. To simplify Solution –8 + (–3) –8 + (–3) = –(8 + 3) = –11. Subtraction of integers Write the following examples of the subtraction of 2: 3–2=1 4–2=2 5 – 2 = 3 Now write examples of the addition of –2: 3 + (–2) = 1 4 + (–2) = 2 5 + (–2) = 3 Comparing the entries in the two columns shows that subtracting 2 gives the same result as adding the opposite of 2. This suggests the following definition of subtraction for all real numbers. The difference a – b is equal to the sum of a + (–b). That is to subtract b, add the opposite of b. 13 3 + (–2) = 1 4 + (–2) = 2 5 + (–2) = 3 3 + (–2) = 1 4 + (–2) = 2 5 + (–2) = 3 a b a b 14 e.g. To simplify 12 – 3 Solution 12 + (–3) 12 – 3 = 9. Multiplication of integers When we multiply any given real number by 1, the product is identical to the given number. e.g. 3 x 1 = 3 1x3=3 When we multiply any given real number by 0, the product is 0: e.g. 3 x 0 = 0 0x3=0 When we multiply any given real number by –1, the product is the opposite of that number. e.g. 3 x –1 = –3 Rules for multiplication of positive and negative numbers 1. The product of a positive and a negative number is a negative number. 2. The product of two positive or two negative numbers is a positive number. Distributive property of multiplication over addition For all real numbers a, b and c: ab + ac = a (b + c) and ba + ca = (b + c) a. e.g. If 4, 5 then 15 and 7 are any three numbers, 4 x 5 + 4 x 7 or = 4 (5 + 7) = 20 + 28 = 48 5 = = = x4+7x4 (5 + 7) 4 20 + 28 48 16 Verifying the distributive property of multiplication over addition e.g. To simplify Solution –5 x (8 + 6) –5 x (8 + 6) = (–5 x 8) + (–5 x 6) –5 x 14 = –40 + (–30) –70 = –70 Division of integers Before going on to division of integers, the terms ‘reciprocal ’ or ‘multiplicative inverse’ must be explained. Two numbers whose product is 1 are called reciprocals or multiplicative inverses of each other. e.g. 3 and 1/3 are reciprocals because 3 x 1/3 = 1. 0 has no reciprocal because the product of 0 and any real number is 0, not 1. The symbol for the reciprocal or multiplicative inverse of a nonzero real number a is ‘1/a’. The reciprocal of a product of non-zero real numbers is the product of the reciprocals of the numbers. That is, for all non-zero real numbers a and b: 1/ab = 1/a x 1/b To divide a by b we multiply a by the reciprocal of b The quotient is often represented as a fraction: e.g. a divided by b = a/b We can use the definition of division to replace any quotient by a product. 17 b a b a 18 e.g. 21/7 = 21 x 1/7 = 3 21/–7 = 21 (–1/7) = –3 –21/7 = –21 x 1/7 = –3 –21/–7 = (–21) (–1/7) = 3 Rules for dividing positive and negative integers 1. The quotient of two positive numbers or two negative numbers is positive. 2. The quotient of a positive number and a negative number is negative. Dividing by zero Dividing by zero would mean multiplying by the reciprocal of zero. But zero has no reciprocal. Therefore, division by zero has no meaning in the set of real numbers. Unit III: Rational Numbers Pages 13-23 Positive integers are called ‘natural numbers’ or ‘counting numbers’. We have learnt earlier that the sum of two natural numbers is always a natural number. Also, when a natural number is subtracted from another natural number, the result is not always a natural number. It could be a whole number or a negative integer. Now let us see what happens when a natural number is divided by another natural number. e.g. when 4 is divided by 2, the result is 2 but when 3 is divided by 4 we get 3/4 which does not belong to any of the number systems that we have already discussed. The number 3/4 is called a ‘rational number ’. 2/3, –5/6, 0, 4, –2, etc. are all rational numbers. So we can define a rational number as ‘any number that can be expressed in the form of a/b, where a and b are integers and b is not equal to 0 ’. 19 20 Positive integers, negative integers, zero and common fractions all belong to the system of rational numbers. Rational numbers can be expressed as a quotient of integers in a number of ways. e.g. 2 can be written as 2, 4/2, 8/4, –12/–6, etc. To determine which of any two rational numbers is greater, we can find a common denominator for the numbers and then compare them. The fraction with the greater numerator will be the greater number. e.g. To find out which is greater, 9/2 or 13/5? Solution: Finding the LCM of 2 and 5. It is 10. Changing the fractions with the new denominator 45/10 and 26/10. Therefore,45/10 is the greater rational number. Order of rational numbers There is a difference between rational numbers and integers in the sense that there is a higher and a lower integer for any given integer. However, a rational number does not have a next higher or previous lower number. a + 1/2 (b – a) is the formula for calculating the number in the middle of two said rational numbers such as a and b. Follow the example given in book. Representing rational numbers on the number line We divide the number line into as many parts as indicated by the denominator of the fraction. Each part then represents one part of the fraction. 21 22 e.g. To represent 2/5 on the number line we must divide each segment into 5 equal parts. Each part will then represent 1/5. 2/5 will be the second mark lying to the right of the zero. The density property of rational numbers Between every pair of different rational numbers there is another rational number. If a and b are rational numbers and a < b then: the number halfway from a to b is c = 1/2 (b – a) and the number one-third of the way from a to b is c = 1/3 (b – a). Operations on rational numbers Addition Adding rational numbers with a common denominator is easy. If a, b, and c are integers and c is not equal to 0 then a/c + b/c = a + b . c We can extend this rule to rational numbers with unequal denominators. e.g. 3/4 + 2/5 The LCM is 20. Renaming the fractions: 15/20 + 8/20 = 23/20. Note that the sum of a rational number is also a rational number. Adding three or more rational numbers: When we add three or more rational numbers, the order in which we add does not affect the result. 23 a/c + b/c = a + b c 24 e.g. Add 1/2, 2/3, 3/4. (1/2 + 2/3) + 3/4 The LCM is 12. Renaming the fractions: (6/12 + 8/12) + 9/12 = (14/12) + 9/12 = 23/12. By changing the order of the fractions: 1/2 + (2/3 + 3/4) = 6/12 + (8/12 + 9/12) = 6/12 + 17/12 = 23/12. We see that the result is the same in both cases. Additive inverse For any rational number a/b, there is a number –a/b such that a/b + (–a/b) = 0. We say that –a/b is the additive inverse of a/b. Similarly a/b is the additive inverse of –a/b. For integers we have seen that 5 – 2 = 5 + (–2). Here –2 is the additive inverse of 2. Subtraction Subtracting 2 from 5 is the same as adding the additive inverse of 2 with 5. As each rational number has an additive inverse, the concept of subtraction of integers can also be extended to rational numbers. The difference of rational numbers is also a rational number. 25 26 e.g. Subtract 1/2 from 3/4. Solution: 3/4 – 1/2 The LCM is 4. Renaming the fractions: 3/4 – 2/4 = 1/4. To subtract three rational numbers: e.g. 3/4, 1/3, 1/2 Solution: (3/4 – 1/3) – 1/2 The LCM is 12. Renaming the fractions: (9/12 – 4/12 ) – 6/12 = 5/12 – 6/12 = –1/12. Changing the order of the fractions: 3/4 – (1/3 – 1/2) Solution: 3/4 – (2/6 – 3/6) 3/4 – (–1/6) 3/4 + 1/6 LCM is 24 18/24 + 4/24 22/24 = 11/12. Note that by changing the order of subtraction of the numbers we do not get the same result. Multiplication The product of two rational numbers is also a rational number. For any two rational numbers a/b and c/d: a x c = a x c = ac bxd b d bd ac/bd is a rational number. 27 a x c = a x c = ac bxd b d bd 28 e.g. 2/5 x 3/4 Solution: 2/5 x 3/4 = 3/10. If we change the order of the rational numbers, the product is the same. e.g. 3/4 x 2/5 Solution: 3/4 x 2/5 = 3/10. The same is the case when we multiply three or more rational numbers. e.g. Multiply –5/8 x 4/15 x –3/4. We can multiply them in different orders (–5/8 x 4/15) x –3/4 or –5/8 x (4/15 x –3/4) Solution: –1/6 x –3/4 –5/8 x –1/5 = 1/8. = 1/8. We can see that the order of the rational numbers does not affect the product. Multiplicative identity The product of 1 and any rational number is equal to the same rational number. 1 is called the ‘multiplicative identity’ for rational numbers. Division The dividend is multiplied by the multiplicative inverse of the divisor for dividing one rational number by another. e.g. Divide 2/7 Solution: by 4/3 2/7 ÷ 4/3 = 2/7 x 3/4 = 6/28. Verifying commutative and associative properties of addition of rational numbers The commutative property states that the order of adding rational numbers does not alter the result. 29 30 e.g. 2 + 1 = 1 + 2 3 4 4 3 8+3 = 3+8 12 12 11 = 11 12 12 Whereas, the associative property implies that the order of grouping of rational numbers does not alter the result. ( ( ( ( e.g. 1 + 1 + 2 = 1 + 1 + 2 3 4 3 4 3 3 1+ 4+6 = 1+1 + 2 3 12 3 4 1 + 10 = 2 + 2 3 12 3 4 4 + 10 = 8 + 6 12 12 14 = 14 12 12 Verifying commutative and associative properties of multiplication of rational numbers The commutative property of multiplication states that if two rational numbers are multiplied, irrespective of the order in which they are multiplied; the products will be the same. e.g. 1/2 x 1/4 = 1/4 x 1/2 Solution: 1/8 = 1/8 The associative property says that when multiplying, the order of rational numbers does not affect the product. ( ( e.g. 1 x 3 1 2 x 9 31 2 x 5 = 1 x 6 3 3 5 = 1 x 5 63 3 9 5 5 = 27 27 ( ( 2+5 3 6 2 + 1 = 1 + 2 3 4 4 3 8+3 = 3+8 12 12 11 = 11 12 12 1+ 3 ( ( ( ( 1 + 2 = 1 + 1 + 2 4 3 4 3 3 1+ 4+6 = 1+1 + 2 3 12 3 4 1 + 10 = 2 + 2 3 12 3 4 4 + 10 = 8 + 6 12 12 14 = 14 12 12 ( ( 1x 2 x 5 = 1 x 6 3 3 3 1 2 x 5 = 1 x 5 9 63 3 9 5 5 = 27 27 ( ( 2+5 3 6 32 Verifying distributive property of multiplication over addition and subtraction Distributive property of multiplication over addition: For any three rational numbers we can see that multiplication is distributive over addition. e.g. Simplify the expression 1/2 x (2/3 + 2/5). Solution: (1/2 x 2/3) + (1/2 x 2/5) = 1/3 + 1/5 = 8/15. Distributive property of multiplication over subtraction: For any three rational numbers we can see that multiplication is distributive over subtraction. e.g. Simplify the expression1/2 x (1/4 – 1/3). Solution: (1/2 x 1/4) – (1/2 x 1/3) = 1/8 – 1/6 = –1/24. Properties of rational numbers The explanation and examples pertaining to rational numbers discussed above can be summarised as the properties of rational numbers. These are: 1. 2. Commutative property of addition of rational numbers. The order of adding rational numbers does not affect the result. Associative property of addition of rational numbers. The order of grouping of rational numbers does not affect the result. 3. Commutative property of multiplication of rational numbers. The order of multiplication of rational numbers does not affect the product. 4. Associative property of multiplication of rational numbers. The grouping of rational numbers does not affect the product. 33 34 5 Distributive property of multiplication over addition and subtraction. Multiplication is distributed over addition and subtraction. Non‑recurring or terminating decimals To write a rational number as a decimal we can divide the numerator by the denominator. e.g. Express the rational number 1/2 as a decimal. Solution: To express the rational number as a decimal we can express the denominator as a power of 10. Such as: or 1x5 = 5 = 0.5 1/2 = 2x5 10 We can divide 1 by 2. Since 1 is less than 2, we cannot perform division so we take 1 as 10 tenths. 1/2 = 10 tenths = 5 tenths = 0.5 2 We can see that the process of division has ended and there is no remainder. Such a decimal is called a ‘terminating decimal ’. Recurring decimal or non-terminating decimal Now express the following rational number by division: 1/3 Solution: 1 divided by 3. 1/3 = 0.333… We can see that the process of division is never ending. The remainder is not a zero. 35 1x5 = 5 = 0.5 1/2 = 2x5 10 10 2 36 Such a decimal is called a ‘non-terminating decimal ’. All terminating and non-terminating decimals represent rational numbers that can be written in the form n/d where n is an integer and d is a positive integer. Unit IV: Square Roots Pages 24-28 We have learnt that subtracting a number is the inverse of adding that number, and that dividing by a non-zero number is the inverse of multiplying by that number. The inverse of squaring a number is finding the ‘square root’ of that number. Explain the symbol used to denote the square root of a positive number, which is called the ‘radical sign’. Often it is convenient to use the + or – notations with radicals. An expression written beneath the radical sign is called the ‘radicand ’. It is interesting to note that zero has only one square root and that is zero itself. The values of certain square roots can be seen at a glance; e.g. the square root of 49 is 7. We may be able to find other square roots by expressing them as a product of square roots familiar to us. e.g. To find the square root of 144. Solution: Factorizing 144 = 9 x 16 The square root of 9 is 3 and that of 16 is 4. Therefore, the square root of 144 is 3 x 4 = 12. To find square roots by simple factors We keep dividing the number by prime numbers till we get zero as the remainder. Then by pairing off the factors and selecting one factor from each group we get a set of factors that when multiplied together give us the required square root. 37 38 e.g. Find the square root of 324. Solution: 2324 2162 381 327 39 33 1 Grouping two equal factors: 324 = 2 x 2 x 3 x 3 x 3 x 3 Selecting one factor from each pair we get: 2 x 3 x 3 = 18. The square root of 324 is 18. Square root of a fraction We find the square roots of the numerator and the denominator separately, then write them as a fraction. e.g. Find the square root of 25/36. Solution: √25/36 The square root of 25 is 5 and that of 36 is 6. Writing the square roots as a fraction, we get 5/6. The square root of 100 and its powers Give the pupils a tip to find out the square root of 100 and its powers. Give them the following example: e.g.10,0002 = 100,000,000 Then tell them to convert two zeros into one: 1 00 00 00 00 39 2324 2162 381 327 39 33 1 324 = 2 x 2 x 3 x 3 x 3 x 3 √25/36 10,0002 = 100,000,000 1 00 00 00 00 40 Now, each pair should be reduced to one zero to find out its square root. Therefore, the square root of 100,000,000 is 10,000. The square root of a decimal number The square root of 100 and its powers can be shown by finding their prime factors. e.g. Find the square root of 100. Solution: The prime factors of 100 are 10 x 10. So, the square root of 100 is 10. We can use two methods for finding the square root of a decimal number. First, we convert the decimal into a fraction and then we find the square root. e.g. Find the square root of 0.16. By changing it into a decimal fraction we get: 16/100. Finding the square root of the numerator and the denominator we get: 4/10. Changing it back into a decimal number we get: 4/10 = 0.4. A decimal fraction is a perfect square if it can be converted into a perfect square fraction. Word Problems Read the word problems carefully and discuss them thoroughly before attempting to solve them. 41 10,000 100,000,000 42 Unit V: Percentage Pages 30-37 Discuss the term ‘percentage’ as applied to the marks that pupils get in their tests and exams. When a pupil gets 6/10 marks in a test he has actually got 60 marks out of 100 and that means 60%. The ratio of a number can be expressed as a per cent. The word ‘per cent ’ (usually denoted by %) means ‘hundredths’ or ‘divided by 100 ’. e.g. 29/100 is called 29 per cent and is written as 29%. To express per cent as a common fraction A percentage can be changed into a common fraction or a decimal fraction. e.g. To express 20% as a common fraction: 20% = 20/100 = 1/5. To convert a per cent to a decimal When finding a per cent of a number, it is often convenient to express the per cent as a decimal. e.g. 15% = 15/100 = 0.15 There are two types of fractions Fractions whose denominators are divisors of 100 are multiplied by 100 and then reduced to their lowest terms. e.g. 1/10 = 1/10 x 100 = 10%. Fractions whose denominators are not divisors of 100 are also multiplied by 100 and then reduced to their lowest terms. e.g. 1/15 = 1/15 x 100 = 20/3 = 6 2 = 6.66% 3 43 2 3 44 To express a decimal fraction as a percentage Change the decimal fraction into a common fraction and then multiply by 100. e.g.0.19 Changing it into a common fraction: 0.19 = 19/100 Multiplying by 100: 19/100 x 100 = 19% Practical applications of percentage Discuss the word problems and explain what the statements imply before attempting to solve them. Profit and loss Discuss the meaning of ‘profit’ and ‘loss’ in terms of business transactions. When something is bought for a lower price and sold at a higher price, it is called ‘profit’. When something is bought for a higher price and sold for a lower price, it is called ‘loss’. Profit or loss percentage Profit or loss can be calculated in terms of percentage: Loss % = loss/cost price x 100 Profit % = profit/cost price x 100 Selling price when cost price and profit or loss per cent are given From the profit or loss per cent we can calculate the selling price. e.g.Ahmed bought a radio for Rs 640. At what price must he sell it to gain 15%? 45 0.19 0.19 = 19/100 100 19/100 x 100 = 19% 46 A gain of 15% means that when the cost price is Rs 100 the selling price should be Rs 100+15. When the cost price is Rs 640, the selling price should be: 115/100 x 640 = Rs 736. Cost price, when selling price and profit or loss per cent are given From the profit or loss per cent we can calculate the cost price. e.g. By selling a calculator for Rs 475, a shopkeeper lost 5%. What is the cost price of the calculator? A loss of 5% means that when the cost price is Rs100 the selling price is Rs 100 – 5 = Rs 95. When the selling price is Rs 475, the cost price would be: 100/95 x 475 = Rs 500. Unit VI: Ratio and Proportion Pages 38-45 This is a continuation of the unit as given in Book 6. A ratio is a relation that one quantity bears to another quantity of the same kind with regard to their magnitudes. The comparison is made by considering what multiple, part or parts the first quantity is of the second. If we say that the ratio of two quantities is ‘5 is to 6’ we mean that the magnitude of the two quantities has been compared, that is if one quantity has the magnitude 5, then the other has a magnitude of 6. A ratio is an abstract number given by a fraction. The numerator denotes the magnitude of the first quantity and the denominator gives the magnitude of the second quantity. If the quantities to be compared are in different units then it is essential to express them in the same unit to find their ratio. 47 115/100 x 640 = Rs. 736 100/95 x 475 Rs. 500 48 e.g. If two sticks of length 2m and 40 cm are to be compared; then the lengths expressed to the same units are 200 cm and 40 cm respectively and the ratio of their lengths is 200 : 40 i.e. 5 : 1. The colon inserted between the two quantities denotes the ratio and is read as ‘5 is to 1’. A ratio remains unaltered when both the terms are multiplied or divided by the same number. A ratio should be expressed in its simplest form. To find the ratio of two quantities of the same kind: Express the measures in the same unit and then divide them. The ratio 9/6 can be expressed in its simplest form as 3/2. Algebraic method for direct proportion An equation which states that two ratios are equal is called a proportion. We can write proportions in several different ways. e.g. 2 : 3 = 4 : 6 We read it as: ‘2 is to 3 as 4 is to 6’. x/12 = 9/4 We read it as ‘x divided by 12 is equal to 9 divided by 4’. In the proportion 2 : 3 = 4 : 6 2 and 6 are called the ‘extremes’ and 3 and 4 are called the ‘means’. We can use the multiplication property of equalities to show that in any proportion: 49 50 The product of the extremes is equal to the product of the means. e.g. If 15 pens cost Rs 45, find the cost of 25 pens? Solution:Let x be the cost of 25 pens. Hence the proportion: 15 : 45 : : 25 : x 15/45 = 25/x Product of the extremes: 15 multiplied by x. Product of the means: 45 x 25. Product of the extremes = product of the means 15x = 45 x 25 x = 45 x 25 15 x = Rs 75. Proportional division Proportional division means dividing a given quantity in a specified ratio. The number in the ratio is considered to be the total number of units that the given quantity is to be divided into. From this we find the quantity per unit. e.g. Divide Rs Solution: 5,000 in the ratio 2:3:5. The sum of the ratios is 2 + 3 + 5 = 10. Quantity per unit is Rs 5,000 divided by 10. One unit is equal to Rs 500. According to the given ratio: 2 units will be equal to 500 x 2 = Rs 1,000; 3 units will be equal to 500 x 3 = Rs 1,500; and 5 units will be equal to 500 x 5 = Rs 2,500. This is the required proportional division. Quick reckoning This is a simpler method of calculating proportions. Teachers should explain the formula to the pupils so that they can retain it better. 51 45 x 25 15 52 e.g. A’s share = the ratio of A x total quantity. total ratio Continued ratio Three quantities of the same kind are said to be in a continued proportion when the ratio of the first to the second is equal to the ratio of the second to the third. e.g. In the proportion 1 : 3 = 3 : 9; 3 constitutes the second as well as the third terms of the proportion, i.e. the means are the same. In this case, 3 is called the ‘mean proportional’ between 1 and 9. Also, the three numbers 1, 3 and 9 are said to be in continued proportion. The third quantity is called the ‘third proportional’. To convert ratios into continued ratios: Consider the proportion 2 : 3 = 5 : 7 This can be written as: A:B=2:3 B:C=5:7 To change the ratios into continued ratios; we multiply the first ratio by 5 and the second by 3 to make the second proportional B the same. 2 : 3 = 10 : 15 (multiplying by 5) 5 : 7 = 15 : 21 (multiplying by 3) Now B is the same in both the ratios. So the continued proportion will be A : B : C. 10 : 15 : 21 We can divide a given quantity into the given ratio by changing the ratios into continued ratio and then dividing. 53 A:B=2:3 B:C=5:7 54 e.g. Divide Rs 7,000 amongst A, B, C in the following ratios: A:B=2:3 B:C=4:5 Solution: For changing the ratios into continued ratio, multiply the first ratio by 4 and the second ratio by 3. We get 8 : 12, and 12 : 15. So the continued ratio is: A: B: C 8 : 12 : 15 The sum of the ratios is 8 + 12 + 15 = 35. Hence, the shares are: A’s share is 8/35 x 7,000 = Rs 1,600, B’s share is 12/35 x 7,000 = Rs 2,400 and C’s share is 15/35 x 7,000 = Rs 3,000. Partnership In a business partnership, the profit or loss depends upon the amount of money invested (capital) and the period for which each partner has invested his share. If the period of investment is the same for all the partners, the profit is divided amongst them in the same way as for proportional division. But if the period of investment is different for all the partners, we have to calculate the ratio of the partners in continued proportion. e.g. The investments of A, B and C in a business are Rs 2,500 for 6 months, Rs 4,000 for 3 months and Rs 5,000 for 4 months. Calculate their shares if the profit is Rs 4,700. Solution: The ratio of their investments is: 55 A’s investment is Rs 2,500 x 6 = 15,000, B’s investment is Rs 4,000 x 3 = 12,000 and C’s investment is Rs 5,000 x 4 = 20,000. A:B=2:3 B:C=4:5 A: B: C 8 : 12 : 15 56 Ratio of investments is: A: B: C 15,000: 12,000: 20,000 Reducing the ratio to its simplest form: 15: 12: 20 The sum of ratios is 15 + 12 + 20 = 47. The profit is Rs 4,700. A’s share is 15/47 x 4,700 = Rs 1,500, B’s share is 12/47 x 4,700 = Rs 1,200 and C’s share is 20/47 x 4,700 = Rs 2,000. Unit VII: Algebra Pages 46-63 Algebraic expressions A variable is a symbol that is used to represent one or more numbers. e.g. a, b, c . . . . . . are ‘variables’. To solve problems using algebra, we must often translate word phrases of numbers into numerical or variable expressions. An algebraic expression is a combination of numbers and variables connected by one or more symbols such as ‘+’ or ‘–’. Coefficient, base, exponents In the expression: 2 a3, 2 is called the ‘coefficient’, ‘a’ is the base and 3 is the exponent. Polynomial expressions An algebraic expression having one or more variables whose exponents are positive integers, is called a polynomial expression. e.g. 8x, 8x + 9, 8x2 + 2x + 1 A ‘monomial’ is considered a polynomial of one term, which is an expression that is either a numeral, 57 58 a variable, or a product of a numeral and one or more variables. A numeral such as 7, is called a ‘constant monomial’ or a ‘constant’. e.g. 7, a, 3c, 8x2y Binomial expressions have two terms: e.g. 4x + 9, 6b2 – 7a Trinomial expressions have three terms: e.g. x2 – 3x – 2, 5b2 + 3ab – a2 Degree of polynomials The ‘degree of a monomial in a variable’ is the number of times that variable occurs as a factor in the monomial. e.g.3xy2z3 is of degree 1 in x, 2 in y and 3 in z. The degree of any non-zero constant monomial is ‘0’ (It has no degree). The ‘degree of a polynomial’ is the greatest of the degrees of its terms. e.g. 4x3 + 5x2 + 7. The greatest degree of the variable x is 3. So the degree of the polynomial is 3. Ascending and descending order It is often helpful to rearrange the terms of a polynomial so that their degrees in a particular variable are in either increasing or decreasing order. e.g. Consider the expression: 59 – 4x3 + 3x2 + 3x5 60 The lowest power of x is 2, and the highest power of x is 5. To arrange the terms in ascending order we start from the lowest power i.e. 3x2 – 4x3 + 3x5. To arrange the terms in descending order we start from the highest power i.e. 3x5 – 4x3 + 3x2. Exponential expressions The number 9 can be written as: 3 x 3 or 32 and is called a ‘power’ of 3 (It is read as: ‘3 to the second power’ or ‘three squared’). The numbers 27 and 81 are also powers of 3. 27 can be written as 3 x 3 x 3 or 33 (It is read as: ‘three to the third power’ or ‘three-cubed’). In the expression a3, a is called the ‘base’ and the small raised symbol 3 is called the ‘exponent’. The exponent indicates the number of times the base occurs as a factor. Addition of algebraic expressions The terms that have the same variables and exponents are called ‘like terms’. The coefficients of ‘like terms’ may be different. e.g. a, 2a, 3a The terms that have different variables and exponents are called ‘unlike terms’, even if their coefficients are the same. e.g. 2a, 2a2, 2a3 Note: Only ‘like terms’ can be added or subtracted. To add two polynomials, we write the sum and simplify by adding ‘like terms’. 61 62 e.g. 2x + 5x = 7x. e.g. 2a + 3b and 2b + a. Write in a vertical form: 2a + 3b a + 2b 3a + 5b Addition of positive and negative terms e.g.3a – 6a + 9a – 4a First, group the terms with similar signs: 3a + 9a – 6a – 4a Then, add ‘like terms’: 12a – 10a = 2a. Addition of mixed expressions Mixed expressions can be added by associating the ‘like terms’ horizontally or vertically. e.g. Add: 5x2y + 3x2 – 8 + 4x2y + 2x2 + 9. Solution:(i) Associating the terms horizontally: 5x2y + 4x2y + 3x2 + 2x2 – 8 + 9 9x2y + 5x2 + 1. (ii) Associating the terms vertically: 5x2y + 3x2 – 8 4x2y + 2x2 + 9 9x2y + 5x2 + 1 Subtraction of algebraic expressions Subtracting polynomials is very much like subtracting real numbers. To subtract a number, you add the opposite of that number. 63 2x + 5x = 7x. 2a + 3b and 2b + a. 2a + 3b a + 2b 3a + 5b 3a – 6a + 9a – 4a 3a + 9a – 6a – 4a 12a – 10a = 2a 5x2y + 3x2 – 8 + 4x2y + 2x2 + 9 5x2y + 4x2y + 3x2 + 2x2 – 8 + 9 9x2y + 5x2 + 1 5x2y + 3x2 – 8 4x2y + 2x2 + 9 9x2y + 5x2 + 1 64 To subtract a polynomial, you add the opposite of ‘each’ term of the polynomial that you are subtracting and then simplify. e.g. Subtract: –5a2 + 2ab + 3b2 – 4 from 7a2 + 6ab – b2 – 9. Solution: (i) Associating the terms horizontally: (7a2 + 6ab – b2 – 9) – (–5a2 + 2ab + 3b2 – 4) = 7a2 + 6ab – b2 – 9 + 5a2 – 2ab – 3b2 + 4 = (7 + 5) a2 + (6 – 2) ab + (–1 – 3) b2 + (–9 + 4) = 12a2 + 4ab – 4b2 – 5. (ii) Associating the terms vertically: 7a2 + 6ab – b2 – 9 – 5a2 + 2ab + 3b2 – 4 + – – + (changing to the opposite) 12a2 + 4ab – 4b2 – 5(adding) Multiplication of polynomials When we multiply two monomials, we use the rule of exponents along with the commutative and associative properties for multiplication. e.g. (i) Multiply 2x2 by 3. Solution: 2x2 x 3 = 6x2. (ii) Multiply –5x2 by –3. Solution: –5x2 x –3 = + 15x2. Multiplying polynomials When we multiply two powers having the same base, we add the exponents. e.g.(i) x2 x x5 = x2+5 = x7. e.g.(ii) 3x3y4 x –7xy5 = (3 x –7) (x3 x x) (y4 x y5) = (–21) (x3+1) (y4+5) = –21x4y 9. 65 –5a2 + 2ab + 3b2 – 4 7a2 + 6ab – b2 – 9 (7a2 + 6ab – b2 – 9) – (–5a2 + 2ab + 3b2 – 4) = 7a2 + 6ab – b2 – 9 + 5a2 – 2ab – 3b2 + 4 = (7 + 5) a2 + (6 – 2) ab + (–1 – 3) b2 + (–9 + 4) = 12a2 + 4ab – 4b2 – 5 7a2 + 6ab – b2 – 9 – 5a2 + 2ab + 3b2 – 4 + – – + 12a2 + 4ab – 4b2 – 5 66 Multiplying a polynomial by a monomial We use the distributive property and the rules of exponents to multiply. e.g. Multiply 3a2 – 4a + 3 by 4a. We can multiply horizontally or vertically. (i)4a (3a2 – 4a + 3) = 12a3 – 16a2 + 12a (ii)3a2 – 4a + 3 4a 12a3 – 16a2 + 12a (multiplying horizontally) (multiplying vertically) Multiplying two polynomials We can use the distributive property of multiplication to multiply two polynomials. e.g. Multiply 4x + 3 by 3x + 4. (i) Multiplying horizontally: (4x + 3) (3x + 4) = 4x (3x + 4) + 3 (3x + 4) = 12x2 + 16x + 9x + 12 = 12x2 + 25x + 12. (ii) Multiplying vertically: 4x + 3 3x + 4 12x2 + 9x + 16x + 12 2 12x + 25x + 12 Division of polynomials When we divide two monomials, we use the laws of exponents. When the base of the divisor and the dividend is the same then the power of the quotient is the difference of their exponents. 67 4a 3a2 – 4a + 3 4a (3a2 – 4a + 3) = 12a3 – 16a2 + 12a 3a2 – 4a + 3 4a 12a3 – 16a2 + 12a (4x + 3) (3x + 4) = 4x (3x + 4) + 3 (3x + 4) = 12x2 + 16x + 9x + 12 = 12x2 + 25x + 12. 4x 3x 12x2 12x2 +3 +4 +9x +16x + 12 +25x + 12 68 e.g. Divide 5x2 by x. Solution:5x2/x= 5x2-1 = 5x. Note: any number raised to the power ‘0’ is 1. e.g.5x/x = 5x1-1 = 5x0 = 5 x 1 = 5. e.g.15y3/3y = 15/3 x y3-1 = 5y2. Dividing a polynomial by a monomial 2a2x – 10a3x2 + 12a5 x3 ÷ 2ax Divide each term of the polynomial by the monomial: 2 3 2 5 3 = 2a x – 10a x + 12a x 2ax 2ax 2ax Add the quotients: = a – 5a2x + 6a4x2. Substitution The meaning of substitution is to ‘replace’ or ‘put in place of’. Changing the numeral by which a number is named in our expression does not change the value of the expression. Replacing each variable in an expression by a given value and simplifying the result is called ‘evaluating the expression’ or 'finding the value of the expression.’ e.g. Evaluate the expression: a + b, when a = 5 and b = –2. Solution: a+b = 5 + (–2) (putting the values of a & b) =5–2 = 3. 69 5x/x = 5x1-1 = 5x0 = 5 x 1 = 5. 15y3/3y = 15/3 x y3-1 = 5y2. 2a2x – 10a3x2 + 12a5 x3 ÷ 2ax 2 3 2 5 3 = 2a x – 10a x + 12a x 2ax 2ax 2ax = a – 5a2x + 6a4x2. a b a b a b a b 70 Algebraic sentences An algebraic sentence is formed by placing =, >, <, ≠ signs between two numerical or variable expressions. Sentences containing variables such as 5x – 1 = 9, and 3 + c = c + 3, are called ‘open sentences’. When we replace each variable in an open sentence by a given number, we obtain a ‘true’ sentence or a ‘false’ sentence. e.g. 5 + 2 = 7 is a true sentence. 5 > 10 is a false sentence. 3+ =10 is an open sentence. The last sentence is an open sentence because it is impossible to judge whether it is a true or a false sentence, unless the blank space is replaced by a number. If we replace the blank space by 7, the sentence becomes a true sentence. If we replace the blank space by 6, the sentence becomes a false sentence. Equations A polynomial equation has polynomials on both sides and where both the sides have an equality sign in the middle. A simple linear equation An equation is formed by placing an ‘is equal to’ sign between two numerical variable expressions called the ‘sides’ of the equation. In an equation, the sign of equality between the sides shows that the two sides are equal. e.g. x – 5 = 7. An equation in which the exponent of the variable is 1 is called a linear equation. 71 3 + c = c + 3 5x–1=9 x–5=7 72 Equivalent equations Equations having the same solution are called equivalent equations. Transforming equations To solve an equation we usually try to change or ‘transform’ it into a simple equivalent equation, whose solution can be seen at a glance. This transformation into a simple equivalent equation can be done by substitution, addition, or subtraction. Addition and subtraction properties of an equality 1. If the same number is added to equal numbers, the sums are equal. 2. If the same number is subtracted from equal numbers, the differences are equal. We can use these properties to solve some equations. e.g.Solve x – 5 = 7. x – 5 + 5 = 7 + 5 (5 is added to both sides) x = 12. e.g.Solve x + 10 = 20. x + 10 – 10 = 20 – 10 (10 is subtracted from both sides) x = 10. Because errors may occur in transforming equations, we should check our work by substituting the value of the variable found, so as to show that the transformed equation satisfies the original equation. e.g. x + 8 = 3. x+8–8=3–8 x = –5. 73 x–5=7 x–5+5=7+5 x = 12 x + 10 = 20 10 x + 10 – 10 = 20 – 10 x = 10 x + 8 = 3. x+8–8=3–8 x = –5. 74 Substituting the value of x = –5 in the given equation: x + 8 = 3. –5 + 8 = 3 3 = 3. Multiplication and division properties of an equality 1. If equal numbers are multiplied by the same number, the products are equal. 2. If equal numbers are divided by the same non-zero number, the quotients are equal. Transformation by multiplication Multiply each side of a given equation by the same non-zero real number. e.g. x/2 = 14. Multiplying both sides by 2: (x/2)(2) = (14)(2) x = 28. Transformation by division Divide each side of a given equation by the same non-zero real number. e.g.2x = 10. Dividing both sides by 2: 2x/2 = 10/2 x = 5. Using several transformations to solve an equation We know that subtraction is the inverse of addition and that division is the inverse of multiplication. In transforming equations, we often use inverse operations. 75 x + 8 = 3. –5 + 8 = 3 3 = 3. x/2 = 14. (x/2)(2) = (14)(2) x = 28. 2x = 10. 2x/2 = 10/2 x = 5. 76 e.g. Solve 4y + 43 = 19. Solution:4y + 43 = 19 Subtracting 43 from each side: 4y + 43 – 43 = 19 – 43 4y = – 24 Dividing each side by 4: 4y/4 = –24/4 y = –6. To check: Substituting the value of y = –6 in the given equation: 4y + 43 = 19 (4) (–6) + 43 = 19 –24 + 43 = 19 19 = 19 The following steps are usually helpful when we are solving an equation in which all the variables are on the same side. 1. Simplify each side of the equation. 2. If there are indicated additions or subtractions, use the inverse operations to undo them. 3. If there are indicated multiplications or divisions, use the inverse operations to undo them. Solving word problems Follow the steps given below for solving word problems involving linear equations: 1. Read the problem carefully a few times. Decide what numbers are asked for and what information is given. Making a sketch may be helpful. 2. Choose a variable and use it with the given facts to represent the number(s) described in the problem. 3. Reread the problems. Then write an open sentence that represents the relationship amongst the numbers in the problem. 4. Solve the open sentence and find the required numbers. 5. Check your results with the words of the problem. Give the answer. 77 4y + 43 = (4) (–6) + 43 = –24 + 43 = 19 = 19 19 19 19 78 UNIT VIII: Geometry Pages 64-78 Parallel lines Discuss parallel lines with examples of the railway line, the horizontal and vertical sides of rectangular doors and windows. Draw a square and rectangle on the board and explain the definition of parallel lines. Complementary and supplementary angles Lay a large book flat on the desk and gradually lift its cover. Explain the formation of angles by it. Discuss a right angle and also complementary angles. Open the book and lay it flat on the desk. Lift one page and hold it near the center. Discuss that supplementary angles are equal to two right angles. Angles of a triangle Draw a triangle on the board and measure its angles. Explain that the sum of the angles of a triangle is equal to 180°. Construction of triangles Following the steps of construction for the different types of triangles, draw the triangles on the board and give the pupils practice in constructing triangles. Quadrilaterals Discuss the various types of quadrilaterals. Following the steps of construction for the different types of quadrilaterals, draw them on the board. Give the pupils practice in constructing quadrilaterals. Area Discuss the meaning of area. Use a squared paper to explain it. Help the pupils to find the area of different geometrical shapes by using formulae. 79 80 Volume Discuss the meaning of volume. Use solid figures to explain it. Help the pupils find the volume of solid figures by using formulae. Unit IX: Information Handling Pages 79-82 Discuss the meaning of statistics. Explain that we get numerical information from various sources. Numerical information is called ‘data’. Ask the pupils to collect data regarding some information about their class, their school, etc. Discuss the different kinds of data. Write ungrouped data on the board and explain the method of arranging it into grouped data. Classification It is the method of arranging data into certain groups or classes without losing valuable information. Class interval Data is classified on the basis of values or quantities and is divided into a number of classes, each of which is called a ‘class interval’. Classification according to this method involves: (i) number of classes and their sizes, (ii) choice of class limits and (iii) counting the numbers in each class. Class limits The maximum and minimum values within which a class interval lies are known as the ‘class limits’. 81 82 Class size The difference between two class limits is known as the ‘class size’. The difference between the maximum and the minimum number of items of a class is known as the ‘range’ of the data. The class size of the class intervals depends on the range of the data and the number of classes. e.g. Marks obtained by 20 students in science: 42, 38, 37, 75, 45, 49, 52, 78, 84, 34, 41, 65, 73, 32 , 87, 34, 37, 56, 59, 89 The lowest score is 32 and the highest is 89. So the range is 89 – 32 = 57. Suppose we want to make 6 classes, the number nearest to 57, which is divisible by 6 is 60. Hence, the class size will be: 60/6 = 10. Thus, the classes may be taken as: 30 – 39, 40 – 49, 50 – 59, 60 – 69, 70 – 79, 80 – 89. Pie chart Diagrams and graphs help in giving us an overall view of the data under consideration. They present the facts in the form of pictures by which data or information can easily be compared. 83 42, 38, 37, 75, 45, 49, 52, 78, 84, 34, 41, 65, 73, 32 , 87, 34, 37, 56, 59, 89 89 – 32 = 57 60/6 = 10 30 – 39, 40 – 49, 50 – 59, 60 – 69, 70 – 79, 80 – 89. 84 One kind of chart called the ‘pie chart’ is constructed by dividing a circle into different sectors. Each sector corresponds to the percentage of a category of data under consideration. The angle of each sector is proportional to the number the sector represents. To obtain the sectors in a pie chart we need to find the angles at the center of the circle. Since the total area of the circle corresponds to the total number of degrees in the circle, i.e. 360, we can find the angle of the circle by dividing each item by the total number of items and multiplying by 360°. To draw a pie chart First find out the angles of each sector. Then draw the radius of the circle and with the help of a protractor, draw the required angle. Using the arm of the angle drawn, draw the next angle corresponding to the next sector and so on. Fill till all the sectors have been represented. Write the item in the sector it represents. Colour each sector in a different shade. e.g. Draw a pie chart to represent the monthly expenditure of a family. House rent = 30% Food =50% Education =10% Health = 2% Savings = 8% 85 30%= 50%= 10%= 2%= 8%= 86 (i) Adding all the items: =100% (ii) Finding the angles of each sector (a) House rent = 30/100 x 360° = 108° (b) Food = 50/100 x 360° = 180° (c) Education = 10/100 x 360° = 36° (d) Health = 2/100 x 360° = 7.2° (e) Savings = 8/100 x 360° = 28.8° (a) 108° draw the circle draw the radius (i) (ii) (i) (b) 180° (i) (a) 108° 87 (a) 108° (ii) (b) 180° (iii) draw the third sector (v) (a) 108° (e) 28.8° ➝ (iv) ➝ ➝ draw the second sector (iv) (c) 36° (iii) (i) ➝ (ii) (b) 180° draw the first sector (c) 36° (ii) (iii) (d) 7.2° then, draw the last two sectors (vi) 30/100 x 360° = 108° = 50/100 x 360° = 180° = 10/100 x 360° = 36° = 2/100 x 360° = 7.2° = 8/100 x 360° = 28.8° = (a) 108° (i) (b) 180° (i) (a) 108° (a) 108° (c) 36° (ii) (b) 180° (iii) (a) 108° (e) 28.8° ➝ (iv) ➝ ➝ ➝ (ii) (b) 180° (i) (c) 36° (ii) (iii) (d) 7.2° 88 NOTE 89 NOTE 90 NOTE 91