Exponents and Powers Chapter 11 PDF

170 MATHEMATICS Chapter 11 Exponents and Powers 11.1 INTRODUCTION Do you know what the mass of earth is? It is...

170 MATHEMATICS Chapter 11 Exponents and Powers 11.1 INTRODUCTION Do you know what the mass of earth is? It is 5,970,000,000,000,000,000,000,000 kg! Can you read this number? Mass of Uranus is 86,800,000,000,000,000,000,000,000 kg. Which has greater mass, Earth or Uranus? Distance between Sun and Saturn is 1,433,500,000,000 m and distance between Saturn and Uranus is 1,439,000,000,000 m. Can you read these numbers? Which distance is less? These very large numbers are difficult to read, understand and compare. To make these numbers easy to read, understand and compare, we use exponents. In this Chapter, we shall learn about exponents and also learn how to use them. 11.2 EXPONENTS We can write large numbers in a shorter form using exponents. Observe 10, 000 = 10 × 10 × 10 × 10 = 104 The short notation 104 stands for the product 10×10×10×10. Here ‘10’ is called the base and ‘4’ the exponent. The number 104 is read as 10 raised to the power of 4 or simply as fourth power of 10. 104 is called the exponential form of 10,000. We can similarly express 1,000 as a power of 10. Note that 1000 = 10 × 10 × 10 = 103 Here again, 103 is the exponential form of 1,000. Similarly, 1,00,000 = 10 × 10 × 10 × 10 × 10 = 105 105 is the exponential form of 1,00,000 In both these examples, the base is 10; in case of 103, the exponent is 3 and in case of 105 the exponent is 5. 2024-25  EXPONENTS AND POWERS 171 We have used numbers like 10, 100, 1000 etc., while writing numbers in an expanded form. For example, 47561 = 4 × 10000 + 7 × 1000 + 5 × 100 + 6 × 10 + 1 This can be written as 4 × 104 + 7 ×103 + 5 × 102 + 6 × 10 + 1. Try writing these numbers in the same way 172, 5642, 6374. In all the above given examples, we have seen numbers whose base is 10. However the base can be any other number also. For example: 81 = 3 × 3 × 3 × 3 can be written as 81 = 34, here 3 is the base and 4 is the exponent. Some powers have special names. For example, 102, which is 10 raised to the power 2, also read as ‘10 squared’ and 103, which is 10 raised to the power 3, also read as ‘10 cubed’. Can you tell what 53 (5 cubed) means? 53 = 5 × 5 × 5 = 125 So, we can say 125 is the third power of 5. What is the exponent and the base in 53? Similarly, 25 = 2 × 2 × 2 × 2 × 2 = 32, which is the fifth power of 2. In 25, 2 is the base and 5 is the exponent. In the same way, 243 = 3 × 3 × 3 × 3 × 3 = 35 64 = 2 × 2 × 2 × 2 × 2 × 2 = 26 625 = 5 × 5 × 5 × 5 = 54 TRY THESE Find five more such examples, where a number is expressed in exponential form. Also identify the base and the exponent in each case. You can also extend this way of writing when the base is a negative integer. What does (–2)3 mean? It is (–2)3 = (–2) × (–2) × (–2) = – 8 Is (–2)4 = 16? Check it. Instead of taking a fixed number let us take any integer a as the base, and write the numbers as, a × a = a2 (read as ‘a squared’ or ‘a raised to the power 2’) a × a × a = a3 (read as ‘a cubed’ or ‘a raised to the power 3’) a × a × a × a = a4 (read as a raised to the power 4 or the 4th power of a).............................. a × a × a × a × a × a × a = a7 (read as a raised to the power 7 or the 7th power of a) and so on. a × a × a × b × b can be expressed as a3b2 (read as a cubed b squared) 2024-25  172 MATHEMATICS TRY THESE a × a × b × b × b × b can be expressed as a2b4 (read as a Express: squared into b raised to the power of 4). (i) 729 as a power of 3 EXAMPLE 1 Express 256 as a power 2. (ii) 128 as a power of 2 S OLUTION We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2. (iii) 343 as a power of 7 So we can say that 256 = 28 E XAMPLE 2 Which one is greater 23 or 32? S OLUTION We have, 23 = 2 × 2 × 2 = 8 and 32 = 3 × 3 = 9. Since 9 > 8, so, 32 is greater than 23 E XAMPLE 3 Which one is greater 82 or 28? S OLUTION 82 = 8 × 8 = 64 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 Clearly, 28 > 82 E XAMPLE 4 Expand a3 b2, a2 b3, b2 a3, b3 a2. Are they all same? S OLUTION a3 b2 = a3 × b2 = (a × a × a) × (b × b) = a×a×a×b×b a b = a2 × b3 2 3 = a×a×b×b×b b a = b2 × a3 2 3 = b×b×a×a×a b3 a2 = b3 × a2 = b×b×b×a×a 3 2 2 3 Note that in the case of terms a b and a b the powers of a and b are different. Thus a3 b2 and a2 b3 are different. On the other hand, a3 b2 and b2 a3 are the same, since the powers of a and b in these two terms are the same. The order of factors does not matter. Thus, a3 b2 = a3 × b2 = b2 × a3 = b2 a3. Similarly, a2 b3 and b3 a2 are the same. E XAMPLE 5 Express the following numbers as a product of powers of prime factors: (i) 72 (ii) 432 (iii) 1000 (iv) 16000 2 72 S OLUTION 2 36 (i) 72 = 2 × 36 = 2 × 2 × 18 2 18 =2×2×2×9 3 9 = 2 × 2 × 2 × 3 × 3 = 23 × 32 Thus, 72 = 23 × 32 (required prime factor product form) 3 2024-25  EXPONENTS AND POWERS 173 (ii) 432 = 2 × 216 = 2 × 2 × 108 = 2 × 2 × 2 × 54 = 2 × 2 × 2 × 2 × 27 = 2 × 2 × 2 × 2 × 3 × 9 =2×2×2×2×3×3×3 or 432 = 24 × 33 (required form) (iii) 1000 = 2 × 500 = 2 × 2 × 250 = 2 × 2 × 2 × 125 = 2 × 2 × 2 × 5 × 25 = 2 × 2 × 2 × 5 × 5 × 5 or 1000 = 23 × 53 Atul wants to solve this example in another way: 1000 = 10 × 100 = 10 × 10 × 10 = (2 × 5) × (2 × 5) × (2 × 5) (Since10 = 2 × 5) =2×5×2×5×2×5=2×2×2×5×5×5 or 1000 = 23 × 53 Is Atul’s method correct? (iv) 16,000 = 16 × 1000 = (2 × 2 × 2 × 2) ×1000 = 24 ×103 (as 16 = 2 × 2 × 2 × 2) = (2 × 2 × 2 × 2) × (2 × 2 × 2 × 5 × 5 × 5) = 24 × 23 × 53 (Since 1000 = 2 × 2 × 2 × 5 × 5 × 5) = (2 × 2 × 2 × 2 × 2 × 2 × 2 ) × (5 × 5 × 5) or, 16,000 = 27 × 53 EXAMPLE 6 Work out (1)5, (–1)3, (–1)4, (–10)3, (–5)4. SOLUTION (i) We have (1)5 = 1 × 1 × 1 × 1 × 1 = 1 In fact, you will realise that 1 raised to any power is 1. (–1) odd number = –1 (ii) (–1)3 = (–1) × (–1) × (–1) = 1 × (–1) = –1 (–1)even number =+1 (iii) (–1)4 = (–1) × (–1) × (–1) × (–1) = 1 ×1 = 1 You may check that (–1) raised to any odd power is (–1), and (–1) raised to any even power is (+1). (iv) (–10)3 = (–10) × (–10) × (–10) = 100 × (–10) = – 1000 (v) (–5)4 = (–5) × (–5) × (–5) × (–5) = 25 × 25 = 625 E XERCISE 11.1 1. Find the value of: (i) 26 (ii) 9 3 (iii) 112 (iv) 54 2. Express the following in exponential form: (i) 6 × 6 × 6 × 6 (ii) t × t (iii) b × b × b × b (iv) 5 × 5× 7 × 7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d 2024-25 174 MATHEMATICS 3. Express each of the following numbers using exponential notation: (i) 512 (ii) 343 (iii) 729 (iv) 3125 4. Identify the greater number, wherever possible, in each of the following? (i) 43 or 34 (ii) 53 or 35 (iii) 28 or 82 (iv) 1002 or 2100 (v) 210 or 102 5. Express each of the following as product of powers of their prime factors: (i) 648 (ii) 405 (iii) 540 (iv) 3,600 6. Simplify: (i) 2 × 103 (ii) 72 × 22 (iii) 23 × 5 (iv) 3 × 44 (v) 0 × 102 (vi) 52 × 33 (vii) 24 × 32 (viii) 32 × 104 7. Simplify: (i) (– 4)3 (ii) (–3) × (–2)3 (iii) (–3)2 × (–5)2 (iv) (–2)3 × (–10)3 8. Compare the following numbers: (i) 2.7 × 1012 ; 1.5 × 108 (ii) 4 × 1014 ; 3 × 1017 11.3 LAWS OF EXPONENTS 11.3.1 Multiplying Powers with the Same Base (i) Let us calculate 22 × 23 22 × 23 = (2 × 2) × (2 × 2 × 2) = 2 × 2 × 2 × 2 × 2 = 25 = 22+3 Note that the base in 22 and 23 is same and the sum of the exponents, i.e., 2 and 3 is 5 (ii) (–3)4 × (–3)3 = [(–3) × (–3) × (–3)× (–3)] × [(–3) × (–3) × (–3)] = (–3) × (–3) × (–3) × (–3) × (–3) × (–3) × (–3) = (–3)7 = (–3)4+3 Again, note that the base is same and the sum of exponents, i.e., 4 and 3, is 7 (iii) a2 × a4 = (a × a) × (a × a × a × a) = a × a × a × a × a × a = a6 (Note: the base is the same and the sum of the exponents is 2 + 4 = 6) Similarly, verify: 42 × 42 = 42+2 32 × 33 = 32+3 2024-25  EXPONENTS AND POWERS 175 Can you write the appropriate number in the box. TRY THESE (–11)2 × (–11)6 = (–11) Simplify and write in 2 3 b × b =b (Remember, base is same; b is any integer). exponential form: c3 × c4 =c (c is any integer) (i) 25 × 23 (ii) p3 × p2 d 10 × d 20 = d (iii) 43 × 42 From this we can generalise that for any non-zero integer a, where m (iv) a3 × a2 × a7 and n are whole numbers, (v) 53 × 57 × 512 am × an = am + n (vi) (– 4)100 × (– 4)20 Caution! Consider 23 × 32 Can you add the exponents? No! Do you see ‘why’? The base of 23 is 2 and base of 32 is 3. The bases are not same. 11.3.2 Dividing Powers with the Same Base Let us simplify 37 ÷ 34? 37 3×3×3× 3× 3× 3× 3 37 ÷ 34 = 4 = 3 3×3×3×3 = 3 × 3 × 3 = 33 = 37 – 4 Thus 37 ÷ 34 = 37 – 4 (Note, in 37 and 34 the base is same and 37 ÷ 34 becomes 37–4) Similarly, 56 5 × 5 × 5 × 5 × 5 × 5 5 ÷5 = 2 = 6 2 5 5×5 = 5 × 5 × 5 × 5 = 54 = 56 – 2 or 56 ÷ 52 = 56 – 2 Let a be a non-zero integer, then, a4 a × a × a × a a ÷a = 2 = = a × a = a 2 = a4 4 2 2 a a×a or a4 ÷ a2 = a4 – 2 Now can you answer quickly? 108 ÷ 103 = 108 – 3 = 105 79 ÷ 76 = 7 a8 ÷ a5 = a 2024-25  176 MATHEMATICS For non-zero integers b and c, TRY THESE b10 ÷ b5 = b Simplify and write in exponential c100 ÷ c90 = c form: (eg., 116 ÷ 112 = 114) In general, for any non-zero integer a, (i) 29 ÷ 23 (ii) 108 ÷ 104 am ÷ an = am – n (iii) 911 ÷ 97 (iv) 2015 ÷ 2013 (v) 713 ÷ 710 where m and n are whole numbers and m > n. 11.3.3 Taking Power of a Power Consider the following ( ) ( ) 2 4 Simplify 23 ; 32 Now, ( 2 ) means 2 is multiplied two times with itself. 3 2 3 (2 ) = 2 × 2 3 2 3 3 = 23 + 3 (Since am × an = am + n) = 26 = 23 × 2 Thus (2 )3 2 = 23×2 Similarly (3 )2 4 = 32 × 32 × 32 × 32 = 32 + 2 + 2 + 2 = 38 (Observe 8 is the product of 2 and 4). = 32 × 4 Can you tell what would ( 7 2 ) would be equal to? 10 So (2 )3 2 = 23 × 2 = 26 (3 )2 4 = 32 × 4 = 38 = 72 × 10 = 720 TRY THESE (a )2 3 = a 2 × 3 = a6 Simplify and write the answer in = am × 3 = a3m exponential form: From this we can generalise for any non-zero integer ‘a’, where ‘m’ and ‘n’ are whole numbers, (i) ( 6 2 ) 4 (ii) (2 )2 100 (a )m n = amn (iii) (7 ) 50 2 (iv) (5 )3 7 2024-25  EXPONENTS AND POWERS 177 E XAMPLE 7 Can you tell which one is greater (52) × 3 or (52 ) ? 3 S OLUTION (52) × 3 means 52 is multiplied by 3 i.e., 5 × 5 × 3 = 75 but (52 ) means 52 is multiplied by itself three times i.e. , 3 52 × 52 × 52 = 56 = 15,625 Therefore (52)3 > (52) × 3 11.3.4 Multiplying Powers with the Same Exponents Can you simplify 23 × 33? Notice that here the two terms 23 and 33 have different bases, but the same exponents. Now, 23 × 33 = (2 × 2 × 2) × (3 × 3 × 3) = (2 × 3) × (2 × 3) × (2 × 3) =6×6×6 = 63 (Observe 6 is the product of bases 2 and 3) Consider 44 × 34 = (4 × 4 × 4 × 4) × (3 × 3 × 3 × 3) = (4 × 3) × (4 × 3) × (4 × 3) × (4 × 3) = 12 × 12 × 12 × 12 = 12 4 TRY THESE Put into another form using Consider, also, 32 × a2 = (3 × 3) × (a × a) am × bm = (ab)m: = (3 × a) × (3 × a) (i) 43 × 23 (ii) 25 × b5 2 = (3 × a) 2 (iii) a2 × t2 (iv) 56 × (–2)6 = (3a) (Note: 3×a = 3a ) 4 4 (v) (–2)4 × (–3)4 Similarly, a × b = (a × a × a × a) × (b × b × b × b) = (a × b) × (a × b) × (a × b) × (a × b) = (a × b)4 = (ab)4 (Note a × b = ab) In general, for any non-zero integer a am × bm = (ab)m (where m is any whole number) E XAMPLE 8 Express the following terms in the exponential form: (i) (2 × 3)5 (ii) (2a)4 (iii) (– 4m)3 S OLUTION (i) (2 × 3)5 = (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) = (2 × 2 × 2 × 2 × 2) × (3 × 3× 3 × 3 × 3) = 25 × 35 2024-25  178 MATHEMATICS (ii) (2a)4 = 2a × 2a × 2a × 2a = (2 × 2 × 2 × 2) × (a × a × a × a) = 24 × a4 3 (iii) (– 4m) = (– 4 × m)3 = (– 4 × m) × (– 4 × m) × (– 4 × m) = (– 4) × (– 4) × (– 4) × (m × m × m) = (– 4)3 × (m)3 11.3.5 Dividing Powers with the Same Exponents TRY THESE Observe the following simplifications: Put into another form 4 24 2 × 2 × 2 × 2 2 2 2 2  2  (i) 4 = = × × × =  3 3 × 3× 3 × 3 3 3 3 3  3  m m a m using a ÷ b =   :  b a3 a × a × a a a a  a  3 (i) 5 4 ÷3 5 (ii) 3 = = × × =  (ii) 25 ÷ b5 b b × b ×b b b b  b  (iii) (– 2)3 ÷ b3 From these examples we may generalise (iv) p4 ÷ q4 m am  a  (v) 56 ÷ (–2)6 a ÷ b = m =   where a and b are any non zero integers m m b  b and m is a whole number. 5 4  −4  E XAMPLE 9 Expand: (i)   3 (ii)    5  7 What is a0? Obeserve the following pattern: S OLUTION 26 = 64 25 = 32  3 4 34 3× 3× 3× 3 (i)   = 4 = 24 = 16  5 5 5× 5×5× 5 23 = 8 22 = ? 5  −4  ( − 4)5 21 = ? (ii)   = =  7 75 20 = ? You can guess the value of 20 l Numbers with exponent zero by just studying the pattern! You find that 20 = 1 35 Can you tell what equals to? 35 If you start from 36 = 729, and proceed as shown above 35 3× 3× 3× 3× 3 finding 35, 34, 33,... etc, what 5 = =1 3 3× 3× 3× 3× 3 will be 30 = ? by using laws of exponents 2024-25  EXPONENTS AND POWERS 179 35 ÷ 35 = 35 – 5 = 30 So 30 = 1 Can you tell what 70 is equal to? 73 ÷73 = 73 – 3 = 70 73 7×7×7 And 3 = =1 7 7×7×7 Therefore 70 = 1 Similarly a3 ÷ a3 = a3–3 = a0 a3 a × a × a And a ÷a = 3 = 3 3 =1 a a×a×a Thus a0 = 1 (for any non-zero integer a) So, we can say that any number (except 0) raised to the power (or exponent) 0 is 1. 11.4 M ISCELLANEOUS E XAMPLES USING THE L AWS OF E XPONENTS Let us solve some examples using rules of exponents developed. EXAMPLE 10 Write exponential form for 8 × 8 × 8 × 8 taking base as 2. SOLUTION We have, 8 × 8 × 8 × 8 = 84 But we know that 8 = 2 × 2 × 2 = 23 Therefore 84 = (23)4 = 23 × 23 × 23 × 23 = 23 × 4 [You may also use (am)n = amn] = 2 12 EXAMPLE 11 Simplify and write the answer in the exponential form.  37  5 (i)  32  × 3 (ii) 23 × 22 × 55 (iii) (62 × 64) ÷ 63   (iv) [(22)3 × 36] × 56 (v) 82 ÷ 23 S OLUTION  37  5 (i)  32  × 3 = (3 ) × 3 7 −2 5   = 35×35 = 35+5 = 310 2024-25 180 MATHEMATICS (ii) 23 × 22 × 55 = 23+2 × 55 = 25 × 55 = (2 × 5)5 = 105 (iii) (6 2 × 64 ) ÷ 63 = 6 2 + 4 ÷ 6 3 66 6−3 = 3 =6 =6 3 6 (iv) ( 2 ) × 3  × 5 = [26 × 36] × 56 2 3 6 6   ( 2 × 3) 6 = × 56 ( 2 × 3 × 5) 6 6 = = 30 (v) 8 = 2 × 2 × 2 = 23 Therefore 82 ÷ 23 = (23)2 ÷ 23 = 26 ÷ 23 = 26 – 3 = 23 E XAMPLE 12 Simplify: 124 × 93 × 4 2 × 34 × 25 (i) (ii) 23 × a3 × 5a4 (iii) 63 × 82 × 27 9 × 42 S OLUTION (i) We have 124 × 93 × 4 (22 ×3) 4 ×(32 )3×22 = 63 × 82 × 27 (2×3)3×(23 )2 ×33 = ( 2 ) × (3) 2 4 4 × 32 ×3 × 22 28 × 22 × 34 × 36 = 3 23 × 33 × 22×3 × 33 2 × 26 × 33 × 33 28 + 2 ×34 + 6 210 ×310 = 3+ 6 3+ 3 = 9 6 2 ×3 2 ×3 = 210 – 9 × 310 – 6 = 21 × 34 = 2 × 81 = 162 (ii) 23 × a3 × 5a4 = 23 × a3 × 5 × a4 = 23 × 5 × a3 × a4 = 8 × 5 × a3 + 4 = 40 a7 2024-25  EXPONENTS AND POWERS 181 2 × 34 × 25 2×34 ×25 2×25 ×34 (ii) = 2 2 = 9 × 42 3 × 22 ( ) 32 ×22 × 2 21+ 5 ×34 26 ×34 = = = 26 − 4 ×34 − 2 24 ×32 24 ×32 = 22 × 32 = 4 × 9 = 36 Note: In most of the examples that we have taken in this Chapter, the base of a power was taken an integer. But all the results of the chapter apply equally well to a base which is a rational number. EXERCISE 11.2 1. Using laws of exponents, simplify and write the answer in exponential form: (i) 32 × 34 × 38 (ii) 615 ÷ 610 (iii) a3 × a2 ( ) 3 (iv) 7x ×72 (v) 52 ÷ 53 (vi) 25 × 55 (viii) (34 ) (2 ÷ 215 ) × 23 3 20 (vii) a4 × b4 (ix) (x) 8t ÷ 82 2. Simplify and express each of the following in exponential form: (i) 23 × 34 × 4 3 × 32 (ii) (( 5 ) × 5 ) ÷ 5 2 3 4 7 (iii) 254 ÷ 53 3 × 7 2 × 118 37 (iv) (v) 4 3 (vi) 20 + 30 + 40 21× 113 3 ×3 0 0 0 0 0 0 28 × a 5 (vii) 2 ×3 ×4 (viii) (3 + 2 ) × 5 (ix) 3 3 4 ×a  a5  4 5 × a 8b 3 (x)  3  × a (2 × 2) 8 3 2 (xi) 5 (xii) a  4 × a 5b 2 3. Say true or false and justify your answer: (i) 10 × 1011 = 10011 (ii) 23 > 52 (iii) 23 × 32 = 65 (iv) 30 = (1000)0 2024-25 182 MATHEMATICS 4. Express each of the following as a product of prime factors only in exponential form: (i) 108 × 192 (ii) 270 (iii) 729 × 64 (iv) 768 5. Simplify: (i) (2 ) 5 2 × 73 (ii) 25 × 52 × t 8 (iii) 35 × 105 × 25 83 × 7 103 × t 4 57 × 65 11.5 DECIMAL NUMBER SYSTEM Let us look at the expansion of 47561, which we already know: 47561 = 4 × 10000 + 7 × 1000 + 5 × 100 + 6 × 10 + 1 We can express it using powers of 10 in the exponent form: Therefore, 47561 = 4 × 104 + 7 × 103 + 5 × 102 + 6 × 101 + 1 × 100 (Note 10,000 = 104, 1000 = 103, 100 = 102, 10 = 101 and 1 = 100) Let us expand another number: 104278 = 1 × 100,000 + 0 × 10,000 + 4 × 1000 + 2 × 100 + 7 × 10 + 8 × 1 = 1 × 105 + 0 × 104 + 4 × 103 + 2 × 102 + 7 × 101 + 8 × 100 = 1 × 105 + 4 × 103 + 2 × 102 + 7 × 101 + 8 × 100 Notice how the exponents of 10 start from a maximum value of 5 and go on decreasing by 1 at a step from the left to the right upto 0. 11.6 EXPRESSING LARGE NUMBERS IN THE STANDARD FORM Let us now go back to the beginning of the chapter. We said that large numbers can be conveniently expressed using exponents. We have not as yet shown this. We shall do so now. 1. Sun is located 300,000,000,000,000,000,000 m from the centre of our Milky Way Galaxy. 2. Number of stars in our Galaxy is 100,000,000,000. 3. Mass of the Earth is 5,976,000,000,000,000,000,000,000 kg. These numbers are not convenient to write and read. To make it convenient we use powers. TRY THESE Expand by expressing Observe the following: powers of 10 in the 59 = 5.9 × 10 = 5.9 × 101 exponential form: 590 = 5.9 × 100 = 5.9 × 102 (i) 172 5900 = 5.9 × 1000 = 5.9 × 103 (ii) 5,643 (iii) 56,439 59000 =5.9 × 10000 = 5.9 × 104 and so on. (iv) 1,76,428 2024-25  EXPONENTS AND POWERS 183 We have expressed all these numbers in the standard form. Any number can be expressed as a decimal number between 1.0 and 10.0 including 1.0 multiplied by a power of 10. Such a form of a number is called its standard form. Thus, 5,985 = 5.985 × 1,000 = 5.985 × 103 is the standard form of 5,985. Note, 5,985 can also be expressed as 59.85 × 100 or 59.85 × 102. But these are not the standard forms, of 5,985. Similarly, 5,985 = 0.5985 × 10,000 = 0.5985 × 104 is also not the standard form of 5,985. We are now ready to express the large numbers we came across at the beginning of the chapter in this form. The, distance of Sun from the centre of our Galaxy i.e., 300,000,000,000,000,000,000 m can be written as 3.0 × 100,000,000,000,000,000,000 = 3.0 × 1020 m Now, can you express 40,000,000,000 in the similar way? Count the number of zeros in it. It is 10. So, 40,000,000,000 = 4.0 × 1010 Mass of the Earth = 5,976,000,000,000,000,000,000,000 kg = 5.976 × 1024 kg Do you agree with the fact, that the number when written in the standard form is much easier to read, understand and compare than when the number is written with 25 digits? Now, Mass of Uranus = 86,800,000,000,000,000,000,000,000 kg = 8.68 × 1025 kg Simply by comparing the powers of 10 in the above two, you can tell that the mass of Uranus is greater than that of the Earth. The distance between Sun and Saturn is 1,433,500,000,000 m or 1.4335 × 1012 m. The distance betwen Saturn and Uranus is 1,439,000,000,000 m or 1.439 × 1012 m. The distance between Sun and Earth is 149, 600,000,000 m or 1.496 × 1011m. Can you tell which of the three distances is smallest? E XAMPLE 13 Express the following numbers in the standard form: (i) 5985.3 (ii) 65,950 (iii) 3,430,000 (iv) 70,040,000,000 S OLUTION (i) 5985.3 = 5.9853 × 1000 = 5.9853 × 103 (ii) 65,950 = 6.595 × 10,000 = 6.595 × 104 (iii) 3,430,000 = 3.43 × 1,000,000 = 3.43 × 106 (iv) 70,040,000,000 = 7.004 × 10,000,000,000 = 7.004 × 1010 2024-25 184 MATHEMATICS A point to remember is that one less than the digit count (number of digits) to the left of the decimal point in a given number is the exponent of 10 in the standard form. Thus, in 70,040,000,000 there is no decimal point shown; we assume it to be at the (right) end. From there, the count of the places (digits) to the left is 11. The exponent of 10 in the standard form is 11 – 1 = 10. In 5985.3 there are 4 digits to the left of the decimal point and hence the exponent of 10 in the standard form is 4 – 1 = 3. E XERCISE 11.3 1. Write the following numbers in the expanded forms: 279404, 3006194, 2806196, 120719, 20068 2. Find the number from each of the following expanded forms: (a) 8 ×104 + 6 ×103 + 0×102 + 4×101 + 5×100 (b) 4 ×105 + 5×103 + 3×102 + 2×100 (c) 3 ×104 + 7×102 + 5×100 (d) 9 ×105 + 2×102 + 3×101 3. Express the following numbers in standard form: (i) 5,00,00,000 (ii) 70,00,000 (iii) 3,18,65,00,000 (iv) 3,90,878 (v) 39087.8 (vi) 3908.78 4. Express the number appearing in the following statements in standard form. (a) The distance between Earth and Moon is 384,000,000 m. (b) Speed of light in vacuum is 300,000,000 m/s. (c) Diameter of the Earth is 1,27,56,000 m. (d) Diameter of the Sun is 1,400,000,000 m. (e) In a galaxy there are on an average 100,000,000,000 stars. (f) The universe is estimated to be about 12,000,000,000 years old. (g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m. (h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm. (i) The earth has 1,353,000,000 cubic km of sea water. (j) The population of India was about 1,027,000,000 in March, 2001. 2024-25  EXPONENTS AND POWERS 185 W HAT HAVE WE DISCUSSED? 1. Very large numbers are difficult to read, understand, compare and operate upon. To make all these easier, we use exponents, converting many of the large numbers in a shorter form. 2. The following are exponential forms of some numbers? 10,000 = 104 (read as 10 raised to 4) 243 = 35, 128 = 27. Here, 10, 3 and 2 are the bases, whereas 4, 5 and 7 are their respective exponents. We also say, 10,000 is the 4th power of 10, 243 is the 5th power of 3, etc. 3. Numbers in exponential form obey certain laws, which are: For any non-zero integers a and b and whole numbers m and n, (a) am × an = am + n (b) am ÷ an = am – n, m>n (c) (am)n = amn (d) am × bm = (ab)m m  a (e) a ÷ b =   m m b (f) a0 = 1 (g) (–1)even number = 1 (–1)odd number = – 1 2024-25  160 MATHEMATICS Chapter 10 Algebraic Expressions 10.1 INTRODUCTION We have already come across simple algebraic expressions like x + 3, y – 5, 4x + 5, 10y – 5 and so on. In Class VI, we have seen how these expressions are useful in formu- lating puzzles and problems. We have also seen examples of several expressions in the chapter on simple equations. Expressions are a central concept in algebra. This Chapter is devoted to algebraic expressions. When you have studied this Chapter, you will know how algebraic expressions are formed, how they can be combined, how we can find their values and how they can be used. 10.2 HOW ARE EXPRESSIONS FORMED? We now know very well what a variable is. We use letters x, y, l, m,... etc. to denote variables. A variable can take various values. Its value is not fixed. On the other hand, a constant has a fixed value. Examples of constants are: 4, 100, –17, etc. We combine variables and constants to make algebraic expressions. For this, we use the operations of addition, subtraction, multiplication and division. We have already come across expressions like 4x + 5, 10y – 20. The expression 4x + 5 is obtained from the variable x, first by multiplying x by the constant 4 and then adding the constant 5 to the product. Similarly, 10y – 20 is obtained by first multiplying y by 10 and then subtracting 20 from the product. The above expressions were obtained by combining variables with constants. We can also obtain expressions by combining variables with themselves or with other variables. Look at how the following expressions are obtained: x2, 2y2, 3x2 – 5, xy, 4xy + 7 (i) The expression x2 is obtained by multiplying the variable x by itself; x × x = x2 Just as 4 × 4 is written as 42, we write x × x = x2. It is commonly read as x squared. 2024-25  ALGEBRAIC EXPRESSIONS 161 (Later, when you study the chapter ‘Exponents and Powers’ you will realise that x2 may also be read as x raised to the power 2). In the same manner, we can write x × x × x = x3 Commonly, x3 is read as ‘x cubed’. Later, you will realise that x3 may also be read as x raised to the power 3. x, x2, x3,... are all algebraic expressions obtained from x. (ii) The expression 2y 2 is obtained from y: 2y 2 = 2 × y × y Here by multiplying y with y we obtain y 2 and then we multiply y2 by the constant 2. (iii) In (3x2 – 5) we first obtain x2, and multiply it by 3 to get 3x2. From 3x2, we subtract 5 to finally arrive at 3x2 – 5. TRY THESE (iv) In xy, we multiply the variable x with another variable y. Thus, Describe how the x × y = xy. following expressions are obtained: (v) In 4xy + 7, we first obtain xy, multiply it by 4 to get 4xy and add 7 to 4xy to get the expression. 7xy + 5, x2y, 4x2 – 5x 10.3 TERMS OF AN EXPRESSION We shall now put in a systematic form what we have learnt above about how expressions are formed. For this purpose, we need to understand what terms of an expression and their factors are. Consider the expression (4x + 5). In forming this expression, we first formed 4x separately as a product of 4 and x and then added 5 to it. Similarly consider the expression (3x2 + 7y). Here we first formed 3x2 separately as a product of 3, x and x. We then formed 7y separately as a product of 7 and y. Having formed 3x2 and 7y separately, we added them to get the expression. You will find that the expressions we deal with can always be seen this way. They have parts which are formed separately and then added. Such parts of an expression which are formed separately first and then added are known as terms. Look at the expression (4x2 – 3xy). We say that it has two terms, 4x2 and –3xy. The term 4x2 is a product of 4, x and x, and the term (–3xy) is a product of (–3), x and y. Terms are added to form expressions. Just as the terms 4x and 5 are added to form the expression (4x + 5), the terms 4x2 and (–3xy) are added to give the expression (4x2 – 3xy). This is because 4x2 + (–3xy) = 4x2 – 3xy. Note, the minus sign (–) is included in the term. In the expression 4x2 –3xy, we took the term as (–3xy) and not as (3xy). That is why we do not need to say that terms are ‘added or subtracted’ to form an expression; just ‘added’ is enough. Factors of a term We saw above that the expression (4x2 – 3xy) consists of two terms 4x2 and –3xy. The term 4x2 is a product of 4, x and x; we say that 4, x and x are the factors of the term 4x2. A term is a product of its factors. The term –3xy is a product of the factors –3, x and y. 2024-25  162 MATHEMATICS We can represent the terms and factors of the terms of an expression conveniently and elegantly by a tree diagram. The tree for the expression (4x 2 – 3xy) is as shown in the adjacent figure. Note, in the tree diagram, we have used dotted lines for factors and continuous lines for terms. This is to avoid mixing them. Let us draw a tree diagram for the expression 5xy + 10. The factors are such that they cannot be further factorised. Thus we do not write 5xy as 5 × xy, because xy can be further factorised. Similarly, if x3 were a term, it would be written as x × x × x and not x2 × x. Also, remember that 1 is not taken as a separate factor. Coefficients TRY THESE We have learnt how to write a term as a product of factors. 1. What are the terms in the One of these factors may be numerical and the others algebraic following expressions? (i.e., they contain variables). The numerical factor is said to be Show how the terms are the numerical coefficient or simply the coefficient of the term. formed. Draw a tree diagram It is also said to be the coefficient of the rest of the term (which for each expression: is obviously the product of algebraic factors of the term). Thus 8y + 3x2, 7mn – 4, 2x2y. in 5xy, 5 is the coefficient of the term. It is also the coefficient of xy. In the term 10xyz, 10 is the coefficient of xyz, in the term 2. Write three expression each –7x 2y 2, –7 is the coefficient of x 2y 2. having 4 terms. When the coefficient of a term is +1, it is usually omitted. For example, 1x is written as x; 1 x2y2 is written as x2y2 and so on. Also, the coefficient (–1) is indicated only by the minus sign. Thus (–1) x is written as – x; (–1) x 2 y 2 is written as – x2 y2 and so on. Sometimes, the word ‘coefficient’ is used in a more general way. Thus TRY THESE we say that in the term 5xy, 5 is the coefficient of xy, x is the coefficient of 5y and y is the coefficient of 5x. In 10xy 2, 10 is the coefficient of xy 2, x is the Identify the coefficients coefficient of 10y 2 and y 2 is the coefficient of 10x. Thus, in this more general of the terms of following way, a coefficient may be either a numerical factor or an algebraic factor or expressions: a product of two or more factors. It is said to be the coefficient of the 4x – 3y, a + b + 5, 2y + 5, 2xy product of the remaining factors. E XAMPLE 1 Identify, in the following expressions, terms which are not constants. Give their numerical coefficients: xy + 4, 13 – y2, 13 – y + 5y2, 4p2q – 3pq2 + 5 2024-25  ALGEBRAIC EXPRESSIONS 163 S OLUTION S. No. Expression Term (which is not Numerical a Constant) Coefficient (i) xy + 4 xy 1 (ii) 13 – y2 – y2 –1 (iii) 13 – y + 5y2 –y –1 5y2 5 (iv) 4p2q – 3pq2 + 5 4p2q 4 – 3pq2 –3 E XAMPLE 2 (a) What are the coefficients of x in the following expressions? 4x – 3y, 8 – x + y, y2x – y, 2z – 5xz (b) What are the coefficients of y in the following expressions? 4x – 3y, 8 + yz, yz2 + 5, my + m SOLUTION (a) In each expression we look for a term with x as a factor. The remaining part of that term is the coefficient of x. S. No. Expression Term with Factor x Coefficient of x (i) 4x – 3y 4x 4 (ii) 8–x+y –x –1 2 2 (iii) yx–y yx y2 (iv) 2z – 5xz – 5xz – 5z (b) The method is similar to that in (a) above. S. No. Expression Term with factor y Coefficient of y (i) 4x – 3y – 3y –3 (ii) 8 + yz yz z (iii) yz2 + 5 yz2 z2 (iv) my + m my m 10.4 LIKE AND UNLIKE TERMS When terms have the same algebraic factors, they are like terms. When terms have different algebraic factors, they are unlike terms. For example, in the expression 2xy – 3x + 5xy – 4, look at the terms 2xy and 5xy. The factors of 2xy are 2, x and y. The factors of 5xy are 5, x and y. Thus their algebraic (i.e., those which contain variables) factors are the same and 2024-25  164 MATHEMATICS hence they are like terms. On the other hand the TRY THESE terms 2xy and –3x, have different algebraic factors. Group the like terms together from the They are unlike terms. Similarly, the terms, 2xy following: and 4, are unlike terms. Also, the terms –3x and 4 12x, 12, – 25x, – 25, – 25y, 1, x, 12y, y are unlike terms. 10.5 MONOMIALS , B INOMIALS, TRINOMIALS AND POLYNOMIALS An expression with only one term is called a monomial; for example, 7xy, – 5m, 3z2, 4 etc. An expression which contains two unlike terms is called a binomial; for example, x + y, m – 5, mn + 4m, a2 – b2 are TRY THESE binomials. The expression 10pq is not a binomial; it is a Classify the following monomial. The expression (a + b + 5) is not a binomial. expressions as a monomial, It contains three terms. a binomial or a trinomial: a, An expression which contains three terms is called a a + b, ab + a + b, ab + a trinomial; for example, the expressions x + y + 7, ab + a +b, + b – 5, xy, xy + 5, 3x2 – 5x + 2, m + n + 10 are trinomials. The expression 5x2 – x + 2, 4pq – 3q + 5p, ab + a + b + 5 is, however not a trinomial; it contains four 7, 4m – 7n + 10, 4mn + 7. terms and not three. The expression x + y + 5x is not a trinomial as the terms x and 5x are like terms. In general, an expression with one or more terms is called a polynomial. Thus a monomial, a binomial and a trinomial are all polynomials. E XAMPLE 3 State with reasons, which of the following pairs of terms are of like terms and which are of unlike terms: (i) 7x, 12y (ii) 15x, –21x (iii) – 4ab, 7ba (iv) 3xy, 3x (v) 6xy2, 9x2y (vi) pq2, – 4pq2 (vii) mn2, 10mn SOLUTION S. Pair Factors Algebraic Like/ Remarks No. factors same Unlike or different terms (i) 7x 7, x Different Unlike The variables in the 12y 12, y terms are different. (ii) 15x 15, x Same Like –21x –21, x (iii) – 4ab – 4, a, b Same Like Remember 7 ba 7, a, b ab = ba 2024-25  ALGEBRAIC EXPRESSIONS 165 (iv) 3xy 3, x, y Different Unlike The variable y is only 3x 3, x in one term. (v) 6xy2 6, x, y, y Different Unlike The variables in the two 9x 2y 9, x, x, y terms match, but their powers do not match. (vi) pq2 1, p, q, q Same Like Note, numerical 2 – 4pq – 4, p, q, q factor 1 is not shown Following simple steps will help you to decide whether the given terms are like or unlike terms: (i) Ignore the numerical coefficients. Concentrate on the algebraic part of the terms. (ii) Check the variables in the terms. They must be the same. (iii) Next, check the powers of each variable in the terms. They must be the same. Note that in deciding like terms, two things do not matter (1) the numerical coefficients of the terms and (2) the order in which the variables are multiplied in the terms. E XERCISE 10.1 1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations. (i) Subtraction of z from y. (ii) One-half of the sum of numbers x and y. (iii) The number z multiplied by itself. (iv) One-fourth of the product of numbers p and q. (v) Numbers x and y both squared and added. (vi) Number 5 added to three times the product of numbers m and n. (vii) Product of numbers y and z subtracted from 10. (viii) Sum of numbers a and b subtracted from their product. 2. (i) Identify the terms and their factors in the following expressions Show the terms and factors by tree diagrams. (a) x – 3 (b) 1 + x + x2 (c) y – y3 (d) 5xy2 + 7x2y (e) – ab + 2b2 – 3a2 (ii) Identify terms and factors in the expressions given below: (a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y2 (d) xy + 2x2y2 (e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a 2024-25 166 MATHEMATICS 3 1 (g) x+ (h) 0.1 p2 + 0.2 q2 4 4 3. Identify the numerical coefficients of terms (other than constants) in the following expressions: (i) 5 – 3t2 (ii) 1 + t + t2 + t3 (iii) x + 2xy + 3y 2 2 (iv) 100m + 1000n (v) – p q + 7pq (vi) 1.2 a + 0.8 b 2 (vii) 3.14 r (viii) 2 (l + b) (ix) 0.1 y + 0.01 y2 4. (a) Identify terms which contain x and give the coefficient of x. (i) y2x + y (ii) 13y2 – 8yx (iii) x + y + 2 (iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy2 + 25 (vii) 7x + xy2 (b) Identify terms which contain y2 and give the coefficient of y2. (i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2 5. Classify into monomials, binomials and trinomials. (i) 4y – 7z (ii) y2 (iii) x + y – xy (iv) 100 2 2 (v) ab – a – b (vi) 5 – 3t (vii) 4p q – 4pq (viii) 7mn 2 2 2 2 (ix) z – 3z + 8 (x) a + b (xi) z + z (xii) 1 + x + x2 6. State whether a given pair of terms is of like or unlike terms. 5 (i) 1, 100 (ii) –7x, x (iii) – 29x, – 29y 2 (iv) 14xy, 42yx (v) 4m2p, 4mp2 (vi) 12xz, 12x2z2 7. Identify like terms in the following: (a) – xy2, – 4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, – 11yx, 20x2y, – 6x2, y, 2xy, 3x (b) 10pq, 7p, 8q, – p2q2, – 7qp, – 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp, 13p2q, qp2, 701p2 10.6 FINDING THE VALUE OF AN EXPRESSION We know that the value of an algebraic expression depends on the values of the variables forming the expression. There are a number of situations in which we need to find the value of an expression, such as when we wish to check whether a particular value of a variable satisfies a given equation or not. We find values of expressions, also, when we use formulas from geometry and from everyday mathematics. For example, the area of a square is l2, where l is the length of a side of the square. If l = 5 cm., the area is 52 cm2 or 25 cm2; if the side is 10 cm, the area is 102 cm2 or 100 cm2 and so on. We shall see more such examples in the next section. 2024-25  ALGEBRAIC EXPRESSIONS 167 E XAMPLE 4 Find the values of the following expressions for x = 2. (i) x + 4 (ii) 4x – 3 (iii) 19 – 5x2 (iv) 100 – 10x3 S OLUTION Putting x = 2 (i) In x + 4, we get the value of x + 4, i.e., x+4=2+4=6 (ii) In 4x – 3, we get 4x – 3 = (4 × 2) – 3 = 8 – 3 = 5 (iii) In 19 – 5x2, we get 19 – 5x2 = 19 – (5 × 22) = 19 – (5 × 4) = 19 – 20 = – 1 (iv) In 100 – 10x3, we get 100 – 10x3 = 100 – (10 × 23) = 100 – (10 × 8) (Note 23 = 8) = 100 – 80 = 20 E XAMPLE 5 Find the value of the following expressions when n = – 2. (i) 5n – 2 (ii) 5n2 + 5n – 2 (iii) n3 + 5n2 + 5n – 2 S OLUTION (i) Putting the value of n = – 2, in 5n – 2, we get, 5(– 2) – 2 = – 10 – 2 = – 12 (ii) In 5n2 + 5n – 2, we have, for n = –2, 5n – 2 = –12 and 5n2 = 5 × (– 2)2 = 5 × 4 = 20 [as (– 2)2 = 4] Combining, 5n2 + 5n – 2 = 20 – 12 = 8 (iii) Now, for n = – 2, 5n2 + 5n – 2 = 8 and n3 = (–2)3 = (–2) × (–2) × (–2) = – 8 Combining, n3 + 5n2 + 5n – 2 = – 8 + 8 = 0 We shall now consider expressions of two variables, for example, x + y, xy. To work out the numerical value of an expression of two variables, we need to give the values of both variables. For example, the value of (x + y), for x = 3 and y = 5, is 3 + 5 = 8. 2024-25 168 MATHEMATICS E XAMPLE 6 Find the value of the following expressions for a = 3, b = 2. (i) a + b (ii) 7a – 4b (iii) a2 + 2ab + b2 (iv) a3 – b3 S OLUTION Substituting a = 3 and b = 2 in (i) a + b, we get a+b=3+2=5 (ii) 7a – 4b, we get 7a – 4b = 7 × 3 – 4 × 2 = 21 – 8 = 13. (iii) a2 + 2ab + b2, we get a2 + 2ab + b2 = 32 + 2 × 3 × 2 + 22 = 9 + 2 × 6 + 4 = 9 + 12 + 4 = 25 (iv) a3 – b3, we get a3 – b3 = 33 – 23 = 3 × 3 × 3 – 2 × 2 × 2 = 9 × 3 – 4 × 2 = 27 – 8 = 19 E XERCISE 10.2 1. If m = 2, find the value of: (i) m – 2 (ii) 3m – 5 (iii) 9 – 5m 5m (iv) 3m2 – 2m – 7 (v) 4 2 2. If p = – 2, find the value of: (i) 4p + 7 (ii) – 3p2 + 4p + 7 (iii) – 2p3 – 3p2 + 4p + 7 3. Find the value of the following expressions, when x = –1: (i) 2x – 7 (ii) – x + 2 (iii) x2 + 2x + 1 (iv) 2x2 – x – 2 4. If a = 2, b = – 2, find the value of: (i) a2 + b2 (ii) a2 + ab + b2 (iii) a2 – b2 5. When a = 0, b = – 1, find the value of the given expressions: (i) 2a + 2b (ii) 2a2 + b2 + 1 (iii) 2a2b + 2ab2 + ab (iv) a2 + ab + 2 6. Simplify the expressions and find the value if x is equal to 2 (i) x + 7 + 4 (x – 5) (ii) 3 (x + 2) + 5x – 7 (iii) 6x + 5 (x – 2) (iv) 4(2x – 1) + 3x + 11 7. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2. (i) 3x – 5 – x + 9 (ii) 2 – 8x + 4x + 4 2024-25  ALGEBRAIC EXPRESSIONS 169 (iii) 3a + 5 – 8a + 1 (iv) 10 – 3b – 4 – 5b (v) 2a – 2b – 4 – 5 + a 8. (i) If z = 10, find the value of z3 – 3(z – 10). (ii) If p = – 10, find the value of p2 – 2p – 100 9. What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0? 10. Simplify the expression and find its value when a = 5 and b = – 3. 2(a2 + ab) + 3 – ab W HAT HAVE WE DISCUSSED? 1. Algebraic expressions are formed from variables and constants. We use the operations of addition, subtraction, multiplication and division on the variables and constants to form expressions. For example, the expression 4xy + 7 is formed from the variables x and y and constants 4 and 7. The constant 4 and the variables x and y are multiplied to give the product 4xy and the constant 7 is added to this product to give the expression. 2. Expressions are made up of terms. Terms are added to make an expression. For example, the addition of the terms 4xy and 7 gives the expression 4xy + 7. 3. A term is a product of factors. The term 4xy in the expression 4xy + 7 is a product of factors x, y and 4. Factors containing variables are said to be algebraic factors. 4. The coefficient is the numerical factor in the term. Sometimes anyone factor in a term is called the coefficient of the remaining part of the term. 5. Any expression with one or more terms is called a polynomial. Specifically a one term expression is called a monomial; a two-term expression is called a binomial; and a three-term expression is called a trinomial. 6. Terms which have the same algebraic factors are like terms. Terms which have different algebraic factors are unlike terms. Thus, terms 4xy and – 3xy are like terms; but terms 4xy and – 3x are not like terms. 7. In situations such as solving an equation and using a formula, we have to find the value of an expression. The value of the expression depends on the value of the variable from which the expression is formed. Thus, the value of 7x – 3 for x = 5 is 32, since 7(5) – 3 = 35 – 3 = 32. 2024-25  144 MATHEMATICS Chapter 9 Perimeter and Area 9.1 AREA OF A PARALLELOGRAM We come across many shapes other than squares and rectangles. How will you find the area of a land which is a parallelogram in shape? Let us find a method to get the area of a parallelogram. Can a parallelogram be converted into a rectangle of equal area? Draw a parallelogram on a graph paper as shown in Fig 9.1(i). Cut out the parallelogram. Draw a line from one vertex of the parallelogram perpendicular to the opposite side [Fig 9.1(ii)]. Cut out the triangle. Move the triangle to the other side of the parallelogram. (i) (ii) (iii) Fig 9.1 What shape do you get? You get a rectangle. Is the area of the parallelogram equal to the area of the rectangle formed? Yes, area of the parallelogram = area of the rectangle formed What are the length and the breadth of the rectangle? Fig 9.2 2024-25  PERIMETER AND AREA 145 We find that the length of the rectangle formed is equal to the base of the parallelogram and the breadth of the rectangle is equal to the height of the parallelogram (Fig 9.2). Now, Area of parallelogram = Area of rectangle = length × breadth = l × b But the length l and breadth b of the rectangle are exactly the base b and the height h, respectively of the parallelogram. Thus, the area of parallelogram = base × height = b × h. Any side of a parallelogram can be chosen as base of the D C parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height (altitude). In the parallelogram ABCD, DE is height perpendicular to AB. Here AB is the A C E B D base and DE is the height of the base F parallelogram. base In this parallelogram ABCD, BF is the A B perpendicular to opposite side AD. Here AD is the base and BF is the height. height Consider the following parallelograms (Fig 9.2). Fig 9.3 Find the areas of the parallelograms by counting the squares enclosed within the figures and also find the perimeters by measuring the sides. 2024-25 146 MATHEMATICS Complete the following table: Parallelogram Base Height Area Perimeter (a) 5 units 3 units 15 sq units (b) (c) (d) (e) (f) (g) You will find that all these parallelograms have equal areas but different perimeters. Now, consider the following parallelograms with sides 7 cm and 5 cm (Fig 9.4). Fig 9.4 Find the perimeter and area of each of these parallelograms. Analyse your results. You will find that these parallelograms have different areas but equal perimeters. To find the area of a parallelogram, you need to know only the base and the corresponding height of the parallelogram. 2024-25  PERIMETER AND AREA 147 TRY THESE Find the area of following parallelograms: (i) (ii) (iii) In a parallelogram ABCD, AB = 7.2 cm and the perpendicular from C on AB is 4.5 cm. 9.2 AREA OF A TRIANGLE A gardener wants to know the cost of covering the whole of a triangular garden with grass. In this case we need to know the area of the triangular region. Let us find a method to get the area of a triangle. Draw a scalene triangle on a piece of paper. Cut out the triangle. B Place this triangle on another piece of paper and cut out another triangle of the same size. So now you have two scalene triangles of the same size. Are both the triangles congruent? E Superpose one triangle on the other so that they match. C You may have to rotate one of the two triangles. A Now place both the triangles such that a pair of corresponding sides is joined as shown in Fig 9.5. Is the figure thus formed a parallelogram? F D Compare the area of each triangle to the area of the parallelogram. Compare the base and height of the triangles with the base and height of the parallelogram. You will find that the sum of the areas of both the triangles is equal to the area of the parallelogram. The base and the height of the triangle are the same as the base and the height of the parallelogram, respectively. 1 Area of each triangle = (Area of parallelogram) 2 Fig 9.5 1 = (base × height) (Since area of a parallelogram = base × height) 2 1 1 = (b × h) (or bh , in short) 2 2 2024-25  148 MATHEMATICS TRY THESE 1. Try the above activity with different types of triangles. 2. Take different parallelograms. Divide each of the parallelograms into two triangles by cutting along any of its diagonals. Are the triangles congruent? In the figure (Fig 9.6) all the triangles are on the base AB = 6 cm. What can you say about the height of each of the triangles corresponding to the base AB? Can we say all the triangles are equal in area? Yes. Are the triangles congruent also? No. 6 cm We conclude that all the congruent triangles are equal in area but the triangles equal in area Fig 9.6 need not be congruent. Consider the obtuse-angled triangle ABC of base 6 cm (Fig 9.7). A Its height AD which is perpendicular from the vertex A is outside the triangle. 4 cm Can you find the area of the triangle? EXAMPLE 1 One of the sides and the corresponding height of a D B 6 cm C parallelogram are 4 cm and 3 cm respectively. Find the area of the parallelogram (Fig 9.8). Fig 9.7 SOLUTION Given that length of base (b) = 4 cm, height (h) = 3 cm Area of the parallelogram = b × h = 4 cm × 3 cm = 12 cm2 E XAMPLE 2 Find the height ‘x’ if the area of the parallelogram is 24 cm2 and the base is 4 cm. Fig 9.8 SOLUTION Area of parallelogram = b × h Therefore,24 = 4 × x (Fig 9.9) 24 or = x or x = 6 cm 4 Fig 9.9 So, the height of the parallelogram is 6 cm. 2024-25  PERIMETER AND AREA 149 E XAMPLE 3 The two sides of the parallelogram ABCD are 6 cm and 4 cm. The height corresponding to the base CD is 3 cm (Fig 9.10). Find the (i) area of the parallelogram. (ii) the height corresponding to the base AD. S OLUTION (i) Area of parallelogram = b × h = 6 cm × 3 cm = 18 cm2 (ii) base (b) = 4 cm, height = x (say), Area = 18 cm2 A B Area of parallelogram = b × x x 18 = 4 × x m 3 cm 4c 18 =x 4 D C 6 cm Therefore, x = 4.5 cm Fig 9.10 Thus, the height corresponding to base AD is 4.5 cm. E XAMPLE 4 Find the area of the following triangles (Fig 9.11). S (i) Fig 9.11 (ii) S OLUTION 1 1 (i) Area of triangle = bh = × QR × PS 2 2 1 = × 4 cm × 2 cm = 4 cm2 2 1 1 (ii) Area of triangle = bh = × MN × LO 2 2 1 = × 3 cm × 2 cm = 3 cm2 2 2024-25 150 MATHEMATICS E XAMPLE 5 Find BC, if the area of the triangle ABC is 36 cm2 and the height AD is 3 cm (Fig 9.12). S OLUTION Height = 3 cm, Area = 36 cm2 Fig 9.12 1 Area of the triangle ABC = bh 2 1 36 × 2 or 36 = × b × 3 i.e., b= = 24 cm 2 3 So, BC = 24 cm E XAMPLE 6 In ∆PQR, PR = 8 cm, QR = 4 cm and PL = 5 cm (Fig 9.13). Find: (i) the area of the ∆PQR (ii) QM S OLUTION Fig 9.13 (i) QR = base =4 cm, PL = height = 5 cm 1 Area of the triangle PQR = bh 2 1 = × 4 cm × 5 cm = 10 cm2 2 (ii) PR = base = 8 cm QM = height = ? Area = 10 cm2 1 1 Area of triangle = ×b×h i.e., 10 = ×8× h 2 2 10 5 h= = = 2.5. So, QM = 2.5 cm 4 2 2024-25  PERIMETER AND AREA 151 E XERCISE 9.1 1. Find the area of each of the following parallelograms: (a) (b) (c) (d) (e) 2. Find the area of each of the following triangles: (a) (b) (c) (d) 3. Find the missing values: S.No. Base Height Area of the Parallelogram a. 20 cm 246 cm2 b. 15 cm 154.5 cm2 c. 8.4 cm 48.72 cm2 d. 15.6 cm 16.38 cm2 2024-25 152 MATHEMATICS 4. Find the missing values: Base Height Area of Triangle 15 cm ______ 87 cm2 _____ 31.4 mm 1256 mm2 22 cm ______ 170.5 cm2 Fig 9.14 5. PQRS is a parallelogram (Fig 9.14). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find: (a) the area of the parallegram PQRS (b) QN, if PS = 8 cm 6. DL and BM are the heights on sides AB and AD Fig 9.15 respectively of parallelogram ABCD (Fig 9.15). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL. 7. ∆ABC is right angled at A (Fig 9.16). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ∆ABC. Also find the length of AD. Fig 9.16 Fig 9.17 8. ∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 9.17). The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE? 9.3 CIRCLES A racing track is semi-circular at both ends (Fig 9.18). Can you find the distance covered by an athlete if he takes two rounds of a racing track? We need to find a method to find the distances around when a shape is Fig 9.18 circular. 2024-25  PERIMETER AND AREA 153 9.3.1 Circumference of a Circle Tanya cut different cards, in curved shape from a cardboard. She wants to put lace around to decorate these cards. What length of the lace does she require for each? (Fig 9.19) (a) (b) (c) Fig 9.19 You cannot measure the curves with the help of a ruler, as these figures are not “straight”. Fig 9.20 What can you do? Here is a way to find the length of lace required for shape in Fig 9.19(a). Mark a point on the edge of the card and place the card on the table. Mark the position of the point on the table also (Fig 9. 20). Now roll the circular card on the table along a straight line till Fig 9.21 the marked point again touches the table. Measure the distance along the line. This is the length of the lace required (Fig 9.21). It is also the distance along the edge of the card from the marked point back to the marked point. You can also find the distance by putting a string on the edge of the circular object and taking all round it. The distance around a circular region is known as its circumference. D O THIS Take a bottle cap, a bangle or any other circular object and find the circumference. Now, can you find the distance covered by the athlete on the track by this method? Still, it will be very difficult to find the distance around the track or any other circular object by measuring through string. Moreover, the measurement will not be accurate. So, we need some formula for this, as we have for rectilinear figures or shapes. Let us see if there is any relationship between the diameter and the circumference of the circles. Consider the following table: Draw six circles of different radii and find their circumference by using string. Also find the ratio of the circumference to the diameter. Circle Radius Diameter Circumference Ratio of Circumference to Diameter 22 1. 3.5 cm 7.0 cm 22.0 cm = 3.14 7 2024-25 154 MATHEMATICS 44 2. 7.0 cm 14.0 cm 44.0 cm = 3.14 14 66 3. 10.5 cm 21.0 cm 66.0 cm = 3.14 21 132 4. 21.0 cm 42.0 cm 132.0 cm = 3.14 42 32 5. 5.0 cm 10.0 cm 32.0 cm = 3.2 10 94 6. 15.0 cm 30.0 cm 94.0 cm = 3.13 30 What do you infer from the above table? Is this ratio approximately the same? Yes. Can you say that the circumference of a circle is always more than three times its diameter? Yes. 22 This ratio is a constant and is denoted by π (pi). Its approximate value is or 3.14. 7 C So, we can say that = π , where ‘C’ represents circumference of the circle and ‘d ’ d its diameter. or C = πd We know that diameter (d) of a circle is twice the radius (r) i.e., d = 2r So, C = πd = π × 2r or C = 2πr. TRY THESE In Fig 9.22, (a) Which square has the larger perimeter? (b) Which is larger, perimeter of smaller square or the circumference of the circle? Fig 9.22 D O THIS Take one each of quarter plate and half plate. Roll once each of these on a table-top. Which plate covers more distance in one complete revolution? Which plate will take less number of revolutions to cover the length of the table-top? 2024-25  PERIMETER AND AREA 155 E XAMPLE 7 What is the circumference of a circle of diameter 10 cm (Take π = 3.14)? S OLUTION Diameter of the circle (d) = 10 cm Circumference of circle = πd = 3.14 × 10 cm = 31.4 cm So, the circumference of the circle of diameter 10 cm is 31.4 cm. EXAMPLE 8 What is the circumference of a circular disc of radius 14 cm?  22   Use π =  7 S OLUTION Radius of circular disc (r) = 14 cm Circumference of disc = 2πr 22 = 2× × 14 cm = 88 cm 7 So, the circumference of the circular disc is 88 cm. E XAMPLE 9 The radius of a circular pipe is 10 cm. What length of a tape is required to wrap once around the pipe (π = 3.14)? S OLUTION Radius of the pipe (r) = 10 cm Length of tape required is equal to the circumference of the pipe. Circumference of the pipe = 2πr = 2 × 3.14 × 10 cm = 62.8 cm Therefore, length of the tape needed to wrap once around the pipe is 62.8 cm. 22 E XAMPLE 10 Find the perimeter of the given shape (Fig 9.23) (Take π = ). 7 S OLUTION In this shape we need to find the circumference of semicircles on each side of the square. Do you need to find the perimeter of the square also? No. The outer boundary, of this figure is made up of semicircles. Diameter of each semicircle is 14 cm. We know that: Circumference of the circle = πd 1 14 cm Circumference of the semicircle = πd 2 14 cm 1 22 = × × 14 cm = 22 cm 2 7 Circumference of each of the semicircles is 22 cm Therefore, perimeter of the given figure = 4 × 22 cm = 88 cm Fig 9.23 2024-25  156 MATHEMATICS E XAMPLE 11 Sudhanshu divides a circular disc of radius 7 cm in two equal parts. 22 What is the perimeter of each semicircular shape disc? (Use π = ) 7 SOLUTION To find the perimeter of the semicircular disc (Fig 9.24), we need to find (i) Circumference of semicircular shape (ii) Diameter Given that radius (r) = 7 cm. We know that the circumference of circle = 2πr 1 So, the circumference of the semicircle = × 2πr = πr 2 22 × 7 cm = 22 cm = 7 Fig 9.24 So, the diameter of the circle = 2r = 2 × 7 cm = 14 cm Thus, perimeter of each semicircular disc = 22 cm + 14 cm = 36 cm 9.3.2 Area of Circle Consider the following: l A farmer dug a flower bed of radius 7 m at the centre of a field. He needs to purchase fertiliser. If 1 kg of fertiliser is required for 1 square metre area, how much fertiliser should he purchase? l What will be the cost of polishing a circular table-top of radius 2 m at the rate of ` 10 per square metre? Can you tell what we need to find in such cases, Area or Perimeter? In such cases we need to find the area of the circular region. Let us find the area of a circle, using graph paper. Draw a circle of radius 4 cm on a graph paper (Fig 9.25). Find the area by counting the number of squares enclosed. As the edges are not straight, we get a rough estimate of the area of circle by this method. There is another way of finding the area of a circle. Fig 9.25 Draw a circle and shade one half of the circle [Fig 9.26(i)]. Now fold the circle into eighths and cut along the folds [Fig 9.26(ii)]. (i) (ii) Fig 9.27 Fig 9.26 Arrange the separate pieces as shown, in Fig 9.27, which is roughly a parallelogram. The more sectors we have, the nearer we reach an appropriate parallelogram. 2024-25  PERIMETER AND AREA 157 As done above if we divide the circle in 64 sectors, and arrange these sectors. It gives nearly a rectangle (Fig 9.28). Radius Circumference Fig 9.28 What is the breadth of this rectangle? The breadth of this rectangle is the radius of the circle, i.e., ‘r’. As the whole circle is divided into 64 sectors and on each side we have 32 sectors, the length of the rectangle is the length of the 32 sectors, which is half of the circumference. (Fig 9.28) Area of the circle = Area of rectangle thus formed = l × b 1  = (Half of circumference) × radius =  × 2πr  × r = πr2 2  So, the area of the circle = πr 2 TRY THESE Draw circles of different radii on a graph paper. Find the area by counting the number of squares. Also find the area by using the formula. Compare the two answers. E XAMPLE 12 Find the area of a circle of radius 30 cm (use π = 3.14). S OLUTION Radius, r = 30 cm Area of the circle = πr2 = 3.14 × 302 = 2,826 cm2 E XAMPLE 13 Diameter of a circular garden is 9.8 m. Find its area. S OLUTION Diameter, d = 9.8 m. Therefore, radius r = 9.8 ÷ 2 = 4.9 m 22 22 Area of the circle = πr2 = × (4.9) 2 m2 = × 4.9 × 4.9 m2 = 75.46 m2 7 7 E XAMPLE 14 The adjoining figure shows two circles with the same centre. The radius of the larger circle is 10 cm and the radius of the smaller circle is 4 cm. Find: (a) the area of the larger circle (b) the area of the smaller circle (c) the shaded area between the two circles. (π = 3.14) 2024-25 158 MATHEMATICS S OLUTION (a) Radius of the larger circle = 10 cm So, area of the larger circle = πr 2 = 3.14 × 10 × 10 = 314 cm2 (b) Radius of the smaller circle = 4 cm Area of the smaller circle = πr 2 = 3.14 × 4 × 4 = 50.24 cm2 (c) Area of the shaded region = (314 – 50.24) cm2 = 263.76 cm2 E XERCISE 9.2 22 1. Find the circumference of the circles with the following radius: (Take π = ) 7 (a) 14 cm (b) 28 mm (c) 21 cm 2. Find the area of the following circles, given that: 22 (a) radius = 14 mm (Take π = ) (b) diameter = 49 m 7 (c) radius = 5 cm 3. If the circumference of a circular sheet is 154 m, find its radius. Also find the area of 22 the sheet. (Take π = ) 7 4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the 22 rope, if it costs ` 4 per meter. (Take π = ) 7 5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14) 6. Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ` 15. (Take π = 3.14) 7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter. 8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ` 15/m2. (Take π = 3.14) 9. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more