Ex 9.5, 6 - Using identities, evaluate (i) 71^2 (ii) 99^2 (iii) 102^2

Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3 Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4 Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 5 Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 6 Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 7 Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 8 Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 9 Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 10 Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 11 Ex 9.5, 6 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 12

  1. Chapter 9 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

Transcript

Ex 9.5, 6 Using identities, evaluate. (i) ใ€–71ใ€—^2 ใ€–71ใ€—^2 =(70+1)^2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 70 & ๐‘ = 1 = (70)^2+(1)^2+2(70)(1) = 4900+1+140 = ๐Ÿ“๐ŸŽ๐Ÿ’๐Ÿ Ex 9.5, 6 Using identities, evaluate. (ii) ใ€–99ใ€—^2 ใ€–99ใ€—^2 =(100โˆ’1)^2 (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 100 & ๐‘ = 1 = (100)^2+(1)^2โˆ’2(100)(1) = 10000+1โˆ’200 = 10001โˆ’200 = ๐Ÿ—๐Ÿ–๐ŸŽ๐Ÿ Ex 9.5, 6 Using identities, evaluate. (iii) ใ€–102ใ€—^2 ใ€–102ใ€—^2 =(100+2)^2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 100 & ๐‘ = 2 = (100)^2+(2)^2+2(100)(2) = 10000+4+400 = ๐Ÿ๐ŸŽ404 Ex 9.5, 6 Using identities, evaluate. (iv) ใ€–998ใ€—^2 ใ€–998ใ€—^2 =(1000โˆ’2)^2 (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 1000 & ๐‘ = 2 = (1000)^2+(2)^2โˆ’2(1000)(2) = 1000000+4โˆ’4000 = 1000004โˆ’4000 = ๐Ÿ—๐Ÿ—๐Ÿ”๐ŸŽ๐ŸŽ๐Ÿ’ Ex 9.5, 6 Using identities, evaluate. (v) ใ€–5.2ใ€—^2 ใ€–5.2ใ€—^2 =(5+0.2)^2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 5 & ๐‘ = 0.2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 5 & ๐‘ = 0.2 = 27+ 4/100 = 27+0.04 = ๐Ÿ๐Ÿ•.๐ŸŽ๐Ÿ’ Ex 9.5, 6 Using identities, evaluate. (vi) 297 ร— 303 297 ร— 303 = (300โˆ’3)ร—(300+3) (๐‘Žโˆ’๐‘)(๐‘Ž+๐‘)=๐‘Ž^2โˆ’๐‘^2 Putting ๐‘Ž = 300 & ๐‘ = 3 = (300)^2โˆ’(3)^2 = 90000โˆ’9 = ๐Ÿ–๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ Ex 9.5, 6 Using identities, evaluate. (vii) 78 ร— 82 78 ร— 82 = (80โˆ’2)ร—(80+2) (๐‘Žโˆ’๐‘)(๐‘Ž+๐‘)=๐‘Ž^2โˆ’๐‘^2 Putting ๐‘Ž = 300 & ๐‘ = 3 = (80)^2โˆ’(2)^2 = 6400โˆ’4 = ๐Ÿ”๐Ÿ‘๐Ÿ—๐Ÿ” Ex 9.5, 6 Using identities, evaluate. (viii) ใ€–8.9ใ€—^2 ใ€–8.9ใ€—^2 = (9โˆ’0.1)^2 (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 9 & ๐‘ = 0.1 = (9)^2+(0.1)^2โˆ’2(9)(0.9) = 81+(1/10)^2โˆ’(2ร—9ร—1/10) = 81 + 1/100 โˆ’ 18/10 = (81 ร— 100 + 1 โˆ’ 180)/100 = (8100 + 1 โˆ’ 18)/100 = (8101 โˆ’ 18)/100 = 7921/100 = 79.21 Ex 9.5, 6 Using identities, evaluate. (ix) 1.05ร—9.5 1.05ร—9.5 = 105/100ร—95/10 = (105 ร— 95)/1000 = ((100 + 5) ร— (100 โˆ’ 5))/1000 (๐‘Ž+๐‘)(๐‘Žโˆ’๐‘)=๐‘Ž^2โˆ’๐‘^2 Putting ๐‘Ž = 100 & ๐‘ = 5 = ((100)^2 โˆ’ (5)^2)/1000 = (10000 โˆ’ 25)/1000 = 9975/1000 = 9.975

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.