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

Transcript

Ex 9.5, 6 Using identities, evaluate. (i) γ€–71γ€—^2 γ€–71γ€—^2 =(70+1)^2 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 70 & 𝑏 = 1 = (70)^2+(1)^2+2(70)(1) = 4900+1+140 = πŸ“πŸŽπŸ’πŸ Ex 9.5, 6 Using identities, evaluate. (ii) γ€–99γ€—^2 γ€–99γ€—^2 =(100βˆ’1)^2 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 100 & 𝑏 = 1 = (100)^2+(1)^2βˆ’2(100)(1) = 10000+1βˆ’200 = 10001βˆ’200 = πŸ—πŸ–πŸŽπŸ Ex 9.5, 6 Using identities, evaluate. (iii) γ€–102γ€—^2 γ€–102γ€—^2 =(100+2)^2 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 100 & 𝑏 = 2 = (100)^2+(2)^2+2(100)(2) = 10000+4+400 = 𝟏𝟎404 Ex 9.5, 6 Using identities, evaluate. (iv) γ€–998γ€—^2 γ€–998γ€—^2 =(1000βˆ’2)^2 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 1000 & 𝑏 = 2 = (1000)^2+(2)^2βˆ’2(1000)(2) = 1000000+4βˆ’4000 = 1000004βˆ’4000 = πŸ—πŸ—πŸ”πŸŽπŸŽπŸ’ Ex 9.5, 6 Using identities, evaluate. (v) γ€–5.2γ€—^2 γ€–5.2γ€—^2 =(5+0.2)^2 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 5 & 𝑏 = 0.2 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 5 & 𝑏 = 0.2 = 27+ 4/100 = 27+0.04 = πŸπŸ•.πŸŽπŸ’ Ex 9.5, 6 Using identities, evaluate. (vi) 297 Γ— 303 297 Γ— 303 = (300βˆ’3)Γ—(300+3) (π‘Žβˆ’π‘)(π‘Ž+𝑏)=π‘Ž^2βˆ’π‘^2 Putting π‘Ž = 300 & 𝑏 = 3 = (300)^2βˆ’(3)^2 = 90000βˆ’9 = πŸ–πŸ—πŸ—πŸ—πŸ Ex 9.5, 6 Using identities, evaluate. (vii) 78 Γ— 82 78 Γ— 82 = (80βˆ’2)Γ—(80+2) (π‘Žβˆ’π‘)(π‘Ž+𝑏)=π‘Ž^2βˆ’π‘^2 Putting π‘Ž = 300 & 𝑏 = 3 = (80)^2βˆ’(2)^2 = 6400βˆ’4 = πŸ”πŸ‘πŸ—πŸ” Ex 9.5, 6 Using identities, evaluate. (viii) γ€–8.9γ€—^2 γ€–8.9γ€—^2 = (9βˆ’0.1)^2 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 9 & 𝑏 = 0.1 = (9)^2+(0.1)^2βˆ’2(9)(0.9) = 81+(1/10)^2βˆ’(2Γ—9Γ—1/10) = 81 + 1/100 βˆ’ 18/10 = (81 Γ— 100 + 1 βˆ’ 180)/100 = (8100 + 1 βˆ’ 18)/100 = (8101 βˆ’ 18)/100 = 7921/100 = 79.21 Ex 9.5, 6 Using identities, evaluate. (ix) 1.05Γ—9.5 1.05Γ—9.5 = 105/100Γ—95/10 = (105 Γ— 95)/1000 = ((100 + 5) Γ— (100 βˆ’ 5))/1000 (π‘Ž+𝑏)(π‘Žβˆ’π‘)=π‘Ž^2βˆ’π‘^2 Putting π‘Ž = 100 & 𝑏 = 5 = ((100)^2 βˆ’ (5)^2)/1000 = (10000 βˆ’ 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 9 years. He provides courses for Maths and Science at Teachoo.