# Ex.9.5 Q6 Algebraic Expressions and Identities Solution - NCERT Maths Class 8

## Question

Using identities, evaluate.

(i) \begin{align}{{71^2}} \end{align}

(ii) \begin{align}{{99^2}}\end{align}

(iii) \begin{align}{{102^2}}\end{align}

(iv) \begin{align}{{998^2}}\end{align}

(v) \begin{align}{{\left( {5.2} \right)^2}}\end{align}

(vi) \begin{align}297\times 303\end{align}

(vii) \begin{align} 78\times 82\end{align}

(viii) \begin{align} {{8.9^2}}\end{align}

(ix) \begin{align} 1.05\times 9.5\end{align}

Video Solution
Algebraic Expressions & Identities
Ex 9.5 | Question 6

## Text Solution

What is known?

Expressions

What is unknown?

Values of the expressions

Reasoning:

\begin{align}{(a + b)^2} &= {a^2} + 2ab + {b^2}\\{(a - b)^2} &= {a^2} - 2ab + {b^2}\a + b)(a - b) &= {a^2} - {b^2}\end{align} Steps: (i) \(\begin{align} \quad {{71^2}} \end{align}

\begin{align}&= {{\left( {70 + 1} \right)}^2}\\&= {{\left( {70} \right)}^2} + 2\left( {70} \right)\left( 1 \right) + {{\left( 1 \right)}^2} \\ & \quad [ \left( {a + b} \right)^2 = {a^2} + 2ab + {b^2} ]\\&= 4900 + 140 + 1\\ &= 5041\end{align}

(ii) \begin{align} \quad {{99^2}}\end{align}

\begin{align} &={{(100-1)}^{2}} \\&=\left( 100 \right)^{2}-2\left( 100 \right)\left( 1 \right)+{{\left( 1 \right)}^{2}} \\ & \quad [ \left( a-b \right)^{2} ={{a}^{2}} -2ab+{{b}^{2}} ]\\& =10000-200+1 \\& =9801\end{align}

(iii) \begin{align} \quad {{102^2}}\end{align}

\begin{align} &= {{\left( {100 + 2} \right)}^2}\\&= {{\left( {100} \right)}^2} + 2\left( {100} \right)\left( 2 \right) + {{\left( 2 \right)}^2} \\ & \quad [\left( {a + b} \right)^2 = {a^2} + 2ab + {b^2} ] \\&= 10000 + 400 + 4\\&= 10404\end{align}

(iv) \begin{align} \quad {{998^2}}\end{align}

\begin{align}&= \left( {1000 - 2} \right)^2\\&= \left( {1000} \right)^2 - 2\left( {1000} \right) \!\left( 2 \right) \!\!+ \!\!\left( 2 \right)^2 \\ & \quad [\left( {a - b} \right)^2 = {a^2} - 2ab + {b^2}] \\&= 1000000 - 4000 + 4\\& = 996004\end{align}

(v) \begin{align} \quad {{\left( {5.2} \right)^2}}\end{align}

\begin{align}&= \left( {5.0 + 0.2}\right)^2\\&= \left( {5.0} \right)^2 \!+ 2\left( {5.0} \right) \! \left( {0.2} \right) \! + \left( \! {0.2} \! \right)^2 \\ & \quad [ \left( {a + b} \right)^2 = {a^2} + 2ab + {b^2}] \\&= 25 + 2 + 0.04\\&= 27.04\end{align}

(vi) \begin{align} \quad 297\times 303\end{align}

\begin{align}&= \left( {300 - 3} \right) \times \left( {300 + 3} \right)\\ &= {{\left( {300} \right)}^2} - {{\left( 3 \right)}^2} \\ & \quad [ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}] \\&= 90000 - 9\\&= 89991\end{align}

(vii) \begin{align} \quad 78\times 82\end{align}

\begin{align}&= \left( {80 - 2} \right)\left( {80 + 2} \right)\\ &= {{\left( {80} \right)}^2} - {{\left( 2 \right)}^2} \\ & \quad[ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}] \\&= 6400 - 4\\&= 6396\end{align}

(viii) \begin{align} \quad {{8.9^2}}\end{align}

\begin{align}&= \left( {9.0 - 0.1} \right)^2 \\&= \left( {9.0} \right)^2 - \!2\left( {9.0}\! \right) \!\left( {0.1} \!\right) \!+ \left( {0.1} \!\right)^2 \\ & \quad [\left( {a - b} \right)^2 = {a^2} - 2ab + {b^2} ] \\&= 81 - 1.8 + 0.01\\&= 79.21\end{align}

(ix) \begin{align} \quad 1.05\times 9.5\end{align}

\begin{align}&= 1.05 \times 0.95 \times 10\\&= \left( {1 + 0.05} \right)\left( {1 - 0.05} \right) \times 10\\&= \left[ {{{\left( 1 \right)}^2} - {{\left( {0.05} \right)}^2}} \right] \times 10\\&= \left[ {1 - 0.0025} \right] \times 10 \\ & \quad [ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}] \\&= 0.9975 \times 10\\&= 9.975\end{align}

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