# Ex.9.5 Q7 Algebraic Expressions and Identities Solution - NCERT Maths Class 8

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## Question

Using $${a^2}{\rm{ }} - {\rm{ }}{b^2} = \left( {a{\rm{ }} + {\rm{ }}b} \right){\rm{ }}\left( {a{\rm{ }} - {\rm{ }}b} \right),$$ find

(i) \begin{align} \quad{{51}^2} -{{49}^2}\end{align}

(ii) \begin{align} \quad {{\left( {1.02} \right)}^2} - {{\left( {0.98} \right)}^2}\end{align}

(iii) \begin{align}\quad {{153}^2} - {{147}^2}\end{align}

(iv) \begin{align} \quad {{12.1}^2} - {{7.9}^2}\end{align}

Video Solution
Algebraic Expressions & Identities
Ex 9.5 | Question 7

## Text Solution

What is known?

$${a^2}{\rm{ }} - {\rm{ }}{b^2} = \left( {a{\rm{ }} + {\rm{ }}b} \right){\rm{ }}\left( {a{\rm{ }} - {\rm{ }}b} \right),$$

What is unknown?

Results of the given expression with their corresponding values

Steps:

(i) \begin{align} \quad{{51}^2} -{{49}^2}\end{align}

\begin{align}&= \left( {51 + 49} \right)\left( {51 - 49} \right)\\&= \left( {100} \right)\left( 2 \right) = 200\end{align}

(ii) \begin{align} \quad {{\left( {1.02} \right)}^2} - {{\left( {0.98} \right)}^2}\end{align}

\begin{align}&= \left( {1.02 + 0.98} \right)\left( {1.02 - 0.98} \right)\\&= \left( 2 \right)\left( {0.04} \right)\\&= 0.08\end{align}

(iii) \begin{align}\quad {{153}^2} - {{147}^2}\end{align}

\begin{align}&= \left( {153 + 147} \right)\left( {153 - 147} \right)\\&= \left( {300} \right)\left( 6 \right)\\&= 1800\end{align}

(iv) \begin{align} \quad {{12.1}^2} - {{7.9}^2}\end{align}

\begin{align}&= \left( {12.1 + 7.9} \right)\left( {12.1 - 7.9} \right) \\&= \left( {20.0} \right)\left( {4.2} \right) \\&= 84\end{align}

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