Question

Using identities, evaluate the following:
(i) ${{71}^{2}}$
(ii) ${{99}^{2}}$
(iii) ${{102}^{2}}$
(iv) ${{998}^{2}}$
(v) ${{5.2}^{2}}$
(vi) $297\times 303$
(vii) $78\times 82$
(viii) ${{8.9}^{2}}$
(ix) $1.05\times 9.5$

Answer 1

Hint: Use different algebraic identities in different questions to simplify the problem and for ease in calculation. Use:

& {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ & \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \\\ \end{aligned}$$ “Complete step-by-step answer:” An algebraic identity is an equality that holds for any value of its variable. For example, the identity ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$ holds for all values of $x$ and $y$. $$\begin{aligned} & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\text{ }................\text{(1)} \\\ & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\text{ }................\text{(2)} \\\ & \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\text{ }................\text{(3)} \\\ \end{aligned}$$ Now, we come to the question one by one. (i) ${{71}^{2}}$ This can be solved by using identity (1). $$71\text{ }=\text{ }70+1$$, therefore, $\begin{aligned} & {{71}^{2}}={{\left( 70+1 \right)}^{2}}={{70}^{2}}+{{1}^{2}}+2\times 70\times 1 \\\ & =4900+1+140 \\\ & =5041 \\\ \end{aligned}$ (ii) ${{99}^{2}}$ This can be solved by using identity (2). $99=100-1$, therefore, $\begin{aligned} & {{99}^{2}}={{\left( 100-1 \right)}^{2}}={{100}^{2}}+{{1}^{2}}-2\times 100\times 1 \\\ & =10000+1-200 \\\ & =9801 \\\ \end{aligned}$ (iii) ${{102}^{2}}$ $102=100+2$ Using identity (1) we have, $\begin{aligned} & {{\left( 102 \right)}^{2}}={{\left( 100+2 \right)}^{2}}={{100}^{2}}+{{2}^{2}}+2\times 100\times 2 \\\ & =10000+4+400 \\\ & =10404 \\\ \end{aligned}$ (iv) ${{998}^{2}}$ $$\begin{aligned} & {{\left( 998 \right)}^{2}}={{\left( 1000-2 \right)}^{2}}={{1000}^{2}}+{{2}^{2}}-2\times 1000\times 2 \\\ & =1000000+4-4000 \\\ & =996004 \\\ \end{aligned}$$ (v) ${{5.2}^{2}}$ $$\begin{aligned} & {{\left( 5.2 \right)}^{2}}={{\left( 5+0.2 \right)}^{2}}={{5}^{2}}+{{0.2}^{2}}-2\times 5\times 0.2 \\\ & =25+0.04-2 \\\ & =23.04 \\\ \end{aligned}$$ (vi) $297\times 303$ This can be written as, $297\times 303=(300-3)\times (300+3)$. Now, using identity (3) we get, $\begin{aligned} & (300-3)\times (300+3) \\\ & ={{300}^{2}}-{{3}^{2}} \\\ & =90000-9 \\\ & =89991 \\\ \end{aligned}$ (vii) $78\times 82$ $78\times 82=(80-2)\times (80+2)$ Again, using identity (3), $\begin{aligned} & (80-2)\times (80+2) \\\ & ={{80}^{2}}-{{2}^{2}} \\\ & =6400-4 \\\ & =6396 \\\ \end{aligned}$ (viii) ${{8.9}^{2}}$ $$8.9=(9-0.1)$$, therefore using identity (2) we have, $\begin{aligned} & {{8.9}^{2}}={{(9-0.1)}^{2}} \\\ & ={{9}^{2}}+{{0.1}^{2}}-2\times 9\times 0.1 \\\ & =81+0.01-1.8 \\\ & =79.21 \\\ \end{aligned}$ (ix) $1.05\times 9.5$ $1.05\times 9.5=\left( 1+0.05 \right)\times \left( 1-0.5 \right)$ We can clearly see that this problem is not based on any identity. But it can be solved easily when we will break the decimal part and then multiply the two terms. Therefore, $\begin{aligned} & \left( 1+0.05 \right)\times \left( 1-0.5 \right) \\\ & =1\times 1-1\times 0.5+1\times 0.05-0.05\times 0.5 \\\ & =1-0.5+0.05-0.025 \\\ & =0.525 \\\ \end{aligned}$ Note: We have applied identity (1) in question (i), (iii) and (iv); identity (2) in question (ii), (iv) and (viii) and identity (3) in (vi) and (vii), while no identity has been used in question (ix). They are used in their respective places to make our calculation easier. Otherwise it would be difficult for us to directly multiply and calculate the value like in question number (iv) and (vi).