Question

Using identities, evaluate
(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

i) ${71}^2=(70+1{)}^2$

$=(70{)}^2+1^2+2(70\left)\right(1)$ $\because \left[\right(a+b{)}^2$$=a^2+b^2+2ab]$

$=4900+1+140$

$=5041$

ii) ${99}^2=(100-1{)}^2$

$=(100{)}^2+1^2-2(100\left)\right(1)$ $\because \left[\right(a-b{)}^2$$=a^2+b^2-2ab]$

$=10000+1-200$

$=9801$

iii) ${102}^2=(100+2{)}^2$

$=(100{)}^2+2^2+2(100\left)\right(2)$ $\because \left[\right(a+b{)}^2$$=a^2+b^2+2ab]$

$=10000+4+400$

$=10404$

iv) ${998}^2=(1000-2{)}^2$

$=(1000{)}^2+2^2-2(1000\left)\right(2)$ $\because \left[\right(a-b{)}^2$$=a^2+b^2-2ab]$

$=1000000+4-4000$

$=996004$

v) ${5.2}^2=(5+0.2{)}^2$

$=(5{)}^2+(0.2{)}^2+2\left(5\right)\left(0.2\right)$ $\because \left[\right(a+b{)}^2$$=a^2+b^2+2ab]$

$=25+0.04+2$

$=27.04$

vi) $297\times 303=(300-3)(300+3)$

$=(300{)}^2-(3{)}^2$ $\because \left[\right(a^2-b^2)$$=(a+b)(a-b)]$

$=90000-9$

$=89991$

vii) $78\times 82=(80-2)(80+2)$

$=(80{)}^2-(2{)}^2$ $\because \left[\right(a^2-b^2)$$=(a+b)(a-b)]$

$=6400-4$

$=6396$

viii) $(8.9{)}^2=(9-0.1{)}^2$

$=(9{)}^2+(0.1{)}^2-2\left(9\right)\left(0.1\right)$ $\because \left[\right(a-b{)}^2$$=a^2+b^2-2ab]$

$=81+0.01-1.8$

$=79.21$

vii) $1.05\times 9.5=\frac{105}{100}\times \frac{95}{10}$

$=\frac{1}{1000}\times (105\times 95)$

$=\frac{1}{1000}\times (100+5)(100-5)$

$=\frac{1}{1000}\times \left[\right(100{)}^2-\left(5{)}^2\right]$

$=\frac{1}{1000}\times (10000-25)$

$=\frac{9975}{1000}=9.975$