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) $10.5\times 9.5$

(i) Given that, ${71}^2$

We know that ${\left(a-b\right)}^2=a^2-2ab+b^2 , {\left(a+b\right)}^2=a^2+2ab+b^2 and a^2-b^2=(a-b)(a+b)$

Using identities to evaluate the following,

${71}^2 = {(70+1)}^2 = {70}^2 + 140 + 12 = 4900 + 140 +1 = 5041$

Hence, ${71}^2=5401$.

(ii) Given that , ${99}^2$

Using identities to evaluate the following,

${99}^2 = {(100 -1)}^2 = {100}^2 – 200 + 12 = 10000 – 200 + 1 = 9801$

Hence, ${99}^2=9801$.

(iii) Given that , ${102}^2$

Using identities to evaluate the following,

${102}^2 = {(100 + 2)}^2 = {100}^2 + 400 + 2^2 = 10000 + 400 + 4 = 10404$

Hence, ${102}^2=10404$.

(iv) Given that , ${998}^2$

Using identities to evaluate the following,

${998}^2 = {(1000 – 2)}^2 = {1000}^2 – 4000 + 2^{ }2 = 1000000 – 4000 + 4 = 996004$

Hence, ${998}^2=996004$.

(v) Given that , $5.2^2$

Using identities to evaluate the following,

$5.2^2 = {(5 + 0.2)}^2 = 5^2 + 2 + 0.2^2 = 25 + 2 + 0.04 = 27.04$

Hence, $5.2^2=27.04$.

(vi) Given that , $297\times 303$

Using identities to evaluate the following,

$297\times 303 = (300 – 3 )(300 + 3) = {300}^2 – 3^2 = 90000 – 9 = 89991$

Hence, $297\times 303=89991$.

(vii) Given that , $78\times 82$

Using identities to evaluate the following,

$78\times 82 = (80 – 2)(80 + 2) = {80}^2 – 2^2 = 6400 – 4 = 6396$

Hence, $78\times 82=6396$.

(viii) Given that , $8.9^2$

Using identities to evaluate the following,

$8.9^2 = {(9 – 0.1)}^2 = 9^2 – 1.8 + 0.1^2 = 81 – 1.8 + 0.01 = 79.21$

Hence, $8.9^2=79.21$.

(ix) Given that , $10.5\times 9.5$

Using identities to evaluate the following,

$10.5 x 9.5 = (10 + 0.5)(10 – 0.5) = {10}^2 – 0.5^2 = 100 – 0.25 = 99.75$

Hence, $10.5\times 9.5=99.75$.