6.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) $$1.05 \times 9.5$$

(i)To find: $$71^2$$

$$=(70+1)^2$$

$$=(70)^2+2(70)(1)+1^2\quad$$[∵$$(a+b)^2=a^2+2ab+b^2$$]

$$=4900+140+1$$

$$=5041$$

Thus, we have $$71^2=5041$$

(ii)To find: $$99^2$$

$$=(100-1)^2$$

$$=(100)^2-2(100)(1)+1^2\quad$$[∵$$(a-b)^2=a^2-2ab+b^2$$]

$$=10000-200+1$$

$$=9801$$

Thus, we have $$99^2=9801$$

(iii)To find: $$102^2$$

$$=(100+2)^2$$

$$=(100)^2+2(100)(2)+2^2\quad$$[∵$$(a+b)^2=a^2+2ab+b^2$$]

$$=10000+400+4$$

$$=10404$$

Thus, we have $$102^2=10404$$

(iv)To find: $$998^2$$

$$=(1000-2)^2$$

$$=(1000)^2-2(1000)(2)+2^2\quad$$[∵$$(a-b)^2=a^2-2ab+b^2$$]

$$=1000000-4000+4$$

$$=996004$$

Thus, we have $$998^2=996004$$

(v)To find: $$5.2^2$$

$$=(5+0.2)^2$$

$$=(5)^2+2(5)(0.2+(0.2)^2\quad$$[∵$$(a+b)^2=a^2+2ab+b^2$$]

$$=25+2+0.04$$

$$=27.04$$

Thus, we have $$(5.2)^2=27.04$$

(vi)To find:$$297\times 303$$

$$(300-3)(300+3)=(300)^2-(3)^2\quad$$[using $$(a-b)(a+b)=a^2-b^2$$]>

$$=90000-9$$

$$=89991$$

Hence we have,$$297\times303=89991$$

(vii)To find:$$78\times 82$$

$$(80-2)(80+2)=(80)^2-(2)^2\quad$$[using $$(a-b)(a+b)=a^2-b^2$$]>

$$=6400-4$$

$$=6396$$

Hence we have,$$78\times82=6396$$

(viii)To find: $$8.9^2$$

$$=(9-0.1)^2$$

$$=(9)^2-2(9)(0.1)+(0.1)^2\quad$$[∵$$(a-b)^2=a^2-2ab+b^2$$]

$$=81-1.8+0.01$$

$$=79.21$$

Thus, we have $$(8.9)^2=79.21$$

(ix)To find:$$1.05\times 9.5$$

$$(1+0.05)(10-0.5)=1(10-0.5)+0.05(10-0.5)$$>

$$=10-0.5+0.05\times10-0.05\times0.5$$

$$=10-0.5+0.5-0.025$$

$$=10.5-0.525$$

$$=9.975$$

Hence we have,$$1.05\times9.5=9.975$$