Evaluate each of the following using identities:
(i) $(399)^{2}$
(ii) $(0.98)^{2}$
(iii) $991 \times 1009$
(iv) $117 \times 83$
(i) We have,
$399^{2}=(400-1)^{2}$
$=(400)^{2}+(1)^{2}-2 \times 400 \times 1 \quad\left[(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
Where, a = 400 and b = 1
= 160000 + 1 - 8000
= 159201
Therefore, $(399)^{2}=159201$.
(ii) We have,
$(0.98)^{2}=(1-0.02)^{2}$
$=(1)^{2}+(0.02)^{2}-2 \times 1 \times 0.02$
= 1 + 0.0004 - 0.04 [Where, a = 1 and b = 0.02]
= 1.0004 - 0.04
= 0.9604
Therefore, $(0.98)^{2}=0.9604$
(iii) 991 × 1009
Solution:
We have,
991 × 1009
= (1000 - 9)(1000 + 9)
$=(1000)^{2}-(9)^{2} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$
= 1000000 - 81 [Where a = 1000 and b = 9]
= 999919
Therefore, 991 × 1009 = 999919
(iv) We have,
117 × 83
= (100 + 17)(100 - 17)
$=(100)^{2}-(17)^{2} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$
= 10000 - 289 [Where a = 100 and b = 17]
= 9711
Therefore, 117 × 83 = 9711