Using suitable identity, evaluate the following.
(a) $ (102)^{2} $
(b) $ (98)^{2} $
(c) $ 104 \times 105 $
(d) $ 215^{2}-205^{2} $
(e) $ 100.4 \times 99.6 $
(f) $ \frac{5.27 \times 5.27-0.27 \times 0.27}{5.54} $

Given:

(a) \( (102)^{2} \)
(b) \( (98)^{2} \)
(c) \( 104 \times 105 \)
(d) \( 215^{2}-205^{2} \)
(e) \( 100.4 \times 99.6 \)
(f) \( \frac{5.27 \times 5.27-0.27 \times 0.27}{5.54} \)

To do:

We have to evaluate the given expressions using suitable identities.

Solution:

We know that,

$(a-b)^2=a^2-2ab+b^2$

$(a-b)^2=a^2-2ab+b^2$

$a^2-b^2=(a+b)(a-b)$

Therefore,

(a) $(102)^2=(100+2)^2$

$=(100)^2+2\times 100\times2 +(2)^2$

$=10000+400+4$

$=10404$

(b) $(98)^2=(100-2)^2$

$=(100)^2-2\times 100\times2 +(2)^2$

$=10000-400+4$

$=9604$

(c) $(104\times105)=(100+4)(100+5)$

$=(100)^2+100\times5+ 4\times100 +4\times5$

$=10000+500+400+20$

$=10920$

(d) $(215)^2-(205)^2=(215+205)(215-205)$

$=420\times10$

$=4200$

(e) $100.4\times99.6=(100+0.4)\times(100-0.4)$

$=(100)^2-(0.4)^2$

$=10000-0.16$

$=9999.84$

(f) $\frac{5.27 \times 5.27-0.27 \times 0.27}{5.54}=\frac{(5.27)^2-(0.27)^2}{5.54}$

$=\frac{(5.27+0.27)(5.27-0.27)}{5.54}$

$=\frac{(5.54)\times5}{5.54}$

$=5$

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Updated on: 10-Oct-2022

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