Find the squares of the following numbers using the identity $(a + b)^2 = a^2 + 2ab + b^2$
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605.

To find: 

We have to find the squares of the given numbers using the identity $(a + b)^2 = a^2 + 2ab + b^2$.

Solution:

We know that,

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

(i) $405$ can be written as,

$=400+5$

Therefore,

$(405)^2=(400+5)^2$

$=(400)^2+2\times400\times5+(5)^2$

$=160000+4000+25$

$=    164025$

(ii) $510$ can be written as,

$=500+10$

Therefore,

$(510)^2=(500+10)^2$

$=(500)^2+2\times500\times10+(10)^2$

$=250000+10000+100$

$=    260100$

(iii) $1001$ can be written as,

$=1000+1$

Therefore,

$(1001)^2=(1000+1)^2$

$=(1000)^2+2\times1000\times1+(1)^2$

$=1000000+2000+1$

$=  1002001  $ 

(iv) $209$ can be written as,

$=200+9$

Therefore,

$(209)^2=(200+9)^2$

$=(200)^2+2\times200\times9+(9)^2$

$=40000+3600+81$

$=  43681$  

(v)  $605$ can be written as,

$=600+5$

Therefore,

$(605)^2=(600+5)^2$

$=(600)^2+2\times600\times5+(5)^2$

$=360000+6000+25$

$=  366025$   

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Updated on: 10-Oct-2022

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