# Find the squares of the following numbers using the identity $(a - b)^2 = a^2 - 2ab + b^2$:(i) 395(ii) 995(iii) 495(iv) 498(v) 99(vi) 999(vii) 599.

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) $395$ can be written as,

$=400-5$

Therefore,

$(395)^2=(400-5)^2$

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

$=160000-4000+25$

$= 156025$

(ii) $995$ can be written as,

$=1000-5$

Therefore,

$(995)^2=(1000-5)^2$

$=(1000)^2-2\times1000\times5+(5)^2$

$=1000000-10000+25$

$= 990025$

(iii) $495$ can be written as,

$=500-5$

Therefore,

$(495)^2=(500-5)^2$

$=(500)^2-2\times500\times5+(5)^2$

$=250000-5000+25$

$= 245025$

(iv) $498$ can be written as,

$=500-2$

Therefore,

$(498)^2=(500-2)^2$

$=(500)^2-2\times500\times2+(2)^2$

$=250000-2000+4$

$= 248004$

(v) $99$ can be written as,

$=100-1$

Therefore,

$(99)^2=(100-1)^2$

$=(100)^2-2\times100\times1+(1)^2$

$=10000-200+1$

$= 9801$

(vi) $999$ can be written as,

$=1000-1$

Therefore,

$(999)^2=(1000-1)^2$

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

$=1000000-2000+1$

$= 998001$

(vii) $599$ can be written as,

$=600-1$

Therefore,

$(599)^2=(600-1)^2$

$=(600)^2-2\times600\times1+(1)^2$

$=360000-1200+1$

$= 358801$

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Updated on: 10-Oct-2022

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