# Using the formula for squaring a binomial, evaluate the following:(i) $(102)^2$(ii) $(99)^2$(iii) $(1001)^2$(iv) $(999)^2$(v) $(703)^2$

To do:

We have to evaluate the given expressions using the formula for squaring a binomial.

Solution:

Here, we have to find the squares of some large numbers. We can find the squares of multiples of $10^n$ easily. So, express the given numbers as the sum of multiples of $10^n$ and other numbers. We can then find the squares of the given numbers by expanding the squares using the algebraic expressions:

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

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

(i) The given expression is $(102)^2$.

$102$ can be written as $100+2$

We know that,

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

Here, $a=100$ and $b=2$

Therefore,

$(100+2)^2=(100)^2+2\times100\times2+2^2$

$(100+2)^2=10000+400+4$

$(100+2)^2=10404$

Hence, $(102)^2=10404$.

(ii) The given expression is $(99)^2$.

$99$ can be written as $100-1$

We know that,

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

Here, $a=100$ and $b=1$

Therefore,

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

$(100-1)^2=10000-200+1$

$(100-1)^2=10001-200$

$(100-1)^2=9801$

Hence, $(99)^2=9801$.

(iii) The given expression is $(1001)^2$.

$1001$ can be written as $1000+1$

We know that,

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

Here, $a=1000$ and $b=1$

Therefore,

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

$(1000+1)^2=1000000+2000+1$

$(1000+1)^2=1002001$

Hence, $(1001)^2=1002001$.

(iv) The given expression is $(999)^2$.

$999$ can be written as $1000-1$

We know that,

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

Here, $a=1000$ and $b=1$

Therefore,

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

$(1000-1)^2=1000000-2000+1$

$(1000-1)^2=1000001-2000$

$(1000-1)^2=998001$

Hence, $(999)^2=998001$.

(v) The given expression is $(703)^2$.

$703$ can be written as $700+3$

We know that,

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

Here, $a=700$ and $b=3$

Therefore,

$(700+3)^2=(700)^2+2\times700\times3+3^2$

$(700+3)^2=490000+4200+9$

$(700+3)^2=494209$

Hence, $(703)^2=494209$.