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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, $(703)^2=494209$.
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