Using an identity- find the square of 101-10201104011020110204

The correct option is A 10201To find: (101)^2 nbsp; (101)^2 can be written as (100+1)^2. Using the identity, (a+b)^2=a^2+2ab+b^2, we get, (100+1)^2=100^2+ ...

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Standard VIII

Mathematics

Algebraic identity for (a+b)^2

Using an iden...

Question

Using an identity, find the square of 101.

10201

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Solution

The correct option is A $10201$
$To find: (101)^{2}$

$(101)^{2} can be written as (100 + 1)^{2} .$

Using the identity,
$(a + b)^{2} = a^{2} + 2 a b + b^{2},$

we get,
$(100 + 1)^{2} = 100^{2} + 2 \times 100 \times 1 + 1^{2}$
$= 10000 + 200 + 1$
$= 10201$

Suggest Corrections

Using an identity, find the square of 101.

The correct option is A 10201To find: (101)2 (101)2 can be written as (100+1)2. Using the identity, (a+b)2=a2+2ab+b2, we get, (100+1)2=1002+2×100×1+12 =10000+200+1 =10201

The correct option is A $10201$
$To find: (101)^{2}$

$(101)^{2} can be written as (100 + 1)^{2} .$

Using the identity,
$(a + b)^{2} = a^{2} + 2 a b + b^{2},$

we get,
$(100 + 1)^{2} = 100^{2} + 2 \times 100 \times 1 + 1^{2}$
$= 10000 + 200 + 1$
$= 10201$