⇒ 21^2 nbsp;=(20+1)^2 ...(i) By comparing the above equation with the nbsp;algebraic identity (a+b)^2 = a^2+2ab+b^2 ...(ii) we get, a=20, b = 1 Substitute ...

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Standard VIII

Mathematics

Interesting Patterns

212 = ?

Question

$21^{2} = ?$

441

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221

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481

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411

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Solution

The correct option is A
441

$\Rightarrow 21^{2} = (20 + 1)^{2} . . . (i)$

By comparing the above equation with the algebraic identity
$(a + b)^{2} = a^{2} + 2 a b + b^{2} . . . (i i)$
we get,
$a = 20, b = 1$

Substitute these value in eq (ii), we get
$= (20 + 1)^{2}$
$= (20^{2} + 2 \times 20 \times 1 + 1^{2})$
$= (400 + 40 + 1)$
$= 441$

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Similar questions

212×424+212×366+212×240

Explain With Solution

Q. Which of the following transmutations demonstrate(s) beta decay?
i.

B u - 212 \to P o - 212

ii.

P b - 212 \to B i - 212

iii.

R a - 228 \to A c - 228

cos (\frac{π}{2^{2}}) . . . . . . cos (\frac{π}{2^{12}}) . sin (\frac{π}{2^{12}}) =

{(\frac{- 1 + i \sqrt{3}}{2})}^{12} - {(\frac{- 1 - i \sqrt{3}}{2})}^{12} =

Q.
The value of

{(\frac{- 1 + i \sqrt{3}}{2})}^{12} + {(\frac{- 1 - i \sqrt{3}}{2})}^{12}

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212= ?

The correct option is A 441 ⇒212=(20+1)2...(i) By comparing the above equation with the algebraic identity (a+b)2 = a2+2ab+b2...(ii) we get, a=20,b=1 Substitute these value in eq (ii), we get =(20+1)2 =(202+2×20×1+12) =(400+40+1) =441

$21^{2} = ?$