Question

hii please give answer to my question
please give answer of (iv)​

Answer 1

${(101)}^{2} \\ = {(100 + 1)}^{2} \\ = {(100)}^{2} + {(1)}^{2} + 2 \times 1001 \\ = 10000 + 1 + 200 \\ = 10201$

Answer 2

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Explanation:

solving part (iv)

${(203)}^{2}$

Here ,this identity is used:

$\bold{\boxed{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

Now break the value 203 into 200+3

${(200 + 3)}^{2} = {(200)}^{2} + {(3)}^{2} + 2(200)(3)$

$= 40000 + 9 + 1200 = 40009 + 1200 = 41209$

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Now solving part (v)

${(400 + 10)}^{2} = {400)}^{2} + {(10)}^{2} + 2(100)(10)$

$= 160000 + 100 + 8000 = 168000 + 100 = 168100$