Compute the squares of these numbers mentally.

(i) 102

(ii) 202

(iii) 1001

(iv) 2002

(v) 205

(vi) 10.3

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Solution

We know that (x + y)² = x² + y² + 2xy

(i)

102 = 100 + 2

⇒ (102)² = (100 + 2)²

= (100)² + 2 × 100 × 2 + 2²

= 10000 + 400 + 4

=10404

(ii)

202 = 200 + 2

⇒ (202)² = (200 +2)²

= (200)² + 2 × 200 × 2 + 2²

= 40000 + 800 + 4

= 40804

(iii)

1001 = 1000 + 1

⇒ (1001)² = (1000 + 1)²

= (1000)² + 2 × 1000 × 1 + 1²

= 1000000 + 2000 + 1

= 1002001

(iv)

2002 = 2000 + 2

⇒ (2002)² = (2000 + 2)²

= (2000)² + 2 × 2000 × 2 + 2²

= 4000000 + 8000 + 4

= 4008004

(v)

205 = 200 + 5

⇒ (205)² = (200 + 5)²

= (200)² + 2 × 200 × 5 + 5²

= 40000 + 2000 + 25

= 42025

(vi)

10.3 = 10 + 0.3

⇒ (10.3)² = (10 + 0.3)²

= (10)² + 2 × 10 × 0.3 + (0.3)²

= 100 + 6 + 0.09

=106.09

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Similar questions

Mentally compute the squares of the numbers given below:

(i) 98

(ii) 198

(iii) 999

(iv) 1998

(v) 195

(vi) 9.7

Find the squares of the following numbers using the identity $a + b)^{2} = a^{2} + 2 a b + b^{2}$

(i) 405 (ii) 510
(iii) 1001 (iv) 209
(v) 605

Compute the following difference

$(11.3 \times 10.7) - (11.7 \times 10.3)$

Compute the following differences $(10.3 \times 9.7) - (10.7 \times 9.3)$

Now compute the square root of these by factorization.

(i) 3025

(ii) 2304

(iii) 9604

(iv) 6561

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