Ncert solutions

# Question 1

Q1) Find the square of the following numbers.

(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

Solution:

(i)

\left(32\right)^2=\left(30+2\right)^2=\left(30\right)^2+2\times30\times2+\left(2\right)^2

[\because\left(a+b\right)^2=a^2+2ab+b^2]

= 900 + 120 + 4 = 1024

(ii) \left(35\right)^2=\left(30+5\right)^2=\left(30\right)^2+2\times30\times5+\left(5\right)^2

[\because\left(a+b\right)^2=a^2+2ab+b^2

= 900 + 300 + 25 = 1225

(iii) \left(86\right)^2=\left(80+6\right)^2=\left(80\right)^2+2\times80\times6+\left(6\right)^2

[\because\left(a+b\right)^2=a^2+2ab+b^2

= 8100 + 540 + 9 = 8649

(iv) \left(93\right)^2=\left(90+3\right)^2=\left(90\right)^2+2\times90\times3+\left(3\right)^2

[\because\left(a+b\right)^2=a^2+2ab+b^2]

= 8100 + 540 + 9 = 8649

(v) \left(71\right)^2=\left(70+1\right)^2=\left(70\right)^2+2\times70\times1+\left(1\right)^2

[\because\left(a+b\right)^2=a^2+2ab+b^2

= 4900 + 140 + 1 = 5041

(vi) \left(46\right)^2=\left(40+6\right)^2=\left(40\right)^2+2\times40\times6+\left(6\right)^2

[\because\left(a+b\right)^2=a^2+2ab+b^2]

= 1600 + 480 + 36 = 2116

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