The square root exercise 1.2.1 is related to the chapter – Square Root – Class IX – Karnataka State Syllabus mathematics Notes.
Square Root Exercise 1.2.1

Find the square root of the following numbers by the factorization method
(i) 82944
Solution:
82944 = 2^{10 } x 3^{4}
= (2^{5})^{2} x (3^{2})^{2}
√(82944) = √((2^{5})^{2} x (3^{2})^{2 }) = 2^{5 }x 3^{2} = 288
(ii) 155236
Solution:
155236 = (2)^{2} x (197)^{2}
√(155236) = √((2)^{2} x (197)^{2 }) = 2^{ }x 197 = 394
(iii) 19881
Solution:
19881 = (3)^{2} x (47)^{2}
√(19881) = √((3)^{2} x (47)^{2 }) = 2^{ }x 47 = 141
2. Find the square root of the following numbers.
(i) 184.96
Solution:
^{= (184.96X100)}⁄_{100 = (18496)⁄}_{100 }
= ^{((2³)²x(17)²)}⁄_{10²}
_{√(184.96) = √ ((2³)²x(17)²⁄10²) = (2³)x(17)⁄10 }
= ^{(136)}⁄_{10}
= 13.6
(iii) 19.5364
Solution:
19.5364 = ^{(19.5364×10000)}⁄_{10000}
= ^{(195364)}⁄_{10000}
_{√(195364/10000) = √ ((2)²x(221)²⁄100²) = (2)x(221)⁄100 }
= ^{(442)}⁄_{100}
= 4.42
3. Exploration:
Find the squares of the numbers 9, 10, 99, 100, 1000, and 9999.Tabulate these numbers. How many digits are there in the squares of a numbers when it has even number of digits and odd number of digits?
Solution:
9² = 81; digits2
10² = 100; digits3
99² = 9801; digits4
100² = 10000; digits5
990² = 998001; digits6
1000² = 1000000; digits7
9999² = 99980001; digits8
They follow the rule
2n if the number has even digits
2n1 if the number has odd digits.
4. If a perfect square A has A digits, how many digits to you expect in √𝐀
Solution:
If a perfect square A has n digits then, √A has
^{(n)}⁄_{2} Digits if n is even
^{(n+1)}⁄_{2} Digits if n is odd.