Square Root Exercise 1.2.1 – Square Root – Class IX

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

  1. Find the square root of the following numbers by the factorization method

(i) 82944

Solution:

82944 = 210  x 34

= (25)2   x (32)2

Square Root Exercise 1.2.1

√(82944) = √((25)2   x (32)) = 2x 32 = 288


(ii) 155236

Solution:

Square Root Exercise 1.2.1

155236 = (2)2 x  (197)2

√(155236) = √((2)2   x (197)) = 2 x 197 = 394


(iii) 19881

Solution:

3

19881 = (3)2 x  (47)2

√(19881) = √((3)2   x (47)) = 2 x 47 = 141


2. Find the square root of the following numbers.

(i) 184.96

Solution:

= (184.96X100)100 = (18496)100 

Square Root Exercise 1.2.1

= ((2³)²x(17)²)10²

√(184.96) = √ ((2³)²x(17)²10²) = (2³)x(17)10  
(136)10

= 13.6


(iii) 19.5364

Solution:

19.5364 = (19.5364×10000)10000
= (195364)10000

Square Root Exercise 1.2.1

√(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; digits-2
10² = 100; digits-3
99² = 9801; digits-4
100² = 10000; digits-5
990² = 998001; digits-6
1000² = 1000000; digits-7
9999² = 99980001; digits-8

They follow the rule
2n if the number has even digits
2n-1 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.