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; 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.