# Square root of a perfect square by factorization

Example 1: Find the square root of 5929.

Solution:

Thus, 7 x 7 x 11 x 11 = 5929

Therefore, [group it]

(7 x 11) x (7 x 11) = 77 x 77 = 772 = 5929.

Example 2: Find the square root of 6724.

Solution:

Thus, 6724 = 2 x 2 x 41 x 41

6724 = (2 x 41) x (2 x 41)

= 82 x 82

Example 3: Find the smallest positive integer with which one has to divide 336 to get a perfect square.

Solution:

We observe that 336 = 2 x 2 x 2 x 2 x 3 x 7. Here both 3 and 7 occur only once. Hence we have to remove them to get a perfect square.

We divide 336 by 3 and 7

42.

The required least number is 21.

Exercise 1.2.5

1. Find the square roots of the following numbers by factorization:

i) 196

Solution:

196 = 2 x 2 x 7 x 7

= (2 x 7) x (2 x 7)

= 14 x 14

196 = 142

ii) 256

Solution:

256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

= (2 x 2 x 2 x 2) x (2 x 2 x 2 x 2)

= 16 x 16

= 162

iii) 10404

Solution:

10404 = 2 x 2 x 3 x 3 x 17 x 17

= (2 x 3 x 17) x (2 x 3 x 17)

= (102) x (102)

= 1022

iv) 1156

Solution:

1156 = 2 x 2 x 17 x 17

= (2 x 17) x (2 x 17)

= 34 x 34

= 342

v) 13225

Solution:

13225 = 5 x 5 x 23 x 23

= (5 x 23) x (5 x 23)

= 115 x 115

= 1152

2. Simplify:

i) √100 + √36

Solution:

√100 + √36 = 10 + 6

= 16

ii) √(1360 + 9)

Solution:

√(1360 + 9) = √1369 = √(37 x 37) = 37

iii) √2704 + √144 + √289

Solution:

√2704 + √144 + √289 = √(52 x 52) + √(12 x 12) + √(17 x 17)

= 52 + 12 + 17

= 81

iv) √225 – √25

Solution:

√225 – √25 = √(15 x 15) – √(5 x 5)

= 15 – 5

= 10.

v) √1764 – √1444

Solution:

√1764 – √1444 = √(42 x 42) – √(38 x 38)

= 42 – 38

= 4

vi) √169 x √361

√169 x √361 = √(13 x 13) x √(19 x 19)

= 13 x 19

= 247

1. A square yard has area 1764m2. From a corner of this yard, other square part of area 784m2 is taken out for public utility. The remaining portion is divided in to 5 equal parts. What is the perimeter of each of these equal parts?

Solution:

Area of square = a2

Area of given square yard = 1764 m2

Area used for public utility = 784 m2

∴ Area of remaining portion = 1764 – 784 = 980 m2

If the area is divided into 5 parts

Then, the area of each Square = 980/5 = 196 m2

Length of each side = √ 196, a = 14m

Perimeter of square = 4a = 4 x 14 = 56m

4. Find the smallest positive integer with which one has to multiply each of the following numbers to get a perfect square:

i) 847

Solution:

847 = 11 x 11 x 7

Here 7 occur only once. Hence we have to multiply by 7 them to get a perfect square.

ii) 450

Solution:

450 = 5 x 5 x 3 x 3 x 2

Here 2 occur only once. So we have multiply by 2 to get a perfect square.

iii) 1445

Solution:

1445 = 17 x 17 x 5

Here 5 occur only once. So we have multiply by 5 to get a perfect square.

iv) 1352

Solution:

1352 = 2 x 2 x 13 x 13 x 2

Here 2 occur only three times. So we have multiply by 2 to get a perfect square.

5. Find the largest perfect square factor of each of the following numbers:

i) 48

Solution:

48 = 2 x 2 x 2 x 2 x 3 = (2 x 2) x (2 x 2) x 3

= 4 x 4 x 3 = 16 x 3 = 42 + 3

Therefore, 16 is the largest perfect square factor of 48.

ii) 11280

Solution:

11280 = 2 x 2 x 2 x 2 x 3 x 47

Therefore, 16 is the largest perfect square factor of 11280

iii) 729

Solution:

729 = 3 x 3 x 3 x 3 x 3 x 3

= (27) x (27)

= 272

Therefore, 729 is the largest perfect square factor of 729.

iv) 1352

Solution:

1352 = 2 x 2 x 2 x 13 x 13

= (2 x 13) x (2 x 13) x 2

= 26 x 26 x 2

Therefore, 676 is the largest perfect square of 1352.

1. Find a proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs?

Solution:

The factor of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Multiples of 48 are = 48, 96 , 144, 192 , 240, 280 ….

1) 48 + 1 = 49 = 72

2 ) 192 + 4 = 196 = 142

3)288 + 1 = 289 = 172

4 ) 96 + 4 = 100 = 102

5)96 + 48 = 144 = 122

6) 48 + 16 = 64 = 82

7)240 + 16 = 256 = 162

Yes, we can prove that there are infinitely many such fairs.