#### NCERT Solutions for Class 8th: Ch 6 Squares and Square Roots Maths Part-2

**Exercise 6.4**

1.Find the square root of each of the following numbers by Division method.

(i) 2304

(ii) 4489

(iii) 3481

(iv) 529

(v) 3249

(vi) 1369

(vii) 5776

(viii) 7921

(ix) 576

(x) 1024

(xi) 3136

(xii) 900

**Solution**

(i)

âˆ´ âˆš2304 = 48

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

âˆ´ âˆš1024 = 32

(xi)Â

âˆ´Â âˆš3136 = 56

(xii)Â

âˆ´ âˆš900 = 30

2.Â Find the number of digits in the square root of each of the following numbers (without any calculation).Â

(i) 64

(ii) 144

(iii) 4489

(iv) 27225

(v) 390625

**Solution**

(i)Â

âˆ´Â âˆš64 = 8

Hence, the square root of the number 64 has 1 digit.

(ii)Â

âˆ´Â âˆš144 = 12

Hence, the square root of the number 144 has 2 digits.

(iii)Â

âˆ´ âˆš4489 = 67

Hence, the square root of the number 4489 has 2 digits.

(iv)Â

âˆ´ âˆš27225 = 165

Hence, the square root of the number 27225 has 3 digits.

(v)Â

âˆ´Â âˆš390625Â = 625

Hence, the square root of the number 390625 has 3 digits.

3. Find the square root of the following decimal numbers.Â

(i) 2.56

(ii) 7.29

(iii) 51.84

(iv) 42.25

(v) 31.36

**Solution**

(i)Â

âˆ´ âˆš2.56Â = 1.6

(ii)Â

âˆ´ âˆš7.29 = 2.7

(iii)Â

âˆ´ âˆš51.84 = 7.2

(iv)Â

âˆ´ âˆš42.25Â = 6.5

(v)Â

âˆ´ âˆš31.36 = 5.6

4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.Â

(i) 402

(ii) 1989

(iii) 3250

(iv) 825

(v) 4000

**Sol:**Â (i)

**Â**

âˆ´ We must subtracted 2 from 402 to get a perfect square.

New number = 402 â€“ 2 = 400

âˆ´ âˆš400 = 20

(ii)Â

âˆ´ We must subtracted 53 from 1989 to get a perfect square.

New number = 1989 â€“ 53 = 1936

âˆ´ âˆš1936 = 44

(iii)Â

âˆ´ We must subtracted 1 from 3250 to get a perfect square.

New number = 3250 â€“ 1 = 3249

âˆ´ âˆš3249 = 57

(iv)

âˆ´ We must subtracted 41 from 825 to get a perfect square.

New number = 825 â€“ 41 = 784

âˆ´ âˆš784 = 28

(v)Â

âˆ´ We must subtracted 31 from 4000 to get a perfect square.

New number = 4000 â€“ 31 = 3961

âˆ´ âˆš3136 = 56

5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

(i) 525

(ii) 1750

(iii) 252

(iv) 1825

(v) 6412

**Solution**

(i)Â

Here, (22)

^{2Â }Â < 525 > (23)^{2}
We can say 525 is ( 129 â€“ 125 ) 4 less than (23)

^{2}.
âˆ´ If we add 4 to 525, it will be perfect square.

New number = 525 + 4 = 529

âˆ´ âˆš529 = 23

(ii)Â

Here, (41)

^{2}Â < 1750 > (42)^{2}
We can say 1750 is ( 164 â€“ 150 ) 14 less than (42)

^{2}.
âˆ´ If we add 14 to 1750, it will be perfect square.

New number = 1750 + 14 = 1764

âˆ´âˆš1764Â = 42

(iii)Â

Here, (15)

^{2Â }Â < 252 > (16)^{2}
We can say 252 is ( 156 â€“ 152 ) 4 less than (16)

^{2}.
âˆ´ If we add 4 to 252, it will be perfect square.

New number = 252 + 4 = 256

âˆ´ âˆš256 = 16

(iv)Â

Here, (42)

^{2Â }Â < 1825 > (43)^{2}
We can say 1825 is ( 249 â€“ 225 ) 24 less than (43)

^{2}.
âˆ´ If we add 24 to 1825, it will be perfect square.

New number = 1825 + 24 = 1849Â Â

âˆ´ âˆš1849 = 43

(v)

Here, (80)

^{2Â }Â < 6412 > (81)^{2}
We can say 6412 is ( 161 â€“ 12 ) 149 less than (81)

^{2}.
âˆ´ If we add 149 to 6412, it will be perfect square.

New number = 6412 + 149 = 6561Â

âˆ´ âˆš6561 = 81

6. Find the length of the side of a square whose area is 441 m

^{2}.**Solution**

Let the length of each side of the field = a

Then, area of the field = 441 m

^{2}
â‡’ a

^{2}Â = 441Â m^{2}
â‡’a = âˆš441 m

âˆ´ The length of each side of the field = a m = 21 m.

7. In a right triangle ABC, âˆ B = 90Â°.

(a) If AB = 6 cm, BC = 8 cm, find AC

(b) If AC = 13 cm, BC = 5 cm, find AB

**Solution**

(a)Â

Given, AB = 6 cm,

BC = 8 cm

Let AC be x cm.

âˆ´ AC

^{2}Â = AB^{2}Â + BC^{2}
Hence, AC = 10 cm.

(b)

8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.

Let the number of rows and column be, x.

âˆ´ Total number of row and column= xÃ—x = x

As per question, x

â‡’ x = âˆš1000

Here, (31)

We can say 1000 is ( 124 â€“ 100 ) 24 less than (32)

âˆ´ 24 more plants is needed.

9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.

(b)

Given, AC = 13 cm,

BC = 5 cm

Let AB be x cm.

âˆ´ AC

^{2Â }= AB^{2Â }+ BC^{2}
â‡’ AC

^{2Â }- BC^{2Â }= AB^{2}
Hence, AB = 12 cm

8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.

**Solution**Let the number of rows and column be, x.

âˆ´ Total number of row and column= xÃ—x = x

^{2}As per question, x

^{2}Â = 1000â‡’ x = âˆš1000

Here, (31)

^{2}Â < 1000 > (32)^{2}We can say 1000 is ( 124 â€“ 100 ) 24 less than (32)

^{2}.âˆ´ 24 more plants is needed.

9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.

**Solution**
Let the number of rows and column be, x.

âˆ´ Total number of row and column= x Ã— x = x

As per question, x

x = âˆš500

âˆ´ Total number of row and column= x Ã— x = x

^{2}As per question, x

^{2}Â = 500x = âˆš500

Hence, 16 children would be left out in the arrangement.