ncert solutions for class 8 maths chapter 6 deals with Squares and Square Roots.

Exercises and solutions of Squares and Square Roots available here so students can read online view.

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Find solutions to exercise 6.1, exercise 6.2, 6.3 and 6.4 in chapter 6 in class 8 maths.

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Find solutions to ncert maths class 8 chapter 6 exercise 6.

class 8 maths chapter 6 exercise 6.1

class 8 maths chapter 6 exercise 6.1 (1)

What will be the unit digit of the squares of the following numbers?

(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555 

Solution to class 8 maths chapter 6 exercise 6.1 (1).

(i) Unit’s place digit of the number 81 is 1. So, square of 1 is 1. Hence, unit’s digit of square of 81 is 1.

(ii) Unit’s place digit of the number 272 is 2. So, square of 2 is 4. Hence, unit’s digit of square of 272 is 4.

(iii)Unit’s place digit of the number 799 is 9. So, square of 9 is 81. Hence, unit’s digit of square of 799 is 1.

(iv) Unit’s place digit of the number 3853 is 3. So, square of 3 is 9. Hence, unit’s digit of square of 3853 is 9.

(v) Unit’s place digit of the number 1234 is 4. So, square of 4 is 16. Hence, unit’s digit of square of 1234 is 6.

(vi) Unit’s place digit of the number 26387 is 7. So, square of 7 is 49. Hence, unit’s digit of square of 26387 is 9.

(vii) Unit’s place digit of the number 52698 is 8. So, square of 8 is 64. Hence, unit’s digit of square of 52698 is 4.

(viii) Unit’s place digit of the number 99880 is 0. So, square of 0 is 0. Hence, unit’s digit of square of 99880 is 0.

(ix)Unit’s place digit of the number 12796 is 6. So, square of 6 is 36. Hence, unit’s digit of square of 12796 is 6.

(x) Unit’s place digit of the number 55555 is 5. So, square of 5 is 25. Hence, unit’s digit of square of 55555 is 5.

class 8 maths chapter 6 exercise 6.1 (2)

The following numbers are obviously not perfect squares. Give reason.

(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050

Solution to class 8 maths chapter 6 exercise 6.1 (2).

(i) Numbers ending with 2,3,7 and 8 are not perfect squares.

1057 is ends with 7. So 1057 is not a perfect square.

(ii) Numbers ending with 2,3,7 and 8 are not perfect squares.

23453 is ends with 3. So 23453 is not a perfect square.

(iii) Numbers ending with 2,3,7 and 8 are not perfect squares.

7928 is ends with 8. So 7928 is not a perfect square.

(iv) Numbers ending with 2,3,7 and 8 are not perfect squares.

222222 is ends with 2. So 222222 is not a perfect square.

(v) A number ending with an odd number of zeroes is never a perfect square. 

64000 is ends with odd number of zeroes. So 64000 is not a perfect square.

(vi)  Numbers ending with 2,3,7 and 8 are not perfect squares.

89722 is ends with 2. So 89722 is not a perfect square.

(vii) A number ending with an odd number of zeroes is never a perfect square.

222000 is ends with odd number of zeroes. So 222000 is not a perfect square.

(viii) A number ending with an odd number of zeroes is never a perfect square.

505050 is ends with odd number of zeroes. So 505050 is not a perfect square.

class 8 maths chapter 6 exercise 6.1 (3)

3  The squares of which of the following would be odd numbers?

(i) 431 (ii) 2826 (iii) 7779 (iv) 82004.

Solution to class 8 maths chapter 6 exercise 6.1 (3).

(i)431 – Unit’s digit of given number is 1 and square of 1 is 1.

∴ Square of 431 would be an odd number

(ii) 2826 – Unit’s digit of given number is 6  square of 6 is 36.

∴  Square of 2826 would not be an odd number.

(iii) 7779- Unit’s digit of given number is 9 and square of 9 is 81.

∴ square of 7779 would be an odd number.

(iv) 82004 – Unit’s digit of given number is 4 and square of 4 is 16. 

 ∴ square of 82004 would not be an odd number.

class 8 maths chapter 6 exercise 6.1 (4)

4  Observe the following pattern and find the missing digits.

                  11²=121

                101²=10201

             1001²=1002001

          100001²=1…….2…….1 

       10000001²=……………….

Solution to class 8 maths chapter 6 exercise 6.1 (4).

11²=121 

101²=10201

1001²=1002001   

100001²=10000200001 

10000001²=100000020000001 

class 8 maths chapter 6 exercise 6.1 (5)

5  Observe the following pattern and supply the missing numbers.

11²=121

101²=10201

10101²=102030201

1010101²=…………….

………………²=10203040504030201

Solution to class 8 maths chapter 6 exercise 6.1 (5).

11²=121

101²=10201

10101²=102030201

1010101²=1020304030201

101010101²=10203040504030201

class 8 maths chapter 6 exercise 6.1 (6)

6  Using the given pattern, find the missing numbers.

1²+2²+2²=3²

2²+3²+6²=7²

3²+4²+12²=13²

4²+5²+ _ ²=21²

5²+ _ ²+30²=31²

6²+7²+ _ ²=  _ ²

Solution to class 8 maths chapter 6 exercise 6.1 (6).

1²+2²+2²=3²

2²+3²+6²=7²

3²+4²+12²=13²

4²+5²+ 20²=21²

5²+ 6²+30²=31²

6²+7²+ 42²= 43²

class 8 maths chapter 6 exercise 6.1 (7)

7  Without adding, find the sum.

(i) 1 + 3 + 5 + 7 + 9

(ii) 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Solution to class 8 maths chapter 6 exercise 6.1 (7).

We know that the sum of first ‘n’ odd natural numbers is equal to ‘n²’.

(i) Sim of first 5 odd natural numbers=

1 + 3 + 5 + 7 + 9 =5²=25

(ii) Sum of first 10 odd natural numbers=

1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19 =10²=100

(iii) Sum of first 12 odd natural numbers=

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23=12²=144.

class 8 maths chapter 6 exercise 6.1 (8) 

8  (i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11 odd numbers.

Solution to class 8 maths chapter 6 exercise 6.1 (8).

(i) 49 = 7².

7 is an odd number.

49 we can write as sum of first 7 odd numbers.

49=1+3+5+7+9+11+13 

(ii) 121=11².

11 is an odd number.

121 we can write as sum of first 11 odd numbers.

121=1+3+5+7+9+11+13+15+17+19+21

class 8 maths chapter 6 exercise 6.1 (9)

9  How many numbers lie between squares of the following numbers?

(i) 12 and 13  (ii) 25 and 26  (iii) 99 and 100

Solution to class 8 maths chapter 6 exercise 6.1 (9).

There are ‘2n’ non-perfect square numbers between n² and (n+1)².

(i) There are 2(12)=24  non-perfect square numbers between 12 and 13.

(ii) There are 2(25)=50  non-perfect square numbers between 25 and 26.

(iii) There are 2(99)=198  non-perfect square numbers between 99 and 100.

 

class 8 maths chapter 6 exercise 6.2

class 8 maths chapter 6 exercise 6.2 (1)

Find the square of the following numbers.

(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

Solution to class 8 maths chapter 6 exercise 6.2 (1)

(i) (32)²= (30+2)²= (30)²+2×30×2+ (2)²     [ ∵ (a+b)²= a²+2ab+b²]

=900+120+4=1024

(ii) (35)²= (30+5)²= (30)²+2×30×5+ (5)²     [ ∵(a+b)²= a²+2ab+b²]

=900+300+25=1225

(iii) (86)²= (80+6)²= (80)²+2×80×6+ (6)²     [ ∵(a+b)²= a²+2ab+b²]

=6400+960+36=7386

(iv) (93)²= (90+3)²= (90)²+2×90×3+ (3)²     [ ∵ (a+b)²= a²+2ab+b²]

=8100+540+9=8649

(v) (71)²= (70+1)²= (70)²+2×70×1+ (1)²     [ ∵ (a+b)²= a²+2ab+b²]

=4900+140+1=5041

(vi) (46)²= (40+6)²= (40)²+2×40×6+ (6)²     [ ∵ (a+b)²= a²+2ab+b²]

=1600+480+36=2116

class 8 maths chapter 6 exercise 6.2 (2)

2  Write a Pythagorean triplet whose one number is.

(i) 6              (ii) 14                 (iii) 16            (iv) 18

Solution to class 8 maths chapter 6 exercise 6.2 (2)

There are three numbers 2m, m²-1 and m²+1 in a Pythagorean Triplet.

(i) Here, 2m=6  ⇒  m=3

∴  Second number=m²-1=3²-1=9-1=8

Third number =m²+1=3²+1=9+1=10

Pythagorean triplet is (6,8,10)

(ii) Here, 2m=14  ⇒  m=7

∴  Second number=m²-1=7²-1=49-1=48

Third number =m²+1=7²+1=49+1=50

Pythagorean triplet is (14,48,50)

(iii) Here, 2m=16  ⇒  m=8

∴  Second number=m²-1=8²-1=64-1=63

Third number =m²+1=8²+1=64+1=65

Pythagorean triplet is (16,63,65)

(iv) Here, 2m=18  ⇒  m=9

∴  Second number=m²-1=9²-1=81-1=80

Third number =m²+1=9²+1=81+1=82

Pythagorean triplet is (18,80,82)

 

class 8 maths chapter 6 exercise 6.3

class 8 maths chapter 6 exercise 6.3 (1)

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025

Solution to class 8 maths chapter 6 exercise 6.3 (1)

(i) Possible units digits of square root of 9801 is 1 or 9. Since 9²=81 and 1²=1.

(ii) Possible units digits of square root of 99856 is 4 or 6. Since 4²=16 and 6²=36.

(iii) Possible units digits of square root of 998001 is 1 or 9. Since 9²=81 and 1²=1.

(iv) Possible units digits of square root of 657666025 is 5. Since 5²=25.

class 8 maths chapter 6 exercise 6.3 (2)

2. Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257 (iii) 408 (iv) 441

Solution to class 8 maths chapter 6 exercise 6.3 (2)

The numbers which ends with 2,3,7,8 and odd number of zeroes are not  perfect squares.

(i) 153 is not a perfect square. Because it is ending with 3.

(ii) 257 is nat a perfect square. Because it is ending with 7.

(iii) 408 is not a perfect square. Because it is ending with 8.

(iv) 441 is ending with 1. So it would be a perfect square.

class 8 maths chapter 6 exercise 6.3 (3)

3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Solution to class 8 maths chapter 6 exercise 6.3 (3)

By subtraction of  successive odd numbers from 100.

100-1=99         99-3=96            96-5=91

91-7=84           84-9=75             75-11=64

64-13=51        51-15=36             36-17=19

19-19=0 

The successive subtraction completed in 10 steps.

√100 =10.

By subtraction of  successive odd numbers from 169.

169-1=168         168-3=165            165-5=160

160-7=153           153-9=144            144-11=133

133-13= 120         120-15=105           105-17=88

88-19=69              69-21=48            48-23=25

25-25=0

The successive subtraction completed in 13 steps.

∴√169 =13.

class 8 maths chapter 6 exercise 6.3 (4)

4. Find the square roots of the following numbers by the Prime Factorisation Method.

(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v)7744 (vi) 9604 (vii) 5929 (viii) 9216(ix) 529 (x) 8100

Solution to class 8 maths chapter 6 exercise 6.3 (4)

(i) 729

$\begin{array}{|cccc|}\hline 3& & & 729\\ 3& & & 243\\ 3& & & 81\\ 3& & & 27\\ 3& & & 9\\ 3& & & 3\\ & & & 1\\ \hline\end{array}$

 

√(729) =√(3 ×3 ×3 ×3 ×3 ×3 )  =√(3 ×3 ×3 )² 

      =3 ×3 ×3  =27

(ii) 400

$\begin{array}{|cccc|}\hline 2& & & 400\\ 2& & & 200\\ 2& & & 100\\ 2& & & 50\\ 5& & & 25\\ 5& & & 5\\ & & & 1\\ \hline\end{array}$

 

√(400) =√(2×2×2×2×5×5)=√(2×2×5)²

=2 ×2 ×5=20

(iii) 1764 

$\begin{array}{|cccc|}\hline 2& & & 1764\\ 2& & & 882\\ 3& & & 441\\ 3& & & 147\\ 7& & & 49\\ 7& & & 7\\ & & & 1\\ \hline\end{array}$

 

√(1764) =√(2×2×3×3×7×7) =√(2×3×7)²

=2×3×7 =42.

(iv) 4096

$\begin{array}{|cccc|}\hline 2& & & 4096\\ 2& & & 2048\\ 2& & & 1024\\ 2& & & 512\\ 2& & & 256\\ 2& & & 128\\ 2& & & 64\\ 2& & & 32\\ 2& & & 16\\ 2& & & 8\\ 2& & & 4\\ 2& & & 2\\ & & & 1\\ \hline\end{array}$

 

√(4096) =√(2×2×2×2×2×2×2×2×2×2×2×2)

=√(2×2×2×2×2×2)²   =2×2×2×2×2×2

=64

(v) 7744

$\begin{array}{|cccc|}\hline 2& & & 7744\\ 2& & & 3872\\ 2& & & 1936\\ 2& & & 968\\ 2& & & 484\\ 2& & & 242\\ 11& & & 121\\ 11& & & 11\\ & & & 1\\ \hline\end{array}$

 

√(7744) = √(2×2×2×2×2×2×11×11)

=√(2×2×2×11)²  =2×2×2×11 =88.

(vi) 9604 

$\begin{array}{|cccc|}\hline 2& & & 9604\\ 2& & & 4802\\ 7& & & 2401\\ 7& & & 343\\ 7& & & 49\\ 7& & & 7\\ & & & 1\\ \hline\end{array}$

 

√(9604) = √(2×2×7×7×7×7)

√(2×7×7)²= 2×7×7=98

(vii) 5929

$\begin{array}{|cccc|}\hline 7& & & 5929\\ 7& & & 847\\ 11& & & 121\\ 11& & & 11\\ & & & 1\\ \hline\end{array}$

 

√(5929) =√(7×7×11×11)

=√(7×11)²  =7 ×11=77

(viii) 9216 

$\begin{array}{|cccc|}\hline 2& & & 9216\\ 2& & & 4608\\ 2& & & 2304\\ 2& & & 1152\\ 2& & & 576\\ 2& & & 288\\ 2& & & 144\\ 2& & & 72\\ 2& & & 36\\ 2& & & 18\\ 3& & & 9\\ 3& & & 3\\ & & & 1\\ \hline\end{array}$

 

√(9216) = √(2×2×2×2×2×2×2×2×2×2×3×3)

=√(2×2×2×2×2×3)²  =2×2×2×2×2×3=96

(ix) 529

$\begin{array}{|cccc|}\hline 23& & & 529\\ 23& & & 23\\ & & & 1\\ \hline\end{array}$

 

√(529) =√(23 ×23)  =√(23)² =23.

(x) 8100 

$\begin{array}{|cccc|}\hline 2& & & 8100\\ 2& & & 4050\\ 3& & & 2025\\ 3& & & 675\\ 3& & & 225\\ 3& & & 75\\ 5& & & 25\\ 5& & & 5\\ & & & 1\\ \hline\end{array}$

 

√(8100) =√2×2×3×3×3×3×5×5)

=√(2×3×3×5)² =2×3×3×5=90.

class 8 maths chapter 6 exercise 6.3 (5)

5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768

Solution to class 8 maths chapter 6 exercise 6.3 (5)

(1)252

$\begin{array}{|cccc|}\hline 2& & & 252\\ 2& & & 126\\ 3& & & 63\\ 3& & & 21\\ 7& & & 7\\ & & & 1\\ \hline\end{array}$

 

252 = 2 x 2 x 3 x 3 x 7

Here, prime factor 7 has no pair.

∴ 252 must be multiplied by 7 to make it a perfect square. 

⇒ 252 x 7 = 1764

And √(1764)=√( 2 × 3 × 7)² = 2×3×7=42

(ii)180

$\begin{array}{|cccc|}\hline 2& & & 180\\ 2& & & 90\\ 3& & & 45\\ 3& & & 15\\ 5& & & 5\\ & & & 1\\ \hline\end{array}$

 

180 = 2 ×2 × 3 × 3 ×5

Here, prime factor 5 has no pair.

∴ 180 must be multiplied by 5 to make it a perfect square

⇒ 180 x 5 = 900

And √(900) =√(2×3×5)² =2×3×5 =30.

(iii) 1008

$\begin{array}{|cccc|}\hline 2& & & 1008\\ 2& & & 504\\ 2& & & 252\\ 2& & & 126\\ 3& & & 63\\ 3& & & 21\\ 7& & & 7\\ & & & 1\\ \hline\end{array}$

 

1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7

Here, prime factor 7 has no pair.

∴ 1008 must be multiplied by 7 to make it a perfect square.

1008 x 7 = 7056

And √(7056) = √(2×2×3×7)² =2×2×3×7

=84.

(iv)2028

$\begin{array}{|cccc|}\hline 2& & & 2028\\ 2& & & 1014\\ 3& & & 507\\ 13& & & 169\\ 13& & & 13\\ & & & 1\\ \hline\end{array}$

 

2028 = 2 x 2 x 3 x 13 x 13

Here, prime factor 3 has no pair.

∴ 2028 must be multiplied by 3 to make it a perfect square.

2028 x 3 = 6084

And √(6084) =√(2×2×3×3×13×13)²

=2×2×3×3×13×13=78.

(v)

 

$\begin{array}{|cccc|}\hline 2& & & 1458\\ 3& & & 729\\ 3& & & 243\\ 3& & & 81\\ 3& & & 27\\ 3& & & 9\\ 3& & & 3\\ & & & 1\\ \hline\end{array}$

Prime factor 2 has no pair. 1458 = 2x 3 x 3 x 3x 3 x 3 x 3 

∴1458 must be multiplied by 2 to make it a perfect square.

1458×2=2916

And √(2916)= 2x 3 x 3 x 3=54

(vi)

$\begin{array}{|cccc|}\hline 2& & & 768\\ 2& & & 384\\ 2& & & 192\\ 2& & & 96\\ 2& & & 48\\ 2& & & 24\\ 2& & & 12\\ 2& & & 6\\ 3& & & 3\\ & & & 1\\ \hline\end{array}$

 

768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2  x 3 

Prime factor 3 has no pair.

∴768 must be multiplied by 3 to make it a perfect square.

768 x3=2304

√(2304) = 2 x 2 x 2 x 2  x 3 =48

class 8 maths chapter 6 exercise 6.3 (6)

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

(i) 252 (ii) 2925 (iii) 396 (iv) 2645(v) 2800 (vi) 1620

Solution to class 8 maths chapter 6 exercise 6.3 (6)

(i) 252

$\begin{array}{|cccc|}\hline 2& & & 252\\ 2& & & 126\\ 3& & & 63\\ 3& & & 21\\ 7& & & 7\\ & & & 1\\ \hline\end{array}$

 

252=2× 2×  3 × 3× 7 

Prime factor 7 has no pair.

∴252 must be divided by 7 to make it a perfect square.

252÷7=36

And √(36) =2×3=6

(ii) 2925

$\begin{array}{|cccc|}\hline 3& & & 2925\\ 3& & & 975\\ 5& & & 325\\ 5& & & 65\\ 13& & & 13\\ & & & 1\\ \hline\end{array}$

 

2925 =3× 3 × 5× 5× 13

Prime factor 13 has no pair.

∴2925 must be divided by 13 to make it a perfect square.

2925÷13=225

√(225) =3×5=15

(iii) 396

$\begin{array}{|cccc|}\hline 2& & & 396\\ 2& & & 198\\ 3& & & 99\\ 3& & & 33\\ 11& & & 11\\ & & & 1\\ \hline\end{array}$

 

396 =2× 2 × 3 × 3 × 11

Prime factor 11has no pair.

∴396 must be divided by 11 to make it a perfect square.

396÷11=36

√(36) =2×3=6

(iv) 2645

$\begin{array}{|cccc|}\hline 5& & & 2645\\ 23& & & 529\\ 23& & & 23\\ & & & 1\\ \hline\end{array}$

 

2645 = 5× 23 × 23

Prime factor 13 has no pair.

∴2645 must be divided by 5 to make it a perfect square.

2645÷5=529

√(529) =23

(v) 2800

$\begin{array}{|cccc|}\hline 2& & & 2800\\ 2& & & 1400\\ 2& & & 700\\ 2& & & 350\\ 5& & & 175\\ 5& & & 35\\ 7& & & 7\\ & & & 1\\ \hline\end{array}$

 

2800 =2× 2× 2× 2× 5 × 5 × 7

Prime factor 7 has no pair.

∴2800 must be divided by 7 to make it a perfect square.

2800÷7=400

√(400) =2×2×5=20

(vi) 1620

$\begin{array}{|cccc|}\hline 2& & & 1620\\ 2& & & 810\\ 3& & & 405\\ 3& & & 135\\ 3& & & 45\\ 3& & & 15\\ 5& & & 5\\ & & & 1\\ \hline\end{array}$

 

1620 =2× 2× 3× 3× 3 × 3 × 5

Prime factor 5 has no pair.

∴1620 must be divided by 5 to make it a perfect square.

1620÷5=324

√(324) =2×3×3=18

class 8 maths chapter 6 exercise 6.3 (7)

7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Solution to class 8 maths chapter 6 exercise 6.3 (7)

Student donated money = Rs 2401 

$\begin{array}{|cccc|}\hline 7& & & 2401\\ 7& & & 343\\ 7& & & 49\\ 7& & & 7\\ & & & 1\\ \hline\end{array}$

 

Let the number of students = x

Then donated money =(x)(x)=x²

According to problem,

x²=2401

x=√(2401)=√(7×7×7×7)

x=7×7=49

∴ Number of students = 49

class 8 maths chapter 6 exercise 6.3 (8)

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Solution to class 8 maths chapter 6 exercise 6.3 (8)

Given Number of plants = 2025

Let number of rows of planted plants are =x

Then Each row contains number of plants =x

According to problem,

$\begin{array}{|cccc|}\hline 3& & & 2025\\ 3& & & 675\\ 3& & & 225\\ 3& & & 75\\ 5& & & 25\\ 5& & & 5\\ & & & 1\\ \hline\end{array}$

x=√(2025)=√(3×3×3×3×5×5)

x=3×3×5=45

∴ Each row contains 45 plants.

class 8 maths chapter 6 exercise 6.3 (9)

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution to class 8 maths chapter 6 exercise 6.3 (9)

L.C.M. of 4, 9 and 10 is 180.

$\begin{array}{|cccc|}\hline 2& & & 180\\ 2& & & 90\\ 3& & & 45\\ 3& & & 15\\ 5& & & 5\\ & & & 1\\ \hline\end{array}$

 

Prime factors of 180 = 2 x 2 x 3 x 3 x 5

Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square, 180 x 5 = 900

∴ The smallest square number which is divisible by 4, 9 and 10 is 900

class 8 maths chapter 6 exercise 6.3 (10)

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Solution to class 8 maths chapter 6 exercise 6.3 (10)

L.C.M. of 8, 15 and 20 is 120. Prime factors of 120 =2 x 2 x 2 x 3 x 5 

$\begin{array}{|cccc|}\hline 2& & & 120\\ 2& & & 60\\ 2& & & 30\\ 3& & & 15\\ 5& & & 5\\ & & & 1\\ \hline\end{array}$

 

Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be multiplied by 2×3×5 to make it a perfect square.

120 ×2 × 3 × 5 = 3600

∴ The smallest square number which is divisible by 8, 15 and 20 is 3600

class 8 maths chapter 6 exercise 6.4

class 8 maths chapter 6 exercise 6.4(1)

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 to class 8 maths chapter 6 exercise 6.4 (1)

(i) 2304

√(2304)  =48

(ii) 4489

√(4489)  = 67

(iii) 3481

√(3481)  = 59

(iv) 529 

√(529)  =23

(v) 3249 

√(3249)  =57

(vi) 1369

√(1369) =37

(vii) 5776 

√(5776) =76

(viii) 7921 

√(7921) =89

(xi) 576 

√(576) =24

(x) 1024 

√(1024) =32

(xi) 3136 

√(3136) =56

(xii) 900 

√(900) =30

class 8 maths chapter 6 exercise 6.4 (2)

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

(i) 64 (ii) 144 (iii) 4489 (iv) 27225

(v) 390625

Solution to class 8 maths chapter 6 exercise 6.4 (2)

(i)  64 contains two digits which is even

∴ Number of digits in square root = ⁿ/₂=²/₂= 1

(ii) 144 contains three digits which is odd.

∴  Number of digits in square root = ⁽ⁿ⁺¹⁾/₂= ⁽³⁺¹⁾/₂=⁴/₂=2

(iv)  4489 contains four digits which is even

∴ Number of digits in square root =ⁿ/₂ =⁴/₂=2

(v) 390625 contains six digits which is even.

 ∴ Numberof digits in square root = ⁿ/₂ =⁶/₂=3

class 8 maths chapter 6 exercise 6.4 (3)

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

(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25

(v) 31.36

Solution to class 8 maths chapter 6 exercise 6.4 (3)

√(2.56) =1.6

(ii) 7.29

√(7.29) =2.7

 

(iii) 51.84

√(51.84)  =7.2

 

(iv) 42.25

√(42.25) =6.5

 

(v) 31.36 

√(31.36) =5.6

 

 

class 8 maths chapter 6 exercise 6.4 (4)

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. 

(i) 402 (ii) 1989 (iii) 3250 (iv) 825

(v) 4000

Solution to class 8 maths chapter 6 exercise 6.4 (4)

(i) 402

We get remainder 2 

∴ 2 must be subtracted from 402 to get a perfect square.

402-2=400

Hence, √(400) =20

(ii) 1989

We get remainder 53.

∴ 53 must be subtracted from 1989 to get a perfect square.

1989-53=1936

Hence √(1936) =44

(iii) 3250

We get remainder 1.

∴ 1 must be subtracted from 3250 to get perfect square.

3250-1=3249

√(3249) =57

(iv) 825

We get remainder 41′

∴ 41 must be subtracted from 825

825-41=784

Hence, √(784) = 28

(v) 

We get remainder 31.

∴ 31 must be subtracted from 4000 to get a perfect square.

4000-31=3969

√(3969) =63

 

class 8 maths chapter 6 exercise 6.4 (5)

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 to class 8 maths chapter 6 exercise 6.4 (5)

(i) 525

Remainder is 41.

∴ 22²<525

Next perfect square number 23²=529.

Hence number to be added=529-525=4

∴525+4=529

√(529)=23

(ii) 1750

 

Remainder is 69.

∴ 41²<1750

Next perfect square number 42²=1764.

Hence number to be added=1764-1750=14

∴1750+14=1764

√(1764)=42

(iii)252

Remainder is 27.

∴ 15²<252

Next perfect square number 16²=256.

Hence number to be added=256-252=4

∴252+4=256

√(256)=16

(iv) 1825

 

Remainder is 61.

∴ 42²<1825

Next perfect square number 43²=1849.

Hence number to be added=1849-1825=24

∴1825+24=1849

√(1849)=43

(v) 6412

 

Remainder is 12.

∴ 80²<6412

Next perfect square number 81²=6561.

Hence number to be added=6561-6412=149

∴6412+149=6561

√(6561)=81

 

class 8 maths chapter 6 exercise 6.4 (6)

6. Find the length of the side of a square whose area is 441 m²

Solution to class 8 maths chapter 6 exercise 6.4 (6)

Let the length of side of square be x meter.

Area of square =(side)²=x²

According to problem,

x=√(441)=√(3×3×7×7) =3×7

x=21 m

∴ Length of side of a square is 21 m.

class 8 maths chapter 6 exercise 6.4 (7)

7. In a right triangle ABC, ∠B = 90°.

(a) If AB = 6 cm, BC = 8 cm, find AC (b) If AC =13 cm, BC = 5 cm, find AB

Solution to class 8 maths chapter 6 exercise 6.4 (7)

(i) Using Pythagoras theorem,

AC²=AB²+BC²

⇒ AC²=(6)²+(8)²

⇒AC²=36+64=100

⇒AC=10

(ii) Using Pythagoras theorem,

 

AC²=AB²+BC²

⇒ (13)²=(AB)²+(5)²

⇒169=(AB)²+25

⇒AB²=169-25=144

AB=12 cm

class 8 maths chapter 6 exercise 6.4 (8)

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 to class 8 maths chapter 6 exercise 6.4 (8)

Given plants =1000

Since remainder is 39.

∴ 31²<1000

Next perfect square number 32²=1024

Hence number to be added =1024-1000=24

1000+24=1024

∴ The garden required 24 more plants.

class 8 maths chapter 6 exercise 6.4 (9)

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 to class 8 maths chapter 6 exercise 6.4 (9)

Number of children =500

By getting the square root of this number,

We get 

In each row,  the number of children is 22.

And left out children are 16.

Squares

here we can learn Squares 

Square Roots

here we can learn Square Roots