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If you are in search of **NCERT solutions for Class 8 Maths Chapter** **5 Squares and Square Roots**, then you are at the right place. When it comes to **Mathematics**, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed **NCERT solutions for Class 8 Maths**Â of all the chapters,Â **exercise wise**Â withÂ **updated syllabus by NCERT**. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmarkÂ **ncertforclass8.com**Â to boost your preparation.

## NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.3

**Class 8 Maths Chapter** **5 Squares and Square Roots** **Exercise 5.**3

**Exercise 5.3**

**1. What could be the possible â€˜oneâ€™sâ€™ digits of the square root of each of the following numbers?**

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

Solution:

(i) The unitâ€™s digit of the square of a number having digit as unitâ€™s place 1 is 1 and also 9 is 1 [ 9 square = 81 and 1 square = 1]

âˆ´ Unitâ€™s digit of the square root of number **9801** is equal to **1 or 9.**

(ii) The unitâ€™s digit of the square of a number having digit as unitâ€™s place of 6 is 6 and also 4 is 6 [6 square =36 and 4 square =16, both the squares have unit digit 6].

âˆ´ Unitâ€™s digit of the square root of number **99856** is equal to **4 or 6.**

(iii) The unitâ€™s digit of the square of a number having digit as unitâ€™s place 1 is 1 and also 9 is 1 [9 square =81 whose unit place is 1].

âˆ´ Unitâ€™s digit of the square root of number **998001** is equal to **1 or 9.**

(iv) The unitâ€™s digit of the square of a number having digit as unitâ€™s place 5 is 5.

âˆ´ Unitâ€™s digit of the square root of number **657666025** is equal to **5.**

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

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

Solution:

Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.

(i) But given number 153 has its unit digit 3. So it is **not a perfect square** number.

(ii) Given number 257 has its unit digit 7. So it is **not a perfect square** number.

(iii) Given number 408 has its unit digit 8. So it is **not a perfect square** number.

(iv) Given number 441 has its unit digit 1. So it would be a **perfect square** number

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

Solution:

**Square root of 100 = **

100 â€“ 1 = 99

99 â€“ 3 = 96

96 â€“ 5 = 91

91 â€“ 7 = 84

84 â€“ 9 = 75

75 â€“ 11 = 64

64 â€“ 13 = 51

51 â€“ 15 = 36

36 â€“ 17 = 19

19 â€“ 19 = 0

Here, we have performed subtraction ten times.**âˆ´ âˆš100 = 10**

**Square root of 169 =**

169 â€“ 1 = 168

168 â€“ 3 = 165

165 â€“ 5 = 160

160 â€“ 7 = 153

153 â€“ 9 = 144

144 â€“ 11 = 133

133 â€“ 13 = 120

120 â€“ 15 = 105

105 â€“ 17 = 88

88 â€“ 19 = 69

69 â€“ 21 = 48

48 â€“ 23 = 25

25 â€“ 25 = 0

Here, we have performed subtraction thirteen times.**âˆ´ âˆš169 = 13**

**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

(v) 7744 (vi) 9604

**(vii) 5929 (viii) 9216**

(ix) 529 (x) 8100

(ix) 529 (x) 8100

Solution:

**(i) 729** = 3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—1

â‡’ 729 = (3Ã—3Ã—3)Ã—(3Ã—3Ã—3)

`\[â‡’729=(3\times3\times3)^{^2}\]`

`\[\sqrt{729\ }=\ 3\times3\times3\ =\ 27\]`

**(ii) 400** = 2Ã—2Ã—2Ã—2Ã—5Ã—5Ã—1

â‡’ 400 = (2Ã—2Ã—5)Ã—(2Ã—2Ã—5)

`\[â‡’400=(2\times2\times5)^2\]`

`\[\sqrt{400}\ =\ 2\times2\times5=20\]`

**(iii) 1764** = 2Ã—2Ã—3Ã—3Ã—7Ã—7

â‡’ 1764 = (2Ã—3Ã—7)Ã—(2Ã—3Ã—7)

`\[â‡’1764=(2\times3\times7)^2\ \]`

`\[â‡’\ \sqrt{1764}=2\times3\times7=42\]`

**(iv) 4096** = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2

â‡’ 4096 = (2Ã—2Ã—2Ã—2Ã—2Ã—2)Ã—(2Ã—2Ã—2Ã—2Ã—2Ã—2)

`\[â‡’4096=(2\times2\times2\times2\times2\times2)^2\]`

`\[\sqrt{4096\ }\ =\ 2\times2\times2\times2\times2\times2=64\]`

**(v) 7744** = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—11Ã—11Ã—1

â‡’ 7744 = (2Ã—2Ã—2Ã—11)Ã—(2Ã—2Ã—2Ã—11)

`\[â‡’7744=(2\times2\times2\times11)^2\]`

`\[\sqrt{7744}=\ 2\times2\times2\times11=88\]`

**(vi) 9604** = 62 Ã— 2 Ã— 7 Ã— 7 Ã— 7 Ã— 7

â‡’ 9604 = ( 2 Ã— 7 Ã—7 ) Ã— ( 2 Ã— 7 Ã—7 )

`\[â‡’9604=(2\times7\times7)^2\]`

`\[\sqrt{9604}\ =\ 2\times7\times7=98\]`

**(vii) 5929** = 7Ã—7Ã—11Ã—11

â‡’ 5929 = (7Ã—11)Ã—(7Ã—11)

`\[â‡’5929=(7\times11)^2\]`

`\[\sqrt{5929\ }\ =\ 7\times11=77\]`

**(viii) 9216 **= 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—1

â‡’ 9216 = ( 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3) Ã— ( 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3)

â‡’ 9216 = 96 Ã— 96

`\[â‡’9216=(96)^2\]`

`\[\sqrt{9216\ }=\ 96\]`

**(ix) 529** = 23Ã—23

`\[529=(23)^2\]`

`\[\sqrt{529}\ =\ 23\]`

**(x) 8100** = 2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—5Ã—5Ã—1

â‡’ 8100 = (2Ã—3Ã—3Ã—5)Ã—(2Ã—3Ã—3Ã—5)

â‡’ 8100 = 90Ã—90

`\[â‡’8100=(90)^2\]`

`\[\sqrt{8100}\ =\ 90\]`

**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:

(i) **252** = 2 Ã— 2 Ã— 3 x 3 x 7

Here, prime factor 7 has no pair.

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

âˆ´ 252 x 7 = 1764

And, **âˆš1764 = 2 Ã— 3 Ã— 7= 42**

**(ii) 180** = 2 Ã— 2 Ã— 3 x 3x 5

Here, prime factor 5 has no pair.

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

âˆ´ 180 x 5 = 900

And, **âˆš900 = 2 x 3 x 5 = 30**

**(iii) 1008** = 2 Ã— 2 x 2 x 2 x 3 x 3 x 7

Here, prime factor 7 has no pair.

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

âˆ´ 1008 x 7 = 7056

And, **âˆš7056 = 2x2x3x7= 84**

**(iv) 2028** = 2 Ã— 2 Ã— 3 Ã— 13 x 13

Here, prime factor 3 has no pair.

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

âˆ´ 2028 x 3 = 6084

And, **âˆš6084 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 13Ã—13=78 **

**(v) 1458** = 2 x 3 x 3 x 3 x 3 x 3 x 3

Here, prime factor 2 has no pair.

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

âˆ´1458 x 2 = 2916

And, **âˆš2916 = 2x3x3x3= 54**

**(vi) 768** = 2 Ã— 2 Ã— 2Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3

Here, prime factor 3 has no pair.

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

âˆ´ 768 x 3 = 2304

And,** âˆš2304 = 2 x 2 x 2 x 2 x 3 = 48**

**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:

**(i) 252** = 2Ã—2Ã—3Ã—3Ã—7

Here, prime factor 7 has no pair.

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

âˆ´ 252 Ã· 7= 36

And, **âˆš36 = 2 x 3 = 6**

**(ii) 2925** = 3 Ã— 3 Ã— 5 Ã— 5 Ã— 13

Here, prime factor 13 has no pair.

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

âˆ´ 2925 Ã· 13 = 225

And, **âˆš225= 3 x 5 = 15**

**(iii) 396** = 2 Ã— 2Ã—3Ã—3Ã—11

Here, prime factor 11 has no pair.

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

âˆ´ 396 Ã· 11 = 36

And, **âˆš36 = 2×3=6**

**(iv) 2645 =** 5 x 23 x 23

Here, prime factor 5 has no pair.

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

âˆ´ 2645 Ã· 5 = 529

And, **âˆš529 = 23 x 23 = 23**

**(v) 2800** = 2 Ã— 2x 2 x 2 x 5 x 5 x 7

Here, prime factor 7 has no pair.

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

âˆ´ 2800 Ã· 7 = 400

And, **âˆš400 = 2 x 2 x 5=20**

**(vi) 1620** = 2 x 2 x3 x 3 x 3 x 3 x 5

Here, prime factor 5 has no pair.

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

âˆ´1620 Ã· 5 = 324

And, **âˆš324 = 2 x 3 x 3 = 18**

**7. The students of Class VIII of a school donated ` 2401 in all, for Prime Ministerâ€™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:

Here, Donated money = â‚¹2401

Let the number of students be x.

Therefore donated money = *x* **x** *x *

According to question,

`\[x^{2\ }=2401\ \]`

`\[â‡’x\ =\ \sqrt{2401}=\ \sqrt{7\ x\ 7\ x\ 7\ x\ 7}\]`

â‡’x= 7 x7 = 49

Hence, the number of students is** 49**.

**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:

Here, Number of plants = 2025

Let the number of rows of planted plants be x.

And each row contains number of plants = x

According to question,

`\[x^2=2025\]`

`\[â‡’x=\sqrt{2025}\ =\ \sqrt{3\ x\ 3\ x\ 3\ x\ 3\ x\ 5\ x\ 5}\]`

â‡’x= 3x3x5= 45

Hence, each row contains **45 plants.**

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

Solution:

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

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.

âˆ´ 180 x 5 = 900

Hence, the smallest square number which is divisible by 4, 9 and 10 is **900.**

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

Solution:

L.C.M. of 8, 15 and 20 is 120.

Prime factors of 120 = 2 x 2 x 2 x 3 x 5

Here, prime factor 2, 3 and 5 has no pair.

Therefore 120 must be multiplied by 2 Ã— 3 Ã— 5 to make it a perfect square.

âˆ´ 120 x 2 x 3 x 5 = 3600

Hence, the smallest square number which is divisible by 8, 15 and 20 is **3600.**

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