NCERT Solutions For Class 8 Maths Chapter 5 Ex 5.4
ncert solutions for class 8 maths chapter 5, Exercise 5.4 involve complete answers for each question in the exercise 5.4. The solutions provide students a strategic methods to prepare for their exam. Class 8 Maths Chapter 8 Squares and Square Roots Exercise 5.4 questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. NCERT Solutions for Class 8 Maths Chapter 8 Squares and Square Roots Exercise 5.4 prepared by our subject matter experts in very delicate, easy and creative way.1. Find the square root of each of the following numbers by the Division method.
i. 2304
Solution:
∴ √2304 = 48
∴ √900 = 30
2. Find the number of digits in the square root of each of the following numbers (without any
calculation).
i. 144
Solution:
∴ √144 = 12
Hence, the square root of the number 144 has 2 digits.
ii. 4489
√27225 = 165
Hence, the square root of the number 27225 has 3 digits.
iv. 390625
Solution
∴ √390625 = 625
Hence, the square root of the number 390625 has 3 digits.
3. Find the square root of the following decimal numbers.
Solution:
i. 2.56
∴ √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
4. Find the least number which must be subtracted from each of the following numbers to get a perfect square. Also, find the square root of the perfect square so obtained.
i. 402
∴ √402 = 20
∴ we must subtract 2 from 402 to get a perfect square.
New number = 402 – 2 = 400
∴ √400 = 20
ii. 1989
∴ We must subtract 53 from 1989 to get a perfect square. New number = 1989 – 53 = 1936
∴ √1936 = 44
iii. 3250
∴ We must subtract 1 from 3250 to get a perfect square.
New number = 3250 – 1 = 3249
∴ √3249 = 57
iv. 825
We must subtract 41 from 825 to get a perfect square.
New number = 825 – 41 = 784
∴ √784 = 28
v. 4000
∴ we must subtract 31 from 4000 to get a perfect square.
∴ √2304 = 48
ii. 4489
Solution:
∴ √4489 = 67
iii. 3481
Solution:
∴ √3481 = 59
iv. 529
Solution:
∴ √529 = 23
v. 3249
Solution:
∴ √3249 = 57
vi. 1369
Solution:
∴ √1369 = 37
vii. 5776
Solution:
∴ √5776 = 76
viii. 7921
Solution:
∴ √7921 = 89
ix. 576
Solution:
∴ √576 = 24
x. 1024
Solution:
∴ √1024 = 32
xi. 3136
Solution:
∴ √3136 = 56
xii. 900
Solution:
Solution:
∴ √4489 = 67
Solution:
∴ √3481 = 59
Solution:
∴ √529 = 23
Solution:
∴ √3249 = 57
Solution:
∴ √1369 = 37
Solution:
∴ √5776 = 76
Solution:
∴ √7921 = 89
Solution:
∴ √576 = 24
Solution:
∴ √1024 = 32
Solution:
∴ √3136 = 56
Solution:
∴ √900 = 30
2. Find the number of digits in the square root of each of the following numbers (without any
calculation).
i. 144
Solution:
∴ √144 = 12
Hence, the square root of the number 144 has 2 digits.
ii. 4489
Solution
∴ √4489 = 67
Hence, the square root of the number 4489 has 2 digits.
∴ √4489 = 67
Hence, the square root of the number 4489 has 2 digits.
iii. 27225
Solution
Solution
√27225 = 165
Hence, the square root of the number 27225 has 3 digits.
iv. 390625
Solution
∴ √390625 = 625
Hence, the square root of the number 390625 has 3 digits.
3. Find the square root of the following decimal numbers.
Solution:
i. 2.56
∴ √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
4. Find the least number which must be subtracted from each of the following numbers to get a perfect square. Also, find the square root of the perfect square so obtained.
i. 402
∴ √402 = 20∴ we must subtract 2 from 402 to get a perfect square.
New number = 402 – 2 = 400
∴ √400 = 20
ii. 1989
∴ We must subtract 53 from 1989 to get a perfect square. New number = 1989 – 53 = 1936
∴ √1936 = 44
iii. 3250
∴ We must subtract 1 from 3250 to get a perfect square.
New number = 3250 – 1 = 3249
∴ √3249 = 57
iv. 825
We must subtract 41 from 825 to get a perfect square.
New number = 825 – 41 = 784
∴ √784 = 28
v. 4000
∴ we must subtract 31 from 4000 to get a perfect square.
New number = 4000 – 31 = 3969
∴ √3969 = 63
5. Find the least number which must be added to each of the following numbers to get a perfect square. Also, find the square root of the perfect square so obtained.
(i) 525
Here, (22)2 < 525 > (23)2
We can say 525 is ( 129 – 125 ) 4 less than (23)2.
∴ if we add 4 to 525, it will be a perfect square. New number = 525 + 4 = 529
∴ √529 = 23
(ii) 1750
Here, \((41)^2 < 1750 > (42)^2\)
We can say 1750 is ( 164 – 150 ) 14 less than \((42)^2\).
∴ If we add 14 to 1750, it will be a perfect square.
New number = 1750 + 14 = 1764
∴√1764 = 42
(iii) 252
Here, \((15)^2 < 25^2 > (16)^2\)
We can say 252 is ( 156 – 152 ) 4 less than \((16)^2\).
∴ if we add 4 to 252, it will be a perfect square.
New number = 252 + 4 = 256
∴ √256 = 16
(iv)1825
Here, \((42)^2 < 1825 > (43)^2\)
We can say 1825 is ( 249 – 225 ) =24 less than \((43)^2\).
∴ if we add 24 to 1825, it will be a perfect square.
New number = 1825 + 24 = 1849
∴ √1849 = 43
(v)6412
Here, \((80)^2 < 6412 > (81)^2\)
We can say 6412 is ( 161 – 12 )= 149 less than \((81)^2\).
∴ if we add 149 to 6412, it will be a perfect square.
New number = 6412 + 149 = 656
∴ √6561 = 81
6. Find the length of the side of a square whose area is \(441 \text{m}^2\).
Solution:
Let the length of each side of the field = a Then, the area of the field = \(441 \text{m}^2\).
\(a^2 = 441\)
\(a = \sqrt{441}\) m
∴ The length of each side of the field = \(a = 21\) m.
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:
a.
Given, AB = 6 cm, BC = 8 cm
Let AC be x cm.
\(∴ AC^2 = AB^2 + BC^2\)
Hence, AC = 10 cm.
b.
Given, AC = 13 cm, BC = 5 cm
Let AB be x cm.
∴ \(AC^2 = AB^2 + BC^2\)
\( AC^2 – BC^2 = AB^2\)
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 remains the same. Find the minimum number of plants he needs for this.
Solution:
Let the number of rows and columns be x.
∴ The total number of rows and columns\(= x× x = x^2 \)
∴ √3969 = 63
5. Find the least number which must be added to each of the following numbers to get a perfect square. Also, find the square root of the perfect square so obtained.
(i) 525
Here, (22)2 < 525 > (23)2
We can say 525 is ( 129 – 125 ) 4 less than (23)2.
∴ if we add 4 to 525, it will be a perfect square. New number = 525 + 4 = 529
∴ √529 = 23
(ii) 1750
Here, \((41)^2 < 1750 > (42)^2\)
We can say 1750 is ( 164 – 150 ) 14 less than \((42)^2\).
∴ If we add 14 to 1750, it will be a perfect square.
New number = 1750 + 14 = 1764
∴√1764 = 42
(iii) 252
Here, \((15)^2 < 25^2 > (16)^2\)
We can say 252 is ( 156 – 152 ) 4 less than \((16)^2\).
∴ if we add 4 to 252, it will be a perfect square.
New number = 252 + 4 = 256
∴ √256 = 16
(iv)1825
Here, \((42)^2 < 1825 > (43)^2\)
We can say 1825 is ( 249 – 225 ) =24 less than \((43)^2\).
∴ if we add 24 to 1825, it will be a perfect square.
New number = 1825 + 24 = 1849
∴ √1849 = 43
(v)6412
Here, \((80)^2 < 6412 > (81)^2\)
We can say 6412 is ( 161 – 12 )= 149 less than \((81)^2\).
∴ if we add 149 to 6412, it will be a perfect square.
New number = 6412 + 149 = 656
∴ √6561 = 81
6. Find the length of the side of a square whose area is \(441 \text{m}^2\).
Solution:
Let the length of each side of the field = a Then, the area of the field = \(441 \text{m}^2\).
\(a^2 = 441\)
\(a = \sqrt{441}\) m
∴ The length of each side of the field = \(a = 21\) m.
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:
a.
Given, AB = 6 cm, BC = 8 cm
Let AC be x cm.
\(∴ AC^2 = AB^2 + BC^2\)
Hence, AC = 10 cm.
b.
Given, AC = 13 cm, BC = 5 cm
Let AB be x cm.
∴ \(AC^2 = AB^2 + BC^2\)
\( AC^2 – BC^2 = AB^2\)
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 remains the same. Find the minimum number of plants he needs for this.
Solution:
Let the number of rows and columns be x.
∴ The total number of rows and columns\(= x× x = x^2 \)
As per the question, \(x^2 = 1000\)
⇒ x = √1000
Here, \((31)^2 < 1000 > (32)^2\)
We can say 1000 is ( 124 – 100 )=24 less than \((32)^2\).
∴ 24 more plants are needed.
9. There are 500 children in a school. For a P.T. drill, they have to stand so that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?
Solution:
Let the number of rows and columns be x.
∴ The total number of rows and columns \(= x × x = x^2\)
⇒ x = √1000
Here, \((31)^2 < 1000 > (32)^2\)
We can say 1000 is ( 124 – 100 )=24 less than \((32)^2\).
∴ 24 more plants are needed.
9. There are 500 children in a school. For a P.T. drill, they have to stand so that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?
Solution:
Let the number of rows and columns be x.
∴ The total number of rows and columns \(= x × x = x^2\)
As per question, \(x^2 = 500\)
\(x = \sqrt{500}\)
Hence, 16 children would be left out of the arrangement.
\(x = \sqrt{500}\)
Hence, 16 children would be left out of the arrangement.
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