Chapter 5 Squares and Square Roots Ex 5.4 NCERT Solutions for Class 8 Maths

Latest Updated : October 2023

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 of all the chapters, exercise wise with 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 bookmark ncertforclass8.com to boost your preparation.

NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.4

NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.4

Class 8 Maths Chapter 5 Squares and Square Roots Exercise 5.4

Exercise 5.3

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) 2304

\[∴\ \sqrt{2304}\ =\ 48\]

(ii) 4429

\[∴\ \sqrt{4429\ }=\ 67\]

(iii) 3481

\[∴\ \sqrt{3481}=\ 59\]

(iv) 529

\[∴\ \sqrt{529}\ =\ 23\]

(v) 3249

\[∴\ \sqrt{3249}=\ 57\]

(vi) 1369

\[∴\ \sqrt{1369}=\ 37\]

(vii) 5776

\[∴\ \sqrt{5776\ }=\ 76\]

(viii) 7921

\[∴\ \sqrt{7921\ }=\ 89\]

(ix) 576

\[∴\ \sqrt{576}=\ 24\]

(x) 1024

\[∴\ \sqrt{1024}=\ 32\]

(xi) 3136

\[∴\ \sqrt{3136}=\ 56\]

(xii) 900

\[∴\ \sqrt{900}=\ 30\]

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:

(i) Here, 64 contains two digits which is even.

\[Therefore,\ number\ of\ digits\ in\ square\ root=\frac{n}{2}=\ \frac{2}{2}=\ 1\]

(ii) Here, 144 contains three digits which is odd.

\[Therefore,\ number\ of\ digits\ in\ square\ root\ =\frac{\left(\ n+1\right)}{2}=\ \frac{\left(3+1\right)}{2}=\frac{4}{2}=\ 2\]

(iii) Here, 4489 contains four digits which is even.

\[Therefore,\ number\ of\ digits\ in\ square\ root\ =\frac{n}{2}=\ \frac{4}{2}=\ 2\]

(iv) Here, 27225 contains five digits which is odd.

\[Therefore,\ number\ of\ digits\ in\ square\ root\ =\frac{\left(\ n+1\right)}{2}=\ \frac{\left(5+1\right)}{2}=\ \frac{6}{2}=\ 3\]

(v) Here, 390625 contains six digits which is even.

\[Therefore,\ number\ of\ digits\ in\ square\ root\ =\ \frac{n}{2}=\frac{6}{2}=3\]

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:

(i) 2.56

\[∴\ \sqrt{2.56}=\ 1.6\]

(ii) 7.29

\[∴\ \sqrt{7.29}=\ 2.7\]

(iii) 51.84

\[∴\ \sqrt{51.84}=\ 7.2\]

(iv) 42.25

\[∴\ \sqrt{42.25}=\ 6.5\]

(v) 31.36

\[∴\ \sqrt{31.36}=\ 5.6\]

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.

(i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000

Solution:

(i) 402

∴ We must subtract 2 from 402 to get a perfect square.
New number = 402 – 2 = 400

\[∴\ \sqrt{400}=\ 20\]

(ii) 1989

∴ We must subtract 53 from 1989 to get a perfect square.
New number = 1989 – 53 = 1936

\[∴\ \sqrt{1936}=\ 44\]

(iii) 3250

∴ We must subtract 1 from 3250 to get a perfect square.
New number = 3250 – 1 = 3249

\[∴\ \sqrt{3249}=\ 57\]

(iv) 825

∴ We must subtract 41 from 825 to get a perfect square.
New number = 825 – 41 = 784

\[∴\ \sqrt{784}=\ 28\]

(v) 4000

∴ We must subtract 31 from 4000 to get a perfect square.
New number = 4000 – 31 = 3969

\[∴\ \sqrt{3969}=\ 63\]

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) 525

Since remainder is 41.

\[Therefore\ 22^{2\ }\ <\ 525\]

\[Next\ perfect\ square\ number\ =\ 23^2\ =\ 529\]

Hence, number to be added = 529-525 = 4
∴ 525 + 4 = 529

\[∴\ \sqrt{529}=\ 23\]

(ii) 1750

Since remainder is 69.

\[Therefore,\ \ 41^{2\ }<\ 1750\]

\[Next\ perfect\ square\ number\ =\ 42^2\ =\ 1764\]

Hence, number to be added = 1764 – 1750 = 14
∴ 1750+14=1764

\[∴\ \sqrt{1764}=\ 42\]

(iii) 252

Since remainder is 27.

\[Therefore,\ 15^{2\ }<\ 252\]

\[Next\ perfect\ square\ number\ =\ 16^2=\ 256\]

Hence, number to be added = 256-252 = 4.
∴ 252 + 4 = 256

\[∴\ \sqrt{256}=\ 16\]

(iv) 1825

Since remainder is 61.

\[Therefore,\ 42^2\ <\ 1825\]

\[Next\ perfect\ square\ number\ =43^2=\ 1849\]

Hence, number to be added = 1849-1825 = 24
∴ 1825 + 24 = 1849

\[∴\ \sqrt{1849}=\ 43\]

(v) 6412

Since remainder is 12.

\[Therefore,\ 80^2<\ 6412\]

\[Next\ perfect\ square\ number\ =81^2=\ 6561\]

Hence, number to be added = 6561-6412 = 149
∴ 6412 + 149 = 6561

\[∴\ \sqrt{6561}=\ 81\]

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

Solution:

\[Let\ the\ length\ of\ each\ side\ of\ the\ square\ be\ x\ metre.\]

\[Area\ of\ a\ square\ =\ side\ x\ side\ =\ x^2\]

\[According\ to\ the\ question,\ x^{2\ }=\ 441\ m^2\]

\[x\ =\ \sqrt{441}=\ \sqrt{3\ X\ 3\ X\ 7\ X\ 7}\ =\ 3\ X\ 7\ =\ 21\]

\[x\ =\ 21\ m\]

\[Therefore,\ length\ of\ the\ side\ of\ square\ is\ 21m.\]

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:

\[\left(a\right)\ U\sin g\ Pythagoras\ theorem,\]

\[AC^2=\ AB^2+\ BC^2\]

\[AC^2\ =\ 6^2+8^2\]

\[AC^2=\ 36\ +\ 64=\ 100\]

\[AC\ =\ \sqrt{100}=\ 10\ cm.\]

\[\]

\[\left(b\right)\ U\sin g\ Pythagoras\ theorem,\]

\[AC^2=\ AB^2+\ BC^2\]

\[13^2\ =\ AB^2+5^2\]

\[169=\ AB^2+\ 25\]

\[AB^2=\ 169-25\ =\ 144\]

\[AB\ =\ \sqrt{144}=\ 12cm.\]

\[\]

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:

No. of plants = 1000
Since remainder is 39.

\[Therefore,\ 31^2<\ 1000\]

\[Next\ Perfect\ square\ number\ =\ 32^2=1024\]

Hence, number to be added = 1024 – 1000 = 24
∴ 1000+24=1024

Hence, the gardener required 24 more plants

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:

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.

Using Class 8 NCERT solutions offers several key advantages for students:

  1. Accurate Answers: NCERT solutions for Class 8 Maths are prepared by expert educators and thoroughly reviewed to ensure quality and accurate answers.
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