NCERT Solutions Class 8 Maths Chapter 5 Squares and Square Root Exercise 5.4

Exercise 5.4 of Class 8 Maths Chapter 5 demonstrates how to estimate square roots of non-perfect square values. In this exercise you'll identify how to estimate square roots by a simple method. This is an important skill to learn as it helps in quick mental calculations and prepares you for advanced methods you will study in later classes.

The NCERT Solutions in this exercise aligns with the latest NCERT Class 8 syllabus and completing this exercise can help you develop familiarity with non-perfect square root numbers, which is a common question in school examinations and in some competitive exams like the Olympiads.

The NCERT Solutions provide a step-by-step explanation for each question. Regularly practicing these types of problems will allow you to improve your accuracy, speed and confidence in dealing with numbers in a smart manner.

1.0Download NCERT Solutions Class 8 Maths Chapter 5 Squares and Square Root Exercise 5.4: Free PDF

Discover how to quickly calculate square roots by using the NCERT Solutions for Class 8 Maths Chapter 5. The PDF provides all the solutions to Exercise 5.4 in very simple steps. Click below to download the PDF for FREE:

2.0Key Concepts in Exercise 5.4 of Class 8 Maths Chapter 5

This exercise aims to estimate the square roots of non-perfect squares. In this exercise, some of the key concepts include:

• Estimating square roots of non-perfect squares.
• Finding the range between two perfect squares.
• Using the number line to determine a close square root estimate.
• Assessing the accuracy of estimates in relation to other estimates.

3.0NCERT Class 8 Maths Chapter 5: Other Exercises

4.0NCERT Class 8 Maths Chapter 5 Exercise 5.4: Detailed Solutions

• 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 Sol. (i) By long division, we have
  
  $\therefore \sqrt{2304}=48$ (ii) By long division, we have
  
  $\therefore \sqrt{4489}=67$ (iii) By long division, we have
  
  $\therefore \sqrt{3481}=59$ (iv) 529 By long division, we have
  
  $\therefore \sqrt{529}=23$ (v) 3249 By long division, we have
  
  $\therefore \sqrt{3249}=57$ (vi) 1369 By long division, we have
  
  $\therefore \sqrt{1369}=37$ (vii) 5776
  
  $\therefore \sqrt{5776}=76$ (viii) 7921 By long division, we have
  
  $\therefore \sqrt{7921}=89$ (ix) 576
  
  By long division, we have $\therefore \sqrt{576}=24$ (x) 1024 By long division, we have
  
  $\therefore \sqrt{1024}=32$ (xi) 3136 By long division, we have
  
  $\therefore \sqrt{3136}=56$ (xii) 900
  
• 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 Sol. We know that if a perfect square is of $n$ digits, then its square root will have $\frac{n}{2}$ digits, if n is even $\frac{(\mathrm{n}+1)}{2}$ digits, if n is odd. (i) Given number is 64 . It is a 2 -digit number, i.e., even number of digits. $\therefore$ The number of digits in $\sqrt{64}$ is $\frac{2}{2}$, i.e., 1 . (ii) Given number is 144. It is a 3-digit number, i.e., odd number of digits. $\therefore \sqrt{144}$ contains $\frac{3+1}{2}$, i.e., 2-digits. (iii) Given number is 4489. It is a 4-digit number, i.e. even number of digits. $\therefore$ The number of digits in $\sqrt{4489}$ is $\frac{4}{2}$, i.e., 2 digits. (iv) Given number is 27225. It is a 5 -digit number, i.e., odd number of digits $\therefore$ The number of digits in $\sqrt{27225}$ is $\frac{5+1}{2}$, i.e., $\frac{6}{2}=3$ digits. (v) Given number has 6-digits i.e., even number of digits. $\therefore \sqrt{390625}$ contains $\frac{6}{2}$, i.e., 3-digits.
• Find the square root of the following decimal numbers. (i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36 Sol. (i) 2.56 Here, the number of decimal places is already even. So, mark off periods and proceed as under
  
  $\therefore \sqrt{2.56}=1.6$ (ii) 7.29 Here, the number of decimal places is already even. So, mark off periods and proceed as under
  
  $\therefore \sqrt{7.29}=2.7$ (iii) 51.84 Here, the number of decimal places are already even. So, mark off periods and proceed as under
  
  $\therefore \sqrt{51.84}=7.2$ (iv) 42.25 Here, the number of decimal places is already even. So, mark off periods and proceed as under
  
  $\therefore \sqrt{42.25}=6.5$ (v) 31.36 Here, the number of decimal places is already even. So, mark off periods and proceed as under
  
  $\therefore \sqrt{31.36}=5.6$
• 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 Sol. (i) Let us try to find the square root of 402. This shows that $(20{)}^2$ is less than 402 by 2 . So, in order to get a perfect square, 2 must be subtracted from the given number.
  
  $\therefore$ Required perfect square number $=402-2=400$ Also, $\sqrt{400}=20$ (ii) Let us try to find the square root of 1989. This shows that (44) ${}^2$ is less than 1989 by 53. So, in order to get a perfect square, 53 must be subtracted from the given number.
  
  $\therefore$ Required perfect square number = 1989-53 = 1936 Also, $\sqrt{1936}=44$ (iii) Let us try to find the square root of 3250 . This shows that (57) ${}^2$ is less than 3250 by 1 . So, in order to get a perfect square, 1 must be subtracted from the given number.
  
  $\therefore$ Required perfect square number $=3250-1=3249$ Also, $\sqrt{3249}=57$ (iv) Let us try to find the square root of 825 . This shows that $(28{)}^2$ is less than 825 by 41 . So, in order to get a perfect square, 41 must be subtracted from the given number.
  
  $\therefore$ Required perfect square number = $825-41=784$ Also, $\sqrt{784}=28$ (v) Let us try to find the square root of 4000.This shows that $(63{)}^2$ is less than 4000 by 31 . So in order to get a perfect square, 31 must be subtracted from the given number.
  
  $\therefore$ Required perfect square number. $=4000-31=3969$ Also, $\sqrt{3969}=63$
• 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 Sol. (i) We try to find out the square root of 525.
  
  We observe that $(22{)}^2<525<(23{)}^2$ The required number to be added $=(23{)}^2-525$ = 529-525 = 4 $\therefore$ Required perfect square number $=525+4=529$ Clearly, $\sqrt{529}=23$ (ii) We try to find out the square root of 1750
  
  We observe that $(41{)}^2<1750<(42{)}^2$. The required number to be added $=(42{)}^2-1750$ = 1764-1750 $=14$ $\therefore$ Required perfect square number $=(1750+14)=1764$ Clearly, $\sqrt{1764}=42$ (iii) We try to find the square root of 252 .
  
  we observe that $(15{)}^2<252<(16{)}^2$ The required number to be added $=(16{)}^2-252$ $=256-252=4$ $\therefore$ Required perfect square number $=252+4=256$ Clearly, $\sqrt{256}=16$ (iv) We try to find out the square root of 1825
  
  we observe that $(42{)}^2<1825<(43{)}^2$ The required number to be added. $=(43{)}^2-1825$ $=1849-1825=24$ $\therefore$ Required perfect square number. $=1825+24=1849$ Clearly, $\sqrt{1849}=43$ (v) We try to find out the square root of 6412
  
  we observe that $(80.7{)}^2<6412<(81{)}^2$ The required number to be added $=(81{)}^2-6412$ $=6561-6412=149$ $\therefore$ Required perfect square number. $=6412+149=6561$ Clearly, $\sqrt{6561}=81$
• Find the length of the side of a square whose area is $441{\mathrm{m}}^2$. Sol. Area $=(\text{ side }{)}^2$ $=441{\mathrm{m}}^2$
  
  $\therefore$ Side $=\sqrt{441}\mathrm{m}=21\mathrm{m}$
• In a right triangle $\mathrm{ABC},\angle \mathrm{B}=90^{\circ }$ (a) If $AB=6\mathrm{cm},\mathrm{BC}=8\mathrm{cm}$, find AC . (b) If $\mathrm{AC}=13\mathrm{cm},\mathrm{BC}=5\mathrm{cm}$, find AB . Sol. (a) In right angled $△\mathrm{ABC}$, using Pythagoras theorem, we have
  
  $$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ \Rightarrow A{C}^2=6^2+8^2 \\ =36+64=100 \\ \therefore AC=\sqrt{100}\mathrm{cm}=10\mathrm{cm} \end{gathered}$$ (b) In right angled $△\mathrm{ABC}$, using Pythagoras theorem, we have
  
  $$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2 \\ \Rightarrow 13^2={\mathrm{AB}}^2+5^2 \\ {\mathrm{AB}}^2=169-25=144 \\ \therefore AB=\sqrt{144}\mathrm{cm}=12\mathrm{cm} \end{gathered}$$
• 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. Sol. Let us find out the square root of 1000 . Since number of rows and the number of columns of plants are same. So number of plants should be perfect square. We observe that
  
  $(31{)}^2<1000<(32{)}^2$ The required plants to be added. $(32{)}^2-1000$ $\therefore 1024-1000=24$ plants
• There are 500 children in a school. For a P.T. practice 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? Sol. Let us find the square root of 500 .
  
  This shows that $(22{)}^2=484$ is less than 500 by 16 . $\therefore 16$ students have to go out for others to do the P.T. practice as per condition.

5.0Key Features and Benefits of Class 8 Maths Chapter 5 Exercise 5.4

• The exercise emphasizes estimation methods for square roots using simple and logical steps.
• Aligns with the style and level of questions in the Class 8 NCERT textbook.
• This section teaches how to handle large and non-perfect square numbers.
• They support the development of number sense and approximations which are also used in Olympiad problems.
• Regular practice increases the speed in solving estimation based questions during the exam.