NCERT Solutions for Class 8 Maths Chapter 1 - A Square and a Cube

The NCERT Solutions for Class 8 Maths Chapter 1—A Square and A Cube has been developed in such a way that they provide a clear and structured way to understand square and cube numbers along with their roots for all Class 8 students. This chapter forms the basis for developing numerical skills in NCERT Class 8 Maths, which helps develop a student’s ability to solve problems accurately and confidently.

These solutions are created based on the NCERT Curriculum, based on the latest syllabus issued by the National Council Of Educational Research And Training (NCERT) and also the patterns of questions as per the CBSE Examination. The concepts covered in these solutions will be explained step by step and will clarify how to study and revise each concept successfully.

1.0NCERT Solutions for Class 8 Maths Chapter 1 PDF Download

Students can download the A Square and a Cube Class 8 PDF with answers from the table below. The NCERT Solutions for Class 8 Maths Chapter 1 – A Square and a Cube PDF are meticulously prepared by ALLEN’s expert faculty members, who possess strong subject expertise and a deep understanding of the latest NCERT and CBSE curriculum. These solutions are designed to help students understand concepts clearly and perform confidently in examinations.

2.0NCERT Solutions for Class 8 Maths Chapter 1 A Square and a Cube : All Exercises

3.0NCERT Questions with Solutions Class 8 Maths Chapter 1 - A Square and a Cube Detailed Solutions

1. Which of the following numbers are not perfect squares?

(i) 2032

(ii) 2048

(iii) 1027

(iv) 1089

Sol. (i) 2032 is not a perfect square, as a number ending with 2 cannot be a perfect square.

(ii) 2048 is not a perfect square, as a number ending with 8 cannot be a perfect square.

(iii) 1027 is not a perfect square, as a number ending with 7 cannot be a perfect square.

(iv) 1089 ends in 9 at the unit's place. Hence, it is a perfect square.

2. Which one among $64^2,108^2,292^2,36^2$ has the last digit 4?

Sol. (i) Unit’s digit of 64 is 4

∴ $4^2=4\times 4=16$ (last digit = 6)

(ii) Unit’s digit of 108 is 8

∴ $8^2=8\times 8=64$ (last digit = 4)

(iii) Unit’s digit of 292 is 2

∴ $2^2=2\times 2=4$ (last digit = 4)

(iv) Unit’s digit of 36 is 6

∴ $6^2=6\times 6=36$ (last digit = 6)

Hence, the numbers whose squares end in 4 are
$108^2$ and $292^2$

3. Given $125^2=15625$, what is the value of $126^2$?

(i) 15625 + 126

(ii) 15625 +$26^2$

(iii) 15625 + 253

(iv) 15625 + 251

(v) 15625 + $51^2$

Sol.

Here,

$126^2=(125+1{)}^2=(125{)}^2+2\times 125\times 1+(1{)}^2$

[Using identity $(a+b{)}^2=a^2+2ab+b^2$]

$=15625+250+1$

$=15625+251$

So, the value of $126^2$ is option (iv), i.e., 15625 + 251.

4. Find the length of the side of a square whose area is $441m^2$.

Sol. Area of square = side × side = 441

$\Rightarrow sid{e}^2=441$

$\Rightarrow side=\surd 441$

$\begin{array}{cc}3& 441\\ 3& 147\\ 7& 49\\ 7& 7\\ & 1\end{array}$

$441=3\times 3\times 7\times 7$

$\surd 441=3\times 7=21$

Side of the square = 21 m

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

Sol. To find the required smallest square number, we first find the least number divisible by 4, 9, and 10, i.e., the LCM of 4, 9, and 10.

$\begin{array}{cccc}2& 4,& 9,& 10\\ 2& 2,& 9,& 5\\ 3& 1,& 9,& 5\\ 3& 1,& 3,& 5\\ 5& 1,& 1,& 5\\ & 1,& 1,& 1\end{array}$

LCM = 2 × 2 × 3 × 3 × 5 = 180

Prime factorisation of 180 = 2 × 2 × 3 × 3 × 5

Here, 5 is not in pairs, so 180 is not a perfect square.

To make it a perfect square, we multiply 180 by 5.

Required smallest square number = 180 × 5 = 900

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

6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.

Sol.

$\begin{array}{cc}2& 9408\\ 2& 4704\\ 2& 2352\\ 2& 1176\\ 3& 588\\ 3& 294\\ 3& 147\\ 7& 49\\ 7& 7\\ & 1\end{array}$

9408 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7

All prime factors of 9408 are arranged in pairs except 3.

So, we multiply 9408 by 3 to make it a perfect square.

Perfect square = 9408 × 3 = 28224

Now,

$\sqrt{28224}=\sqrt{2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 7\times 7}$

= 2 × 2 × 3 × 3 × 7

= 252

7. How many numbers lie between the squares of the following numbers?

(i) 16 and 17

Sol. Numbers lying between$16^2$ and$17^2$
= 2 × 16
= 32

(ii) 99 and 100

Numbers lying between $99^2$ and $100^2$
= 2 × 99
= 198

8. In the following pattern, fill in the missing numbers:

$1^2+2^2+2^2=3^2$

$2^2+3^2+6^2=7^2$

$3^2+4^2+12^2=13^2$

$4^2+5^2+20^2={(}_{)}^2$

$9^2+10^2+{(}_{)}^2={(}_{)}^2$

Sol.

$1^2+2^2+2^2=3^2$

$2^2+3^2+6^2=7^2$

$3^2+4^2+12^2=13^2$

$4^2+5^2+20^2=21^2$

$9^2+10^2+90^2=91^2$

FIGURE IT OUT - 2

1. Find the cube roots of 27000 and 10648.

Sol. Here,

$\begin{array}{cc}2& 27000\\ 2& 13500\\ 2& 6750\\ 3& 3375\\ 3& 1125\\ 3& 375\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$27000=2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 5$

$\sqrt[3]{27000}=2\times 3\times 5=30$

$\begin{array}{cc}2& 10648\\ 2& 5324\\ 2& 2662\\ 11& 1331\\ 11& 121\\ 11& 11\\ & 1\end{array}$

$\begin{array}{rl}& 10648=2\times 2\times 2\times 11\times 11\times 11\\ & \sqrt[3]{10648}=2\times 11=22\end{array}$

2. What number will you multiply by 1323 to make it a cube number?

Sol. Here

$\begin{array}{cc}3& 1323\\ 3& 441\\ 3& 147\\ 7& 49\\ 7& 7\\ & 1\end{array}$

$1323=3\times 3\times 3\times 7\times 7$

To complete the triplet, one more 7 is required. So, 1323 will be multiplied by 7 to make it a cube number.

So, the cube number $=1323\times 7=9261$ Hence, required number $=7$

3. State true or false. Explain your reasoning. (i) The cube of any odd number is even. (ii) There is no perfect cube that ends with 8. (iii) The cube of a 2-digit number may be a 3 -digit number. (iv) The cube of a 2 -digit number may have seven or more digits. (v) Cube numbers have an odd number of factors. Sol.

(i) The cube of any odd number is even. (False) Reason: The cube of an odd number is always odd, as $3^3=27$ $5^3=125$ $7^3=343$

(ii) There is no perfect cube that ends with 8. (False) Reason: The cubes of all the numbers ending with 2 at the unit place end with 8. $2^3=8$ $12^3=1728$ $22^3=10648$

(iii) The cube of a 2 -digit number may be a 3 -digit number. (False) Reason: Cube of a 2 -digit number may have a minimum of 4 digits to a maximum of 6 digits. 10 is the smallest 2-digit number, and $10^3=1000$, which has 4 digits.

(iv) The cube of a 2 -digit number may have seven or more digits. (False) Reason: Cube of a 2 -digit number may have at most 6 digits. 99 is the largest 2-digit number, and $99^3=970299$, which is a 6 -digit number.

(v) Cube numbers have an odd number of factors. (False) Reason: Cube numbers may have an odd as well as an even number of factors. As $27=3\times 3\times 3$ (odd no. of factors) $64=2\times 2\times 2\times 2\times 2\times 2$ (even no. of factors)

4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.

Sol. To find the cube root of 1331, 1331 We divide the given number 1331 into two groups, starting from the right side, taking three digits in group 1. $331\to$ group 1 $1\to$ group 2 $331\to$ unit digit is 1 Hence, the cube roots of one's digit is 1 .....(1) Group 2, i.e., 1 only, which is $1^3$. So, the cube roots of one's digit is 1 . $\because \sqrt[3]{1331}=11$

4913 Group 1-913 Group 2-4 Unit digit of 913 is 3 . We know that 3 comes at the unit's place when its cube root ends in 7, as 7 $\times 7\times 7$ = 343 So the unit digit of the cube root of 4913 $=7$ …..(1) Group 2-4 4 lies between 1 (i.e., $1^3$ ) and $2^3$ (i.e., 8 ) $1^3<4<2^3$ Taking the lower limit, the tens digit of the cube root of 4913 is 1 . .....(2) $\sqrt[3]{4913}=17$ (from (1) & (2))

12167 Group 1-167 Unit digit = 7 So, unit digit of cube root of $12167=3$ as $3\times 3\times 3=27$ Group 2-12 8 < 12 < 27 $2^3<12<3^3$ Taking the lower limit, the ten's digit of cube root $=2$ So, $\sqrt[3]{12167}=23$

32768 Group 1-768 Unit digit = 8 So unit digit of cube root of $32768=2$ .....(1) as $2\times 2\times 2=8$ $\Rightarrow \sqrt[3]{8}=2$ From Group 2-32 27 < 32 < 64 $3^3<32<4^3$ Taking lower limit, ten's digit of the cube root of 32768 is 3 . .....(2) $\therefore \sqrt[3]{32768}=32$ (from (1) & (2))

5. Which of the following is the greatest? Explain your reasoning. (i) $67^3-66^3$ (ii) $43^3-42^3$ (iii) $67^2-66^2$ (iv) $43^2-42^2$ Sol.

(i) $67^3-66^3=1+67\times 66\times 3$ (ii) $43^3-42^3=1+43\times 42\times 3$ (iii) $67^2-66^2=67+66=133$ (iv) $43^2-42^2=43+42=85$ From above we can see that $67^3-66^3$ is the greatest as $(\mathrm{n}+1{)}^3-{\mathrm{n}}^3=1+(\mathrm{n}+1)\times 3\mathrm{n}$ $(\mathrm{n}+1{)}^2-{\mathrm{n}}^2=\mathrm{n}+\mathrm{n}+1=2\mathrm{n}+1$

4.0Quick Insights About the Chapter – A Square and a Cube

This chapter establishes the fundamental concepts related to number patterns and powers, which are essential for developing numerical understanding in mathematics.

• The definition of square numbers and cube numbers, along with an explanation of how they are formed through repeated multiplication.
• Identification of perfect squares and perfect cubes using number patterns and properties.
• Understanding the properties of square numbers, including patterns related to even and odd numbers and the possible last digits of perfect squares.
• Methods to find square roots and cube roots, primarily using the prime factorisation technique.
• Recognition of number patterns to determine whether a given number is a perfect square or a perfect cube.
• Application of square and cube concepts to solve numerical and reasoning-based problems.

5.0Benefits of NCERT Solutions Class 8 Maths Chapter 1

Students can take advantage of NCERT Solutions for Class 8 Maths Chapter 1 - A Square and a Cube to gain a strong mathematical knowledge base to perform better in exams.

1. Conceptual Clarity: - They provide an in-depth explanation of square numbers, cube numbers, and their roots in a simple format to help students grasp these concepts easily.
2. Conformance to NCERT and CBSE Curriculum: - All solutions conform entirely to the latest NCERT syllabus and the CBSE examination formats.
3. Mathematical Problem Solving Skills: - The solutions explain how to properly approach numerical problems and logical reasoning questions through step-by-step examples.
4. Exam Preparation Tools: - By practising NCERT questions with the correct answers, students will feel more at ease with their speed, accuracy, and confidence during exams.
5. Foundation for Future Topics: - Using square and cube numbers as a foundation prepares students for subjects at a more advanced level, such as Algebra, Exponents, and Number Systems.
6. A Resource for Study and Reinforcement: - The well-organized layouts of the solutions create an easy reference for independent study/reviewing when questions arise.