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

NCERT Solutions for Class 8 Maths Chapter 1: Exercise Questions

Question 1. Which of the following numbers are not perfect squares?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089

Answer:

(i)
2032 = 2 × 2 × 2 × 2 × 127
So, 2032 is not a perfect square as 127 has no pair.

(ii)
2048 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
So, 2048 is not a perfect square as the last 2 has no pair.

(iii)
1027 = 13 × 79
So, 1027 is not a perfect square as 13 and 79 have no pairs.

(iv)
1089 = 3 × 3 × 11 × 11
So, 1089 is a perfect square as 3 and 11 both have pairs.

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

Answer:

$\begin{aligned} & 64^2=4096 \rightarrow \text { ends with } 6 \\ & 108^2=11664 \rightarrow \text { ends with } 4 \\ & 292^2=85264 \rightarrow \text { ends with } 4 \\ & 36^2=1296 \rightarrow \text { ends with } 6\end{aligned}$

So, 108² and 292² have 4 as the last digit.

Question 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) $15625+51^2$

Answer:

We know, $(a+1)^2=a^2+2a+1$
Here, $a=125$
So, $(125+1)^2= 125^2+2\times125\times1 +1^2=15625+251$

Hence, the correct answer is option (iv).

Question 4. Find the length of the side of a square whose area is $441 \mathrm{~m}^2$.

Answer:

We know that the Area of a square = Side × Side

Here, 441 = 21 × 21

So, the side of the square is 21 m.

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

Answer:

LCM of 4,9, and $10=180$
Prime factorisation of $180=2^2 \times 3^2 \times 5$
To make a perfect square, 5 needs to appear twice.

So, $180 \times 5=900=(30)^2$

Hence, the correct answer is 30.

Question 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.

Answer:

Prime factorisation of $9408=2^6 \times 3 \times 7^2$
To make a perfect square, we need one more 3.

Number becomes
$= 9408 \times 3=28224=2^6 \times 3^2 \times 7^2$

Hence, the square root of 2560320 $=2^3\times3\times7=168$

Question 7. How many numbers lie between the squares of the following numbers?
(i) 16 and 17
(ii) 99 and 100

Answer:

(i) Between $16^2=256$ and $17^2=289$ :

Numbers lie between $=289-256-1=32$​​​​​​

(ii) Between $99^2=9801$ and $100^2=10000$ :

Numbers lie between $=10000-9801-1=198$

Question 8. In the following pattern, fill in the missing numbers:
$1^2+2^2+2^2=3^2$
$2^2+3^2+6^2=7^2 \\$
$3^2+4^2+12^2=13^2 \\$
$4^2+5^2+20^2=()^2 \\$
$9^2+10^2+()^2=()^2$

Answer:

$4^2+5^2+20^2=16+25+400=441=21^2$
So, $4^2+5^2+20^2=(21)^2$

$9^2+10^2=81+100=181$
If we add $90^2+181=8281=91^2$
So, $9^2+10^2+(90)^2=(91)^2$

Question 9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares

Answer:

There are 81 tiny squares.

Prime factorisation of 81 = 3 × 3 × 3 × 3 = 3⁴

Question 1. Find the cube roots of 27000 and 10648.

Answer:

Prime factorisation of 27000 = 3 × 3 × 3 × 10 × 10 × 10 = 3³ × 10³

Hence, the cube root of 27000 = 3 × 10 = 30

Prime factorisation of 10648 = 2 × 2 × 2 × 11 × 11 × 11 = 2³ × 11³

Hence, the cube root of 10648 = 2 × 11 = 22

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

Answer:

$1323=3^3 \times 7^2$
To make it a perfect cube, we need one more 7 to make 7³.

Hence, the answer is 3.

Question 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.

Answer:

(i) The cube of any odd number is even.
This statement is False.
Odd $\times$ Odd $\times$ Odd $=$ Odd
Example: $3^3=27$ (odd)

(ii) There is no perfect cube that ends with 8.
This statement is False
Example: $2^3=8$, ends with 8

(iii) The cube of a 2-digit number may be a 3-digit number.
This statement is True.
Example: $10^3=1000$, which is a three-digit number

(iv) The cube of a 2-digit number may have seven or more digits.
This statement is False.
The largest two-digit number is 99.
Try $99^3=970299 \rightarrow$, which is 6 digits

(v) Cube numbers have an odd number of factors.

This statement is False.
Only perfect squares have an odd number of factors.
Example: $8=2^3 \rightarrow$ Factors $=1,2,4,8 \rightarrow 4$ factors (even)

Question 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.

Answer:

Cube Root of 1331:

Ends in $1 \rightarrow$ Cube of a number ending in 1 also ends in 1.

Try: $11^3=1331$

So, Cube root of $1331=11$

Cube Root of 4913:

Ends in $3 \rightarrow$ Cube of a number ending in 7 ends in 3.

(because $7^3=343$ )

Try: $17^3=4913$

So, Cube root of $4913=17$

Cube Root of 12167:

Ends in $7 \rightarrow$ Cube of a number ending in 3 ends in 7 (because $3^3=27$ )

Try: $23^3=12167$

So, Cube root of $12167=23$

Cube Root of 32768:

Ends in $8 \rightarrow$ Cube of a number ending in 2 ends in 8.

Try: $32^3=32768$

So, Cube root of $32768=32$

Question 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) $43^2-42^2$

Answer:

We know that $a^3-b^3=(a-b)\left(a^2+a b+b^2\right)$
In the first two options,
(i) $67^3-66^3=(67-66)(67^2+67\times66+66^2)=13267$
(ii) $43^3-42^3=(43-42)(43^2+43\times43+42^2)=5420$

We know that $a^2-b^2=(a+b)(a-b)$
In the last two options,
(iii) $67^2-66^2=(67+66)(67-66)=133$
(iv) $43^2-42^2=(43+42)(43-42)=85$

Hence, $67^3-66^3$ is the greatest.

A Square and A Cube Class 8 Chapter 1: Topics

The topics discussed in the NCERT Solutions for Class 8 Chapter 1 A Square and A Cube are:

• Introduction
• Square Numbers
• Patterns and Properties of Perfect Squares
• Perfect Squares and Odd Numbers
• Perfect Squares and Triangular Numbers
• Square Roots
• Cubic Numbers
• Taxicab Numbers
• Perfect Cubes and Consecutive Odd Numbers
• Cube Roots
• Successive Differences
• A Pinch of History

NCERT Solutions for Class 8 Maths Chapter 1 A Square and A Cube: Notes

Careers360 has prepared these A Square and A Cube Notes to make your revision smoother and faster. Additionally, these notes will help students to understand the A Square and A Cube class 8 solutions and solve them on their own from next time.

Square Numbers and Perfect Squares

A number obtained by multiplying a number by itself is called a square number. Squares of natural numbers are called perfect squares.

Example: 2 × 2 = 4, it is a perfect square.

Properties of Perfect Squares

1. Non-Negative Perfect Squares:

Perfect squares are always non-negative numbers.
If we multiply two negative numbers, we will get a positive number all the time.
(Example: –8 × –8 = +64)
Or, if we multiply two positive numbers, the resultant number is always positive.
(Example: +8 × +8 = +64).

2. Square Root of a Perfect Square:

The square root of a perfect square is always an integer.
Example: $\sqrt9=3$, which is an integer, $\sqrt{121}=11$, which is an integer

3. Last digit of a Perfect Square:

Most of the time, Perfect squares end with the integers 0, 1, 4, 5, 6, or 9, but never in 2, 3, 7, or 8, when expressed in the base 10.
Example: 100(10²), 121(11²), 144(12²), 225(15²), 196(14²), 169(13²),...

Cubes and Perfect Cubes

In general, we heard about the word “Cube” in Geometry. A cube is a solid figure all of whose sides meet at right angles and are equal.

A number obtained by multiplying a number by itself three times is called a cube or a cubic number. A number is a perfect cube if its prime factors can be split into three identical groups.

Example: 2 × 2 × 2 = 8, it is a perfect cube.

Properties of Perfect Cubes

1. Positive and Negative Perfect Cubes

Perfect cubes can be positive or negative integers, unlike perfect squares, which are always non-negative integers.
Example: −27 is a perfect cube as −27 = (−3)3

2. Cube root of a Perfect Cube

The cube root of a perfect cube is always an integer.
Example: $\sqrt[3]{9}=3$, which is an integer, $\sqrt[3]{64}=4$, which is an integer

3. Last digit of a Perfect Cube

In base 10, a perfect cube can end in any digit, unlike perfect squares, which have more restricted ending digits.

Example: 343(7³), 512(8³), 1728(12³), 2197(13³)
In these examples, the perfect cube ending digits are 3, 2, 8, 7. These digits do not generally come up at the end of a perfect square.

Square roots and Cube roots

The symbol $\sqrt{}$ denotes the square root.

For example, $\sqrt{16}=4$

The symbol $\sqrt[3]{}$ denotes the cube root.
For example, $\sqrt[3]{64}=4$.

Prime Factorisation Method

The most common method to find the square root of a perfect square is:

• Prime factorisation method

The prime factorisation of a perfect square provides insight into its factors. All the exponents in its prime factorisation must be even in a perfect square.

Here are the steps to use the prime factorisation method.

Step 1: Factorise the number into its prime factors.

Step 2: Pair the same type of prime factors.

Step 3: Multiply one number from each pair to get the required square root.

Example:
The prime factors of 225 = 3 × 3 × 5 × 5 = 32 × 52

Taking one number from each pair, we get 3 × 5 =15

So, the square root of 225 is 15.

There are two other methods, which we will learn in later classes.

• Long division method
• Estimation method

Below, you can find a list of a few squares and cubes.

Taxicab Numbers

Numbers that can be expressed as the sum of two cubes in two different ways are called taxicab numbers.

1729 is the first taxicab number
$1729=1^3+12^3=9^3+10^3$

Also, 4104 is a taxicab number.
$4104=2^3+16^3=9^3+15^3$

13832 is a taxicab number.
$13832=2^3+24^3=18^3+20^3$

People also know 1729 as the Hardy–Ramanujan Number, and it was named after the renowned Indian mathematician Srinivasa Ramanujan and the British mathematician Godfrey Harold Hardy.

NCERT Solutions for Class 8 Maths Chapter 2 A Square and A Cube: Points to Remember

Here are some points that will help students while practising the solutions of the A Square and A Cube Class 8 chapter.

• The square of a number is denoted as n2, where n is the number.
• A number which is the square of an integer is called a perfect square.
• We can get the square root of a number when multiplied by itself, which gives the original number.
• The prime factorisation method is the best way to check if a number is a perfect square or a perfect cube.
  Factorise the numbers into their prime factors.
  1. If the exponents of the numbers in the factorisation are even, then it's a perfect square.
  2. If the exponents of the numbers in the factorisation are a multiple of 3, then it's a perfect cube.
• A number can be both a perfect square and a perfect cube. For example: 64, 729, etc
• The square of a number ending with 1, 4, 5, 6, or 9 always ends in the same digit as the number.
• The square of a negative number is also positive, but the cube of a negative number is always negative.
• The difference between consecutive perfect squares gives the sequence of odd numbers.

NCERT Solutions for Class 8 Mathematics - Chapter Wise

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NCERT Class 8 Maths Solutions: Subject Wise

Students can use the links below to prepare efficiently in other subjects as well as Mathematics.

• NCERT solutions for class 8 maths
• NCERT solutions for class 8 science

NCERT Books and NCERT Syllabus

The following links will give students access to the latest CBSE syllabus and some reference books.

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8