NCERT Solutions for Class 8 Maths Chapter 3 - Square Square Root & Cube Cube Root

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Answer:

$\begin{array}{c}196=2\times 2\times 7\times 7\\ ={(2\times 7)}^2\\ ={\left(14\right)}^2\end{array}$
$\therefore$ Square root of 196 is 14.
Hence, the correct answer is option C.

Answer:

We know, unit place of square of any even number will always be even.
Therefore, 144 is a square of an even number.
Hence, correct answer is option A.

Answer:

If any number ends in 9 then unit place of its square will be 1, because 9$\times$9 = 81.
Hence, the correct answer is option C.

Answer:

Unit place of 62 is 2, and square of 2 is 4.
Hence, the correct answer is option B.

Answer:

5² = 25
6² = 36
Natural number between 25 and 36 are 26, 27, 28, 29, 30, 31, 32, 33, 34, 35.
Thus, 10 natural numbers lie between 5² and 6².
Hence, the correct answer is option B.

Answer:

Here
$$\begin{gathered} \sqrt{841}=\sqrt{29\times 29}=29 \\ \sqrt{529}=\sqrt{23\times 23}=23 \\ \sqrt{198}=\sqrt{2\times 3\times 3\times 11} \end{gathered}$$
Hence, the correct answer is option C.

Answer:

Unit digit of 23 is 3.
Now, cube of 3 = 27.
Therefore, unit digit of cube of 23 is 7.
Hence, the correct answer is option B.

Answer:

We know,
Area of square board = (Side)²
⇒ 144 = (Side)²
⇒ Side = $\sqrt{144}$
⇒ Side = 12 units.
Hence, the correct answer is option B.

Answer:

Now, $\sqrt{25}=\sqrt{5\times 5}=5$
And, from number line point C represents the location of number 5.
Hence, the correct answer is option C.

Answer:

We know,
For every natural number , such that  from a Pythagorean triples.
Hence, the correct answer is option B.

Answer:

Given: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
Number of terms, $n=8$
$\begin{array}{rcl}\mathrm{Therefore}, \mathrm{sum} & =& n\times n\\ & =& 8\times 8\\ & =& 64\end{array}$
Hence, the correct answer is option B.

Answer:

We know, first $n$ odd number are given by 1, 3, 5, 7,..................$\left(2n-1\right)$
Here first term $\left(a\right)=1$, common difference = 2.
Thus, formula of sum of $n$ term of an AP

$$\begin{gathered} S_n=\frac{n}{2}\left[2\times 1+\left(n-1\right)2\right] \\ =\frac{n}{2}\times 2n \\ =n^2 \end{gathered}$$

Hence, the correct answer is option B.

Answer:

Here
$$\begin{gathered} \left(\mathrm{i}\right) 243=3\times 3\times 3\times 3\times 3={\left(3\right)}^5 \\ \left(\mathrm{ii}\right) 216=2\times 2\times 2\times 3\times 3\times 3=2^3\times 3^3={\left(6\right)}^3 \\ \left(\mathrm{iii}\right) 392=2\times 2\times 2\times 7\times 7=2^3\times 7^2 \\ \left(\mathrm{iv}\right) 8640=2\times 2\times 2\times 2\times 2\times 2\times 5\times 3\times 3\times 3 =2^6\times 5\times 3^3 \end{gathered}$$ 

We know, a perfect cube has multiples of 3 as powers of prime factors.
Hence, the correct answer is option B.

Answer:

We know, 
$\begin{array}{rcl}{\left(\mathrm{Hypotenuse}\right)}^2& =& {\left(3x\right)}^2+{\left(4x\right)}^2\\ & =& 25x^2\end{array}$
$\begin{array}{rcl}\mathrm{Hypotenuse}& =& \sqrt{25x^2}\\ & =& \sqrt{\left(5\times 5\right)\left(x\times x\right)}\\ & =& 5x\end{array}$
Hence, the correct answer is option A.

Answer:

Here, given pattern general term will be
$n^2 \left(\therefore n=1, 2, 3, 4, 5,.............\right)$
Therefore the next two term in series will be 
1, 4, 9, 16, 25, 36, 49
Hence, the correct answer is option B.

Answer:

Unit digit of 59 is 9
Now square of 9 is 81
Thus 59 ends with digit 1.
Hence, the correct answer is option D.

Answer:

We know, a perfect square never ends with digits, 2, 3, 7 and 8.
Hence, the correct answer is option B.

Answer:

Here,
$$\begin{gathered} 216=6\times 6\times 6={\left(6\right)}^3 \\ 567=9\times 9\times 7={\left(9\right)}^2\times {\left(7\right)}^1 \\ 125=5\times 5\times 5={\left(5\right)}^3 \\ 343=7\times 7\times 7={\left(7\right)}^3 \end{gathered}$$
We know, a perfect cube has multiples of 3 as powers of prime factors.
Hence, the correct answer is option B.

Answer:

Here,
$\sqrt[3]{1000}=\sqrt{10\times 10\times 10}=10$
Hence, the correct answer is option A.

Answer:

Given: square of $n=m$
$$\begin{gathered} i.e., {\left(n\right)}^2=m. \\ \Rightarrow n=\sqrt{m}. \end{gathered}$$
Hence, the correct answer is option D.

Answer:

We know, if $\text{'}n\text{'}$ is even then square will contain $\frac{n}{2}$ digits.
Hence, the correct answer is option B.

Answer:

Given: $m$ is cube root of $n$
$$\begin{gathered} i.e., \sqrt[3]{n}=m \\ \Rightarrow n=m^3 \end{gathered}$$
Hence, the correct answer is option A.

Answer:

Here, $\sqrt{248+\sqrt{52+\sqrt{144}}}$
$$\begin{gathered} \Rightarrow \sqrt{248+\sqrt{52+12}} \\ \Rightarrow \sqrt{248+8} \\ \Rightarrow \sqrt{256} \\ =16 \end{gathered}$$
Hence, the correct answer is option C.

Answer:

Given: $\sqrt{4096}=64$
Now, $\sqrt{4096}+\sqrt{40.96}$
$$\begin{gathered} =64+6.4 \\ =70.4 \end{gathered}$$
Hence, the correct answer is option D.

Answer:

Here,
${\text{2}}^2=4,\mathrm{\hspace{0.17em}}{3}^2=9,\mathrm{\hspace{0.17em}}{4}^2=16,\mathrm{\hspace{0.17em}}{5}^2=25,\mathrm{\hspace{0.17em}}{6}^2=36,\mathrm{\hspace{0.17em}}{7}^2=49,\mathrm{\hspace{0.17em}}{8}^2=64,\mathrm{\hspace{0.17em}}{9}^2=81$
Therefore, between 1 and 100, there are 8 perfect squares.
Hence, there are 8 perfect squares between 1 and 100.

Answer:

Here,
$2^3=8, 3^3=27, 4^3=64, 5^3=125, 6^3=216, 7^3=343, 8^3=512, 9^3=729$
Therefore, between 1 and 1000 there are 8 perfect cubes.
Hence, there are 8 perfect cubes.

Answer:

We know, 4 or 6 at the units place ends in 6.
4 × 4 = 16
Hence, the units digit in the square of 1294 is 6.

Answer:

${\left(500\right)}^2=500\times 500=250000$
Thus, the square of 500 will have 4 zeros.

Answer:

Natural number between $n^2$ and ${\left(n+1\right)}^2$
$$\begin{gathered} =\left\{{\left(n+1\right)}^2-n^2\right\}-1 \\ =\left\{n^2+2n+1-n^2-1\right\} \\ =2n \end{gathered}$$
Thus, there are $2n$ natural numbers between $n^2 \mathrm{and} {\left(n+1\right)}^2$

Answer:

Given: 24025
Number of digits = 5
$\begin{array}{rcl}\mathrm{Number} \mathrm{of} \mathrm{digit} \mathrm{in} \mathrm{square} \mathrm{root} & =& \frac{\mathrm{n}+1}{2}\\ & =& \frac{5+1}{2}\\ & =& 3\end{array}$

Hence, the square root of 24025 will have 3 digits.

Answer:

Square of 5.5 = 5.5 $\times$ 5.5 = 30.25

Hence, the square of 5.5 is 30.25.

Answer:

Square root of 5.3 $\times$ 5.3 = $\sqrt{5.3\times 5.3}=5.3$

Hence, the square root of 5.3 × 5.3 is 5.3.

Answer:

$\begin{array}{rcl}\mathrm{Cube} \mathrm{of} 100 & =& {\left(100\right)}^3\\ & =& 100\times 100\times 100\\ & =& 1000000\end{array}$

Hence, the cube of 100 will have 6 zeros.

Answer:

1 m² = (100 cm) $\times$ (100 cm) = 10000 cm²

Hence, 1m² = 10⁴ cm².

Answer:

$\begin{array}{rcl}1 {\mathrm{m}}^3 & =& 100 \times 100 \times 100 {\mathrm{cm}}^3\\ & =& 1000000 {\mathrm{cm}}^3\\ & =& {10}^6 {\mathrm{cm}}^3\end{array}$

Hence, 1 m³ = 10⁶ cm³.

Answer:

Ones digit in the cube of 38 is the ones digit in the cube of 8.

$\begin{array}{rcl}\mathrm{Cube} \mathrm{of} 8& =& 8\times 8\times 8\\ & =& 512\end{array}$

Hence, one digit in cube of 38 is 2.

Answer:

Square of 0.7 = 0.7 $\times$ 0.7 = 0.49

Hence, the square of 0.7 is 0.49.

Answer:

Sum of six odd number = n² = 6 × 6 = 36.

Hence, the sum of first six odd natural numbers is 36.

Answer:

Unit digit of 57 = 7
Square of 7 = 49
Thus, unit digit in square of 57 is 9.

Answer:

We know, For every natural number $\mathrm{m}>1, 2\mathrm{m}, {\mathrm{m}}^2+1$ from a Pythagoras triplet.
Now,
${\left({\mathrm{M}}^2+1\right)}^2=\left({\mathrm{M}}^2-1\right)+{\left(2\mathrm{M}\right)}^2$
$$\begin{gathered} \mathrm{Where} {\mathrm{M}}^2+1=17 \\ \Rightarrow {\mathrm{M}}^2=16 \\ \Rightarrow \mathrm{M}=4 \end{gathered}$$

Thus, 2m = 2 $\times$ 4 = 8
and 
${\mathrm{M}}^2-1={\left(4\right)}^2-1=15$
Hence, the sides of a right triangle whose hypotenuse is 17 cm are 8 and 15.

Answer:

$\begin{array}{rcl}\sqrt{1.96}& =& \sqrt{\frac{196}{100}}=\sqrt{\frac{14\times 14}{10\times 10}}\\ & =& \frac{14}{10}\\ & =& 1.4\end{array}$

Hence, $\sqrt{1.96}=1.4$.

Answer:

$\begin{array}{rcl}{\left(1.2\right)}^3& =& {\left(\frac{12}{10}\right)}^3=\frac{12\times 12\times 12}{10\times 10\times 10}\\ & =& \frac{1728}{1000}\\ & =& 1.728\end{array}$

Hence, (1.2)³ = 1.728.

Answer:

The cube of an odd number is always an odd number.

Answer:

The cube root of a number is denoted by $x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$.

Answer:

125 = 5 × 5 × 5

5 is not in the pairs so we need to multiply 125 by 5 in order to make it a perfect square.

∴ 125 × 5 = 625

And, $\sqrt{625}=\sqrt{5\times 5\times 5\times 5}=5\times 5=25$

Hence, the least number by which 125 must be multiplied is 5.

Answer:

Prime factors of 72 = $\left(3\times 3\right)\times \left(2\times 2\times 2\right)$
Hence, least number by which 72 must be multiplied to make it a perfect cube is 3.

Answer:

Prime factors of 72 = $\left(2\times 2\times 2\right)\times \left(3\times 3\right)$
Clearly, if we divide 72 by 9, the quotient will be a perfect cube.
Hence, the least number by which 72 must be divided to make it a perfect cube is 9.

Answer:

We know, cube of a number ending in digits 3 or 7 ends in digit 7 or 3.
Respectively,
i.e. $7\times 7\times 7=343$
Hence, cube of a number ending in 7 will end in digit 3.

Answer:

True
We know, unit's digit of a square of a number having 4 or 6 as unit digit is 6.

Answer:

False
For example, 9 and 36 are perfect squares but, 9 + 36 = 45 is not a perfect square.

Answer:

True
For example: 16 and 25 are perfect squares and 16 $\times$ 25 is also a perfect square.

Answer:

True
There is no perfect square number between 50 and 60.

Answer:

False
Because, square of 31 = 31 × 31 = 961.

Answer:

True
A perfect cube always be expressed as the product of triplets of prime factors.

Answer:

False
$\begin{array}{rcl}\mathrm{Square} \mathrm{of} 2.8& =& {\left(\frac{28}{10}\right)}^2\\ & =& \frac{784}{100}\\ & =& 7.84\end{array}$

Answer:

True
Cube of 0.4 = 0.4 × 0.4 × 0.4
                   = 0.064
Hence, statement is true.

Answer:

False
Here, square of (0.3) = 0.3 × 0.3
                                  = 0.09
Thus, square root of 0.09 is 0.3.
Hence, statement is false.

Answer:

False
For example: square of 1 = 1 × 1
                                         = 1
Thus, square of 1 is equal to 1
Hence, statement is false.

Answer:

False
Prime factorisation of 8000 = 20 × 20 × 20
         Therefore  $3\sqrt{8000} =3\sqrt{20\times 20\times 20}$
                                          = 20
Hence, given statement is false.

Answer:

False, 
Between 1 and 100, there are 3 perfect cubes
2³ = 2 × 2 × 2 = 8

3³ = 3 × 3 × 3 = 27
4³ = 4 × 4 × 4 = 64
​Hence, given statement is false.

Answer:

Ture,
Natural numbers between a and b = b – a – 1
= (101)² – (100)² – 1
= (101 – 100)(101 + 100) – 1
= 200
Hence, given statement is true.
 

Answer:

True
$$\begin{gathered} \mathrm{Sum} \mathrm{of} \mathrm{odd} \mathrm{numbers}=\sum (2n-1) \\ =\frac{\left(2\times n(n+1)\right)}{2}-n \\ =\frac{2n^2}{2}+\frac{2n}{2}-n \\ =n^2 \end{gathered}$$
Hence, the given statement is true.

Answer:

False
Prime factorization of 1000 = 2³ × 5³ 
Thus, 1000 is not a perfect square
Hence, given statement is false.

Answer:

False 
A perfect square can never have 8 as its units digit.
Hence, given statement is false.

Answer:

False, 
For every natural number m>1,
2m, m² – 1, m² + 1 from a phythagoras triplet 
Hence, given statement is false.

Answer:

False, 
Three natural numbers x, y, z are said to form pythagoras triplet if
c² = a² + b^{2 
For example: 6, 8, 10 form  a pythagoras triplet} 
(10)² = (6)² + (8)²
Here, 6,8 and 10 are even.
Here, given statement is false.

Answer:

False
Let us consider a = –2
$$\begin{gathered} \therefore {\left(-2\right)}^3<{\left(-2\right)}^2 \\ -8<4 (a^3<a^2) \end{gathered}$$
Hence, given statement is false.

Answer:

False
Let us consider two integers –3 and –2
Then
$$\begin{gathered} \Rightarrow {\left(-3\right)}^2>{(-2)}^2 \\ \Rightarrow 9>4 \end{gathered}$$
But
${\left(-3\right)}^3=-27<{(-2)}^3=-8$

Hence, the given statement is false.

Answer:

True.
Since, x and y are natural number
$$\begin{gathered} \therefore \frac{\mathit{y}}{\mathit{x}}=\mathrm{natural} \mathrm{number} \left(\mathrm{given}\right) \\ \mathrm{Then} \frac{y^3}{x^3}=\left(\frac{y}{x}\right)\left(\frac{y}{x}\right)\left(\frac{y}{x}\right) \\ =\mathrm{Natural} \mathrm{number} \end{gathered}$$
Hence, given statement is true.
 

Answer:

Let us consider a number, a = 55
                                     $$\begin{gathered} \therefore a^2=55\times 55 \\ =3025 \end{gathered}$$
3025 ends is 5
Now a³ = (55)³ =166375
166375 does not ends in 25.
Hence, given statement is false.

Answer:

True
Let us consider a number, a = 23
$\therefore a^2=23\times 23=529$
529 ends in 9.
Now, a³ = 23 × 23 × 23
             = 12,167
Thus, 12,167 ends in 7.
​Hence, given statement is true.

Answer:

If n is odd, then square root will contain $\frac{n+1}{2}\mathrm{digits}.$
Hence, the given statement is true.

Answer:

True 
Square root of a number is denoted by $\sqrt{x}$
Hence, given statement is true.

Answer:

False 
17² = 17 × 17 = 289
27² = 27 × 27 = 729
Thus, a number having 7 at its ones place, will have 9 at its units place.
Hence, given statement is false.

Answer:

True
7³ = 7 × 7 × 7 = 343
17³ = 17 × 17 × 17 = 4913
27³ = 27 × 27 × 27 = 19683
Thus, a number having 7 at its ones place will have 3 at ones place in cube.
Hence, given statement is true.

Answer:

False
1³ = 1, 2³ = 8
3³ = 27, 4³ = 64
Thus, the cube of a one digit number can be a two digit number.
Hence, given statement is false

Answer:

False
Cube of  an even number is always even not odd.
Example: (2)³ = 8
                (4)³ = 64
Hence, given statement is true.

Answer:

False
Cube of an odd number is odd
Example: 3³ = 27
                5³ = 125
Hence, give statement is false.

Answer:

True,
Cube of even number is even
Example: 2³ = 8
                4³ = 64
Hence, given statement is true.
 

Answer:

True
Cube of odd number is odd
(3)³ = 27
(5)³ = 125
Hence, given statement is true.

Answer:

False
Prime factorization of 999 = 3 × 3 × 3 × 37
                                           = (3)³ × (37)¹ 
Therefore, 999 is not a perfect cube.
Hence, given statement is false.

Answer:

False, 
Prime factorization = 363 × 81
                                = ((3) × 11 × 11) × (3 × 3 × 3 × 3)
                                = 3⁵ × 11²
Here, 3 and 11 does not occur in triplets.
Thus, (363 × 81) is not a perfect cube.
​Hence, given statement is false.
       

Answer:

False
Cube of 2 = 2 × 2 × 2
                 = 8
Cube 9 – 2 = (–2) × (–2) × (–2)
                  = – 8
Thus, cube root of 8 is 2
Hence, given statement is false.

Answer:

False
Taking RHS, 
$$\begin{gathered} \sqrt[3]{8}+\sqrt[3]{27}=\sqrt[3]{2\times 2\times 2}+\sqrt[3]{3\times 3\times 3} \\ =2+3 \\ =5 \end{gathered}$$
Now, In LHS,
$\sqrt[3]{8+27}=\sqrt[3]{34}$
Since LHS $\ne$RHS.
Hence, the given statement is false.

Answer:

False
Consider (–8)
$$\begin{gathered} 3\sqrt{-8}=3\sqrt{(-2)\times (-2)\times (-2)} \\ =-2 \end{gathered}$$
Hence, given statement is false.

Answer:

False
Example: Let us take a number (–5)
Now, (–5)² = (–5) × (–5)
                   = 25 (Positive)
Also, 
(–5)³ = (–5) × (–5) × (–5)
          = –125 (negative)
Hence, given statement is false.

Answer:

First five square numbers:
(1)² = 1
(2)² = 4
(3)² = 9
(4)² = 16
(5)² = 25

Answer:

Fist three multiplies of 3 are 3, 6, 9
Cube of first three multiples:
(3)³ = 27
(6)³ = 216
(9)³ = 729
​

Answer:

Prime factors of 500 = (2 × 2) × (5 × 5 × 5)
                                  = (2)² × (5)³ 
Since, prime factors of 5 do not form pair.

i.e (2 × 2) × (5 × 5) × (5 × 1)
Thus, 500 is not a perfect square.

Answer:

Fist nine odd consecutive = 1, 3, 5, 7, 9, 11, 13, 15, 17
Now
Sum of first nine consecutive odd numbers.
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81. 

Answer:

(a) 484
Using prime factorization,
484 = 2 × 2 × 11 × 21
       = (2)² × (11)² 
       = (22)² 
Thus, 484 is a perfect square of 22.

(b) 11250
Using prime factorization
11250 = 2 × 3 × 3 × 5 × 5 × 5 × 5
Thus, 11250, is not a perfect square because does not appears in pair.

(c) 841
Using prime factorization,
841 = 29 × 29
Thus, 841 is a perfect square of 29.

(d) 729
Using prime factorization
729 = 3 × 3 × 3 × 3 × 3 × 3
       = 27 × 27
       = (27)² 
​Thus, 729 is a perfect square of 27.

 

Answer:

(a) 128 
Using prime factorization
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
       = (4)³ × (2)¹ 
Since, 2 does not form a triplet,
Thus, 128 does not form a perfect cube.

(b) 343
Using prime factorization
343 = 7 × 7 × 7
       = (7)³  
Thus, 343 is a perfect cube.

(c) 729 
Using prime factorization
729 = 3 × 3 × 3 × 3 × 3 × 3
       = 9 × 9 × 9
       = (9)³ 
Thus, 729 is a perfect cube.

(d) 1331
Using prime factorization
1331 = 11 × 11 × 11
         = (11)³
​Thus, 1331 is a perfect cube. 

 

Answer:

(a) 101
(101)² = 101 × 101
           = 101 × (100 + 1) ( $\because$ Using Distribution Property)
           = 10100 + 101
           = 10201

(b) 72
(72)² = 72 × 70
         = 72 × (70 + 2) ($\because$ Using Distributive Property)
         = 5040 + 144
         = 5184

Answer:

We know for right angle triangle the square of one side must be equal to the sum of the square of the other two sides.
i.e.  c² = a² + b² 
Let, a = 8cm, b = 6cm, c = 10cm
(10)² = (8)² + (16)² 
⇒ 100 = 64 + 36
⇒ 100 = 100
Hence, a right angle triangle can be formed by sides 6cm, 8cm and 10cm.

Answer:

For every natural number m (m>1)
2m, m² + 1, m² – 1 form pythagoras triplet.
Let 2m = 4
    $\Rightarrow$m =2
So, m² + 1 = (2)² + 1
                  = 5
and, m² – 1 = (2)² – 1
                    = 3
Hence, the required pythagoras triplet is 3, 4 and 5.

Answer:

(a) 11025
Using prime factorization
11025 = 3 × 3 × 5 × 5 × 7 × 7
$\therefore \sqrt{11025}=\sqrt{3\times 3\times 5\times 5\times 7\times 7}$
                 = 3 × 5 × 7
                 = 105
Hence, square root of 11025 is 105.

(b) 4761
Using prime factorization
4761 = 3 × 3 × 23 × 23
$$\begin{gathered} \therefore \sqrt{4761} = \sqrt{3\times 3\times 23\times 23} \\ =3 \times 23 \\ =69 \end{gathered}$$
​Hence, square root of 4761 is 69.
 

Answer:

(a) 512
Using prime factorization
512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)
       = 8 × 8 × 8
$\begin{array}{rcl}\therefore \sqrt[3]{512} & =& \sqrt[3]{8\times 8\times 8}\\ & =& 8\end{array}$
Hence, the cube root of 512 is 8.

(b) 2197
Using prime factorization.
2197 = 13 × 13 × 13
$\begin{array}{rcl}\therefore \sqrt[3]{2197} & =& \sqrt[3]{13\times 13\times 13}\\ & =& 13\end{array}$
​Therefore cube root of 2197 is 13.

Answer:

Using prime factorization,
176 = 2 × 2× 2× 2 × 11
       = (2 × 2) × (2 × 2) × (11 × 1)
Now, while grouping, factor 11 has no pair.
Therefore, 176 is not a perfect square.
​
Then, 176 × 11 = 2 × 2× 2× 2 × 11× 11
⇒ 1936 = 2² × 2² × 11² = (44)²

Hence, the smallest number by which it must be multiplied to make a perfect square is 11.

 

Answer:

Using prime factorization method,
9720 = (2 × 2× 2)×(3 × 3 × 3) × (3 × 3 × 5)  
Since, factor 3 and 5 do not form triplet 
Therefore, 9720 is not a perfect cube.
Thus, smallest number it must be divided 
to make perfect cube = 3 × 3 × 5
                                  = 45
$\therefore 9720 ÷ 45=216$
Facture of 216 = (2 × 2 × 2) × (3 × 3 × 3)
                        = 6 × 6 × 6
Hence, 216 is a perfect cube.

Answer:

For every natural number m(m>1)
2m, m² + 1, m² – 1 form pythagoras triplet.
Now m² + 1 = 5(given)
              m² = 4
              m = 2
Then, 2m = 2 × 2
                = 4
Also, m² – 1 = (2)² – 1
                     = 3
Therefore, pythagoras triplet is 3, 4 and 5.
Similarly, other pythagoras triplet is 5, 12 and 13.
 

Answer:

Factors of 216 = (2 × 2 × 2) × (3 × 3 × 3)

                        = (2 × 2) × (3 × 3) × 2 × 3
​ Now, while grouping, factors 2 and 3 has no pair.
​Therefore, 216 is not a perfect square.
​Thus, the smallest number by which it must be divided to get perfect square is 3 × 2 = 6

$$\begin{gathered} \therefore 216 ÷6 \\ =36 \end{gathered}$$

Factors of 36 = 6 × 6

$$\begin{gathered} \sqrt{36}=\sqrt{6\times 6} \\ =6 \end{gathered}$$
Hence, 36 is a perfect square and 6 is the square root of 36.

Answer:

Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = (2 × 2 × 2) × 2 × (3 × 3) × (5 × 5)

Since, while grouping, factors 2, 3 and 5 has no pair.

$\therefore$ 3600 is not a perfect cube.

Thus, the smallest number by which 3600 must be multiplied to make it perfect cube =  2 × 2 × 3 × 5 = 60

Then, 3600 × 60 = 216000

Factors of 216000 = 60 × 60 × 60

$\begin{array}{rcl}\mathrm{Now}, \sqrt[3]{216000}& =& \sqrt[3]{60\times 60\times 60}\\ & =& 60\end{array}$
Hence, the cube root of 216000 is 60. 

Answer:

(a) 1369

$\therefore \sqrt{1369}=37$
Hence, square root of 1369 is 37.

(b) 5625

$\therefore \sqrt{5625}=75$
Hence, square root of 5625 is 75.
 

Answer:

(a) 27.04

$\therefore \sqrt{27.04}=5.2$
Hence, square root of 27.04  is 5.2

(b) 1.44

$\therefore \sqrt{1.44}=1.2$
Hence, square root of 1.44 is 1.2
 

Answer:

Given: 1385
From Long Division Method,

Thus, the least number that should be subtracted from 1385 to get perfect square is 16.
$\therefore$1385 – 16
 =1369 
$$\begin{gathered} \Rightarrow \sqrt{1369}=\sqrt{37\times 37} \\ =37 \end{gathered}$$
Hence, the square root of 1369 is 37.
 

Answer:

Given: 6200
From Long Division Method.

Now, (78)² = 6084 < 6200
Next perfect square
(79)² = 6241
Least number added to make it perfect square = 6241 – 6200 = 41

Hence, 41 is the least number that should be added to 6200 to make it a perfect square.


 

Answer:

We know, 
The least number of four digit = 1000

Now, (32)² = 1024
   
Hence, the smallest four digit number which is a perfect square is 1024.

Answer:

We know, 
Largest 3 digit number = 999.
Now, 
(31)² = 31 × 31
         = 961

Hence, the largest 3 digit number which is perfect is 961.

Answer:

Let us first calculate LCM of 3, 4, 5, 6 and 8.

LCM of 3, 4, 5, 6  and 8 = 2 × 2 × 2 × 3 × 5
                                        = 120
Now, factors of 120
120 = 2 × 2 × 2 × 3 × 5
       = (2 × 2) × (2 × 1) × (3 × 1) × (5 × 1)
So, to make 120 perfect square multiplied 
$\Rightarrow$2 × 3 × 5
$\Rightarrow$30
Then, 120 × 30
       = 3600
Hence, 3600 is the least square number exactly divisible by 3, 4, 5, 6 and 8.

Answer:

Given: Length of diagonal = 10 cm
Let the length of the side of the square be 'a' cm.
By Pythagoras theorem,
(10)² = a² + a²
⇒ 100 = 2a²
⇒ a² = 50
⇒ a = $5\sqrt{2}$
Hence, the length of the side of the square is â€‹$5\sqrt{2}$ units.

Answer:

Let the decimal number be 'a'.
Now,
a × a = 51.84
a² = 51.84
a = $\sqrt{51.84}$
The required number is found out by long division method:

Thus,
a = $\sqrt{51.84}$ = 7.2
Hence, required decimal number is 7.2.

Answer:

Let the decimal fraction be 'x'
Now, 
x × x = 84.64
x² = 84.64
x = $\sqrt{84.64}$
The required number is found out by long division method,

Thus, x = $\sqrt{84.64}$
             = 9.2
Hence, required decimal number is 9.2

Answer:

Length of square feild = 150 m
Now, Area of square feild = 150 × 150
                                          = 22,500 m² 
Therefore, area of square feild is 22, 500 m²

Answer:

Given: Total number of unit squares = 256
Let us assume the number be 'a'.
Now, 
a × a = 256
⇒ a² = $\sqrt{256}$
⇒ a = $\sqrt{256}$
Thus, By long division method.

$\therefore a =\sqrt{16\times 16}$ = 16
Therefore, the number of unit squares are 16.

Answer:

Length of side = 15 m
Volume of cube = (Side)³ 
                          = 15 × 15 × 15
                          = 3375 m³ 
Hence, volume of cube is 3375 m³.

Answer:

Length of rectangular field = 80 m
Breadth of rectangular field = 18 m
 $$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{diogonal}=\sqrt{{\left(80\right)}^2+{\left(18\right)}^2} \\ =\sqrt{6400+324} \\ =\sqrt{6724} \\ =82 \mathrm{m} \end{gathered}$$
Hence, length of diagonal is 82 m.

Answer:

Perimeter of square field = 96
$\therefore$ 4 × Side = 96
⇒ Side = 24 m
Now, area of square field = (Side)² 
                                         = 24 × 24
                                         = 576 m² 
Hence, area of square field is 576 m².
 

Answer:

Given: Volume of cube = 512
$\Rightarrow$(Side of cube)³ = 512
$\Rightarrow$Side of cube = $\sqrt[3]{512}$
                         = 8 cm
Hence, side of cube is 8 cm.

Answer:

Let us assume three numbers be a, 2a, 3a.
Given: Sum of cube of three numbers is 4500
$\Rightarrow$a³ + (2a)³ + (3a)³ = 4500
$\Rightarrow$a³ + 8a³ + 27a³ = 4500
$\Rightarrow$36a³ = 4500
$\Rightarrow$a³ = 125
$\Rightarrow$a = $\sqrt[3]{5\times 5\times 5}$
$\Rightarrow$a = 5
Thus, numbers are 5, 10 and 15.
 

Answer:

Length of square side = 6.5 m
Area of square = Side × Side
                        = 6.5 × 6.5
                        = 42.25 m² 

Answer:

Given: Length of rectangle = 6.4 m
Breadth of rectangle = 2.5 m
Area of rectangle = length × breadth
                             = 6.4 × 2.5
                             = 16 m² 
Let us assume side of square be 'a' m.
Area of square = Area of rectangle (given)

a² = 6.4 × 2.5
⇒ a² = 16
⇒ a = $\sqrt{16}$
⇒ a = 4 m

Hence, the side of the square is 4 m.

Answer:

Let the two numbers be x and y such that (x > y).
Now, 
x³ – y³ = 189
⇒ x³ = 189 + y³           .....(1)
Also, cube root of smaller = 3
i.e. $\sqrt[3]{y}$ = 3
$\Rightarrow$y = 27
From (1)
x³ = 189 + 27
⇒ x³ = 216
⇒ $\sqrt[3]{x}=\sqrt[3]{216}$
⇒ x = 6
Hence, larger number is 6.
 

Answer:

Let the number of plants in each row = x
Number of rows = Number of plants in each row = x
                                
So, total number of plants = 1024
⇒ x² = 1024
⇒ x = $\sqrt{1024}$
⇒ x = 32
​Hence, plants in each row is 32.
 

Answer:

Let the number of seats in each row be x.
Number of rows = Number of seats in each row (given)
$\therefore$ Total number of plants = x² â€‹
⇒ 2704 = x² 
⇒ x = $\sqrt{2704}$
⇒ x = 52
Hence, the number of seats in each row is 52.
 

Answer:

Number of soldiers = 7500
Now, it is between (85)² and (90)² 
(85)² = 7225, (90)² = 8100
(86)² = 7396, (87)² = 7569
Left out = 7500 – 7396
              = 104

Hence, 104 soldiers are left out after the arrangement.

Answer:

Let number of students in each row be '$x$'.
Number of row = number of students in each row = $x$
$\therefore$ Total number of students = $x\times x$
$$\begin{gathered} \Rightarrow x\times x=8649 \\ \Rightarrow x^2=8649 \\ \Rightarrow x=\sqrt{8649} \\ \Rightarrow x=93 \end{gathered}$$
Hence, students in each row is 93.

Answer:

From figure
Distance covered while returning = BO
Therefore, By pythagoras theorem, 
$$\begin{gathered} {\left(\mathrm{BO}\right)}^2={\left(\mathrm{AO}\right)}^2+{\left(\mathrm{AB}\right)}^2 \\ {\left(\mathrm{BO}\right)}^2={\left(12\right)}^2+{\left(35\right)}^2 \\ {\left(\mathrm{BO}\right)}^2=144+1225 \\ {\left(\mathrm{BO}\right)}^2=1369 \\ \mathrm{Bo}=\sqrt{1369} \\ \mathrm{BO}=37 \mathrm{m} \end{gathered}$$
Here, distance traveled while return is 37 m.

Answer:

Let the distance between the wall and the foot of the ladder be '$x$' m.
By Pythagoras theorem,
$$\begin{gathered} {\left(5.5\right)}^2={\left(4.4\right)}^2+x^2 \\ {x}^2={\left(5.5\right)}^2-{\left(4.4\right)}^2 \\ {x}^2=30.35-19.36 \\ {x}^2=10.89 \\ x=\sqrt{10.89} \\ x=3.3 \mathrm{m} \end{gathered}$$
Hence, the distance between ladder foot and wall is3.3 m.

Answer:

From question,
Amount received at end of 30 days = 1 + 3 + 5 + ...
Since, it is an odd number series,
We know, sum of 'n' odd terms = $n^2$
Here, $n=30$.
$\begin{array}{rcl}\therefore \mathrm{Sum} \mathrm{of} \mathrm{odd} \mathrm{natural} \mathrm{number} & =& n^2\\ & =& {\left(30\right)}^2\\ & =& 900\end{array}$
Hence, he will receive 900 coins in that month.

Answer:

Let the numbers be $2x, 3x \mathrm{and} 5x$.

$$\begin{gathered} \therefore {\left(2x\right)}^2+{\left(3x\right)}^2+{\left(5x\right)}^2=608 \left(\mathrm{given}\right) \\ \Rightarrow 4x^2+9x^2+25x^2=608 \\ \Rightarrow 38x^2=608 \\ \Rightarrow x^2=16 \\ \Rightarrow x=4 \end{gathered}$$
Hence, numbers are 8, 12 and 20.

Answer:

LCM of 8, 9, 10 = $2\times 2\times 2\times 3\times 3\times 5$
                           = 360
Since, 2 and 5 have incomplete pair.
$\begin{array}{rcl}\therefore \left(2\times 2\right)\times \left(2\times 2\right)\times \left(3\times 3\right)\times \left(5\times 5\right)& =& 360\times 10\\ & =& 3600\end{array}$
Hence, the required smallest number is 3600.

Answer:

Area of a square plot = $101\frac{1}{400}$ m².
Let the length of the square plot be x m.
$\therefore$Area = $x\times x=101\frac{1}{400}$
$$\begin{gathered} \Rightarrow x^2=\frac{40401}{400} \\ \Rightarrow x=\sqrt{\frac{40401}{400}} \\ \Rightarrow x=\frac{201}{20} \\ \Rightarrow x=10\frac{1}{20} \end{gathered}$$

Hence, the length of the side of the square is $10\frac{1}{20} \mathrm{m}$.

Answer:

By successive subtracting odd numbers, starting from 1.

$$\begin{gathered} 324-1=323 \\ 323-3=320 \\ 320-5=315 \\ 315-7=308 \\ 308-9=299 \\ 299-11=288 \\ 288-13=275 \\ 275-15=260 \\ 260-17=243 \\ 243-19=224 \\ 224-21=203 \\ 203-23=180 \\ 180-25=155 \\ 155-27=128 \\ 128-29=99 \\ 99-31=68 \\ 68-33=35 \\ 35-35=0 \end{gathered}$$

Here, we see 324 reduces after subtracting 18 odd numbers.

Hence, the square root of 324 is 18.
 

Answer:

Let numbers be $2x, 3x \mathrm{and} 4x$.

Now,

${\left(2x\right)}^3+{\left(3x\right)}^3+{\left(4x\right)}^3=0.334125$

$$\begin{gathered} \Rightarrow 8x^3+27x^3+64x^3=0.334125 \\ \Rightarrow 99x^3=0.334125 \\ \Rightarrow x^3=0.334125÷99 \\ \Rightarrow x^3=0.003375 \\ \Rightarrow x=0.15 \end{gathered}$$

Hence, the required numbers are 0.3, 0.45 and 0.6.

Answer:

$\begin{array}{rcl}\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}& =& 3+0.2+0.4\\ & =& 3+0.6\\ & =& 3.6\end{array}$

Answer:

$\begin{array}{rcl}{\left\{\left(5^2+{\left({12}^2\right)}^{\frac{1}{2}}\right)\right\}}^3& =& {\left\{\left(25+{\left(12\right)}^{2\times \frac{1}{2}}\right)\right\}}^3\\ & =& {\left\{\left(25+12\right)\right\}}^3\\ & =& {\left\{37\right\}}^3\\ & =& 50,653\end{array}$

Answer:

$\begin{array}{rcl}{\left\{\left(6^2+{\left(8^2\right)}^{\frac{1}{2}}\right)\right\}}^3& =& {\left\{\left(36+{\left(8\right)}^{2\times \frac{1}{2}}\right)\right\}}^3\\ & =& {\left\{36+8\right\}}^3\\ & =& {\left\{44\right\}}^3\\ & =& 85,184\end{array}$

Answer:

Let the number be WXYZ.
Now, from left to right, even, even, odd, even,
So, number = 8836 $\left(\therefore {\left(94\right)}^2=8836\right)$

Answer:

Let the number be a, b and c
Such that, a = 6, b = 19, c = 30
a + b = 19 + 6 = (5)²  ($\therefore$ Perfect square)
b + c = 19 + 30 = (7)²
c + a = 30 + 6 = (6)²

Answer:

Let the sides of the two squares be x and y respectively and the side of the third square be z.
Now,
The perimeter of first square = 40 m
$$\begin{gathered} 4x=40 \\ \Rightarrow x=10 \mathrm{m} \end{gathered}$$.
Also, the perimeter of second square = 96
$$\begin{gathered} 4y=96 \\ \Rightarrow y=24 \mathrm{m} \end{gathered}$$
Given:
Area of 3^rd square = Area of 1^st square + Area of 2^nd square
$$\begin{gathered} z^2=\left(10\times 10\right)+\left(24\times 24\right) \\ {z}^2=100+5+6 \\ z=\sqrt{676} \\ z=26 \mathrm{m} \end{gathered}$$
Perimeter of 3^rd square = $4\times 26$ = 104 m.

Answer:

The digit 1, 0, 8 remains same when viewed upside down, 
Whereas, 9 become 6 and 6 become 9.
$\therefore$ 3- digit perfect square when viewed upside to down = 196 and 961

Answer:

Given: ${\left(13\right)}^2=169, {\left(31\right)}^2=961$

Similar, two such numbers can be

${\left(12\right)}^2=144, {\left(21\right)}^2=441$

Hence, the required numbers are 12 and 21.