S.chand books class 8 solution maths chapter 3 Square and Square root

Exercise 3 (A)

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Question1

Find the square of the following numbers :

(i) 15

(ii) 48

(iii) ​$\dfrac{6}{7}$​

(iv) ​$\dfrac{21}{25}$​

(v) ​$6\dfrac{3}{8}$​

(vi) 0.9

(vii) 1.1

(viii) 0.018

Sol :

(i)

=15×15

=225

(ii)

=48×48

=2304

(iii)

$=\frac{6}{7} \times \frac{6}{7}$

$=\frac{36}{49}$

(iv)

$=\frac{21}{25} \times \frac{21}{25}$

$=\frac{441}{625}$

(v) 

$=\frac{48+3}{8}$

$=\frac{51}{8} \times \frac{51}{8}$

$=\frac{2601}{64}$

$=40 \frac{41}{64}$

(vi)

=0.9×0.9

=0.81

(vii)

=1.1×1.1

=1.21

(viii)

=0.018×0.018

=0.000324

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Question 2

Determine whether square of the following numbers will be even or odd.(Do not find the square)

Note : Square of even number are even

Square of odd number is odd

(i) 529

Sol :

529 is odd and we know square of odd number is odd

(ii) 30976

Sol :

30976 is even and we know square of even number is even

(iii) 893025

Sol :

893025 is odd and we know square of odd number is odd

(iv) 6990736

Sol :

6990736 is even and we know square of even number is even

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Question3

Just by looking at the following numbers, give reason why they are not perfect squares.

Note : Any number ending with 2,3,7,8 and having odd number of zeros are not perfect square

(i) 4893

Sol : Not a perfect square

(ii) 65000

Sol : Not a perfect square

(iii) 4422

Sol : Perfect square

(iv) 89138

Sol : Perfect square

(v) 150087

Sol : Not a perfect square

 

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Question 4

Using prime factorization method, find which of the following are perfect square numbers.

Note : For a number to be a perfect square it's prime factors should be to a power of 2 or any even number.

(i) 100

Sol :

$\begin{array}{l|l} 2&100 \\ \hline 2&50 \\ \hline 5&25 \\ \hline 5&5 \\ \hline &1 \end{array}$

100=2×2×5×5 =2²×5²

Since the powers of the prime factors are even therefore 100 is a perfect square.

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(ii) 284

Sol :

$\begin{array}{l|l}2&284\\ \hline 2&142 \\ \hline 71& 71 \\ \hline & 1\end{array}$

281=2×2×71 =2²×71¹ 

Since the powers of the prime factors are not even (i.e. 71¹) therefore 284 is not a perfect square.

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(iii) 784

Sol :

$\begin{array}{l|l}2&784 \\ \hline 2&392 \\ \hline 2&196 \\ \hline 2&98 \\ \hline 7&49 \\ \hline 7&7 \\ \hline &1\end{array}$

784=2×2×2×2×7×7 =2⁴×7²

Since the powers of the prime factors are even therefore 784 is a perfect square.

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(iv) 1444

Sol :

$\begin{array}{l|l} 2 & 1444 \\ \hline 2 & 722 \\ \hline 19 & 361 \\ \hline 19 & 19 \\ \hline & 1 \\ \end{array}$

1444=2×2×19×19 =2²×19²

Since the powers of the prime factors are even therefore 1444 is a perfect square.

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(v) 768

Sol :

$\begin{array}{l|l} 2 & 768 \\ \hline 2 & 384 \\ \hline 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

768=2×2×2×2×2×2×2×2×3 =2⁸×3¹

Since the powers of the prime factors are not even therefore 768 is not a perfect square.

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(vi) 4225

Sol :

$\begin{array}{l|l} 5 & 4225 \\ \hline 5 & 845 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

4225=5×5×13×13 =5²×13²

Since the powers of the prime factors are even therefore 4225 is a perfect square.

 

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(vii) 3375

Sol :

$\begin{array}{l|l} 3 & 3375 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

3375=3×3×3×5×5×5 =3³×5³

Since the powers of the prime factors are not even therefore 3375 is not a perfect square.

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(viii) 15625

Sol :

$\begin{array}{l|l} 5 & 15625 \\ \hline 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

15625=5×5×5×5×5×5 =5⁶

Since the powers of the prime factors are even therefore 15625 is a perfect square.

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Question 5

Using prime factorization method, find the square root of the following numbers.

(i) 256

Sol :

$\begin{array}{l|l} 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

√15625=√2×2×2×2×2×2×2×2 =2⁴=16

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(ii) 1936

Sol :

$\begin{array}{l|l} 2 & 1936 \\ \hline 2 & 968 \\ \hline 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

√1936=√2×2×2×2×11×11=2×2×11=44

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(iii) 3364

Sol :

$\begin{array}{l|l} 2 & 3364 \\ \hline 2 & 1682 \\ \hline 29 & 841 \\ \hline 29 & 29 \\ \hline & 1 \\ \end{array}$

√3364=√2×2×29×29=2×29=58

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(iv) 5625

Sol :

$\begin{array}{l|l} 3 & 5625 \\ \hline 3 & 1875 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

√5625=√3×3×5×5×5×5=3×5×5=75

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(v) 7744

Sol :

$\begin{array}{l|l} 2 & 7744 \\ \hline 2 & 3872 \\ \hline 2 & 1936 \\ \hline 2 & 968 \\ \hline 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

√7744=√2×2×2×2×2×2×11×11=2×2×2×11=88

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(vi) 12544

Sol :

$\begin{array}{l|l} 2 & 12544 \\ \hline 2 & 6272 \\ \hline 2 & 3136 \\ \hline 2 & 1568 \\ \hline 2 & 784 \\ \hline 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7&7\\ \hline &1 \end{array}$

√12544=√2×2×2×2×2×2×2×2×7×7=2×2×2×2×7=112

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Question 6

Find the square root of the following fractions.

(i) ​$\dfrac{529}{1225}$​

(ii) $\dfrac{1444}{3249}$​

(iii) $1\dfrac{396}{9604}$​

(iv) 0.01

(v) 0.1764

(vi) 0.00003136

Sol :

(i)

$=\sqrt{\frac{23 \times 23}{5^{2} \times 7^{2}}}$

$=\frac{23}{5 \times 7}$

$=\frac{23}{35}$

(ii)

$=\sqrt{\frac{2^{2} \times 19^{2}}{3^{2} \times 19^{2}}}$

$=\frac{2 \times 19}{3 \times 19}$

$=\frac{2}{3}$

(iii)

$=\sqrt{\frac{9604 \times 1+396}{9604}}$

$=\sqrt{\frac{10000}{9604}}$

$=\sqrt{\frac{2^{4} \times 5^{4}}{2^{2} \times 7^{4}}}$

$=\frac{2 \times 2 \times 5 \times 5}{2 \times 7 \times 1}$

(iv)

$=\sqrt{\frac{0.01}{1 \times 100}}$

$=\frac{1}{100}$

$\frac{1}{\sqrt{2^{2} \times 5^{2}}}$

$=\frac{1}{2 \times 5}$

$=\frac{1}{10}$

(v)

$=\sqrt{\frac{0.1764 \times 10000}{10000}}$

$=\sqrt{\frac{1764}{10000}}$

$=\sqrt{ \frac{2^{2} \times 3^{2} \times 7^{2}}{2^{4} \times 5^{4}}}$

$=\frac{2 \times 3 \times 7}{2 \times 2 \times 5 \times 5}$

$=\frac{21}{50}$

=0.42

(vi)

$=\sqrt{\frac{0.00003136 \times 100000000}{100000000}}$

$=\sqrt{\frac{3136}{100000000}}$

$=\sqrt{\frac{2^{6} \times 7^{2}}{2^{8} \times 5^{8}}}$

$=\frac{2^{3} \times 7}{2^{4} \times 5^{4}}$

$=\frac{7}{2 \times 5^{4}}$

$=\frac{7}{1250}$

=0.0056

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Question 7

Simplify:

(i) ​$\sqrt{10^2-{6^2}}$​

 

(ii) ​$\Bigg(-\sqrt{\dfrac{9}{16}}\Bigg)\Bigg(\sqrt{\dfrac{64}{81}}\Bigg)$​

 

(iii) ​$\sqrt{\dfrac{16}{36}-\dfrac{1}{4}}$​

Sol :

(i)

=√100-36

=√64

=√8×8

=8

(ii)

$=\left(-\sqrt{\frac{3 \times 3}{4 \times 4}}\right)\left(\sqrt{\frac{8 \times 8}{9 \times 9}}\right)$

$=-\frac{3}{4} \times \frac{8}{9}$

$=-\frac{2}{3}$

(iii)

L.C.M of 36 and 4 is 36

$=\sqrt{\frac{16 \times 1+1 \times 9}{36}}$

$=\sqrt{\frac{16+9}{36}}=\sqrt{\frac{25}{36}}$

$=\sqrt{\frac{5 \times 5}{6 \times 6}}=\frac{5}{6}$

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Question 8

Find the square root of 1024. Hence find the value of ​$\sqrt{10.24} + \sqrt{0.1024}+\sqrt{10240000}$​.

Sol :

$\begin{array}{l|l}2&1024\\ \hline 2&512 \\ \hline 2& 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2& 32 \\ \hline 2& 16 \\ \hline 2& 8 \\ \hline 2 & 4\\ \hline 2& 2 \\ \hline &1 \end{array}$

=√2¹⁰

=2⁵

=32

=√10.24+=√0.1024+=√10240000

$=\sqrt{\frac{1024}{100}}+\sqrt{\frac{1024}{10000}}+\sqrt{1024 \times 10000}$

$=\frac{32}{10}+\frac{32}{100}+32 \times 100$

=3.2+0.32+3200

=3203.52

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Question 9

Find the smallest number by which each of the given number s ,ust be multiplied so that the product is a perfect square. Also find the square root of this new number.

(i) 175

Sol :

$\begin{array}{l|l}5&175 \\ \hline 5& 35 \\ \hline 7& 7 \\ \hline & 1\end{array}$

Resolving prime factors

175=5×5×7

In the above product 5 occur in pairs whereas 7 does not exist in pair.So we have to multiply 7 to 175 to make it a perfect square

Perfect square number=1225

Square root=√1225

=√5×5×7×7

=5×7

=35

(ii) 325

Sol :

$\begin{array}{l|l}5&325\\ \hline 5&65 \\ \hline 13& 13\\ \hline &1\end{array}$

Resolving prime factors

325=5×5×13

In the above product 5 occur in pairs whereas 13 does not exist in pair.So we have to multiply 325 by 13 to make it a perfect square

Perfect square number=4225

Square root=√4225

=√5×5×13×13

=5×13

=65

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(iii) 720

Sol :

$\begin{array}{l|l} 2 & 720 \\ \hline 2 & 360 \\ \hline 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline \\ \end{array}$

Resolving Prime factors

720= (2×2)×(2×2)×(3×3)×5

Above ,product of 2 and 3 occur in pairs whereas 5 does not exist in pair so we have to multiply 720 by 5 to make it a perfect square

Perfect square number=3600

Square  root=√(2×2)×(2×2)×(3×3)×(5×5)

=2×2×3×5 =60

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(iv) 3150

Sol :



$\begin{array}{l|l} 2 & 3150 \\ \hline 3 & 1575 \\ \hline 3 & 525 \\ \hline 5 & 175 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

Resolving Prime factors

3150= 2×(3×3)×(5×5)×7

Above ,product of 3 and 5 occur in pairs whereas 2 and 7 does not exist in pair so we have to multiply 3150 by 2 and 7 (i.e. 14) to make it a perfect square

Perfect square number=44100

Square  root=√(2×2)×(3×3)×(5×5)×(7×7)

=2×3×5×7 =210

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Question 10

Find the smallest number by which each of the given numbers must be divided so that the quotient is a perfect square.Also find the square root of this quotient.

(i) 2700

Sol :

$\begin{array}{l|l} 2 & 2700 \\ \hline 2 & 1350 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline \\ \end{array}$

Resolving Prime factors

2700= (2×2)×(3×3)×3×(5×5)

Above ,product of 2 , 3 and 5 occur in pairs but their is a single 3 extra so we have to divide 2700 by 3  to make it a perfect square

Perfect square number=900

Square  root=√(2×2)×(3×3)×(5×5)

=2×3×5 =30

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(ii) 5488

Sol :

$\begin{array}{l|l} 2 & 5488 \\ \hline 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

Resolving Prime factors

5488= (2×2)×(2×2)×(7×7)×7

Above ,product of 2 and 7 occur in pairs but their is a single 7 extra so we have to divide 5488 by 7  to make it a perfect square

Perfect square number=784

Square  root=√(2×2)×(2×2)×(7×7)

=2×2×7 =28

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(iii) 7203

Sol :

$\begin{array}{l|l} 3 & 7203 \\ \hline 7 & 2401 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

Resolving Prime factors

7203= 3×(7×7)×(7×7)

Above ,product of 7 occur in pairs but their is a single 3 extra so we have to divide 7203 by 3  to make it a perfect square

Perfect square number=2401

Square  root=√(7×7)×(7×7)

=7×7 =49

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(iv) 20886

Sol :

$\begin{array}{l|l} 2 & 20886 \\ \hline 3 & 10443 \\ \hline 59 & 3481 \\ \hline 59 & 59 \\ \hline & 1 \\ \end{array}$

Resolving Prime factors

7203= 3×(7×7)×(7×7)

Above ,product of 7 occur in pairs but their is a single 3 extra so we have to divide 7203 by 3  to make it a perfect square

Perfect square number=2401

Square  root=√(7×7)×(7×7)

=7×7 =49

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Question11

Find the smallest number which is completely divisible by each of the numbers 10 , 16 and 24.

Sol :

Here, we need to find the perfect number divisible by square of L.C.M of 10 , 16 and 24

L.C.M of 10 ,16 , 24 is =2×2×2×2×3×5 =240

$\begin{array}{l|l} 2 & 10,16,24 \\ \hline 2 & 5,8,12 \\ \hline 2 & 5,4,6 \\ \hline 2 & 5,2,3 \\ \hline 3 & 5,1,3 \\ \hline 5 & 5,1,1 \\ \hline &1,1,1 \end{array}$

Resolving Prime factors

240= (2×2)×(2×2)×3×5

To make it into a perfect square is should be multiplied by 3×5=15

∴Required square number =240×15=3600

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Question12

The children of class VIII of a school contributed ₹ 3025 for the Prime Minister's National Relief Fund, contributing as much money as there are children in the class.How many children are there in the class ?

Sol :

Let the number of children be x

Then, according to question,
Money donated by individual = number of children = x

Now,
Total money donated = money donated by individual × number of children

⇒x×x=3025

⇒x²=3025

⇒x=√3025

⇒x=√5×5×11×11

⇒x=5×11=55

Hence there are 55 children in the class and all of them donated 55 ₹ each

 

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