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

## Exercise 3 (A)

#### Question 1

Find the square of the following numbers :

(i) 15

(ii) 48

(iii) ​ (iv) ​ (v) ​ (vi) 0.9

(vii) 1.1

(viii) 0.018

Sol : #### Question

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

#### Question 3

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

#### 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 :

 2 100 2 50 5 25 5 5 1

100=2×2×5×5 =22×52

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

(ii) 284

Sol :

 2 284 2 142 71 71 1

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.

(iii) 784

Sol :

 2 784 2 392 2 196 2 98 7 49 7 7 1

784=2×2×2×2×7×7 =24×72

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

(iv) 1444

Sol :

 2 1444 2 722 19 361 19 19 1

1444=2×2×19×19 =22×192

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

(v) 768

Sol :

 2 768 2 384 2 192 2 96 2 48 2 24 2 12 2 6 3 3 1

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.

(vi) 4225

Sol :

 5 4225 5 845 13 169 13 13 1

4225=5×5×13×13 =52×132

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

(vii) 3375

Sol :

 3 3375 3 1125 3 375 5 125 5 25 5 5 1

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.

(viii) 15625

Sol :

 5 15625 5 3125 5 625 5 125 5 25 5 5 1

15625=5×5×5×5×5×5 =56

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

#### Question 5

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

(i) 256

Sol : 2 256 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1

√15625=√2×2×2×2×2×2×2×2 =24=16

(ii) 1936

Sol :

 2 1936 2 968 2 484 2 242 11 121 11 11 1

√1936=√2×2×2×2×11×11=2×2×11=44

(iii) 3364

Sol :

 2 3364 2 1682 29 841 29 29 1

√3364=√2×2×29×29=2×29=58

(iv) 5625

Sol :

 3 5625 3 1875 5 625 5 125 5 25 5 5 1

√5625=√3×3×5×5×5×5=3×5×5=75

(v) 7744

Sol :

 2 7744 2 3872 2 1936 2 968 2 484 2 242 11 121 11 11 1

√7744=√2×2×2×2×2×2×11×11=2×2×2×11=88

(vi) 12544

Sol :

 2 12544 2 6272 2 3136 2 1568 2 784 2 392 2 196 2 98 7 49 7 7 1

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

#### Question 6

Find the square root of the following fractions.

(i) ​ (ii) (iii) (iv) 0.01

(v) 0.1764

(vi) 0.00003136

Sol :

#### Question 7

Simplify:

(i) ​ (ii) ​ (iii) ​ Sol : #### Question 8

Find the square root of 1024. Hence find the value of ​ ​.

Sol : #### 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 : (ii) 325

Sol : (iii) 720

Sol :

 2 720 2 360 2 180 2 90 3 45 3 15 5 5 1

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

(iv) 3150

Sol :

 2 3150 3 1575 3 525 5 175 5 35 7 7 1

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

#### 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 :

 2 2700 2 1350 3 675 3 225 3 75 5 25 5 5 1

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

(ii) 5488

Sol :

 2 5488 2 2744 2 1372 2 686 7 343 7 49 7 7 1

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

(iii) 7203

Sol :

 3 7203 7 2401 7 343 7 49 7 7 1

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

(iv) 20886

Sol :

 2 20886 3 10443 59 3481 59 59 1

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

#### Question 11

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

 2 10 , 16 , 24 2 5 , 8 , 12 2 5 , 4 , 6 2 5 , 2 , 3 3 5 , 1 , 3 5 5 , 1 , 1 1 , 1 , 1

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

#### Question 12

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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