RS Aggarwal solution class 8 chapter 3 Squares and Square roots Exercise 3A

Exercise 3A

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Q1 | Ex-3A | Squares and Square roots | Chapter 3| Class 8 | RS AGGARWAL  | myhelper

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Question 1:

Using the prime factorisation method, find which of the following numbers are perfect squares:
(i) 441
(ii) 576
(iii) 11025
(iv) 1176
(v) 5625
(vi) 9075
(vii) 4225
(viii) 1089

Answer 1:

A perfect square can always be expressed as a product of equal factors.

(i)
Resolving into prime factors:
441=49×9=7×7×3×3=7×3×7×3=21×21=(21)²

Thus, 441 is a perfect square.

(ii)
Resolving into prime factors:
576=64×9=8×8×3×3=2×2×2×2×2×2×3×3=24×24=(24)²

Thus, 576 is a perfect square.

(iii)
Resolving into prime factors:
11025=441×25=49×9×5×5=7×7×3×3×5×5

=7×5×3×7×5×3=105×105=(105)²

Thus, 11025 is a perfect square.

(iv)
Resolving into prime factors:
1176=7×168=7×21×8=7×7×3×2×2×2

1176 cannot be expressed as a product of two equal numbers. Thus, 1176 is not a perfect square.

(v)
Resolving into prime factors:
5625=225×25=9×25×25=3×3×5×5×5×5

=3×5×5×3×5×5=75×75=(75)²

 Thus, 5625 is a perfect square.

(vi)
Resolving into prime factors:
 $9075=25\times 363=5\times 5\times 3\times 11\times 11=55\times 55\times 3$

9075 is not a product of two equal numbers. Thus, 9075 is not a perfect square.

(vii)
Resolving into prime factors:
4225=25×169=5×5×13×13=5×13×5×13=65×65=(65)²

Thus, 4225 is a perfect square.

(viii)
Resolving into prime factors: 
1089=9×121=3×3×11×11=3×11×3×11=33×33=(33)²

Thus, 1089 is a perfect square.

Q2 | Ex-3A | Squares and Square roots | Chapter 3| Class 8 | RS AGGARWAL | myhelper

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Question 2:

Show that each of the following numbers is a perfect square. In each case, find the number whose square is the given number:
(i) 1225
(ii) 2601
(iii) 5929
(iv) 7056
(v) 8281

Answer 2:

A perfect square is a product of two perfectly equal numbers.

(i)
Resolving into prime factors: 

1225=25×49=5×5×7×7=5×7×5×7=35×35=(35)²

Thus, 1225 is the perfect square of 35.

(ii)
Resolving into prime factors:
2601=9×289=3×3×17×17=3×17×3×17=51×51=(51)²

Thus, 2601 is the perfect square of 51.

(iii)
Resolving into prime factors: 

5929=11×539=11×7×77=11×7×11×7=77×77=(77)²

Thus, 5929 is the perfect square of 77.

(iv)
Resolving into prime factors:
7056=12×588=12×7×84=12×7×12×7=(12×7)²=(84)²

Thus, 7056 is the perfect square of 84.

(v)
Resolving into prime factors:

8281=49×169=7×7×13×13=7×13×7×13=(7×13)²=(91)²

Thus, 8281 is the perfect square of 91.

Q3 | Ex-3A | Squares and Square roots | Chapter 3| Class 8 | RS AGGARWAL | myhelper

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Question 3:

By what least number should the given number be multiplied to get a perfect square number? In each case, find the number whose square is the new number.
(i) 3675
(ii) 2156
(iii) 3332
(iv) 2925
(v) 9075
(vi) 7623
(vii) 3380
(viii) 2475

Answer 3:

1. Resolving 3675 into prime factors:
 $3675=3\times 5\times 5\times 7\times 7$ 

Thus, to get a perfect square, the given number should be multiplied by 3.

New number$= (3^2\times 5^2\times 7^2)=(3\times 5\times 7{)}^2=(105{)}^2$

Hence, the new number is the square of 105.

2. Resolving 2156 into prime factors:
$2156=2\times 2\times 7\times 7\times 11=(2^2\times 7^2\times 11)$

Thus to get a perfect square, the given number should be multiplied by 11.

New number $=(2^2\times 7^2\times {11}^2)=(2\times 7\times 11{)}^2=(154{)}^2$

Hence, the new number is the square of 154.

3. Resolving 3332 into prime factors:
​$3332=2\times 2\times 7\times 7\times 17=2^2\times 7^2\times 17$

Thus, to get a perfect square, the given number should be multiplied by 17.

New number $=(2^2\times 7^2\times {17}^2)=(2\times 7\times 17{)}^2=(238{)}^2$

Hence, the new number is the square of 238.

4. Resolving 2925 into prime factors:
​$2925=3\times 3\times 5\times 5\times 13=3^2\times 5^2\times 13$

Thus, to get a perfect square, the given number should be multiplied by 13.

New number $=(3^2\times 5^2\times {13}^2)=(3\times 5\times 13{)}^2=(195{)}^2$

Hence, the number whose square is the new number is 195.

5. Resolving 9075 into prime factors:

 ​$9075=3\times 5\times 5\times 11\times 11=3\times 5^2\times {11}^2$

Thus, to get a perfect square, the given number should be multiplied by 3.

New number $=(3^2\times 5^2\times {11}^2)=(3\times 5\times 11{)}^2=(165{)}^2$

Hence, the new number is the square of 165.

6. Resolving 7623 into prime factors:
 ​$7623=3\times 3\times 7\times 11\times 11=3^2\times 7\times {11}^2$

Thus, to get a perfect square, the given number should be multiplied by 7.

New number $=(3^2\times 7^2\times {11}^2)=(3\times 7\times 11{)}^2=(231{)}^2$

Hence, the number whose square is the new number is 231.

7. Resolving 3380 into prime factors:
 ​$3380=2\times 2\times 5\times 13\times 13=2^2\times 5\times {13}^2$

Thus, to get a perfect square, the given number should be multiplied by 5.

New number $=(2^2\times 5^2\times {13}^2)=(2\times 5\times 13{)}^2=(130{)}^2$

Hence, the new number is the square of 130.

8. Resolving 2475 into prime factors:
 ​$2475=3\times 3\times 5\times 5\times 11=3^2\times 5^2\times 11$

Thus, to get a perfect square, the given number should be multiplied by 11.

New number $=(3^2\times 5^2\times {11}^2)=(3\times 5\times 11{)}^2=(165{)}^2$

Hence, the new number is the square of 165.

Q4 | Ex-3A | Squares and Square roots | Chapter 3| Class 8 | RS AGGARWAL | myhelper

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Question 4:

By what least number should the given number be divided to get a perfect square number? In each case, find the number whose square is the new number.
(i) 1575
(ii) 9075
(iii) 4851
(iv) 3380
(v) 4500
(vi) 7776
(vii) 8820
(viii) 4056

Answer 4:

(i) Resolving 1575 into prime factors:
 $1575=3\times 3\times 5\times 5\times 7=3^2\times 5^2\times 7$

Thus, to get a perfect square, the given number should be divided by 7

New number obtained$=(3^2\times 5^2)=(3\times 5{)}^2=(15{)}^2$

Hence, the new number is the square of 15

(ii) Resolving 9075 into prime factors: 

$9075=3\times 5\times 5\times 11\times 11=3\times 5^2\times {11}^2$

Thus, to get a perfect square, the given number should be divided by 3

New number obtained$=(5^2\times {11}^2)=(5\times 11{)}^2=(55{)}^2$

Hence, the new number is the square of  55

(iii) Resolving 4851 into prime factors:

 $4851=3\times 3\times 7\times 7\times 11=3^2\times 7^2\times 11$

Thus, to get a perfect square, the given number should be divided by 11

New number obtained$=(3^2\times 7^2)=(3\times 7{)}^2=(21{)}^2$

Hence, the new number is the square of 21

(iv) Resolving 3380 into prime factors: 

$3380=2\times 2\times 5\times 13\times 13=2^2\times 5\times {13}^2$

Thus, to get a perfect square, the given number should be divided by 5

New number obtained$=(2^2\times {13}^2)=(2\times 13{)}^2=(26{)}^2$

Hence, the new number is the square of 26

(v) Resolving 4500 into prime factors: 

$4500=2\times 2\times 3\times 3\times 5\times 5\times 5=2^2\times 3^2\times 5^2\times 5$

Thus, to get a perfect square, the given number should be divided by 5

New number obtained$=(2^2\times 3^2\times 5^2)=(2\times 3\times 5{)}^2=(30{)}^2$

Hence, the new number is the square of 30

(vi) Resolving 7776 into prime factors:

7776=2×2×2×2×2×3×3×3×3×3=2²×2²×2×3²×3²×3

Thus, to get a perfect square, the given number should be divided by 6 whish is a product of 2 and 3

New number obtained$=(2^2\times 2^2\times 3^2\times 3^2)=(2\times 2\times 3\times 3{)}^2=(36{)}^2$

Hence, the new number is the square of 36

(vii) Resolving 8820 into prime factors:

 $8820=2\times 2\times 3\times 3\times 5\times 7\times 7=2^2\times 3^2\times 5\times 7^2$

Thus, to get a perfect square, the given number should be divided by 5

New number obtained$=(2^2\times 3^2\times 7^2)=(2\times 3\times 7{)}^2=(42{)}^2$

Hence, the new number is the square of 42

(viii) Resolving 4056 into prime factors: 

$4056=2\times 2\times 2\times 3\times 13\times 13=2^2\times 2\times 3\times {13}^2$

Thus, to get a perfect square, the given number should be divided by 6, which is a product of 2 and 3

New number obtained$=(2^2\times {13}^2)=(2\times 13{)}^2=(26{)}^2$

Hence, the new number is the square of 26

Q5 | Ex-3A | Squares and Square roots | Chapter 3| Class 8 | RS AGGARWAL | myhelper

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Question 5:

Find the largest number of 2 digits which is a perfect square.

Answer 5:

The first three digit number (100) is a perfect square. Its square root is 10.
The number before 10 is 9.
Square of 9 $= (9{)}^2=81$
Thus, the largest 2 digit number that is a perfect square is 81.

Q6 | Ex-3A | Squares and Square roots | Chapter 3| Class 8 | RS AGGARWAL | myhelper

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Question 6:

Find the largest number of 3 digits which is a perfect square.

Answer 6:

The largest 3 digit number is 999.

The number whose square root is 999 is 31.61.

Thus, the square of any number greater than 31.61 will be a 4 digit number.
Therefore, the square of 31 will be the greatest 3 digit perfect square.

${31}^2=31\times 31=961$