In this page we have *Important questions for Square And Square roots Class 8 Maths*
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**Question 1**
Find the square root of each of the following numbers by using the method of prime factorization:

(a) 121

(b) 441

(c) 625

(d) 729

(e) 1521

(f) 2025

(g) 4096

(h) 5776

(i) 8100

(j) 9216

(k) 11236

(l) 15876

(m) 18496

Solution
(a) 121 = 11 x 11

$\sqrt {121} =11$

(b) 441= 3 x 3 x 7 x 7

$\sqrt {441} =3 \times 7=21$

(c) 625= 5 x 5 x 5 x 5

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

(d) 729= 3 x 3 x 3 x 3 x 3 x 3

$\sqrt {729} =3 \times 3 \times 3=27$

(e)1521= 3 x 3 x 13 x 13

$\sqrt {1521} =3 \times 13=39$

(f) 2025= 3 x 3 x 3 x 3 x 5 x 5

$\sqrt {2025} =3 \times 3 \times 5=45$

(g) 4096=2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

$\sqrt {4096} =2 \times 2 \times 2 \times 2 \times 2 \times 2=64$

(h) 5776= 2 x 2 x 2 x 2 x 19 x 19

$\sqrt {5776} =2 \times 2 \times 19=76$

(i) 8100= 2 x 2 x 3 x 3 x 3 x 3 x 5 x 5

$\sqrt {8100} =2 \times 3 \times 3 \times 5=90$

(j) 9216= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

$\sqrt {9216} =2 \times 2 \times 2 \times 2 \times 2 \times 3=96$

(k)11236 =2 x 2 x 53 x 53

$\sqrt {11236} =2 \times 53=106$

(l) 15876= 2 x 2 x 3 x 3 x 3 x 3 x 7 x 7

$\sqrt {15876} =2 \times 3 \times 3 \times 7=126$

(m) 18496=2 x 2 x 2 x 2 x 2 x 2 x 17 x 17

$\sqrt {18496} =2 \times 2 \times 2 \times 17=136$

**Question 2**
Find the smallest number by which following number must be multiplied to get a perfect square. Also, find the square root of the perfect square so obtained.

- 1008
- 1280
- 1875

Solution
(i)1008= 2 x 2 x 2 x 2 x 3 x 3 x 7

We can see number 7 is not in pair, So to make a perfect square, it has to be multiplied by 7

Then number will become=7056

Now

7056= 2 x 2 x 2 x 2 x 3 x 3 x 7 x 7

$\sqrt {7056} =2 \times 2 \times 3 \times 7=84$

(ii)1280=2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5

We can see number 5 is not in pair, So to make a perfect square, it has to be multiplied by 5

Then number will become=6400

Now

6400= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5

$\sqrt {6400} =2 \times 2 \times 2 \times 2 \times 5=80$

(iii)1875= 3 x 5 x 5 x 5 x 5

We can see number 3 is not in pair, So to make a perfect square, it has to be multiplied by 3

Then number will become=5625

Now

5625= 3 x 3 x 5 x 5 x 5 x 5

$\sqrt {5625} =3 \times 5 \times 5 =75$

**Question 3. **
676 students are to be sit in a hall in such a way that each row contains as many students as the number of rows. Find the number of rows and the number of students in each row.

Solution
Here we need to find the square root of the Number 676

676=2 x 2 x 13 x 13

$\sqrt {676} =2 \times 13 =26$

So There are 26 rows and each rows has 26 students

**Question 4.**
What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 1801

(ii) 856

(iii) 1008001

(iv) 6577525

Solution
(i)Last digit is 1 , So one digit can be 1 or 9 as $1^2=1$ and $9^2=81$

(ii)Last digit is 5 , So one digit can be 4 or 6 as $4^2=16$ and $6^2=36$

(i)Last digit is 1 , So one digit can be 1 or 9 as $1^2=1$ and $9^2=81$

(i)Last digit is 5 , So one digit will be 5 as $5^2=25$

**Question 5. **
The students of a class arranged a gift for the class teacher. Each student contributed as many rupees as the number of students in the class. If the total contribution is Rs 1521, find the strength of the class.

Solution
Here we need to find the square root of the Number 1521

1521= 3 x 3 x 13 x 13

$\sqrt {1521} =3 \times 13=39$

So There are 39 students and each contributed has Rs 39

**Question 6. **
Find the least number which when added to 4529 to make it a perfect square?

Solution
Let us find the square root of 4529 using Long division method

So remainder is 40

Therfore $67^2 < 4529$

Next perfect square would be $68^2=4624$

hence the number to be added = 4624 - 4529 = 95

So addition of 95 to 4529 will make it perfect square

**Question 7. **
Find the least number which must be subtracted from 2361 to make it a perfect square?

Solution
Let us find the square root of 2361 using Long division method

So remainder is 57

Therfore $48^2 < 2361$

Now if we subtract the remainder from main number, it will be perfect square
So subtraction of 57 from 2361 will make it perfect square

**Question 8**
Find the smallest number by which following number must be divided to get a perfect square. Also, find the square root of the perfect square so obtained.

(i)600

(ii)2904

Solution
(i) 600= 2 x 2 x 2 x 3 x 5 x 5

We can see number 2 and 3 is not in pair, So to make a perfect square, it has to be divided by 6

Then number will become=100

Now

100= 2 x 2 x 5 x 5

$\sqrt {100} =2 \times 5 =10$

(ii)2904=2 x 2 x 2 x 3 x 11 x 11

We can see number 2 and 3 is not in pair, So to make a perfect square, it has to be divided by 6

Then number will become=484

Now

484= 2 x 2 x 11 x 11

$\sqrt {484} =2 \times 11 =22$

**Question 9**
Find the value of

$ \sqrt {176 + \sqrt {2401}}$

Solution
$ \sqrt {176 + \sqrt {2401}} = \sqrt {176 + 49} = \sqrt {225} =15$

**Question 10**
Find the square root

(i).0151129

(ii) 83.3569

Solution
Square roots for decimal are found using the same long division method. We put bars on both integral part and decimal part.For integral we move fron the unit's place close to the decimal and move towards left. For decimal part, we start from the decimal and move towards right

(i) By long division method

$\sqrt {.0151129}=.123$

(ii) By long division method

$\sqrt {83.3569}=9.13$

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