# Exercise 5.2 class 8 solution

### Exercise 5.2 class 8 solution-Finding Square Roots

Before we learn how to find square root of a number we must know why we need to find square root of a number? Consider the following situations:

1. We know that area of square = (Side)?

If you are given area of square as 3125cm how will you find its side?

1. Suppose you are given with sides of a rectangle, how will you find the diagonal?
2. Suppose you have a right triangle whose adjacent sides are given, how will you calculate its hypotenuse?

In all above situations and many other similar situations we need to calculate square root at some stage.

Before we learn to calculate square root of a number. Let’s connect square and square root of a number through properties already learnt.

1.Square root of even number is even and that of odd number is odd.

square root(196) = 14 as 14 ^ 2 = 196

and square root(225) = 15 as 15 ^ 2 = 225

2.Ones place digit of square root of any perfect square number ending with 1 is either 1 or 9.

i.e √121-11 as 11^2-=121

and √361 19 as 19^2 = 361

3. Ones place digit of square root of any perfect square number ending with 4 is either 2 or 8.

i.e. √144-12 as 12^2 = 144

and √324-18 as 18^2=324

4.Ones place digit of square root of a perfect square ending with 9 is either 3 or 7

i.e. √169=13 as 13^2 = 169

and √729=27 as 27^2-729

5.Ones place digit of square root of any perfect square ending with 5 is 5.

i.e √225 = 15 as 15^2=225

and √625-25 as 25^2=625

.6.Any number which ends with 2, 3, 7, 8 or have odd number of zeros at its end cannot be a perfect square. Square root of these type numbers will not be a natural number. e.g. 232, 407, 1603, 1008 and 1690 can never be a perfect square numbers

### Exercise 5.2 class 8 solution-Finding square root by repeated subtraction .

As we have learnt earlier that the sum of the first n odd natural numbers is n². That is,every square number can be expressed as sum of consecutive odd natural number starting from 1.

from 1. consider square root (49)

Step (1) 49-1-48 Step (ii) 48-3-45

Step (iii) 45-5-40 Step (iv) 40-7-33

Step (v) 33-9-24 Step (vi) 24-11-13

Step (vii) 13 – 13 = 0

We can write 49 = 1 + 3 + 5 + 7 + 5 + 9 + 11 + 13

Se here we have started subtracting successive odd natural numbers starting from 1 and obtained 0 at 7th step. So square root (49) = 7

### Exercise 5.2 class 8 solution-finding square root through prime factorization method

Finding the square root of a number through prime factorization involves breaking down the number into its prime factors and then taking the square root of those factors. Here’s how you can use the prime factorization method to find the square root of a number:

1. Prime Factorization: First, find the prime factorization of the number you want to find the square root of. This involves breaking down the number into its prime factors. For example, let’s find the square root of 144:
• 144 = 2^4 * 3^2
2. Take the Square Root of Each Prime Factor: Now, take the square root of each prime factor. In this case:
• √(2^4) = 2^2 = 4
• √(3^2) = 3
3. Multiply the Square Roots: Finally, multiply the square roots of the prime factors to find the square root of the original number:
• √144 = 4 * 3 = 12

So, the square root of 144 is 12.

This method works for any positive integer, and it’s particularly useful when you’re dealing with numbers that have prime factors with exponents that are multiples of 2 .

To find the square root of 625 using the prime factorization method, you can follow these steps:

Step 1: Find the prime factorization of 625.

625 = 5^4

Step 2: Take the square root of each prime factor.

√(5^4) = √(5^2 * 5^2)

Step 3: Apply the properties of square roots to simplify.

√(5^2 * 5^2) = 5^2

Step 4: Calculate the square root.

5^2 = 25

So, the square root of 625 is 25.

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5.2-

Exercise 5.4 class 8 solution