Class 8 Maths Square Square Root Examples and assignment

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Square /Square  Root Examples and assignment

1 - What will be the unit digit of the squares of the following numbers?

(i) 81

Answer: 1 Since, 1² ends up having 1 as the digit at unit’s place so 81² will have 1 at unit’s place.

(ii) 272

Asnwer: 4 Since, 2² = 4, therefore, square of 272 will have 2 at it's unit place.

 2. The following numbers are obviously not perfect squares. Give reason.

(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050

Answer: (i), (ii), (iii), (iv), (vi) don’t have any of the 0, 1, 4, 5, 6, or 9 at unit’s place, so they are not be perfect squares.

(v), (vii) and (viii) don’t have even number of zeroes at the end so they are not perfect squares.

3. Write a Pythagorean triplet whose one member is:

(i) 6

Solution : As we know 2m, m ² + 1 and m² - 1 form a Pythagorean triplet for any number, m > 1.

Let us assume 2m = 6

Therefore, m = 3

And, m² + 1 = 3 ² + 1= 9 + 1 = 10

And, m ² - 1 = 3 ² - 1 = 9 - 1 = 8

Test: 6 ² + 8 ² = 36 + 64 = 100 = 10²

Hence, the triplet is 6, 8, and 10 Answer

There are 16 primitive Pythagorean triples with c ≤ 100:

(3, 4, 5 )	(5, 12, 13)	(8, 15, 17)	(7, 24, 25)
(20, 21, 29)	(12, 35, 37)	( 9, 40, 41)	(28, 45, 53)
(11, 60, 61)	(16, 63, 65)	(33, 56, 65)	(48, 55, 73)
(13, 84, 85)	(36, 77, 85)	(39, 80, 89)	(65, 72, 97)

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

(i) 9801

Answer: 1 and 9.

Explanation: Since 1² and 9² give 1 at unit’s place, so these are the possible values of unit digit of the square root.

5. For the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252

Solution:

By prime factorisation we get,

252 = 2 x 2 x 3 x 3 x 7

Here, 2 and 3 are in pairs but 7 needs a pair. Thus, 7 can become pair after multiplying 252 with 7.

So, 252 will become a perfect square when multiplied by 7.

Thus, Answer = 7

6. For the following number, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

(i) 252

Solution:

By prime factorisation of 252, we get

252 = 2 x 2 x 3 x 3 x 7

Here, 2 and 3 are in pair, but 7 has no pair, which can be eliminated after dividing 768 by 7.

Hence, 252 needs to be divided by 7 to become a perfect square

Thus, Answer = 7

7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Solution:

We need to calculate the square root of 2401 to get the solution.

By prime factorisation of 2401, we get

2401 = 7 x 7 x 7 x 7

There are 49 students, each contributing 49 rupees

Thus, Answer = 49

8. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution: Let us find LCM of 4, 9 and 10

4 = 2 x 2

9 = 3 x 3

10 = 5 x 2

So, LCM = 2 ² x 3 ² x 5 = 180

Now the LCM gives us a clue that if 180 is multiplied by 5 then it will become a perfect square.

The Required number = 180 x 5 = 900

Examples on square root of a perfect square by using the long division method

1. Find the square root of 784 by the long-division method.
Solution:
Marking periods and using the long-division method,

Therefore, √784 = 28

2. Evaluate: √10609.

Solution:

Marking periods and using the long-division method,

Therefore, √10609 = 103

3. What least number must be subtracted from 7250 to get a perfect square? Also, find the square root of this perfect square.

Solution:

Let us try to find the square root of 7250.

This shows that (85)² is less than 7250 by 25.

So, the least number to be subtracted from 7250 is 25.

Required perfect square number = (7250 - 25) = 7225

And, √7225 = 85.

4. Find the greatest number of four digits which is a perfect square.
Solution
Greatest number of four digits = 9999.
Let us try to find the square root of 9999.

This shows that (99)² is less than 9999 by 198.

So, the least number to be subtracted is 198.

Hence, the required number is (9999 - 198) = 9801.

Unsloved Questions

1.     What least number must be added to 5607 to make the sum a perfect square? Find this perfect square and its square root

2.     Find the least number of six digits which is a perfect square. Find the square root of this number.

3.     Find the smallest number by which 1100 must be divided so that the quotient is a perfect square.

4.     Write a Pythagorean triplet whose one member is:18

5.     There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.

Ans

1.     18

2.     489, 317

3.     11

4.     18, 82, 80

5.     16 children will be left out in the arrangement.