Squares and Square Roots Class 8 Extra Questions with Answers

These Class 8 Maths Extra Questions Chapter 6 Squares and Square Roots will help students prepare well for the exams.

Class 8 Maths Chapter 6 Extra Questions Squares and Square Roots

Squares and Square Roots Extra Questions Class 8

Question 1.
Represent
i) 64 as sum of 8 odd numbers.
ii) 100 as sum of 10 odd numbers.
Solution:
i) 64 as sum of 8 odd numbers
1 + 3 + 5 + 7 – 9 + 11 + 13 + 15 = 8² = 64

ii) 100 as sum of 10 odd numbers
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
= 100 = (10)²

Question 2.
Find the smallest positive integer that should be added to 1500 to make it a perfect quare.
Solution:
1500

Remainder = 21
If we add ’21’ to 1500. we get a perfect square.
∴ The required smallest positive integer that should be added to 1500 to make it a perfect square = 21
(OR)
30² = 900, 40² = 1600
30² < 1500 < 40²
35² = 1225
∴ 35² < 1500 < 40²
1500 is nearer to 40²
39² = 1521
38² = 1444
∴ 38² < 1500 < 39²
∴ The smallest positive integer, that should be added to
1500 = 1521 – 1500 = 21

Squares and Square Roots Class 8 Extra Questions

Question 1.
Find the square of the following numbers without actual multiplication.
i) 39
ii) 42
Solution:
i) 39² – (30 + 9)² = 30(30 + 9) + 9(30 + 9)
= 30² + 30 × 9 + 9 × 30 + 9² = 900 + 270 + 270 + 81 = 1521

ii) 42² = (40 + 2)²2 = 40(40 + 2) + 2(40 + 2)
= 40² + 40 × 2 + 2 × 40 + 2² = 1600 + 80 + 80 + 4 = 1764

Some other interesting patterns in squares:
(A5)² = [A × (A + 1)] hundreds + 25
for example
put A = 2 then (25)² = [2 × (2 + 1)] × 100 + 25 = (2 × 3 × 100) + 25 = 625 ……………. (1)
Put A = 4 then (45)² = 4 × (4 + 1) × 100 + 25 = 4 × 5 × 100 + 25 = 2000 + 25 45² = 2025 ……………… (2)
Put A = 8 then (85)² = 8 × 9 × 100 + 25 = 7200 + 25 = 7225
∴ 85² = 7225 ……………… (3)
Put A = 9 then
(95)² = (9 × 10 × 100) + 25 – 9025
95² = 9025 …………………. (4)
Proof : A 5 = 10 (A) + 1 (5) = (10 A+ 5)
(A5)² = (10 A + 5) (10 A + 5)
= 10 A (10 A + 5) + 5(10 A + 5)
= 10 A. (10 A) + 10 A. 5 + 5 (10 A) + 5.5
= 100 AA + 50 A + 50 A + 25
= 100 A. A + 110 A + 25
= 100 A (A + 1) + 25
= 100A (A + 1) + 25
(A5)² = A (A + 1) (100) + 25

Question 2.
Write a Pythagorean triplet whose smallest member is 8.
Solution:
We can get Pythagorean triplets by using general form 2m, m² – 1, m² + 1.
Let us first take m² – 1 = 8
So, m² = 8 + 1 = 9
which gives m = 3
Therefore, 2m = 6 and m² + 1 = 10
The triplet is thus 6, 8,10. But 8 is not the smallest member of this.
So, let us try 2m = 8
then m = 4
We get m² – 1 = 16 – 1 = 15
and m² + 1 = 16 + 1 = 17
The triplet is 8, 15, 17 with 8 as the smallest member.

Question 3.
Find a Pythagorean triplet in which one member is 12.
Solution:
If we take m² – 1 = 12
Then, m² = 12 + 1 = 13
Then the value of m will not be an integer.
So, we try to take m² + 1 = 12. Again m² =
11 will not give an integer value for m.
So, let us take 2m = 12
then m = 6
Thus, m² – 1 = 36 – 1 = 35 and
m² + 1 = 36 + 1 = 37
Therefore, the required triplet is 12, 35, 37.
Note: All Pythagorean triplets may not be obtained using this form. For example another triplet 5, 12, 13 also has 12 as a member.

Question 4.
Find the square root of 6400.
Solution:
Write
6400 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
Therefore \(\sqrt{6400}\) = 2 × 2 × 2 × 2 × 5 = 80

Question 5.
Is 90 a perfect square ?
Solution:
We have 90 = 2 × 3 × 3 × 5

The prime factors 2 and 5 do not occur in pairs. Therefore, 90 is not a perfect square.
That 90 is not a perfect square can also be seen from the fact that it has only one zero.

Question 6.
Is 2352 a perfect square? If not, find the smallest multiple of 2352 which is a perfect square, Find the square root of the new number.
Solution:
We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.

If 3 gets a pair then the number will become perfect square. So, we multiply
2352 by 3 to get,
2352 × 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 × 3 = 7056 is a perfect square. Thus the required smallest multiple of 2352 is 7056 which is a perfect square.
And, \(\sqrt{7056}\) = 2 × 2 × 3 × 7 = 84

Question 7.
Find the smallest number by which 9408 must be divided so that the quotient is a perfect square. Find the square root of the quotient.
Solution:
We have,
9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7
If we divide 9408 by the factor 3. then 9408 + 3 = 3136
= 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7
which is a perfect square.
Therefore, the required smallest number is 3.
And, \(\sqrt{3136}\) = 2 × 2 × 2 × 7 = 56.

Question 8.
Find the smallest square number which is divisible by each of the numbers 6, 9 and 15.
Solution:
This has to be done in two steps. First find the smallest common multiple and then find the square number needed. The least number divisible by each one of 6. 9 and 15 is their LCM.

The LCM of 6,9 and 15 is 2 × 3 × 3 × 5 = 90.
Prime factorisation of 90 is 90 = 2 × 3 × 3 × 5.
We see that prime factors 2 and 5 are not in pairs. Therefore 90 is not a perfect square.

In order to get a perfect square, each factor of 90 must be paired. So we need to make pairs of 2 and 5. Therefore, 90 should be multiplied by 2 × 5, i.e., 10. Hence, the required square number is 90 × 10 = 900.

Question 9.
Find the square root of: (i) 729 (ii) 1296
Solution:

Question 10.
Find the least number that must be subtracted from 5607 so square. Also find the square root of the perfect square.
Solution:

Let us try to find \(\sqrt{5607}\) by long division method.
We get the remainder 131. It shows that 74² is less than 5607 by 131.
This means if we subtract the remainder from the number, we get a perfect square. Therefore, the required perfect square is 5607 – 131 = 5476.
And, \(\sqrt{5476}\) = 74.

Question 11.
Find the greatest 4 – digit number which is a perfect square.
Solution:

Greatest number of 4 – digits = \(\sqrt{9999}\). We find 79999 by long division method. The remainder is 198. This shows 99² is less than 9999 by 198.
This means if we subtract the remainder from the number, we get a perfect square. Therefore, the required perfect square is 9999 – 198 = 9801.
And, \(\sqrt{9999}\) = 99

Question 12.
Find the least number that must be added to 1300 so as to get a perfect square. Also find the square root of the perfect square.
Solution:

We find \(\sqrt{1300}\) by long division method. The remainder is 4.
This shows that 36² < 1300.
Next perfect square number is 37² = 1369.
Hence, the number to be added is 37² – 1300 = 1369 – 1300 = 69.

Question 13.
Find the square root of 12.25.
Solution:

Question 14.
Area of a square plot is 2304 m². Find the side of the square plot.
Solution:

Area of square plot = 2304 m²
Therefore, side of the square plot = \(\sqrt{2304}\) m
We find that, \(\sqrt{2304}\) = 48
Thus, the side of the square plot is 48 m.

Question 15.
There are 2401 students in a school. P.T. teacher wants them to stand in rows and columns such that the number of rows is equal to the number of columns. Find the number of rows.
Solution:

Let the number of rows be x 4
So, the num ber of columns = x
Therefore, number of students = x × x = x²
Thus, x² = 2401 gives x = \(\sqrt{2401}\) = 49
The number of rows = 49.