Class 8 Maths Chapter 5 Question Answers - Squares and Square Roots

Q1: Using the prime factorisation method, find which of the following numbers are not perfect squares.
(i) 768
(ii) 1296
Ans:

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, 3 is not in pair.
768 is not a perfect square.

1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
Here, there is no number left to make a pair.
1296 is a perfect square.

Q2: Find the square root of the following using successive subtraction of odd numbers starting from 1.
(i) 169
(ii) 81
(iii) 225
Ans: 
(i) 169 – 1 = 168,
168 – 3 = 165,
165 – 5 = 160,
160 – 7 = 153,
153 – 9 = 144,
144 – 11 = 133,
133 – 13 = 120,
120 – 15 = 105,
105 – 17 = 88,
88 – 19 = 69,
69 – 21 = 48,
48 – 23 = 25,
25 – 25 = 0
We have subtracted odd numbers 13 times to get 0.
√169 = 13
(ii) 81 – 1 = 80,
80 – 3 = 77,
77 – 5 = 72,
72 – 7 = 65,
65 – 9 = 56,
56 – 11 = 45,
45 – 13 = 32,
32 – 15 = 17,
17 – 17 = 0
We have subtracted 9 times to get 0.
√81 = 9
(iii) 225 – 1 = 224,
224 – 3 = 221,
221 – 5 = 216,
216 – 7 = 209,
209 – 9 = 200,
200 – 11 = 189,
189 – 13 = 176,
176 – 15 = 161,
161 – 17 = 144,
144 – 19 = 125,
125 – 21 = 104,
104 – 23 = 81,
81 – 25 = 56,
56 – 27 = 29,
29 – 29 = 0
We have subtracted 15 times to get 0.
√225 = 15

Q3: Find the least square number which is divisible by each of the number 4, 8 and 12.
Ans:
LCM of 4, 8, 12 is the least number divisible by each of them.
LCM of 4, 8 and 12 = 24
24 = 2 × 2 × 2 × 3
To make it perfect square multiply 24 by the product of unpaired numbers, i.e., 2 × 3 = 6
Required number = 24 × 6 = 144


Q4: What is the least number that must be subtracted from 3793 so as to get a perfect square? Also, find the square root of the number so obtained.
Ans:
First, we find the square root of 3793 by division method.

Here, we get a remainder 72
612 < 3793
Required perfect square number = 3793 – 72 = 3721 and √3721 = 61

Q5: Simplify: √900 + √0.09 + √0.000009
Ans: 
We know that √(ab) = √a × √b
√900 = √(9 × 100) = √9 × √100 = 3 × 10 = 30
√0.09 = √(0.3 × 0.3) = 0.3
√0.000009 = √(0.003 × 0.003) = 0.003
√900 + √0.09 + √0.000009 = 30 + 0.3 + 0.003 = 30.303

Q6: Which of the following triplets are Pythagorean?
(i) (14, 48, 50)
(ii) (18, 79, 82)
Ans: 
We know that 2m, m² – 1 and m² + 1 make Pythagorean triplets.
(i) For (14, 48, 50),
Put 2m =14 ⇒ m = 7
m² – 1 = (7)² – 1 = 49 – 1 = 48
m² + 1 = (7)² + 1 = 49 + 1 = 50
Hence (14, 48, 50) is a Pythagorean triplet.
(ii) For (18, 79, 82)
Put 2m = 18 ⇒ m = 9
m² – 1 = (9)² – 1 = 81 – 1 = 80
m² + 1 = (9)² + 1 = 81 + 1 = 82
Hence (18, 79, 82) is not a Pythagorean triplet.

Q7: Find the square rootofthe following using prime factorisation
(i) 441
(ii) 2025
(iii) 7056
(iv) 4096
Ans:
(i) 441 = 3 × 3 × 7 × 7
√441 = 3 × 7 = 21

(ii) 2025 = 3 × 3 × 3 × 3 × 5 × 5
√2025 = 3 × 3 × 5 = 45

(iii) 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
√7056 = 2 × 2 × 3 × 7 = 84


Q8: Find the square roots of the following decimal numbers
(i) 1056.25
(ii) 10020.01
Ans: (i)



Q9: Fill in the blanks:
(а) The perfect square number between 60 and 70 is …………
(b) The square root of 361 ends with digit …………..
(c) The sum of first n odd numbers is …………
(d) The number of digits in the square root of 4096 is ………..
(e) If (-3)² = 9, then the square root of 9 is ……….
(f) Number of digits in the square root of 1002001 is …………
(g) Square root of 366/25 is ………..
(h) The value of √(63 × 28) = …………
Ans: 
(a) 64
(b) 9
(c) n²
(d) 2
(e) ±3
(f) 4
(g) 6/25
(h) 42