Class 8 | NCERT Solution Maths Chapter 6 | Squares and Square Roots
Answers of NCERT Class 8 Maths Chapter 6 – Squares and Square Roots
Exercise 6.1 Page: 96
1. What will be the unit digit of the squares of the following numbers?
i. 81
ii. 272
iii. 799
iv. 3853
v. 1234
we. 26387
vii. 52698
viii. 99880
ix. 12796
x. 55555
Solution:
The unit digit of square of a number having ‘a’ at its unit place ends with a×a.
i. The unit digit of the square of a number having digit 1 as unit’s place is 1.
∴ Unit digit of the square of number 81 is equal to 1.
ii. The unit digit of the square of a number having digit 2 as unit’s place is 4.
∴ Unit digit of the square of number 272 is equal to 4.
iii. The unit digit of the square of a number having digit 9 as unit’s place is 1.
∴ Unit digit of the square of number 799 is equal to 1.
iv. The unit digit of the square of a number having digit 3 as unit’s place is 9.
∴ Unit digit of the square of number 3853 is equal to 9.
v. The unit digit of the square of a number having digit 4 as unit’s place is 6.
∴ Unit digit of the square of number 1234 is equal to 6.
vi. The unit digit of the square of a number having digit 7 as unit’s place is 9.
∴ Unit digit of the square of number 26387 is equal to 9.
vii. The unit digit of the square of a number having digit 8 as unit’s place is 4.
∴ Unit digit of the square of number 52698 is equal to 4.
viii. The unit digit of the square of a number having digit 0 as unit’s place is 01.
∴ Unit digit of the square of number 99880 is equal to 0.
ix. The unit digit of the square of a number having digit 6 as unit’s place is 6.
∴ Unit digit of the square of number 12796 is equal to 6.
x. The unit digit of the square of a number having digit 5 as unit’s place is 5.
∴ Unit digit of the square of number 55555 is equal to 5.
2. The following numbers are obviously not perfect squares. Give reason.
i. 1057
ii. 23453
iii. 7928
iv. 222222
v. 64000
we. 89722
vii. 222000
viii. 505050
Solution:
We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.
i. 1057 ⟹ Ends with 7
ii. 23453 ⟹ Ends with 3
iii. 7928 ⟹ Ends with 8
iv. 222222 ⟹ Ends with 2
v. 64000 ⟹ Ends with 0
vi. 89722 ⟹ Ends with 2
vii. 222000 ⟹ Ends with 0
viii. 505050 ⟹ Ends with 0
3. The squares of which of the following would be odd numbers?
i. 431
ii. 2826
iii. 7779
iv. 82004
Solution:
We know that the square of an odd number is odd and the square of an even number is even.
i. The square of 431 is an odd number.
ii. The square of 2826 is an even number.
iii. The square of 7779 is an odd number.
iv. The square of 82004 is an even number.
4. Observe the following pattern and find the missing numbers. 112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 …….2………1
100000012 = ……………………..
Solution:
We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.
∴ 1000012 = 10000200001
100000012 = 100000020000001
5. Observe the following pattern and supply the missing numbers. 112 = 121
1012 = 10201
101012 = 102030201
10101012 = ………………………
…………2 = 10203040504030201
Solution:
We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And, the square is symmetric about the middle digit. If the middle digit is 4, then the number to be squared is 10101 and its square is 102030201.
So, 10101012 =1020304030201
1010101012 =10203040505030201
6. Using the given pattern, find the missing numbers. 12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + _2 = 212
5 + _ 2 + 302 = 312
6 + 7 + _ 2 = _ 2
Solution:
Given, 12 + 22 + 22 = 32
i.e 12 + 22 + (1×2 )2 = ( 12 + 22 -1 × 2 )2
22 + 32 + 62 =72
∴ 22 + 32 + (2×3 )2 = (22 + 32 -2 × 3)2
32 + 42 + 122 = 132
∴ 32 + 42 + (3×4 )2 = (32 + 42 – 3 × 4)2
42 + 52 + (4×5 )2 = (42 + 52 – 4 × 5)2
∴ 42 + 52 + 202 = 212
52 + 62 + (5×6 )2 = (52+ 62 – 5 × 6)2
∴ 52 + 62 + 302 = 312
62 + 72 + (6×7 )2 = (62 + 72 – 6 × 7)2
∴ 62 + 72 + 422 = 432
7. Without adding, find the sum.
i. 1 + 3 + 5 + 7 + 9
Solution:
Sum of first five odd number = (5)2 = 25
ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19
Solution:
Sum of first ten odd number = (10)2 = 100
iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
Sum of first thirteen odd number = (12)2 = 144
8. (i) Express 49 as the sum of 7 odd numbers.
Solution:
We know, sum of first n odd natural numbers is n2 . Since,49 = 72
∴ 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) Express 121 as the sum of 11 odd numbers. Solution:
Since, 121 = 112
∴ 121 = sum of first 11 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9. How many numbers lie between squares of the following numbers?
i. 12 and 13
ii. 25 and 26
iii. 99 and 100
Solution:
Between n2 and (n+1)2, there are 2n non–perfect square numbers.
i. 122 and 132 there are 2×12 = 24 natural numbers.
ii. 252 and 262 there are 2×25 = 50 natural numbers.
iii. 992 and 1002 there are 2×99 =198 natural numbers.
Exercise 6.2 Page: 98
1. Find the square of the following numbers.
i. 32
ii. 35
iii. 86
iv. 93
v. 71
we. 46
Solution:
i. (32)2
= (30 +2)2
= (30)2 + (2)2 + 2×30×2 [Since, (a+b)2 = a2+b2 +2ab]
= 900 + 4 + 120
= 1024
ii. (35)2
= (30+5 )2
= (30)2 + (5)2 + 2×30×5 [Since, (a+b)2 = a2+b2 +2ab]
= 900 + 25 + 300
= 1225
iii. (86)2
= (90 – 4)2
= (90)2 + (4)2 – 2×90×4 [Since, (a+b)2 = a2+b2 +2ab]
= 8100 + 16 – 720
= 8116 – 720
= 7396
iv. (93)2
= (90+3 )2
= (90)2 + (3)2 + 2×90×3 [Since, (a+b)2 = a2+b2 +2ab]
= 8100 + 9 + 540
= 8649
v. (71) 2
= (70+1 )2
= (70)2 + (1)2 +2×70×1 [Since, (a+b)2 = a2+b2 +2ab]
= 4900 + 1 + 140
= 5041
we. (46) 2
= (50 -4 )2
= (50)2 + (4)2 – 2×50×4 [Since, (a+b)2 = a2+b2 +2ab]
= 2500 + 16 – 400
= 2116
2. Write a Pythagorean triplet whose one member is.
i. 6
ii. 14
iii. 16
iv. 18
Solution:
For any natural number m, we know that 2m, m2–1, m2+1 is a Pythagorean triplet.
i. 2m = 6
⇒ m = 6/2 = 3
m2–1= 32 – 1 = 9–1 = 8
m2+1= 32+1 = 9+1 = 10
∴ (6, 8, 10) is a Pythagorean triplet.
ii. 2m = 14
⇒ m = 14/2 = 7
m2–1= 72–1 = 49–1 = 48
m2+1 = 72+1 = 49+1 = 50
∴ (14, 48, 50) is not a Pythagorean triplet.
iii. 2m = 16
⇒ m = 16/2 = 8
m2–1 = 82–1 = 64–1 = 63
m2+ 1 = 82+1 = 64+1 = 65
∴ (16, 63, 65) is a Pythagorean triplet.
iv. 2m = 18
⇒ m = 18/2 = 9
m2–1 = 92–1 = 81–1 = 80
m2+1 = 92+1 = 81+1 = 82
∴ (18, 80, 82) is a Pythagorean triplet.
Exercise 6.3 Page: 102
1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
i. 9801
ii. 99856
iii. 998001
iv. 657666025
Solution:
i. We know that the unit’s digit of the square of a number having digit as unit’s
place 1 is 1 and also 9 is 1[92=81 whose unit place is 1].
∴ Unit’s digit of the square root of number 9801 is equal to 1 or 9.
ii. We know that the unit’s digit of the square of a number having digit as unit’s
place 6 is 6 and also 4 is 6 [62=36 and 42=16, both the squares have unit digit 6].
∴ Unit’s digit of the square root of number 99856 is equal to 6.
iii. We know that the unit’s digit of the square of a number having digit as unit’s
place 1 is 1 and also 9 is 1[92=81 whose unit place is 1].
∴ Unit’s digit of the square root of number 998001 is equal to 1 or 9.
iv. We know that the unit’s digit of the square of a number having digit as unit’s
place 5 is 5.
∴ Unit’s digit of the square root of number 657666025 is equal to 5.
2. Without doing any calculation, find the numbers which are surely not perfect squares.
i. 153
ii. 257
iii. 408
iv. 441
Solution:
We know that natural numbers ending with the digits 0, 2, 3, 7 and 8 are not perfect square.
i. 153⟹ Ends with 3.
∴, 153 is not a perfect square
ii. 257⟹ Ends with 7
∴, 257 is not a perfect square
iii. 408⟹ Ends with 8
∴, 408 is not a perfect square
iv. 441⟹ Ends with 1
∴, 441 is a perfect square.
3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution:
100
100 – 1 = 99
99 – 3 = 96
96 – 5 = 91
91 – 7 = 84
84 – 9 = 75
75 – 11 = 64
64 – 13 = 51
51 – 15 = 36
36 – 17 = 19
19 – 19 = 0
Here, we have performed subtraction ten times.
∴ √100 = 10
169
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
Here, we have performed subtraction thirteen times.
∴ √169 = 13
4. Find the square roots of the following numbers by the Prime Factorisation Method.
i. 729
ii. 400
iii. 1764
iv. 4096
v. 7744
we. 9604
vii. 5929
viii. 9216
ix. 529
x. 8100
Solution:
729 = 3×3×3×3×3×3×1
⇒ 729 = (3×3)×(3×3)×(3×3)
⇒ 729 = (3×3×3)×(3×3×3)
⇒ 729 = (3×3×3)2
⇒ √729 = 3×3×3 = 27
400 = 2×2×2×2×5×5×1
⇒ 400 = (2×2)×(2×2)×(5×5)
⇒ 400 = (2×2×5)×(2×2×5)
⇒ 400 = (2×2×5)2
⇒ √400 = 2×2×5 = 20
1764 = 2×2×3×3×7×7
⇒ 1764 = (2×2)×(3×3)×(7×7)
⇒ 1764 = (2×3×7)×(2×3×7)
⇒ 1764 = (2×3×7)2
⇒ √1764 = 2 ×3×7 = 42
4096 = 2×2×2×2×2×2×2×2×2×2×2×2
⇒ 4096 = (2×2)×(2×2)×(2×2)×(2×2)×(2×2)×(2×2)
⇒ 4096 = (2×2×2×2×2×2)×(2×2×2×2×2×2)
⇒ 4096 = (2×2×2×2×2×2)2
⇒ √4096 = 2×2×2 ×2×2×2 = 64
7744 = 2×2×2×2×2×2×11×11×1
⇒ 7744 = (2×2)×(2×2)×(2×2)×(11×11)
⇒ 7744 = (2×2×2×11)×(2×2×2×11)
⇒ 7744 = (2×2×2×11)2
⇒ √7744 = 2×2×2×11 = 88
we.
9604 = 62 × 2 × 7 × 7 × 7 × 7
⇒ 9604 = ( 2 × 2 ) × ( 7 × 7 ) × ( 7 × 7 )
⇒ 9604 = ( 2 × 7 ×7 ) × ( 2 × 7 ×7 )
⇒ 9604 = ( 2×7×7 )2
⇒ √9604 = 2×7×7 = 98
vii.
5929 = 7×7×11×11
⇒ 5929 = (7×7)×(11×11)
⇒ 5929 = (7×11)×(7×11)
⇒ 5929 = (7×11)2
⇒ √5929 = 7×11 = 77
viii.
9216 = 2×2×2×2×2×2×2×2×2×2×3×3×1
⇒ 9216 = (2×2)×(2×2) × ( 2 × 2 ) × ( 2 × 2 ) × ( 2 × 2 ) × ( 3 × 3 )
⇒ 9216 = ( 2 × 2 × 2 × 2 × 2 × 3) × ( 2 × 2 × 2 × 2 × 2 × 3)
⇒ 9216 = 96 × 96
⇒ 9216 = ( 96 )2
⇒ √9216 = 96
ix.
529 = 23×23
529 = (23)2
√529 = 23
x.
8100 = 2×2×3×3×3×3×5×5×1
⇒ 8100 = (2×2) ×(3×3)×(3×3)×(5×5)
⇒ 8100 = (2×3×3×5)×(2×3×3×5)
⇒ 8100 = 90×90
⇒ 8100 = (90)2
⇒ √8100 = 90
5. For each of 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
ii. 180
iii. 1008
iv. 2028
v. 1458
we. 768
Solution:
i.
252 = 2×2×3×3×7
= (2×2)×(3×3)×7
Here, 7 cannot be paired.
∴ We will multiply 252 by 7 to get perfect square.
New number = 252 × 7 = 1764
1764 = 2×2×3×3×7×7
⇒ 1764 = (2×2)×(3×3)×(7×7)
⇒ 1764 = 22×32×72
⇒ 1764 = (2×3×7)2
⇒ √1764 = 2×3×7 = 42
ii.
180 = 2×2×3×3×5
= (2×2)×(3×3)×5
Here, 5 cannot be paired.
∴ We will multiply 180 by 5 to get perfect square.
New number = 180 × 5 = 900
900 = 2×2×3×3×5×5×1
⇒ 900 = (2×2)×(3×3)×(5×5)
⇒ 900 = 22×32×52
⇒ 900 = (2×3×5)2
⇒ √900 = 2×3×5 = 30
iii.
1008 = 2×2×2×2×3×3×7
= (2×2)×(2×2)×(3×3)×7
Here, 7 cannot be paired.
∴ We will multiply 1008 by 7 to get perfect square.
New number = 1008×7 = 7056
7056 = 2×2×2×2×3×3×7×7
⇒ 7056 = (2×2)×(2×2)×(3×3)×(7×7)
⇒ 7056 = 22×22×32×72
⇒ 7056 = (2×2×3×7)2
⇒ √7056 = 2×2×3×7 = 84
iv.
2028 = 2×2×3×13×13
= (2×2)×(13×13)×3
Here, 3 cannot be paired.
∴ We will multiply 2028 by 3 to get perfect square. New number = 2028×3 = 6084
6084 = 2×2×3×3×13×13
⇒ 6084 = (2×2)×(3×3)×(13×13)
⇒ 6084 = 22×32×132
⇒ 6084 = (2×3×13)2
⇒ √6084 = 2×3×13 = 78
v.
1458 = 2×3×3×3×3×3×3
= (3×3)×(3×3)×(3×3)×2
Here, 2 cannot be paired.
∴ We will multiply 1458 by 2 to get perfect square. New number = 1458 × 2 = 2916
2916 = 2×2×3×3×3×3×3×3
⇒ 2916 = (3×3)×(3×3)×(3×3)×(2×2)
⇒ 2916 = 32×32×32×22
⇒ 2916 = (3×3×3×2)2
⇒ √2916 = 3×3×3×2 = 54
we.
768 = 2×2×2×2×2×2×2×2×3
= (2×2)×(2×2)×(2×2)×(2×2)×3
Here, 3 cannot be paired.
∴ We will multiply 768 by 3 to get perfect square.
New number = 768×3 = 2304
2304 = 2×2×2×2×2×2×2×2×3×3
⇒ 2304 = (2×2)×(2×2)×(2×2)×(2×2)×(3×3)
⇒ 2304 = 22×22×22×22×32
⇒ 2304 = (2×2×2×2×3)2
⇒ √2304 = 2×2×2×2×3 = 48
6. For each of the following numbers, 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
ii. 2925
iii. 396
iv. 2645
v. 2800
we. 1620
Solution:
i.
252 = 2×2×3×3×7
= (2×2)×(3×3)×7
Here, 7 cannot be paired.
∴ We will divide 252 by 7 to get perfect square. New number = 252 ÷ 7 = 36
36 = 2×2×3×3
⇒ 36 = (2×2)×(3×3)
⇒ 36 = 22×32
⇒ 36 = (2×3)2
⇒ √36 = 2×3 = 6
ii.
2925 = 3×3×5×5×13
= (3×3)×(5×5)×13
Here, 13 cannot be paired.
∴ We will divide 2925 by 13 to get perfect square. New number = 2925 ÷ 13 = 225
225 = 3×3×5×5
⇒ 225 = (3×3)×(5×5)
⇒ 225 = 32×52
⇒ 225 = (3×5)2
⇒ √36 = 3×5 = 15
iii.
396 = 2×2×3×3×11
= (2×2)×(3×3)×11
Here, 11 cannot be paired.
∴ We will divide 396 by 11 to get perfect square. New number = 396 ÷ 11 = 36
36 = 2×2×3×3
⇒ 36 = (2×2)×(3×3)
⇒ 36 = 22×32
⇒ 36 = (2×3)2
⇒ √36 = 2×3 = 6
iv.
2645 = 5×23×23
⇒ 2645 = (23×23)×5
Here, 5 cannot be paired.
∴ We will divide 2645 by 5 to get perfect square.
New number = 2645 ÷ 5 = 529
529 = 23×23
⇒ 529 = (23)2
⇒ √529 = 23
v.
2800 = 2×2×2×2×5×5×7
= (2×2)×(2×2)×(5×5)×7
Here, 7 cannot be paired.
∴ We will divide 2800 by 7 to get perfect square. New number = 2800 ÷ 7 = 400
400 = 2×2×2×2×5×5
⇒ 400 = (2×2)×(2×2)×(5×5)
⇒ 400 = (2×2×5)2
⇒ √400 = 20
we.
1620 = 2×2×3×3×3×3×5
= (2×2)×(3×3)×(3×3)×5
Here, 5 cannot be paired.
∴ We will divide 1620 by 5 to get perfect square. New number = 1620 ÷ 5 = 324
324 = 2×2×3×3×3×3
⇒ 324 = (2×2)×(3×3)×(3×3)
⇒ 324 = (2×3×3)2
⇒ √324 = 18
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:
Let the number of students in the school be, x.
∴ Each student donate Rs.x .
Total many contributed by all the students= x×x=x2 Given, x2 = Rs.2401
x2 = 7×7×7×7
⇒ x2 = (7×7)×(7×7)
⇒ x2 = 49×49
⇒ x = √(49×49)
⇒ x = 49
∴ The number of students = 49
8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution
Let the number of rows be, x.
∴ the number of plants in each rows = x.
Total many contributed by all the students = x × x =x2
Given,
x2 = Rs.2025
x2 = 3×3×3×3×5×5
⇒ x2 = (3×3)×(3×3)×(5×5)
⇒ x2 = (3×3×5)×(3×3×5)
⇒ x2 = 45×45
⇒ x = √45×45
⇒ x = 45
∴ The number of rows = 45 and the number of plants in each rows = 45.
9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution:
L.C.M of 4, 9 and 10 is (2×2×9×5) 180.
180 = 2×2×9×5
= (2×2)×3×3×5
= (2×2)×(3×3)×5
Here, 5 cannot be paired.
∴ we will multiply 180 by 5 to get perfect square.
Hence, the smallest square number divisible by 4, 9 and 10 = 180×5 = 900
10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Solution:
L.C.M of 8, 15 and 20 is (2×2×5×2×3) 120.
120 = 2×2×3×5×2
= (2×2)×3×5×2
Here, 3, 5 and 2 cannot be paired.
∴ We will multiply 120 by (3×5×2) 30 to get perfect square.
Hence, the smallest square number divisible by 8, 15 and 20 =120×30 = 3600
Exercise 6.4 Page: 107
1. Find the square root of each of the following numbers by Division method.
i. 2304
ii. 4489
iii. 3481
iv. 529
v. 3249
we. 1369
vii. 5776
viii. 7921
ix. 576
x. 1024
xi. 3136
xii. 900
Solution:
i.
∴ √2304 = 48
ii.
∴ √4489 = 67
iii.
∴ √3481 = 59
iv.
∴ √529 = 23
v.
∴ √3249 = 57
we.
∴ √1369 = 37
vii.
∴ √5776 = 76
viii.
∴ √7921 = 89
ix.
∴ √576 = 24
x.
∴ √1024 = 32
xi.
∴ √3136 = 56
xii.
∴ √900 = 30
2. Find the number of digits in the square root of each of the following numbers (without any
calculation).64
i. 144
ii. 4489
iii. 27225
iv. 390625
Solution:
i.
∴ √144 = 12
Hence, the square root of the number 144 has 2 digits.
ii.
∴ √4489 = 67
Hence, the square root of the number 4489 has 2 digits.
iii.
√27225 = 165
Hence, the square root of the number 27225 has 3 digits.
iv.
∴ √390625 = 625
Hence, the square root of the number 390625 has 3 digits.
3. Find the square root of the following decimal numbers.
i. 2.56
ii. 7.29
iii. 51.84
iv. 42.25
v. 31.36
Solution:
i.
∴ √2.56 = 1.6
ii.
∴ √7.29 = 2.7
iii.
∴ √51.84 = 7.2
iv.
∴ √42.25 = 6.5
v.
∴ √31.36 = 5.6
4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
i. 402
ii. 1989
iii. 3250
iv. 825
v. 4000
Solution:
i.
∴ √400 = 20
∴ We must subtracted 2 from 402 to get a perfect square.
New number = 402 – 2 = 400
∴ √400 = 20
ii.
∴ We must subtracted 53 from 1989 to get a perfect square. New number = 1989 – 53 = 1936
∴ √1936 = 44
iii.
∴ We must subtracted 1 from 3250 to get a perfect square.
New number = 3250 – 1 = 3249
∴ √3249 = 57
iv.
∴ We must subtracted 41 from 825 to get a perfect square.
New number = 825 – 41 = 784
∴ √784 = 28
∴ We must subtracted 31 from 4000 to get a perfect square. New number = 4000 – 31 = 3969
∴ √3969 = 63
5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 525
(ii) 1750
(iii) 252
(iv)1825
(v) 6412
Solution:
(i)
Here, (22)2 < 525 > (23)2
We can say 525 is ( 129 – 125 ) 4 less than (23)2.
∴ If we add 4 to 525, it will be perfect square. New number = 525 + 4 = 529
∴ √529 = 23
(ii)
Here, (41)2 < 1750 > (42)2
We can say 1750 is ( 164 – 150 ) 14 less than (42)2.
∴ If we add 14 to 1750, it will be perfect square.
New number = 1750 + 14 = 1764
∴√1764 = 42
(iii)
Here, (15)2 < 252 > (16)2
We can say 252 is ( 156 – 152 ) 4 less than (16)2.
∴ If we add 4 to 252, it will be perfect square.
New number = 252 + 4 = 256
∴ √256 = 16
(iv)
Here, (42)2 < 1825 > (43)2
We can say 1825 is ( 249 – 225 ) 24 less than (43)2.
∴ If we add 24 to 1825, it will be perfect square.
New number = 1825 + 24 = 1849
∴ √1849 = 43
(v)
Here, (80)2 < 6412 > (81)2
We can say 6412 is ( 161 – 12 ) 149 less than (81)2.
∴ If we add 149 to 6412, it will be perfect square.
New number = 6412 + 149 = 656
∴ √6561 = 81
6. Find the length of the side of a square whose area is 441 m2.
Solution:
Let the length of each side of the field = a Then, area of the field = 441 m2
⇒ a2 = 441 m2
⇒a = √441 m
∴ The length of each side of the field = a m = 21 m.
7. In a right triangle ABC, ∠B = 90°.
a. If AB = 6 cm, BC = 8 cm, find AC
b. If AC = 13 cm, BC = 5 cm, find AB
Solution:
a.
Given, AB = 6 cm, BC = 8 cm
Let AC be x cm.
∴ AC2 = AB2 + BC2
Hence, AC = 10 cm.
b.
Given, AC = 13 cm, BC = 5 cm
Let AB be x cm.
∴ AC2 = AB2 + BC2
⇒ AC2 – BC2 = AB2
Hence, AB = 12 cm
8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows
and the number of columns remain same. Find the minimum number of plants he needs more for this.
Solution:
Let the number of rows and column be, x.
∴ Total number of row and column= x× x = x2 As per question, x2 = 1000
⇒ x = √1000
Here, (31)2 < 1000 > (32)2
We can say 1000 is ( 124 – 100 ) 24 less than (32)2.
∴ 24 more plants are needed.
9. 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.
Solution:
Let the number of rows and column be, x.
∴ Total number of row and column= x × x = x2 As per question, x2 = 500
x = √500
Hence, 16 children would be left out in the arrangement
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