ML Aggarwal Solutions for Class 8 Maths Chapter 3 Squares and Square Roots help students understand the types of methods to be followed in solving problems effortlessly. The solutions can be used by the students, irrespective of time constraints, to boost their annual exam preparation. Students can download ML Aggarwal Solutions for Class 8 Maths Chapter 3 Squares and Square Roots free PDF from the link given below.
Chapter 3 has problems on determining the squares and square roots of fractions, natural numbers and decimals. The various steps are explained in simple language to make it easy for the students during revision. The solutions are designed with the aim of eliminating the fear of students and improving confidence in answering complex problems in a shorter duration.
ML Aggarwal Solutions for Class 8 Maths Chapter 3: Squares and Square Roots Download PDF
Access ML Aggarwal Solutions for Class 8 Maths Chapter 3: Squares and Square Roots
Exercise 3.1
1. Which of the following natural numbers are perfect squares? Give reasons in support of your answer.
(i) 729
(ii) 5488
(iii) 1024
(iv) 243
Solution:
(i) 729
We know that
It can be written as
729 = 3 × 3 × 3 × 3 × 3 × 3
Here
729 is the product of pairs of equal prime factors
Therefore, 729 is a perfect square.
(ii) 5488
We know that
It can be written as
5488 = 2 × 2 × 2 × 2 × 7 × 7 × 7
Here
After pairing the same prime factors, one factor 7 is left unpaired.
Therefore, 5488 is not a perfect square.
(iii) 1024
We know that
It can be written as
1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Here
After pairing the same prime factors, there is no factor left.
Therefore, 1024 is a perfect square.
(iv) 243
We know that
It can be written as
243 = 3 × 3 × 3 × 3 × 3
Here
After pairing the same prime factors, factor 3 is left unpaired.
Therefore, 243 is not a perfect square.
2. Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.
(i) 1296
(ii) 1764
(iii) 3025
(iv) 3969
Solution:
(i) 1296
We know that
It can be written as
1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
Here
After pairing the same prime factors, no factor is left.
Therefore, 1296 is a perfect square of 2 × 2 × 3 × 3 = 36.
(ii) 1764
We know that
It can be written as
1764 = 2 × 2 × 3 × 3 × 7 × 7
Here
After pairing the same factors, no factor is left.
Therefore, 1764 is a perfect square of 2 × 3 × 7 = 42.
(iii) 3025
We know that
It can be written as
3025 = 5 × 5 × 11 × 11
Here
After pairing the same prime factors, no factor is left.
Therefore, 3025 is a perfect square of 5 × 11 = 55.
(iv) 3969
We know that
It can be written as
3969 = 3 × 3 × 3 × 3 × 7 × 7
Here
After pairing the same prime factors, no factor is left.
Therefore, 3969 is a perfect square of 3 × 3 × 7 = 63.
3. Find the smallest natural number by which 1008 should be multiplied to make it a perfect square.
Solution:
We know that
It can be written as
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
Here
After pairing the same kind of prime factors, one factor, 7 is left.
Now, multiplying 1008 by 7
We get a perfect square
Therefore, the required smallest number is 7.
4. Find the smallest natural number by which 5808 should be divided to make it a perfect square. Also, find the number whose square is the resulting number.
Solution:
We know that
It can be written as
5808 = 2 × 2 × 2 × 2 × 3 × 11 × 11
Here
After pairing the same kind of prime factors, factor 3 is left.
Now dividing the number by 3, we get a perfect square.
Therefore, the square root of the resulting number is 2 × 2 × 11 = 44.
Exercise 3.2
1. Write five numbers which you can decide by looking at their one’s digit that they are not square numbers. Solution:
We know that
A number which ends with the digits 2, 3, 7 or 8 at its unit places is not a perfect square.
For example, 111, 372, 563, 978, and 1282 are not square numbers.
2. What will be the unit digit of the squares of the following numbers?
(i) 951
(ii) 502
(iii) 329
(iv) 643
(v) 5124
(vi) 7625
(vii) 68327
(viii) 95628
(ix) 99880
(x) 12796
Solution:
(i) 951
The unit digit of the square is 1.
(ii) 502
The unit digit of the square is 4.
(iii) 329
The unit digit of the square is 1.
(iv) 643
The unit digit of the square is 9.
(v) 5124
The unit digit of the square is 6.
(vi) 7625
The unit digit of the square is 5.
(vii) 68327
The unit digit of the square is 9.
(viii) 95628
The unit digit of the square is 4.
(ix) 99880
The unit digit of the square is 0.
(x) 12796
The unit digit of the square is 6.
3. The following numbers are obviously not perfect. Give reason.
(i) 567
(ii) 2453
(iii) 5298
(iv) 46292
(v) 74000
Solution:
In the given numbers,
If the square of a number does not have 2, 3, 7, 8 or 0 as its unit digit, the squares 567, 2453, 5208, 46292 and 74000 cannot be the perfect squares as they have 7, 2, 8, 2 digits at the unit place.
4. The square of which of the following numbers would be an odd number or an even number? Why?
(i) 573
(ii) 4096
(iii) 8267
(iv) 37916
Solution:
We know that
The square of an odd number is odd, and a square of an even number is even.
So 573 and 8262 are odd numbers, and their squares will be odd numbers.
4096 and 37916 are even numbers, and their square will also be even numbers.
5. How many natural numbers lie between squares of the following numbers?
(i) 12 and 13
(ii) 90 and 91
Solution:
(i) We know that
No. of natural numbers between the squares of 12 and 13 = (132 – 122) – 1
By further calculation
= (13 + 12 – 1)
So we get
= 25 – 1
= 24
(ii) We know that
No. of natural numbers between the squares of 90 and 91 = (912 – 902) – 1
By further calculation
= (91 + 90 – 1)
So we get
= 181 – 1
= 180
6. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29
Solution:
(i) We know that
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = n2
Here n = 8
So the sum = 82 = 64
(ii) We know that
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = n2
Here, n = 15
So, the sum = 152 = 225
7. (i) Express 64 as the sum of 8 odd numbers.
(ii) 121 as the sum of 11 odd numbers.
Solution:
(i) We know that
64 as the sum of 8 odd numbers = 82 = n2
Here n = 8
It can be written as
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
(ii) We know that
121 as the sum of 11 odd numbers = 112 = n2
Here n = 11
It can be written as
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
8. Express the following as the sum of two consecutive integers.
(i) 192
(ii) 332
(iii) 472
Solution:
We know that
n2 = (n2 – 1)/ 2 + (n2 + 1)/ 2 is the sum of two consecutive integers when n is odd
(i) 192 = (192 – 1)/ 2 + (192 + 1)/ 2
We know that
192 = 361
By further calculation
= (361 – 1)/ 2 + (361 + 1)/ 2
So we get
= 180 + 181
(ii) 332 = (332 – 1)/ 2 + (332 + 1)/ 2
We know that
332 = 1089
By further calculation
= (1089 – 1)/ 2 + (1089 + 1)/ 1
So we get
= 1088/2 + 1090/2
= 544 + 545
(iii) 472 = (472 – 1)/ 2 + (472 + 1)/ 2
We know that
472 = 2209
By further calculation
= (2209 – 1)/ 2 + (2209 + 1)/ 1
So we get
= 2208/2 + 2210/2
= 1104 + 1105
9. Find the squares of the following numbers without actual multiplication:
(i) 31
(ii) 42
(iii) 86
(iv) 94
Solution:
We know that
(a + b)2 = a2 + 2ab + b2
(i) 312 = (30 + 1)2
We can write it as
= 302 + 2 × 30 × 1 + 12
By further calculation
= 900 + 60 + 1
= 961
(ii) 422 = (40 + 2)2
We can write it as
= 402 + 2 × 40 × 2 + 22
By further calculation
= 1600 + 160 + 4
= 1764
(iii) 862 = (80 + 6)2
We can write it as
= 802 + 2 × 80 × 6 + 62
By further calculation
= 6400 + 960 + 36
= 7396
(iv) 942 = (90 + 4)2
We can write it as
= 902 + 2 × 90 × 4 + 42
By further calculation
= 8100 + 720 + 16
= 8836
10. Find the squares of the following numbers containing 5 in unit’s place:
(i) 45
(ii) 305
(iii) 525
Solution:
(i) 452 = n52
It can be written as
= n (n + 1) hundred + 52
Substituting the values
= 4 × 5 hundred + 25
By further calculation
= 2000 + 25
= 2025
(ii) 3052 = (30 × 31) hundred + 25
By further calculation
= 93000 + 25
= 93025
(iii) 5252 = (52 × 53) hundred + 25
By further calculation
= 275600 + 25
= 275625
11. Write a Pythagorean triplet whose one number is
(i) 8
(ii) 15
(iii) 63
(iv) 80
Solution:
(i) 8
Take n = 8
So, the triplet will be 2n, n2 – 1, n2 + 1
Here
If 2n = 8, then n = 8/2 = 4
Substituting the values
n2 – 1 = 42 – 1 = 16 – 1 = 15
n2 + 1 = 42 + 1 = 16 + 1 = 17
Therefore, the triplets are 8, 15 and 17.
(ii) 15
Take 2n = 15
So n = n/2 is not possible
n2 – 1 = 15
By further calculation
n2 = 15 + 1 = 16 = 42
n = 4
So we get
2n = 2 × 4 = 8
n2 – 1 = 15
n2 + 1 = 42 + 1 = 16 + 1 = 17
Therefore, the triplets are 8, 15 and 17.
(iii) 63
Take n2 – 1 = 63
By further calculation
n2 = 63 + 1 = 64 = 82
So, n = 8
Here
2n = 2 × 8 = 16
n2 – 1 = 63
n2 + 1 = 82 + 1 = 64 + 1 = 65
Therefore, the triplets are 16, 63 and 65.
(iv) 80
Take 2n = 80
n = 80/2 = 40
Here
n2 – 1 = 402 – 1 = 1600 – 1 = 1599
n2 + 1 = 402 + 1 = 1600 + 1 = 1601
Therefore, the triplets are 80, 1599 and 1601.
12. Observe the following pattern and find the missing digits:
212 = 441
2012 = 40401
20012 = 4004001
200012 = 4—4—1
2000012 = ——
Solution:
212 = 441
2012 = 40401
20012 = 4004001
200012 = 400040001
2000012 = 40000400001
13. Observe the following pattern and find the missing digits:
92 = 81
992 = 9801
9992 = 998001
99992 = 99980001
999992 = 9—8—01
9999992 = 9—0—1
Solution:
92 = 81
992 = 9801
9992 = 998001
99992 = 99980001
999992 = 9999800001
9999992 = 9999998000001
14. Observe the following pattern and find the missing digits:
72 = 49
672 = 4489
6672 = 444889
66672 = 44448889
666672 = 4—-8—–9
6666672 = 4—-8—-8-
Solution:
72 = 49
672 = 4489
6672 = 444889
66672 = 44448889
666672 = 4444488889
6666672 = 4444448888889
Exercise 3.3
1. By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not. If the number is a perfect square, then find its square root:
(i) 121
(ii) 55
(iii) 36
(iv) 90
Solution:
(i) We know that
Square root of 121
121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
So the square root of 121 is 11
Hence, 121 is a perfect square.
(ii) We know that
Square root of 55
55 – 1 = 54
54 – 3 = 51
51 – 5 = 46
46 – 7 = 39
39 – 9 = 30
30 – 11 = 19
19 – 13 = 6
6 – 15 = – 9 is not possible
Hence, 55 is not a perfect square.
(iii) We know that
Square root of 36
36 – 1 = 35
35 – 3 = 32
32 – 5 = 27
27 – 7 = 20
20 – 9 = 11
11 – 11 = 0
Hence, 36 is a perfect square, and its square root is 6.
(iv) We know that
Square root of 90
90 – 1 = 89
89 – 3 = 86
86 – 5 = 81
81 – 7 = 74
74 – 9 = 65
65 – 11 = 54
54 – 13 = 41
41 – 15 = 26
26 – 17 = 9
9 – 19 = – 10, which is not possible
Hence, 90 is not a perfect square.
2. Find the square roots of the following numbers by the prime factorisation method:
(i) 784
(ii) 441
(iii) 1849
(iv) 4356
(v) 6241
(vi) 8836
(vii) 8281
(viii) 9025
Solution:
(i) We know that
Square root of 784
It can be written as
So we get
= 2 × 2 × 7
= 28
(ii) We know that
Square root of 441
It can be written as
So we get
= 3 × 7
= 21
(iii) We know that
Square root of 1849
It can be written as
= 43
(iv) We know that
Square root of 4356
It can be written as
So we get
= 2 × 3 × 11
= 66
(v) We know that
Square root of 6241
It can be written as
= 79
(vi) We know that
Square root of 8836
It can be written as
So we get
= 2 × 47
= 94
(vii) We know that
Square root of 8281
It can be written as
So we get
= 7 × 13
= 91
(viii) We know that
Square root of 9025
It can be written as
So we get
= 5 × 19
= 95
3. Find the square roots of the following numbers by the prime factorisation method:
(i) 9 67/121
(ii) 17 13/36
(iii) 1.96
(iv) 0.0064
Solution:
(i) 9 67/121 = (9 × 121 + 67)/ 121
By further calculation
= (1089 + 67)/ 121
= 1156/121
By squaring, we get
We know that
It can be written as
So we get
= (2 × 17)/ 11
= 34/11
= 3 1/11
(ii) 17 13/36 = (17 × 36 + 13)/ 36
By further calculation
= (612 + 13)/ 36
= 625/36
By squaring we get
We know that
It can be written as
So we get
= (5 × 5)/ (2 × 3)
= 25/6
= 4 1/6
(iii) 1.96 = 196/100
By squaring, we get
We know that
It can be written as
So we get
= (2 × 7)/ (2 × 5)
= 14/10
= 1.4
(iv) 0.0064 = 64/10000
By squaring, we get
We know that
It can be written as
So we get
= (2 × 2 × 2)/ (2 × 2 × 5 × 5)
= 8/100
= 0.08
4. For each of the following numbers, find the smallest natural number by which it should be multiplied so as to get a perfect square. Also, find the square root of the square number so obtained:
(i) 588
(ii) 720
(iii) 2178
(iv) 3042
(v) 6300
Solution:
(i) 588 = 2 × 2 × 3 × 7 × 7
We know that
By pairing the same kind of factors, one factor, 3 is left unpaired.
So to make it a pair, we must multiply it by 3
Required least number = 3
Square root of 588 × 3 = 1764
Here
2 × 3 × 7 = 42
(ii) 720 = 2 × 2 × 2 × 2 × 3 × 3 × 5
We know that
By pairing the same kind of factors, one factor, 5 is left unpaired.
So to make it a pair, we must multiply it by 5
Required least number = 5
Square root of 720 × 5 = 3600
Here
2 × 2 × 3 × 5 = 60
(iii) 2178 = 2 × 3 × 3 × 11 × 11
We know that
By pairing the same kind of factors, one factor, 2 is left unpaired.
So to make it a pair, we must multiply it by 2
Required least number = 2
Square root of 2178 × 2 = 4356
Here
2 × 3 × 11 = 66
(iv) 3042 = 2 × 3 × 3 × 13 × 13
We know that
By pairing the same kind of factors, one factor 2 is left unpaired
So to make it a pair, we must multiply it by 2
Required least number = 2
Square root of 3042 × 2 = 6084
Here
2 × 3 × 13 = 78
(v) 6300 = 2 × 2 × 3 × 3 × 5 × 5 × 7
We know that
By pairing the same kind of factors, one factor 7 is left unpaired
So, to make it a pair, we must multiply it by 7
Required least number = 7
Square root of 6300 × 7 = 44100
Here
2 × 3 × 5 × 7 = 210
5. For each of the following numbers, find the smallest natural number by which it should be divided so that this quotient is a perfect square. Also, find the square root of the square number so obtained:
(i) 1872
(ii) 2592
(iii) 3380
(iv) 16224
(v) 61347
Solution:
(i) 1872 = 2 × 2 × 2 × 2 × 3 × 3 × 13
It can be written as
By pairing the same kind of factors, one factor 13 is left unpaired
Required least number = 13
The number 1872 should be divided by 13 so that the resultant number will be a perfect square
Resultant number = 1872 ÷ 13 = 144
Square root = 2 × 2 × 3 = 12
(ii) 2592 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
It can be written as
By pairing the same kind of factors, one factor 2 is left unpaired
Required least number = 2
The number 2592 should be divided by 2 so that the resultant number will be a perfect square
Resultant number = 2592 ÷ 2 = 1296
Square root = 2 × 2 × 3 × 3 = 36
(iii) 3380 = 2 × 2 × 5 × 13 × 13
It can be written as
By pairing the same kind of factors, one factor 5 is left unpaired
Required least number = 5
The number 3380 should be divided by 5 so that the resultant number will be a perfect square
Resultant number = 3380 ÷ 5 = 676
Square root = 2 × 13 = 26
(iv ) 16224 = 2 × 2 × 2 × 2 × 2 × 3 × 13 × 13
It can be written as
By pairing the same kind of factors, two factors 2 and 3 is left unpaired
Required least number = 2 × 3 = 6
The number 16224 should be divided by 6 so that the resultant number will be a perfect square
Resultant number = 16224 ÷ 6 = 2704
Square root = 2 × 2 × 13 = 52
(v) 61347 = 3 × 11 × 11 × 13 × 13
It can be written as
By pairing the same kind of factors, one factor 3 is left unpaired
Required least number = 3
The number 61347 should be divided by 3 so that the resultant number will be a perfect square
Resultant number = 61347 ÷ 3 = 20449
Square root = 11 × 13 = 143
6. Find the smallest square number that is divisible by each of the following numbers:
(i) 3, 6, 10, 15
(ii) 6, 9, 27, 36
(iii) 4, 7, 8, 16
Solution:
(i) 3, 6, 10, 15
The number which is divisible by
3, 6, 10, 15 = LCM of 3, 6, 10, 15
It can be written as
So we get
= 2 × 3 × 5
= 30
(ii) 6, 9, 27, 36
The number which is divisible by
6, 9, 27, 36 = LCM of 6, 9, 27, 36
It can be written as
So we get
= 3 × 3 × 2 × 2 × 3
= 108
Here the smallest square
= 108 × 3
= 324
(iii) 4, 7, 8, 16
The number which is divisible by
4, 7, 8, 16 = LCM of 4, 7, 8, 16
It can be written as
So we get
= 2 × 2 × 2 × 2 × 7
= 112
Here the smallest square
= 112 × 7
= 784
7. 4225 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:
It is given that
Total number of plants = 4225
Here
Number of rows = Number of plants in each row
So the number of rows = square root of 4225
It can be written as
So, we get
= 5 × 13
= 65
Hence, the number of rows is 65, and the number of plants in each row is 65.
8. The area of a rectangle is 1936 sq. m. If the length of the rectangle is 4 times its breadth, find the dimensions of the rectangle.
Solution:
It is given that
Area of rectangle = 1936 sq. m
Take breadth = x m
Length = 4x m
So we get
4x2 = 1936
By further calculation
x2 = 1936/4 = 484
We know that
It can be written as
By further calculation
= 2 × 11
= 22
Here
Length = 4x = 4 × 22 = 88 m
Breadth = x = 22 m
9. In a school, a P.T. teacher wants to arrange 2000 students in the form of rows and columns for P.T. display. If the number of rows is equal to the number of columns and 64 students cannot be accommodated in this arrangement, find the number of rows.
Solution:
It is given that
Total number of students in a school = 2000
The P.T. teacher arranges in such a way that
No. of rows = no. of students in each row
So 64 students are left
Required number of students = 2000 – 64 = 1936
No. of rows = √1936
We know that
It can be written as
By further calculation
= 2 × 2 × 11
= 44
10. In a school, the students of class VIII collected ₹2304 for a picnic. Each student contributed as many rupees as the number of students in the class. Find the number of students in the class.
Solution:
It is given that
Amount collected for picnic = ₹2304
We know that
No. of students = no. of rupees contributed by each student = √2304
Here
It can be written as
By further calculation
= 2 × 2 × 2 × 2 × 34= 48
Therefore, the number of students in class VIII is 4811.
11. The product of two numbers is 7260. If one number is 15 times the other number, find the numbers.
Solution:
It is given that
Product of two numbers = 7260
Consider one number = x
Second number = 15x
It can be written as
15x × x = 7260
15x2 = 7260
By further calculation
x2 = 7260/15 = 484
So we get
= 2 × 11
= 22
Here
One number = 22
Second number = 22 × 15 = 330
12. Find three positive numbers in the ratio 2: 3: 5, the sum of whose squares is 950.
Solution:
It is given that
Ratio of three positive numbers = 2: 3: 5
Sum of their squares = 950
Consider
First number = 2x
Second number = 3x
Third number = 5x
It can be written as
(2x)2+ (3x)2 + (5x)2 = 950
By further calculation
4x2 + 9x2 + 25x2 = 950
38x2 = 950
So we get
X2 = 950/38 = 25
x = √25 = 5
Here
First number = 2 × 5 = 10
Second number = 3 × 5 = 15
Third number = 5 × 5 = 25
13. The perimeter of two squares is 60 metres and 144 metres, respectively. Find the perimeter of another square equal in area to the sum of the first two squares.
Solution:
It is given that
Perimeter of first square = 60 m
Side = 60/4 = 15 m
Perimeter of second square = 144 m
Side = 144/4 = 36 m
So the sum of perimeters of two squares = 60 + 144 = 204 m
Sum of areas of these two squares = 152 + 362
= 225 + 1296
= 1521 m2
Here
Area of third square = 1521 m2
We know that
So we get
Side = √Area = √1521
It can be written as
= 3 × 13
= 39 m
Here
Perimeter = 4 × side
Substituting the values
= 4 × 39
= 156 m
Exercise 3.4
1. Find the square root of each of the following by division method:
(i) 2401
(ii) 4489
(iii) 106929
(iv) 167281
(v) 53824
(vi) 213444
Solution:
(i) √2401 = 49
By division method
(ii) √4489 = 67
By division method
(iii) √106929 = 327
By division method
(iv) √167281 = 409
By division method
(v) √53824 = 232
By division method
(vi) √213444 = 462
By division method
2. Find the number of digits in the square root of each of the following (without any calculation):
(i) 81
(ii) 169
(iii) 4761
(iv) 27889
(v) 525625
Solution:
(i) 81
We know that
In 81, a group of two’s is 1
Therefore, its square root has one digit.
(ii) 169
We know that
In 169, group of two’s are 2
Therefore, its square root has two digits.
(iii) 4761
We know that
In 4761, group of two’s are 2
Therefore, its square root has two digits.
(iv) 27889
We know that
In 27889, groups of two’s are 3
Therefore, its square root has three digits.
(v) 525625
We know that
In 525625, groups of two’s are 3
Therefore, its square root has three digits.
3. Find the square root of the following decimal numbers by division method:
(i) 51.84
(ii) 42.25
(iii) 18.4041
(iv) 5.774409
Solution:
(i) √51.84 = 7.2
By division method
(ii) √42.25 = 6.5
By division method
(iii) √18.4041 = 4.29
By division method
(iv) √5.774409 = 2.403
By division method
4. Find the square root of the following numbers correct to two decimal places:
(i) 645.8
(ii) 107.45
(iii) 5.462
(iv) 2
(v) 3
Solution:
(i) √645.8 = 25.41
It can be written as
(ii) √107.45 = 10.36
It can be written as
(iii) √5.462 = 2.337 = 2.34
It can be written as
(iv) √2 = 1.41
It can be written as
(v) √3 = 1.73
It can be written as
5. Find the square root of the following fractions by division method:
(i) 841/1521
(ii) 8 257/529
(iii) 16 169/441
Solution:
(i) 841/1521
By squaring
It can be written as
(ii) 8 257/529
By squaring
It can be written as
(iii) 16 169/441
By squaring
It can be written as
6. Find the least number which must be subtracted from each of the following numbers to make them a perfect square. Also, find the square root of the perfect square number so obtained:
(i) 2000
(ii) 984
(iii) 8934
(iv) 11021
Solution:
(i) 2000
We know that
By taking the square root, 64 is left as the remainder
Subtracting 64 from 2000
We get 1936 which is a perfect square, and its square root is 44.
(ii) 984
We know that
By taking the square root, 23 is left as the remainder
Subtracting 23 from 984
We get 961 which is a perfect square, and its square root is 31.
(iii) 8934
We know that
By taking the square root, 98 is left as the remainder
Subtracting 98 from 894
We get 8934 – 98 = 8836 which is a perfect square, and its square root is 94.
(iv) 11021
We know that
By taking the square root, 205 is left as the remainder
Subtracting 205 from 11021
We get 11021 – 205 = 10816 which is a perfect square, and its square root is 104.
7. Find the least number which must be added to each of the following numbers to make them a perfect square. Also, find the square root of the perfect square number so obtained:
(i) 1750
(ii) 6412
(iii) 6598
(iv) 8000
Solution:
(i) 1750
We know that
By taking the square root
412 is less than 1750
So by taking 422
164 – 150 = 14 less
Adding 14, we get a square of 42 which is 1764.
(ii) 6412
We know that
By taking the square root
802 is less than 6412
So by taking 812
161 – 12 = 14 less
Adding 149, we get a square of 81 which is 6561.
(iii) 6598
We know that
By taking the square root
812 is less than 6598
So by taking 822
324 – 198 = 126 less
Adding 126, we get a square of 82 which is 6724.
(iv) 8000
We know that
By taking the square root
892 is less than 8000
So by taking 902
8100 – 8000 = 100 less
Adding 100, we get a square of 90 which is 8100.
8. Find the smallest four-digit number, which is a perfect square.
Solution:
It is given that
Smallest four–digit number = 1000
We know that
By taking the square root, we find that 39 is left.
If we subtract any number from 1000, we get 3 digit number
Take 322 = 1024
Here 1024 – 1000 = 24 is to be added to get a perfect square of least 4 digit number
Therefore, the required 4-digit smallest number is 1024.
9. Find the greatest number of six digits, which is a perfect square.
Solution:
It is given that
Greatest six-digit number = 999999
We know that
By taking the square root, we find that 1998 is left
If we subtract 1998 from 999999, we get 998001 which is a perfect square.
Therefore, the required six-digit greatest number is 998001.
10. In a right triangle ABC, ∠B = 900.
(i) If AB = 14 cm, BC = 48 cm, find AC.
(ii) If AC = 37 cm, BC = 35 cm, find AB.
Solution:
(i) In a right-angled triangle ABC
It is given that
AB = 14 cm and BC = 48 cm
Using Pythagoras theorem
AC2 = AB2 + BC2
Substituting the values
= 142 + 482
By further calculation
= 196 + 2304
= 2500
So we get
AC = √2500 = 50 cm
(ii) In a right triangle, ABC
B = 900, AC = 37 cm, BC = 35 cm
Using Pythagoras Theorem
AC2 = AB2 + BC2
Substituting the values
372 = AB2 + 352
By further calculation
1369 = AB2 + 1225
AB2 = 1369 – 1225 = 144
So we get
AB = √144 = 12 cm
11. A gardener has 1400 plants. He wants to plant these in such a way that the number of rows and columns remains the same. Find the minimum number of plants he needs more for this.
Solution:
It is given that
Total number of plants = 1400
We know that
Here
Number of columns = Number of rows
By taking the square root of 1400
372 < 1400
So take 382 = 1444
We need 1444 – 1400 = 44 plants more
Therefore, the minimum number of plants he needs more for this is 44.
12. There are 1000 children in a school. For a P.T. drill, they have to stand in such a way that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?
Solution:
It is given that
No. of total children in a school = 1000
For a P.T. drill, children have to stand in such a way that
No. of rows = No. of columns
Take the square root of 1000
39 is left as the remainder
Left out children = 39
Hence, 39 children would be left out in this arrangement.
13. Amit walks 16 m south from his house and turns east to walk 63 m to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk while returning?
Solution:
It is given that
Amit walks 16 m south from his house and turns east to walk 63 m to reach his friend’s house
Consider O as the house and A and B as the places
OA = 16 m, AO = 63 m
Using Pythagoras theorem
OB2 = OA2 + AB2
Substituting the values
= 162 + 632
By further calculation
= 256 + 3969
= 4225
So we get
OB = √4225 = 65
Therefore, Amit has to walk 65 m to reach his house.
14. A ladder 6 m long leaned against a wall. The ladder reaches the wall to a height of 4.8 m. Find the distance between the wall and the foot of the ladder.
Solution:
It is given that
Length of ladder = 6 m
The ladder reaches the wall to a height of 4.8 m
Consider AB as the ladder and AC as the height of the wall
AB = 6 m and AC = 4.8 m
The distance between the foot of the ladder and the wall is BC
Using the Pythagoras theorem,
AB2 = AC2 + BC2
Substituting the values
62 = 4.82 + BC2
By further calculation
BC2 = 62 – 4.82
BC2 = 36 – 23.04 = 12.96
So we get
BC = √12.96 = 3.6 m
Hence, the distance between the wall and the foot of the ladder is 3.6 m.
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