Karnataka Class 8 Math Solution Chapter Square and Square root by Experience Math Teacher. Karnataka Board Class 8 Math Part 1 PDF Chapter 5 Exercise 5.1, Exercise 5.2, Exercise 5.3, Exercise 5.4 Solution.
Karnataka Class 8 Math Solution Chapter Square and Square root
EXERCISE 5.1
1.) What will be the unit digit of the squares of the following numbers?
(i) 81
Ans:
Unit digit of the square of 81 has 1 as its unit digit, 1 is the unit digit of its square.
(ii) 272
Ans:
Unit digit of the square of 272 has 2 as its unit digit, 4 is the unit digit of its square.
(iii) 799
Ans:
Unit digit of the square of 799 has 9 as its unit digit, 1 is the unit digit of its square.
(iv) 3853
Ans:
Unit digit of the square of 3853 has 3 as its unit digit, 9 is the unit digit of its square.
(v) 1234
Ans:
Unit digit of the square of 1234 has 4 as its unit digit, 6 is the unit digit of its square.
(vi) 26387
Ans:
Unit digit of the square of 26387 has 7 as its unit digit, 9 is the unit digit of its square.
(vii) 52698
Ans:
Unit digit of the square of 52698 has 8 as its unit digit, 4 is the unit digit of its square.
(viii) 99880
Ans:
Unit digit of the square of 99880 has 0 as its unit digit, 0 is the unit digit of its square.
(ix) 12796
Ans:
Unit digit of the square of 12796 has 6 as its unit digit, 6 is the unit digit of its square.
(x) 55555
Ans:
Unit digit of the square of 55555 has 5 as its unit digit, 5 is the unit digit of its square.
2.) The following numbers are obviously not perfect squares. Give reason.
(i) 1057
Ans:
Unit digit of number 1057 is 7.
We know, Unit digit of numbers are 2,3,7,8 and odd numbers of 0 then these number are not perfect square.
(ii) 23453
Ans:
Unit digit of number 23453 is 3.
We know, Unit digit of numbers are 2,3,7,8 and odd numbers of 0 then these number are not perfect square.
(iii) 7928
Ans:
Unit digit of number 7928 is 8.
We know, Unit digit of numbers are 2,3,7,8 and odd numbers of 0 then these number are not perfect square.
(iv) 222222
Ans:
Unit digit of number 222222 is 2.
We know, Unit digit of numbers are 2,3,7,8 and odd numbers of 0 then these number are not perfect square.
(v) 64000
Ans:
64000 has 000 which is odd number of 0.
We know, Unit digit of numbers are 2,3,7,8 and odd numbers of 0 then these number are not perfect square.
(vi) 89722
Ans:
Unit digit of number 89722 is 2.
We know, Unit digit of numbers are 2,3,7,8 and odd numbers of 0 then these number are not perfect square.
(vii) 222000
Ans:
222000 has 000 which is odd number of 0.
We know, Unit digit of numbers are 2,3,7,8 and odd numbers of 0 then these number are not perfect square.
(viii) 505050
Ans:
505050 has 0 which is odd number of 0.
We know, Unit digit of numbers are 2,3,7,8 and odd numbers of 0 then these number are not perfect square.
3.) The squares of which of the following would be odd numbers?
(i) 431
Ans:
We know,
Square of an odd number is always odd and square of an even number is always even.
Square of 431 is odd number.
(ii) 2826
Ans:
We know,
Square of an odd number is always odd and square of an even number is always even.
Square of 2826 is even number.
(iii) 7779
Ans:
We know,
Square of an odd number is always odd and square of an even number is always even.
Square of 7779 is odd number.
(iv) 82004
Ans:
We know,
Square of an odd number is always odd and square of an even number is always even.
Square of 82004 is even number.
4.) Observe the following pattern and find the missing digits.
11 2 = 121
101 2 = 10201
1001 2 = 1002001
1000012 =1 —–2——1
100000012 = —————–
Ans:
11 2 = 121
101 2 = 10201
1001 2 = 1002001
1000012 =10000200001
100000012 =100000020000001
5.) Observe the following pattern and supply the missing numbers.
11 2 =121
101 2 =10201
10101 2 = 102030201
1010101 2 = ————-
———-2 = 10203040504030201
Ans:
11 2 =121
101 2 =10201
10101 2 = 102030201
1010101 2 = 1020304030201
1010101012 = 10203040504030201
6.) Using the given pattern, find the missing numbers.
1 2 + 2 2 + 2 2 = 3 2
2 2 + 3 2 + 6 2 = 72
3 2 + 4 2 + 12 2 = 13 2
4 2 + 5 2 + — 2 = 21 2
5 2 + —2 + 302 = 31 2
6 2 + 7 2 + —2 = — 2
Ans:
1 2 + 2 2 + 2 2 = 3 2
2 2 + 3 2 + 6 2 = 72
3 2 + 4 2 + 12 2 = 13 2
4 2 + 5 2 + 202 = 21 2
5 2 + 6 2 + 302 = 31 2
6 2 + 7 2 + 422 = 432
7.) Without adding, find the sum.
To find pattern Third number is related to first and second number. How? Fourth number is related to third number. How?
(i) 1 + 3 + 5 + 7 + 9
Ans:
We know, sum of the first n odd natural numbers is n².
Here, number of term (n) is 5 sum = (5)² = 25
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
Ans:
We know, sum of the first n odd natural numbers is n².
Here, number of term (n) is 10 sum = (10)² = 100
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23
Ans:
We know, sum of the first n odd natural numbers is n².
Here, number of term (n) is 12 sum = (12)² = 144
8.) (1) Express 49 as the sum of 7 odd numbers.
Ans:
We know,
49 = (7) 2
49 is the sum of first 7 odd natural numbers
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) Express 121 as the sum of 11 odd numbers.
Ans:
We know,
(121) = (11) 2
121 is the sum of first 11 odd natural numbers
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9.) How many numbers lie between squares of the following numbers?
(1) 12 and 13
Ans:
We know, there are “2n” number between the square of the number n and (n+1)
Here, n = 12
2n =2 x 12 = 24
24 numbers between (12) 2 and (13) 2
(ii) 25 and 26
Ans:
We know, there are “2n” number between the square of the number n and (n+1)
Here, n = 25
2 x 25 = 50
50 numbers between (25) 2 and (26) 2
99 and 100
Ans:
We know, there are “2n” number between the square of the number n and (n+1)
Here, n = 99
2 x 99 = 198
198 numbers between (99) 2 and (100) 2
EXERCISE 5.2
1.) Find the square of the following numbers.
(i) 32
Ans:
32 2 = (30 + 2) 2
= 30(30 + 2) + 2(30 + 2)
Here we use (a + b)2= a2 + 2ab + b2
= 30 2 + 30 x 2 + 2 x 30 + 2 2
= 900 + 60 + 60 + 4
=1024
(ii) 35
Ans:
We know, If a number have its unit digit 5 its square number will be [a (a+1) hundred+25].
Where a is first digit of that number.
Here a = 3
352 = (3 x 4) hundred+25
352 = 1225
(iii) 86
Ans:
86 2 = (80 + 6) 2
Here we use (a + b)2= a2 + 2ab + b2
= 80 2 + 80 x 6 + 6 x 80 + 6 2
= 6400+480+480+36
=7396
(iv) 93
Ans:
93 2 = (90 + 3) 2
Here we use (a + b)2= a2 + 2ab + b2
= 90 2 + 90 x 3 + 3 x 90 + 3 2
= 8100 + 270 + 270 + 9
= 8649
(v) 71
Ans:
71 2 = (70 + 1) 2
Here we use (a + b)2= a2 + 2ab + b2
= 70 2 + 70 x 1 + 1 x 70 + 1 2
= 4900 + 70 + 70 + 1
= 5041
(vi) 46
Ans:
46 2 = (40 + 6) 2
Here we use (a + b)2= a2 + 2ab + b2
= 40 2 + 40 x 6 + 6 x 40 + 6 2
= 1600 + 240 + 240 + 36
= 2116
2.) Write a Pythagorean triplet whose one member is.
(i) 6
Ans:
Pythagorean triplets are 6, 8 and 10.
(ii) 14
Ans:
Pythagorean triplets are 14, 48 and 50.
(iii) 16
Ans:
Pythagorean triplets are 16, 63 and 65.
(iv) 18
Ans:
Pythagorean triplets are 18, 80 and 82.
EXERCISE 5.3
1.) What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
(i) 9801
Ans:
We know, the number ends with 1, the one’s digit of the square root may be 1 or 9.
(ii) 99856
Ans:
We know, the number ends with 6, the one’s digit of the square root may be 4 or 6.
(iii) 998001
Ans:
We know, the number ends with 1, the one’s digit of the square root may be 1 or 9.
(iv) 657666025
Ans:
We know, the number ends with 5, the one’s digit of the square root will be 5
2.) without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153
Ans:
We know, the perfect square of a number can end with 0, 1, 2, 4, 5, 6 or 9 at unit’s place.
The number 153 has 3 at its unit’s place, it is not a perfect square.
(ii) 257
Ans:
We know, the perfect square of a number can end with 0, 1, 2, 4, 5, 6 or 9 at unit’s place.
The number 257 has 7 at its unit’s place, it is not a perfect square.
(iii) 408
Ans:
We know, the perfect square of a number can end with 0, 1, 2, 4, 5, 6 or 9 at unit’s place.
The number 408 has 8 at its unit’s place, it is not a perfect square
(iv) 441
Ans:
We know, the perfect square of a number can end with 0, 1, 2, 4, 5, 6 or 9 at unit’s place.
The number 441 has 1 at its unit’s place, it is a perfect square.
4.) Find the square roots of the following numbers by the Prime Factorisation Method.
(i) 729
Ans:
= 3 x 3 x 3
Square root of 729 = 27
(ii) 400
Ans:
= 2 x 2 x 5
Square root of 400 = 20
(iii) 1764
Ans:
= 2 x 3 x 7
Square root of 1764 = 42
(iv) 4096
Ans:
= 2 x 2 x 2 x 2 x 2 x 2
Square root of 4096 = 64
(v) 7744
Ans:
= 2 x 2 x 2 x 11
Square root of 7744 = 88
(vi) 9604
Ans:
= 2 x 7 x 7
Square root of 9604 = 98
(vii) 5929
Ans:
= 7 x 11
Square root of 5929 = 77
(viii) 9216
Ans:
= 2 x 2 x 2 x 2 x 2 x 3
Square root of 9216 = 96
(ix) 529
Ans:
= 23
Square root of 529 = 23
(x) 8100
Ans:
= 2 x 3 x 3 x 5
Square root of 8100 = 90
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
Ans:
252 = 2 x 2 x 3 x 3 x 7
Here,
Prime factor 7 does not have its pair. If we divide 252 by 7 then rest of the prime factor will be in pairs.
252 / 7 = 36
36 is perfect square.
Square root of 36 is 6.
(ii) 2925
Ans:
2925 = 5 x 5 x 3 x 3 x 13
Here,
Prime factor 13 does not have its pair. If we divide 2925 by 13 then rest of the prime factor will be in pairs.
2925 / 13
= 225
225 is a perfect square.
Square root of 225 is 15.
(iii) 396
Ans:
396 = 3 x 3 x 2 x 2 x 11
Here,
Prime factor 11 does not have its pair. If we divide 396 by 11 then rest of the prime factor will be in pairs.
= 396/ 11
= 36
36 is perfect square.
Square root of 36 is 6.
(iv) 2645
Ans:
2645 = 5 x 23 x 23
Here,
Prime factor 5 does not have its pair. If we divide 2645 by 5 then rest of the prime factor will be in pairs.
2645/ 5 =
529
529 is a perfect square
Square root of 529 is 23.
(v) 2800
Ans:
2800 = 2 x 2 x 2 x 5 x 5 x 7
Here,
Prime factor 7 does not have its pair. If we divide 2800by 7 then rest of the prime factor will be in pairs.
2800 / 7
= 400
400 is a perfect square
Square root of 400 is 20.
(vi) 1620
Ans:
1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5
Here,
Prime factor 5 does not have its pair. If we divide 1620 by 5 then rest of the prime factor will be in pairs.
1620 / 5
= 324
324 is a perfect square
Square root of 324 is 18.
7.) The students of Class VIII of a school donated₹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.
Ans:
Given that,
The students of Class VIII of a school donated ₹2401.
Let, number of students in a class be x
Money donated by each student is also x.
We know, Total amount of donation = Rs. 2401
Total amount of donation = number of students ´ money donated by each student
2401 = XxX
2401 = x2
Here, we have to find square root of 2401.
Square root of 2401 = 7 x 7
Square root of 2401 = 49.
The number of students in class is 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.
Ans:
Given that,
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.
Let number of rows be x
Number of plants in each row is also x
We know,
(Number of rows) x (number of plants in each row) = 2025
X xX = 2025
X2 = 2025
Here, we have to find square root of 2025.
Square root of 2025 = 5 x 3 x 3
Square root of2025= 45
The number of rows and the number of plants in each row is 45.
9.) Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Ans:
We know, the number that will be perfectly divisible by each one of the numbers is their LCM.
We have to find LCM of 4, 9 and 10.
LCM of 4, 9 and 10 = 2 x 2 x 3 x 3 x 5
LCM of 4, 9 and 10 = 180
Here, number 5 does not have a pair.
180 is multiplied by 5 then the number obtained is a perfect square.
180 x 5 = 900
900 is the smallest square that is divisible by 4, 9 and 10
10.) Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Ans:
We know, the number that will be perfectly divisible by each one of the numbers is their LCM.
We have to find LCM of 8, 15 and 20.
2 |
8 | 15 | 20 |
2 | 4 | 15 |
10 |
5 |
2 | 15 | 5 |
2 | 3 |
1 |
LCM of 8, 15 and 20= 2 x 2 x 2 x 3 x 5
LCM of 8, 15 and 20 = 120
Here, number 2, 3, 5 does not have a pair.
120 is multiplied by 2x3x5 then the number obtained is a perfect square.
120 x 2 x 3 x 5 = 3600
3600 is the smallest square that is divisible by 8, 15 and 20.
EXERCISE 5.4
1.) Find the square root of each of the following numbers by Division method.
(i) 2304
Ans:
The square root of 2304 by Division method,
The square root of 2304 is 48.
(ii) 4489
Ans:
The square root of 4489 by Division method,
The square root of 4489 is 67.
(iii) 3481
Ans:
The square root of 3481 by Division method,
The square root of 3481 is 59.
(iv) 529
Ans:
The square root of 529 by Division method,
The square root of 529 is 23.
(v) 3249
Ans:
The square root of 3249 by Division method,
The square root of 3249 is 57.
(vi) 1369
Ans:
The square root of 1369 by Division method,
The square root of 1369 is 37.
(vii) 5776
Ans:
The square root of 5776 by Division method,
The square root of 5776 is 76.
(viii) 7921
Ans:
The square root of 7921 by Division method,
The square root of 7921 is 89.
(ix) 576
Ans:
The square root of 576 by Division method,
The square root of 576 is 24.
(x) 1024
Ans:
The square root of 1024 by Division method,
The square root of 1024 is 32.
3.) Find the square root of the following decimal numbers.
(i) 2.56
Ans:
Square root of 2.56
Square root of 2.56 is 1.6.
(ii) 7.29
Ans:
Square root of 7.29
(iii) 51.84
Ans:
Square root of 51.84
Square root of 51.84 is 7.2
(iv) 42.25
Ans:
Square root of 42.25
Square root of 42.25 is 6.5
(v) 31.36
Ans:
Square root of 31.36
Square root of 31.36 is 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
Ans:
Here, when we divide 402 we get remainder 2.
We know, when we subtract remainder from given number we get perfect square number.
402 – 2 = 400
400 is a perfect square.
Square root of 400 is 20.
(ii) 1989
Ans:
Here, when we divide 1989 we get remainder 53.
We know, when we subtract remainder from given number we get perfect square number.
1989 – 53 = 1936
1936is a perfect square.
Square root of 1936 is 44.
(iii) 3250
Ans:
Here, when we divide 3250 we get remainder 1.
We know, when we subtract remainder from given number we get perfect square number.
3250 – 1 = 3249
3249is a perfect square.
Square root of 3249 is 57.
(iv) 825
Ans:
Here, when we divide 825 we get remainder 41.
We know, when we subtract remainder from given number we get perfect square number.
825 – 41 = 784
784 is a perfect square.
Square root of 784 is 28.
(v) 4000
Ans:
Here, when we divide 4000 we get remainder 31.
4000 – 31 = 3969
3969is a perfect square.
Square root of 3969is 63.
6.) Find the length of the side of a square whose area is 441 m².
Ans:
We know,
Area of square = (Length) 2
Given that,
Area of square is 441 m².
(Length) 2= 441 m².
Here, we have to find square root of 441.
Square root of 441 is 21.
Length of Square is 21m.
7.) In a right triangle ABC, ∠B=90°.
(a) IfAB=6cm, BC = 8 cm, find AC
Ans:
Given that,
AB=6cm, BC = 8 cm and ∠B=90°.
By using Pythagoras theorem,
AC2 = AB + BC
AC2= (6)2 + (8)2
AC2= 100
AC = 10 cm.
(b) IfAC=13 cm, BC=5 cm, find AB
Ans:
Given that,
AC=13 cm, BC=5 cmand ∠B=90°.
By using Pythagoras theorem,
AC2 = AB2 + BC2
(13)2 = AB2 + (5)2
169 = AB2 + 25
AB2 = 169 – 25 = 144
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.
Ans:
Given that,
A gardener has 1000 plants.
The number of rows and the number of columns remain same.
We have to find the minimum number of plants he needs more for this.
First we calculate Square root of 1000
The remainder is 39.
We know Square root of 32 is 1024.
Number to be added to 1000 to make it perfect square.
= 322– 1000
= 1024 – 1000 = 24
The required number of plants is 24.
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.
Ans:
Given that,
There are 500 children in a school.
Number of rows is equal to number of columns.
We have to find the how many children would be left out in this arrangement.
First we calculate Square root of 500
The remainder is 16.
Perfect square can be obtained by subtracting 16 from 500
500 – 16
= 484
Square root of 484 is 22.
Number of children left out in PT drill arrangement is 16.
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