Yearly Archives: 2022

Exercise 3.1

Question 1.
Which of the following numbers are perfect squares ?
(i)484
(ii) 625
(iii) 576
(iv) 941
(v) 961
(vi) 2500
Solution:

Grouping the factors in pairs, we have left no factor unpaired
∴ 484 is a perfect square of 22

∴ Grouping the factors in pairs, we have left no factor unpaired
∴ 625 is a perfect square of 25.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 576 is a perfect square of 24
(iv) 941 has no prime factors

∴ 941 is not a perfect square.
(v) 961 =31 x 31
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 961 is a perfect square of 31

Grouping the factors in pairs, we see that no factor is left impaired
∴ 2500 is a perfect square of 50 .

Question 2.
Show that each of the following* numbers is a perfect square. Also find the number whose square is the given number in each case :
(i) 1156
(ii) 2025
(iii) 14641
(iv) 4761
Solution:

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 1156 is a perfect square of 2 x 17 = 34

Grouping the factors in pairs, we see that no factor is left unpaired
2025 is a perfect square of 3 x 3 x 5 =45

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 14641 is a perfect square of 11×11 = 121

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 4761 is a perfect square of 3 x 23 = 69

Question 3.
Find the smallest number by which the given number must be multiplied so that the product is a perfect square.
(i) 23805
(ii) 12150
(iii) 7688
Solution:

Grouping the factors in pairs of equal factors, we see that 5 is left unpaird
∴ In order to complete the pairs, we have to multiply 23805 by 5, then the product will be the perfect square.
Requid smallest number = 5
(ii) 12150 = 2 x 3 x 3×3 x 3×3 x 5×5

Grouping the factors in pairs of equal factors, we see that factors 2 and 3 are left unpaired
∴ In order to complete the pairs, we have to multiply 12150 by 2 x 3 =6 i.e., then the product will be the complete square.
∴ Required smallest number = 6

Grouping the factors in pairs of equal factors, we see that factor 2 is left unpaired
∴ In order to complete the pairs we have to multiply 7688 by 2, then the product will be the complete square
∴ Required smallest number = 2

Question 4.
Find the smallest number by which the given number must be divided so that the resulting number is a perfect square.
(i) 14283
(ii) 1800
(iii) 2904
Solution:

Grouping the factors in pairs of equal factors, we see that factors we see that 3 is left unpaired
Deviding by 3, the quotient will the perfect square.

Grouping the factors in pair of equal factors, we see that 2 is left unpaired.
∴ Dividing by 2, the quotient will be the perfect square.

Grouping the factors in pairs of equal factors, we see that 2 x 3 we left unpaired
∴ Dividing by 2 x 3 = 6, the quotient will be the perfect square.

Question 5.
Which of the following numbers are perfect squares ?
11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121
Solution:
11 is not a perfect square as 11 = 1 x 11
12 is not a perfect square as 12 = 2×2 x 3
16 is a perfect square as 16 = 2×2 x 2×2
32 is not a perfect square as 32 = 2×2 x 2×2 x 2
36 is a perfect square as 36 = 2×2 x 3×3
50 is not a perfect square as 50 = 2 x 5×5
64 is a perfect square as 64 = 2×2 x 2×2 x 2×2
79 is not a perfect square as 79 = 1 x 79
81 is a perfect square as 81 = 3×3 x 3×3
111 is not a perfect square as 111 = 3 x 37
121 is a perfect square as 121 = 11 x 11
Hence 16, 36, 64, 81 and 121 are perfect squares.

Question 6.
Using prime factorization method, find which of the following numbers are perfect squares ?
∴ 189,225,2048,343,441,2916,11025,3549
Solution:

Grouping the factors in pairs, we see that are 3 and 7 are left unpaired
∴ 189 is not a perfect square

Grouping the factors in pairs, we see no factor left unpaired
∴ 225 is a perfect square

Grouping the factors in pairs, we see no factor left unpaired
∴ 2048 is a perfect square

Grouping the factors in pairs, we see that one 7 is left unpaired
∴ 343 is not a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 441 is a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 2916 is a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 11025 is a perfect square.

Grouping the factors in pairs, we see that 3, no factor 7 are left unpaired
∴ 3549 is a perfect square.

Question 7.
By what number should each of the following numbers be multiplied to get a perfect square in each case ? Also, find the number whose square is the new number.
(i) 8820
(ii) 3675
(iii) 605
(iv) 2880
(v) 4056
(vi) 3468
Solution:

Grouping the factors in pairs, we see that 5 is left unpaired
∴ By multiplying 8820 by 5, we get the perfect square and square root of product will be
= 2 x 3 x 5 x 7 = 210

Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3675 by 3, we get a perfect square and square of the product will be
= 3 x 5 x 7 = 105

Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 605 by 5, we get a perfect square and square root of the product will be
= 5 x 11 =55

Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 2880 by 5, we get the perfect square.
Square rooi of product will be = 2 x 2 * 2 – 3 x 5 = 120

Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 4056 by 2 x 3 i.e., 6, we get the perfect square.
and square root of the product will be
= 2 x 2 x 3 x 13 = 156

Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3468 by 3 we get a perfect square, and square root of the product will be 2 x 3 x 17 = 102

Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 7776 by 2 x 3 or 6 We get a perfect square and square root of the product will be
= 2 x 2 x 2 x 3 x 3 x 3 = 216

Question 8.
By what numbers should each of the following be .divided to get a perfect square in each case ? Also find the number whose square is the new number.
(i) 16562
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575
Solution:

Grouping the factors in pairs, we see that 2 is left unpaired
∴ Dividing by 2, we get the perfect square and square root of the quotient will be 7 x 13 = 91

Grouping the factors in pairs, we see that 2 is left unpaired,
∴ Dividing 3698 by 2, the quotient is a perfect square
and square of quotient will be = 43

Grouping the factors in pairs, we see that 7 is left unpaired
∴ Dividing 5103 by 7, we get the quotient a perfect square.
and square root of the quotient will be 3 x 3 x 3 = 27

Grouping the factors iq pairs, we see that 2 and 3 are left unpaired
∴ Dividing 3174 by 2 x 3 i.e. 6, the quotient will be a perfect square and square root of the quotient will be = 23

Grouping the factors in pairs, we find that 7 is left unpaired i
∴ Dividing 1575 by 7, the quotient is a perfect square
and square root of the quotient will be = 3 x 5 = 15

Question 9.
Find the greatest number of two digits which is a perfect square.
Solution:
The greatest two digit number = 99 We know, 92 = 81 and 102 = 100 But 99 is in between 81 and 100
∴ 81 is the greatest two digit number which is a perfect square.

Question 10.
Find the least number of three digits which is perfect square.
Solution:
The smallest three digit number =100
We know that 92 = 81, 102 = 100, ll2 = 121
We see that 100 is the least three digit number which is a perfect square.

Question 11.
Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.
Solution:
By factorization:

Grouping the factors in pairs, we see that 11 is left unpaired
∴ The least number is 11 by which multiplying 4851, we get a perfect square.

Question 12.
Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.
Solution:
By factorization,

Grouping the factors in pairs, we see that 13 is left unpaired
∴ Dividing 28812 by 3, the quotient will be a perfect square.

Question 13.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number.
Solution:
By factorization,

Grouping the factors in pairs, we see that one 2 is left unpaired.
∴ Dividing 1152 by 2, we get the perfect square and square root of the resulting number 576, will be 2 x 2 x 2 x 3 = 24

Exercise 3.2

Question 1.
The following numbers are not perfect squares. Give reason :
(i) 1547
(ii) 45743
(iii) 8948
(iv) 333333

Solution:
We know that if the units digit is 2, 3, 7 or 8 of a number, then the number is not a perfect square.

(i) ∴ 1547 has 7 as units digit.
∴ It is not a perfect square.

(ii) 45743 has 3 as units digit
∴ It is not a perfect square.

(iii)  ∴ 8948 has 8 as units digit
∴ It is not a perfect square.

(iv)  ∴ 333333 has 3 as units digits
∴ It is not a perfect square.

Question 2.
Show that the following numbers are not perfect squares :
(i) 9327
(ii) 4058
(iii) 22453
(iv) 743522

Solution:
(i) 9327
∴ The units digit of 9327 is 7
∴ This number can’t be a perfect square.

(ii) 4058
∴ The units digit of 4058 is 8
∴ This number can’t be a perfect square.

(iii) 22453
∴ The units digit of 22453 is 3
.∴ This number can’t be a perfect square.

(iv) 743522
∴ The units digit of 743522 is 2
∴ This number can’t be a perfect square.

Question 3.
The square of which of the following numbers would be an odd number ?
(i) 731
(ii) 3456
(iii) 5559
(iv) 42008
Solution:
We know that the square of an odd number is odd and of even number is even. Therefore
(i) Square of 731 would be odd as it is an odd number.
(ii) Square of 3456 should be even as it is an even number.
(iii) Square of 5559 would be odd as it is an odd number.
(iv) The square of 42008 would be an even number as it is an even number.
Therefore suqares of (i) 731 and (ii) 5559 will be odd numbers.

Question 4.
What will be the units digit of the squares of the following numbers ?
(i) 52
(ii) 977
(iii) 4583
(iv) 78367
(v) 52698
(vi) 99880
(vii) 12796
(viii) 55555
(ix) 53924

Solution:

(i) Square of 52 will be 2704 or (2)² = 4
∴ Its units digit is 4.

(ii) Square of 977 will be 954529 or (7)² = 49 .
∴ Its units digit is 9

(iii) Square of 4583 will be 21003889 or (3)² = 9
∴ Its units digit is 9

(iv) IS 78367, square of 7 = 7² = 49
∴ Its units digit is 9

(v) In 52698, square of 8 = (8)² = 64
∴ Its units digit is 4

(vi) In 99880, square of 0 = 0² = 0
∴ Its units digit is 0

(vii) In 12796, square of 6 = 6² = 36
∴ Its units digit is 6

(viii) In In 55555, square of 5 = 5² = 25
∴ Its units digit is 5

(ix) In 53924, square pf 4 = 4² = 16
∴ Its units digit is 6

Question 5.
Observe the following pattern
1 + 3 = 2²
1 + 3 + 5 = 3²
1+34-5 + 7 = 4²
and write the value of 1 + 3 + 5 + 7 + 9 +…………upto n terms.
Solution:
The given pattern is
1 + 3 = 2²
1 + 3 + 5 = 3²
1+3 + 5 + 7 = 4²
1+3 + 5 + 7 + 9 +……………… upto n terms (number of terms)² = n²

Question 6.
Observe the following pattern :
2² – 1² = 2 + 1
3² – 2² = 3 + 2
4² – 3² = 4 + 3
5² – 4² = 5 + 4
Find the value of
(i) 100² – 99²
(ii) 111² – 109²
(iii) 99² – 96²
Solution:
From the given pattern,
2² – 1² = 2 + 1
3² – 2² = 3 + 2
4² – 3² = 4 + 3
₅² – 4² = 5 + 4
Therefore
(i) 100²-99° = 100 + 99

(ii) 111² – 109² = 111² – 110²– 109²
= (111² – 110²) + (110² – 109²)
= (111 + 110) + (110+ 109)
= 221 + 219 = 440

(iii) 99² – 96² = 99² – 98² + 98² – 97² + 97² – 96²
= (99² – 98²) + (98² – 97²) + (97² – 96²)
= (99 + 98) + (98 + 97) + (97 + 96)
= 197 + 195 + 193 = 585

Question 7.
Which of the following triplets are Pythagorean ?
(i) (8, 15, 17)
(ii) (18, 80, 82)
(iii) (14, 48, 51)
(iv) (10, 24, 26)
(vi) (16, 63, 65)
(vii) (12, 35, 38)
Solution:
A pythagorean triplet is possible if (greatest number)² = (sum of the two smaller numbers)

(i) 8, 15, 17
Here, greatest number =17
∴ (17)² = 289
and (8)² + (15)² = 64 + 225 = 289
∴ 8² + 152 = 172
∴ 8, 15, 17 is a pythagorean triplet

(ii) 18, 80, 82
Greatest number = 82
∴ (82)² = 6724
and 18² + 80² = 324 + 6400 = 6724
∴ 18² + 80² = 82²
∴ 18, 80, 82 is a pythagorean

(iii) 14, 48, 51
Greatest number = 51
∴ (51)² = 2601
and 14² + 48² = 196 + 2304 = 25 00
∴ 51²≠ 14² + 48²
∴ 14, 48, 51 is not a pythagorean triplet

(iv) 10, 24, 26
Greatest number is 26
∴ 26² = 676
and 102 + 242 = 100 + 576 = 676
∴ 26² = 10² + 24²
∴ 10, 24, 26 is a pythagorean triplet

(vi) 16, 63, 65
Greatest number = 65
∴ 65² = 4225
and 16² + 63² = 256 + 3969 = 4225
∴ 65² = 16² + 63²
∴ 16, 63, 65 is a pythagorean triplet

(vii) 12, 35, 38
Greatest number = 38
∴ 38² = 1444
and 12² + 35² = 144 + 1225 = 1369
∴ 38² ≠12² + 35²
∴ 12, 35, 38 is not a pythagorean triplet.

Question 8.
Observe the following pattern

Solution:
From the given pattern

Question 9.
Observe the following pattern

and find the values of each of the following :
(i) 1 + 2 + 3 + 4 + 5 +….. + 50
(ii) 31 + 32 +… + 50
Solution:
From the given pattern,

Question 10.
Observe the following pattern

and find the values of each of the following :
(i) 1² + 2² + 3² + 4² +…………… + 10²
(ii) 5² + 6² + 7² + 8² + 9² + 10² + 12²
Solution:
From the given pattern,

Question 11.
Which of the following numbers are squares of even numbers ?
121,225,256,324,1296,6561,5476,4489, 373758
Solution:
We know that squares of even numbers is also are even number. Therefore numbers 256, 324,1296, 5476 and 373758 have their units digit an even number.
∴ These are the squares of even numbers.

Question 12.
By just examining the units digits, can you tell which of the following cannot be whole squares ?

1. 1026
2. 1028
3. 1024 
4. 1022
5. 1023
6. 1027

Solution:
We know that a perfect square cam at ends with the digit 2, 3, 7, or 8
∴ By examining the given number, we can say that 1028, 1022, 1023, 1027 can not be perfect squares.

Question 13.
Write five numbers for which you cannot decide whether they are squares.
Solution:
A number which ends with 1,4, 5, 6, 9 or 0
can’t be a perfect square
2036, 4225, 4881, 5764, 3349, 6400

Question 14.
Write five numbers which you cannot decide whether they are square just by looking at the unit’s digit.
Solution:
A number which does not end with 2, 3, 7 or 8 can be a perfect square
∴ The five numbers can be 2024, 3036, 4069, 3021, 4900

Question 15.
Write true (T) or false (F) for the following statements.
(i) The number of digits in a square number is even.
(ii) The square of a prime number is prime.
(iii) The sum of two square numbers is a square number.
(iv) The difference of two square numbers is a square number.
(vi) The product of two square numbers is a square number.
(vii) No square number is negative.
(viii) There is not square number between 50 and 60.
(ix) There are fourteen square number upto 200. 

Solution:
(i) False : In a square number, there is no condition of even or odd digits.
(ii) False : A square of a prime is not a prime.
(iii) False : It is not necessarily.
(iv) False : It is not necessarily.
(vi) True.
(vii) True : A square is always positive.
(viii) True : As 7² = 49, and 8² = 64.
(ix) True : As squares upto 200 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196 which are fourteen in numbers..

Exercise 3.3

Question 1.
Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication :
(i) 25
(ii) 37
(iii) 54
(iv) 71
(v) 96
Solution:
(i) (25)²

Question 2.
Find the squares of the following numbers using diagonal method :
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171
Solution:

Question 3.
Find the squares of the following numbers :
(i) 127
(ii) 503
(iii) 451
(iv) 862
(v) 265
Solution:
(i) (127)² = (120 + 7)²
{(a + b)² = a² + lab + b²}
= (120)² + 2 x 120 x 7 + (7)²
= 14400+ 1680 + 49 = 16129

(ii) (503)² = (500 + 3)²
{(a + b)² = a² + lab + b¹}
= (500)² + 2 x 500 x 3 + (3)²
= 250000 + 3000 + 9 = 353009

(iii) (451)² = (400 + 51)²
{(a + b)² = a² + lab + b²}
= (400)² + 2 x 400 x 51 + (5l)²
= 160000 + 40800 + 2601 = 203401

(iv) (451)² = (800 + 62)²
{(a + b)² = a² + lab + b²}
= (800)² + 2 x 800 x 62 + (62)²
= 640000 + 99200 + 3844 = 743044

(v) (265)²
{(a + b)² = a² + 2ab + b²}
(200 + 65)² = (200)² + 2 x 200 x 65 + (65)²
= 40000 + 26000 + 4225 = 70225

Question 4.
Find the squares of the following numbers
(i) 425
(ii) 575
(iii) 405
(iv) 205
(v) 95
(vi) 745
(vii) 512
(viii) 995
Solution:
(i) (425)²
Here n = 42
∴ n (n + 1) = 42 (42 + 1) = 42 x 43 = 1806
∴ (425)² = 180625

(ii) (575)²
Here n = 57
∴ n (n + 1) = 57 (57 + 1) = 57 x 58 = 3306
∴ (575)² = 330625

(iii) (405)²
Here n = 40
∴ n (n + 1) = 40 (40 + 1) -40 x 41 = 1640
∴ (405)² = 164025

(iv) (205)²
Here n = 20
∴ n (n + 1) = 20 (20 + 1) = 20 x 21 = 420
∴ (205)² = 42025

(v) (95)²
Here n = 9
∴ n (n + 1) = 9 (9 + 1) = 9 x 10 = 90
∴ (95)² = 9025

(vi) (745)²
Here n = 74
∴ n (n + 1) = 74 (74 + 1) = 74 x 75 = 5550
∴ (745)² = 555025

(vii) (512)²
Here a = 1, b = 2
∴ (5ab)² = (250 + ab) x 1000 + (ab)²
∴ (512)² = (250 + 12) x 1000 + (12)²
= 262 x 1000 + 144
= 262000 + 144 = 262144

(viii) (995)²
Here n = 99
∴ n (n + 1) = 99 (99 + 1) = 99 x 100 = 9900
∴ (995)² = 990025

Question 5.
Find the squares of the following numbers using the identity (a + b)1 = a2 + lab + b1
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605
Solution:
a + b)² = a² + lab + b²

(i) (405)² = (400 + 5)²
= (400)² + 2 x 400 x 5 + (5)²
= 160000 + 4000 + 25 = 164025

(ii) (510)² = (500 + 10)²
= (500)² + 2 x 500 x 10 x (10)²
= 250000 + 10000 + 100
= 260100

(iii) (1001)² = (1000+1)²
= (1000)² + 2 X 1000 x 1 + (1)
= 1000000 + 2000 + 1
=1002001

(iv) (209)² = (200 + 9)²
= (200)² + 2 x 200 x 9 x (9)²
= 40000 + 3600 +81
= 43681

(v) (605)² = (600 + 5)²
= (600)² + 2 x 600 x 5 +(5)²
= 360000 + 6000       25
=366025

Question 6.
Find the squares of the following numbers using the identity (a – b)² = a² – 2ab + b² :
(i) 395
(ii) 995
(iii) 495
(iv) 498
(v) 99
(vi) 999
(vii) 599
Solution:
a – b)² = a² – lab + b²

(i) (395)² = (400 – 5)²
= (400)² – 2 x 400 x 5 + (5)²
= 160000-4000 + 25
= 160025-4000
= 156025

(ii) (995)² = (1000 – 5)²
= (1000)² – 2 x 1000 x 5 + (5)²
= 1000000- 10000 + 25
= 1000025- 10000
= 990025

(iii) (495)² = (500 – 5)²
= (500)² – 2 x 500 x 5 + (5)²
= 250000 – 5000 + 25
= 250025 – 5000
= 245025

(iv) (498)² = (500 – 2)²
= (500)² – 2 x 500 x 2 + (2)²
= 250000 – 2000 + 4
= 250004 – 2000
= 248004

(v) (99)² = (100 – l)²
= (100)² – 2 x 100 x 1 + (1)²
= 10000 – 200 + 1
= 10001 – 200
= 9801

(vi) (999)² = (1000- l)²
= (1000)² – 2 x 1000 x 1+ (1)2
= 1000000-2000+1
= 10000001-2000=998001

(vii) (599)² = (600 – 1)²
= (600)² -2 x 600 X 1+ (1)2
= 360000 -1200+1
= 360001 – 1200 = 358801

Question 7.
Find the squares of the following numbers by visual method :
(i) 52
(ii) 95
(iii) 505
(iv) 702
(v) 99
Solution:
(a + b)² = a² – ab + ab + b²
(i) (52)² = (50 + 2)²
= 2500 + 100 + 100 + 4
= 2704

(ii) (95)² = (90 + 5)²
= 8100 + 450 + 450 + 25
= 9025

(iii) (505)² = (500 + 5)²
= 250000 + 2500 + 2500 + 25
= 255025

(iv) (702)² = (700 + 2)²
= 490000 + 1400+ 1400 + 4
= 492804

(v) (99)² = (90 + 9)²
= 8100 + 810 + 810 + 81
= 9801

Exercise 3.4

Question 1.
Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots ?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i)  In 9801−−−−√ ∴ the units digits is 1, therefore, the units digit of the square root can be 1 or 9
(ii) In 799356−−−−−−√ ∴ the units digit is 6
∴ The units digit of the square root can be 4 or 6
(iii) In 7998001−−−−−−−√ ∴ the units digit is 1
∴ The units digit of the square root can be 1 or 9
(iv) In 657666025
∴ The unit digit is 5
∴ The units digit of the square root can be 5

Question 2.
Find the square root of each of the following by prime factorization.
(i) 441
(ii) 196
(iii) 529
(iv) 1764
(v) 1156
(vi) 4096
(vii) 7056
(viii) 8281
(ix) 11664
(x) 47089
(xi) 24336
(xii) 190969
(xiii) 586756
(xiv) 27225
(xv) 3013696
Solution:

Question 3.
Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.
Solution:
Factorising 180,

180 = 2 x 2 x 3 x 3 x 5
Grouping the factors in pairs we see that factor 5 is left unpaired.
∴ Multiply 180 by 5, we get the product 180 x 5 = 900
Which is a perfect square
and square root of 900 = 2 x 3 x 5 = 30

Question 4.
Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorising 147,

147 = 3 x 7×7
Grouping the factors in pairs of the equal factors, we see that one factor 3 is left unpaired
∴ Multiplying 147 by 3, we get the product 147 x 3 = 441
Which is a perfect square
and its square root = 3×7 = 21

Question 5.
Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.
Solution:
Factorising 3645

3645 = 3 x 3 3 x 3 x 3 x 3 x 5
Grouping the factors in pair of the equal factors, we see t at one factor 5 is left unpaired
∴ Dividing 3645 by 5, the quotient 729 will be the perfect square and square root of 729 = 27

Question 6.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorsing 1152,

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
Grouping the factors in pairs of the equal factors, we see that factor 2 is left unpaired.
∴ Dividing by 2, the quotient 576 is a perfect square .
∴ Square root of 576, it is 24

Question 7.
The product of two numbers is 1296. If one number is 16 times the others find the numbers.
Solution:
Product of two numbers = 1296
Let one number = x
Second number = 16x

∴ First number = 9
and second number = 16 x 9 = 144

Question 8.
A welfare association collected Rs. 202500 as donation from the residents. If each paid as many rupees as there were residents find the number of residents.
Solution:
Total donation collected = Rs. 202500
Let number of residents = x
Then donation given by each resident = Rs. x
∴ Total collection = Rs. x x x

Question 9.
A society collected Rs. 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute?
Solution:
Total amount collected = Rs. 92.16 = 9216 paise
Let the number of members = x
Then amount collected by each member = x
paise

∴ Number of members = 96
and each member collected = 96 paise

Question 10.
A school collected Rs. 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school ?
Solution:
Total fee collected = Rs. 2304
Let number of students = x
Then fee paid by each student = Rs. x
∴ x x x = 2304 => x² = 2304
∴ x = 2304−−−−√

Question 11.
The area of a square field is 5184 m². A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.
Solution:
The area of a square field = 5184 m²
Let side of the square = x

∴ side of square= 72 m
∴ Perimeter, of square field = 72 x 4 m = 288 m
Perimeter of rectangle = 288 m
Let breadth of rectangular field (b) = x
Then length (l) = 2x
∴ Perimeter = 2 (l + b)
= 2 (2x + x) = 2 x 3x = 6x
= 2 (2x + x) = 2 x 3x = 6x

∴ Length of rectangular field = 2x = 2 x 48 = 96 m
and breadth = 48 m
and area = l x b = 96 x 48 m²
= 4608 m²

Question 12.
Find the least square number, exactly divisible by each one of the numbers :
(i) 6, 9,15 and 20
(ii) 8,12,15 and 20
Solution:

LCM of 6, 9, 15, 20 = 2 x 3 x 5 x 3 x 2 = 180
=2 x 2 x 3 x 3 x 5
We see that after grouping the factors in pairs, 5 is left unpaired
∴ Least perfect square = 180 x 5 = 900

We see that after grouping the factors,
factors 2, 3, 5 are left unpaired
∴ Perfect square =120 x 2 x 3 x 5 = 120 x 30 = 3600

Question 13.
Find the square roots of 121 and 169 by the method of repeated subtraction.
Solution:

Question 14.
Write the prime factorization of the following numbers and hence find their square roots. ^
(i) 7744
(ii) 9604
(iii) 5929
(iv) 7056
Solution:
Factorization, we get:
(i) 7744 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11

Grouping the factors in pairs of equal factors,

Question 15.
The students of class VIII of a school donated Rs. 2401 for PM’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:
Total amount of donation = 2401
Let number of students in VIII = x
∴ Amount donoted by each student = Rs. x

Question 16.
A PT teacher wants to arrange maximum possible number of 6000 students in a Held such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.
Solution:
Number of students = 6000
Students left out = 71
∴ Students arranged in a field = 6000 – 71=5929

Exercise 3.5

Question 1.
Find the square root of each of the long division method.
(I) 12544
(ii) 97344
(iii) 286225
(iv) 390625
(v) 363609
(vi) 974169
(vii) 120409
(viii) 1471369
(ix) 291600
(x) 9653449
(xi) 1745041
(xii) 4008004
(xiii) 20657025
(xiv) 152547201
(jcv) 20421361
(xvi) 62504836
(xvii) 82264900
(xviii) 3226694416
(xix)6407522209
(xx) 3915380329
Solution:
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Question 2.
Find the least number which must be subtracted from the following numbers to make them a perfect square :
(i) 2361
(ii) 194491
(iii) 26535
(iv) 16160
(v) 4401624
Solution:
(i) 2361
Finding the square root of 2361

We get 48 as quotient and remainder = 57
∴ To make it a perfect square, we have to subtract 57 from 2361
∴ Least number to be subtracted = 57
(ii) 194491
Finding the square root of 194491

We get 441 as quotient and remainder = 10
∴ To make it a perfect square, we have to subtract 10 from 194491
∴ Least number to be subtracted = 10
(iii) 26535
Finding the square root of 26535

We get 162 as quotient and 291 as remainder
∴ To make it a perfect square, we have to subtract 291 from 26535
∴ Least number to be subtracted = 291
(iv)16160
Finding the square root of 16160

We get 127 as quotient and 31 as remainder
∴ To make it a perfect square, we have to subtract 31 from 16160
∴ Least number to be subtracted = 31
(v) 4401624
Find the square root of 4401624

We get 2098 as quotient and 20 as remainder
∴ To make it a perfect square, we have to subtract 20 from 4401624
∴ Least number to be subtracted = 20

Question 3.
Find the least number which must be added to the following numbers to make them a perfect square :
(i) 5607
(ii) 4931
(iii) 4515600
(iv) 37460
(v) 506900
Solution:
(i) 5607

Finding the square root of 5607, we see that 74² = 5607- 131 =5476 and 75² = 5625
∴ 5476 < 5607 < 5625
∴ 5625 – 5607 = 18 is to be added to get a perfect square
∴ Least number to be added = 18
(ii) 4931

Finding the square root of 4931, we see that 70²= 4900
∴ 71² = 5041 4900 <4931 <5041
∴ 5041 – 4931 = 110 is to be added to get a perfect square.
∴ Least number to be added =110
(iii) 4515600

Finding the square root of 4515600, we see
that 2124² = 4511376
and 2 1 25² = 45 1 56 25
∴ 4511376 <4515600 <4515625
∴ 4515625 – 4515600 = 25 is to be added to get a perfect square.
∴ Least number to be added = 25
(iv) 37460

Finding the square root of 37460
that 193² = 37249, 194² = =37636
∴ 37249 < 37460 < 37636
∴ 37636 – 37460 = 176 is to be added to get a perfect square.
∴ Least number to be added =176
(v) 506900

Finding the square root of 506900, we see that
711² = 505521, 712² = 506944
∴ 505521 < 506900 < 506944
∴ 506944 – 506900 = 44 is to be added to get a perfect square.
∴ Least number to be added = 44

Question 4.
Find the greatest number of 5 digits which is a perfect square.
Solution:
Greatest number of 5-digits = 99999 Finding square root, we see that 143 is left as remainder

∴ Perfect square = 99999 – 143 = 99856 If we add 1 to 99999, it will because a number of 6 digits
∴ Greatest square 5-digits perfect square = 99856

Question 5.
Find the least number of four digits which is a perfect square.
Solution:
Least number of 4-digits = 10000
Finding square root of 1000
We see that if we subtract 39

From 1000, we get three digit number
∴ We shall add 124 – 100 = 24 to 1000 to get a
perfect square of 4-digit number
∴ 1000 + 24 = 1024
∴ Least number of 4-digits which is a perfect square = 1024

Question 6.
Find the least number of six-digits which is a perfect square.
Solution:
Least number of 6-digits = 100000

Finding the square root of 100000, we see that if we subtract 544, we get a perfect square of 5-digits.
So we shall add
4389 – 3900 = 489
to 100000 to get a perfect square
Past perfect square of six digits= 100000 + 489 =100489

Question 7.
Find the greatest number of 4-digits which is a perfect square.
Solution:
Greatest number of 4-digits = 9999

Finding the square root, we see that 198 has been left as remainder
∴ 4-digit greatest perfect square = 9999 – 198 = 9801

Question 8.
A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8160, find the number of soldiers in each row.
Solution:
Total number of soldiers = 8160 Soldiers left after arranging them in a square = 60
∴ Number of soldiers which are standing in a square = 8160 – 60 = 8100

Question 9.
The area of a square field is 60025 m². A man cycle along its boundry at 18 km/hr. In how much time will be return at the starting point.
Solution:
Area of a square field = 60025 m²

Question 10.
The cost of levelling and turfing a square lawn at Rs. 250 per m² is Rs. 13322.50. Find the cost of fencing it at Rs. 5 per metre ?
Solution:
Cost of levelling a square field = Rs. 13322.50
Rate of levelling = Rs. 2.50 per m²

and perimeter = 4a = 4 x 73 = 292 m Rate of fencing the field = Rs. 5 per m
∴ Total cost of fencing = Rs. 5 x 292 = Rs. 1460

Question 11.
Find the greatest number of three digits which is a perfect square.
Solution:
3-digits greatest number = 999

Finding the square root, we see that 38 has been left
∴ Perfect square = 999 – 38 = 961
∴ Greatest 3-digit perfect square = 961

Question 12.
Find the smallest number which must be added to 2300 so that it becomes a perfect square.
Solution:
Finding the square root of 2300

We see that we have to add 704 – 700 = 4 to 2300 in order to get a perfect square
∴ Smallest number to be added = 4

Exercise 3.6

Question 1.
Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots ?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i)  In 9801−−−−√ ∴ the units digits is 1, therefore, the units digit of the square root can be 1 or 9
(ii) In 799356−−−−−−√ ∴ the units digit is 6
∴ The units digit of the square root can be 4 or 6
(iii) In 7998001−−−−−−−√ ∴ the units digit is 1
∴ The units digit of the square root can be 1 or 9
(iv) In 657666025
∴ The unit digit is 5
∴ The units digit of the square root can be 5

Question 2.
Find the square root of each of the following by prime factorization.
(i) 441
(ii) 196
(iii) 529
(iv) 1764
(v) 1156
(vi) 4096
(vii) 7056
(viii) 8281
(ix) 11664
(x) 47089
(xi) 24336
(xii) 190969
(xiii) 586756
(xiv) 27225
(xv) 3013696
Solution:

Question 3.
Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.
Solution:
Factorising 180,

180 = 2 x 2 x 3 x 3 x 5
Grouping the factors in pairs we see that factor 5 is left unpaired.
∴ Multiply 180 by 5, we get the product 180 x 5 = 900
Which is a perfect square
and square root of 900 = 2 x 3 x 5 = 30

Question 4.
Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorising 147,

147 = 3 x 7×7
Grouping the factors in pairs of the equal factors, we see that one factor 3 is left unpaired
∴ Multiplying 147 by 3, we get the product 147 x 3 = 441
Which is a perfect square
and its square root = 3×7 = 21

Question 5.
Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.
Solution:
Factorising 3645

3645 = 3 x 3 3 x 3 x 3 x 3 x 5
Grouping the factors in pair of the equal factors, we see t at one factor 5 is left unpaired
∴ Dividing 3645 by 5, the quotient 729 will be the perfect square and square root of 729 = 27

Question 6.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorsing 1152,

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
Grouping the factors in pairs of the equal factors, we see that factor 2 is left unpaired.
∴ Dividing by 2, the quotient 576 is a perfect square .
∴ Square root of 576, it is 24

Question 7.
The product of two numbers is 1296. If one number is 16 times the others find the numbers.
Solution:
Product of two numbers = 1296
Let one number = x
Second number = 16x

∴ First number = 9
and second number = 16 x 9 = 144

Question 8.
A welfare association collected Rs. 202500 as donation from the residents. If each paid as many rupees as there were residents find the number of residents.
Solution:
Total donation collected = Rs. 202500
Let number of residents = x
Then donation given by each resident = Rs. x
∴ Total collection = Rs. x x x

Question 9.
A society collected Rs. 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute?
Solution:
Total amount collected = Rs. 92.16 = 9216 paise
Let the number of members = x
Then amount collected by each member = x
paise

∴ Number of members = 96
and each member collected = 96 paise

Question 10.
A school collected Rs. 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school ?
Solution:
Total fee collected = Rs. 2304
Let number of students = x
Then fee paid by each student = Rs. x
∴ x x x = 2304 => x² = 2304
∴ x = 2304−−−−√

Question 11.
The area of a square field is 5184 m². A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.
Solution:
The area of a square field = 5184 m²
Let side of the square = x

∴ side of square= 72 m
∴ Perimeter, of square field = 72 x 4 m = 288 m
Perimeter of rectangle = 288 m
Let breadth of rectangular field (b) = x
Then length (l) = 2x
∴ Perimeter = 2 (l + b)
= 2 (2x + x) = 2 x 3x = 6x
= 2 (2x + x) = 2 x 3x = 6x

∴ Length of rectangular field = 2x = 2 x 48 = 96 m
and breadth = 48 m
and area = l x b = 96 x 48 m²
= 4608 m²

Question 12.
Find the least square number, exactly divisible by each one of the numbers :
(i) 6, 9,15 and 20
(ii) 8,12,15 and 20
Solution:

LCM of 6, 9, 15, 20 = 2 x 3 x 5 x 3 x 2 = 180
=2 x 2 x 3 x 3 x 5
We see that after grouping the factors in pairs, 5 is left unpaired
∴ Least perfect square = 180 x 5 = 900

We see that after grouping the factors,
factors 2, 3, 5 are left unpaired
∴ Perfect square =120 x 2 x 3 x 5 = 120 x 30 = 3600

Question 13.
Find the square roots of 121 and 169 by the method of repeated subtraction.
Solution:

Question 14.
Write the prime factorization of the following numbers and hence find their square roots. ^
(i) 7744
(ii) 9604
(iii) 5929
(iv) 7056
Solution:
Factorization, we get:
(i) 7744 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11

Grouping the factors in pairs of equal factors,

Question 15.
The students of class VIII of a school donated Rs. 2401 for PM’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:
Total amount of donation = 2401
Let number of students in VIII = x
∴ Amount donoted by each student = Rs. x

Question 16.
A PT teacher wants to arrange maximum possible number of 6000 students in a Held such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.
Solution:
Number of students = 6000
Students left out = 71
∴ Students arranged in a field = 6000 – 71=5929

Exercise 3.7

Find the square root of the following numbers in decimal form :

Question 1.
84.8241
Solution:

Question 2.
0.7225
Solution:

Question 3.
0.813604
Solution:

Question 4.
0.00002025
Solution:

Question 5.
150.0625
Solution:

Question 6.
225.6004
Solution:

Question 7.
3600.720036
Solution:

Question 8.
236.144689
Solution:

Question 9.
0.00059049
Solution:

Question 10.
176.252176
Solution:

Question 11.
9998.0001
Solution:

Question 12.
0.00038809
Solution:

Question 13.
What is that fraction which when multiplied by itself gives 227.798649 ?
Solution:

Question 14.
square playground is 256.6404 square metres. Find the length of one side of the playground.
Solution:
Area of square playground = 256.6404 sq. m

Question 15.
What is the fraction which when multiplied by itself gives 0.00053361 ?
Solution:

Question 16.
Simplify :


Solution:

Question 17.
Evaluate 50625−−−−−√ and hence find the value of 506.25−−−−−√+5.0625−−−−−√.
Solution:

Question 18.
Find the value of 103.0225−−−−−−−√ and hence And the value of

Solution:

Exercise 3.8

Question 1.
Find the square root of each of the following correct to three places of decimal :
(i) 5
(ii) 7
(iii) 17
(iv) 20
(v) 66
(vi) 427
(vii) 1.7
(vii’) 23.1
(ix) 2.5
(x) 237.615
(xi) 15.3215
(xii) 0.9
(xiii) 0.1
(xiv) 0.016
(xv) 0.00064
(xvi) 0.019
(xvii) 78
(xviii) 512
(xix) 2 12
(xx) 287 88
Solution:

Question 2.
Find the square root of 12.0068 correct to four decimal places.
Solution:

Question 3.
Find the square root of 11 correct to five decimal places.
Solution:
https://googleads.g.doubleclick.net/pagead/ads?client=ca-pub-7601472013083661&outp/www.learninsta.com&dtd=1834

Question 4.
Given that √2, = 1-414, √3 = 1.732, √5 = 2.236 and √7 = 2.646. Evaluate each of the following :

Solution:

Question 5.
Given that √2 = 1-414, √3 = 1-732, √5= 2.236 and √7= 2.646, find the square roots of the following :

Solution:

Exercise 3.9

Using square root table, find the square roots of the following :

Question 1.
7
Solution:

Question 2.
15
Solution:

Question 3.
74
Solution:

Question 4.
82
Solution:

Question 5.
198
Solution:

Question 6.
540
Solution:

Question 7.
8700
Solution:

Question 8.
3509
Solution:

Question 9.
6929
Solution:

Question 10.
25725
Solution:

Question 11.
1312
Solution:

Question 12.
4192
Solution:

Question 13.
4955
Solution:

Question 14.
99144
Solution:

Question 15.
57169
Solution:

Question 16.
101169
Solution:

Question 17.
13.21
Solution:

Question 18.
21.97
Solution:

Question 19.
110
Solution:

Question 20.
1110
Solution:

Question 21.
11.11
Solution:

Question 22.
The area of a square field is 325 m². Find the approximate length of one side of the field.
Solution:

Question 23.
Find the length of a side of a square, whose area is equal to the area bf the rectangle with sides 240 m and 70 m.
Solution:

Exercise 2.1

Question 1.
Express the following numbers in standard form :
(i) 6020000000000000
(ii) 0.00000000000942
(iii) 0.00000000085
(iv) 846 X 10⁷
(v) 3759 x 10^-4
(vi) 0.00072984
(vii) 0.000437 x 10⁴ 
(Viii) 4 + 100000
Solution:

Question 2.
Write the following numbers in the usual form :
(i) 4.83 x 10⁷
(ii) 3.02 x 10^-6
(iii) 4.5 x 10⁴
(iv) 3 x 10^-8
(v) 1.0001 x 10⁹
(vi) 5.8 x 10²
(vii) 3.61492 x 10⁶
(viii) 3.25 x 10^-7
Solution:

Exercise 2.2

Question 1.
Write each of the following in exponential form :

Solution:

Question 2.
Evaluate :

Solution:

Question 3.
Express each of the following as a rational number in the form pq:

Solution:

Question 4.
Simplify :

Solution:

Question 5.
Express each of the following rational numbers with a negative exponent :

Solution:

Question 6.
Express each of the following rational numbers with a positive exponent :

Solution:

Question 7.
Simplify :

Solution:

Question 8.
By what number should 5^-1 be multiplied so that the product may be equal to (-7)^-1 ?
Solution:

Question 9.
By what number should (12)−1 be multiplied so that the product may be equal to (−47)−1 ?
Solution:

Question 10.
By what number should (-15)^-1 be divided so that the quotient may be equal to (-5)^-1 ?
Solution:

Question 11.
By what number should (53)−2 be multiplied so that the product may be (73)−1 ?
Solution:

Question 12.
Find x, if

Solution:

Question 13.

Solution:

Question 14.
Find the value of x for which 5^2x + 5^-3 = 5⁵.
Solution:

Exercise 2.3

Question 1.
Express the following numbers in standard form :
(i) 6020000000000000
(ii) 0.00000000000942
(iii) 0.00000000085
(iv) 846 X 10⁷
(v) 3759 x 10^-4
(vi) 0.00072984
(vii) 0.000437 x 10⁴ 
(Viii) 4 + 100000
Solution:

Question 2.
Write the following numbers in the usual form :
(i) 4.83 x 10⁷
(ii) 3.02 x 10^-6
(iii) 4.5 x 10⁴
(iv) 3 x 10^-8
(v) 1.0001 x 10⁹
(vi) 5.8 x 10²
(vii) 3.61492 x 10⁶
(viii) 3.25 x 10^-7
Solution:

Exercise 1.1

Question 1.
Add the following rational numbers:

Solution:

Question 2.
Add the following rational numbers:

Solution:

Question 3.
Simplify:

Solution:

Question 4.
Add and Express the sum as mixed fraction:

Solution:

Exercise 1.2

Question 1.
Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers :

Solution:

Question 2.
Verify associativity of addition of rational numbers i.e., (A: + y) + z = x + (y + z), when :

Solution:

Question 3.
Write the additive inverse of each of the following rational numbers :

Solution:

Question 4.
Write the negative (additive inverse) of each of the following :

Solution:

Question 5.
Using commutativity and associativity of addition of rational numbers, express ‘iach of the following as a rational number :

Solution:

Question 6.
Re-arrange suitably and find the sum in each of the following :


Solution:

Exercise 1.3

Question 1.
Subtract the first rational number from the second in each of the following :



Solution:

Question 2.
Evaluate each of the following :

Solution:

Question 3.
The sum of two numbers is 59. If one of the numbers is 13, find the other.
Solution:

Question 4.
The sum of two numbers is −13. If one of the numbers is −123, find the other.
Solution:

Question 5.
The sum of two numbers is −43. If one of the number is -5, find the
Solution:

Question 6.
The sum of two rational numbers is -8. If one of the numbers is −157 find the other.
Solution:

Question 7.
What should be added to so as to −78 get 59 ?
Solution:

Question 8.
What number should be added to −511 so as to get 263 ?
Solution:

Question 9.
What number should be added to −57 to get −23 ?
Solution:

Question 10.
What number should be subtracted from −53 to get 56 ?
Solution:

Question 11.
What number should be subtracted from 37 to get 54 ?
Solution:

Question 12.
What should be added to (23+35) to get −1215 ?
Solution:

Question 13.
What should be added to (12+13+15) to get 3 ?
Solution:

Question 14.
What should be subtracted from (34−23) to get −16 ?
Solution:

Question 15.
Fill in the blanks :

Solution:

Exercise 1.4

Question 1.
Subtract the first rational number from the second in each of the following :



Solution:

Question 2.
Evaluate each of the following :

Solution:

Question 3.
The sum of two numbers is 59. If one of the numbers is 13, find the other.
Solution:

Question 4.
The sum of two numbers is −13. If one of the numbers is −123, find the other.
Solution:

Question 5.
The sum of two numbers is −43. If one of the number is -5, find the
Solution:

Question 6.
The sum of two rational numbers is -8. If one of the numbers is −157 find the other.
Solution:

Question 7.
What should be added to so as to −78 get 59 ?
Solution:

Question 8.
What number should be added to −511 so as to get 263 ?
Solution:

Question 9.
What number should be added to −57 to get −23 ?
Solution:

Question 10.
What number should be subtracted from −53 to get 56 ?
Solution:

Question 11.
What number should be subtracted from 37 to get 54 ?
Solution:

Question 12.
What should be added to (23+35) to get −1215 ?
Solution:

Question 13.
What should be added to (12+13+15) to get 3 ?
Solution:

Question 14.
What should be subtracted from (34−23) to get −16 ?
Solution:

Question 15.
Fill in the blanks :

Solution:

Exercise 1.5

Question 1.
Multiply:

Solution:

Question 2.
Multiply:

Solution:

Question 3.
Simplify each of the following and express the result as a rational number in standard form :

Solution:

Question 4.
Simplify :


Solution:

Question 5.
Simplify :

Solution:

Exercise 1.6

Question 1.
Verify the property : x x y = y x x by taking :

Solution:

Question 2.
Verify the property : x x (y x z) = (x x y) x z by taking :

Solution:

Question 3.
Verify the property :xx(y + 2) = xxy + x x z by taking :

Solution:

Question 4.
Use the distributivity of multiplication of rational numbers over their addition to simplify :

Solution:

Question 5.
Find the multiplicative inverse (reciprocal) of each of the following rational numbers :


Solution:

Question 6.
Name the property of multiplication of rational numbers illustrated by the following statements :

Solution:

Exercise 1.7

Question 1.
Divide :

Solution:

Question 2.
Find the value and express as a rational number in standard form :

Solution:

Question 3.
The product of two rational numbers is15. If one of the numbers is -10, find the other.
Solution:

Question 4.
The product of two rational numbers is−98 if one of the numbers is −415, find other.
Solution:

Question 5.
By what number should we multiply −16 so that the product may be −239 ?
Solution:

Question 6.
By what number should we multiply −1528 so that the product may be −57 ?
Solution:

Question 7.
By what number should we multiply −813 so that the product may be 24 ?
Solution:

Question 8.
Bv what number should −34 multiplied in order to produce 23 ?
Solution:

Question 9.
Find (x +y) + (x – y), if

Solution:

Question 10.
The cost of 7 23 metres of rope is Rs 12 34.Find its cost per metre.
Solution:

Question 11.
The cost of 2 13 metres of cloth is Rs. 75 14Find the cost of cloth per metre.
Solution:

Question 12.
By what number should −3316 be divided to get −114 ?
Solution:

Question 13.
Divide the sum of −135 and 127 by the product of −317 and −12.
Solution:

Question 14.
Divide the sum of 6512 and 127 bv their difference.
Solution:

Question 15.
If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser ?
Solution:

Exercise 1.8

Question 1.
Find a rational number between -3 and 1.
Solution:

Question 2.
Find any five rational number less than 1.
Solution:

Question 3.
Find four rational numbers between −29 and 59 .
Solution:

Question 4.
Find two rational numbers between 15 and 12 .
Solution:

Question 5.
Find ten rational numbers between 14 and 12 .
Solution:

Question 6.
Find ten rational numbers between −25 and 12 .
Solution:

Question 7.
Find ten rational numbers between 35 and 34 .
Solution:

Exercise 23 A

Page No 238:

Question 1:

The number of students who absented from the class during a week are given below:

Day	Monday	Tuesday	Wednesday	Thursday	Friday
No. of absentees	6	2	4	2	8

Take the scale Figure = 2 absentees.
Draw the pictograph.

ANSWER:

Day	Number of absentees
Monday
Tuesday
Wednesday
Thursday
Friday

Page No 238:

Question 2:

The number of stools in five rooms of a school are given below:

Room number	I	II	III	IV	V
Number of stools	30	40	60	50	20

Taking the scale Figure = 10 stools, draw the pictograph.

ANSWER:

Room Number	Number of stools
I
II
III
IV
V

Page No 238:

Question 3:

In a class test, the number of students passed in various subjects are given below.

Subject	English	Mathematics	Hindi	Drawing
Number of students passed	15	25	10	20

Taking the scale Figure = 5 successful students, draw the pictograph.

ANSWER:

Subject	Number of students passed
English
Mathematics
Hindi
Drawing

Page No 238:

Question 4:

The number of fans sold by a shopkeeper during 6 months are given below:

Month	March	April	May	June	July	August
Number of fans sold	30	40	60	50	20	30

Taking the scale Figure = 10 fans sold, draw the pictograph.

ANSWER:

Month	Number of fans sold
March
April
May
June
July
August

Page No 239:

Question 5:

The following pictograph shows different kinds of trees planted in a park. Each symbol represents 8 trees. Look at the pictograph adn answer the questions given below.

Banyan tree	Figure
Neem tree	Figure
Mango tree	Figure

(i) How many mango trees are there?
(ii) How many banyan trees are there?
(iii) How many neem trees are there?
(iv) How many  trees are there in all?

ANSWER:

(i) Number of mango trees = 3××8 = 24

(ii) Number of banyan trees = 4××8 = 32

(iii) Number of neem trees = 5××8 = 40

(iv) Total number of trees = Number of mango trees + Number of banyan trees + Number of neem trees
                                        = 24+32+40
                                        = 96

Page No 239:

Question 6:

The following pictograph shows the number of scooters sold by a company during a week.

Scale used is Figure = 6 scooters sold.

Day	Figure
Monday	Figure
Tuesday	Figure
Wednesday	Figure
Thursday	Figure
Friday	Figure
Saturday	Figure

Study the pictograph carefully and answer the questions given below.
(i) How many scooters were sold on Monday?
(ii) How many scooters were sold on Tuesday?
(iii) On what day of the week was the sale of the scooters maximum? How many scooters were sold on that day?
(iv) On what day of the week was the sale of the scooters minimum? How many scooters were sold on that day?

ANSWER:

(i) Number of scooters sold on Monday = 5××6 = 30

(ii) Number of scooters sold on Tuesday = 4××6 = 24

(iii) The sale was maximum on Friday.

      Number of scooters sold on Friday = 7××6 = 42

(iv) The sale was minimum on Saturday.

      Number of scooters sold on Saturday = 2××6 = 12
     

Exercise 24A

Page No 243:

Question 1:

Look at the bar graph given below.
Figure
Read it carefully and answer the questions given below:
(i) What information does the bar graph give?
(ii) In which subject is the student poorest?
(iii) In which subject is the student best?
(iv) In which subjects did he get more than 40 marks?

ANSWER:

(i) The given bar graph provides information about the marks obtained by a student in an examinations in four different subjects.

(ii) The student is poorest in science because the height of the bar representing the marks obtained in science is the lowest.

(iii) The student is best in mathematics because the height of the bar representing the marks obtained in mathematics is the highest.

(iv) In Hindi and mathematics he got more than 40 marks. He scored 55 marks in Hindi and 70 marks in mathematics.

Page No 243:

Question 2:

In a survey of 60 families of a colony, the number of members in each family was recorded and the data has been represented by the bar graph given below:
Figure
Read the bar graph carefully and answer the following questions:
(i) What information does the bar graph give?
(ii) How many families have 3 members?
(iii) How many couples have no child?
(iv) Which type of family is the most common?

ANSWER:

(i) The given bar graph provides information about the number of members in 60 families of a colony.

(ii) There are 10 families that have 3 members.

(iii) There are 5 couples that do not have children.

(iv) The families having two children is most common because the bar of the families having four members has the highest height.

Page No 244:

Question 3:

Look at the bar graph given below:
Figure
Study the bar graph carefully and answer the questions given below:
(i) In which week was the production maximum?
(ii) In which week was the production minimum?
(iii) What is the average production during these five weeks?
(iv) How many cycles were produced in the first 3 weeks?

ANSWER:

(i) Production was maximum in the 2nd week.

(ii) Production was minimum in the 4th week.

(iii) Average production of these five weeks =Total production of all weeksNumber of weeksTotal production of all weeksNumber of weeks
     
                                                                            =1000 + 600 + 800 + 500 + 7005=36005=720=1000 + 600 + 800 + 500 + 7005=36005=720

(iv) Number of cycles produced in the first 3 week = Production in the 1st week + Production in the 2nd week + Production in the 3rd week

                                                                                 = 600 + 1000 + 800
                                                                                 = 2400 cycles

Page No 244:

Question 4:

51 students from a locality use different modes of transport to school, as shown by the bar graph given below:
Figure
Look at the bar graph given above and answer the questions given below:
(i) What does the above bar graph show?
(ii) Which mode of transport is used by maximum number of students?
(iii) How many students use bus for going to school?
(iv) How many students of the locality do not use bus for going to school?

ANSWER:

(i) The given bar graph gives information about the different modes of transport used by the students for going to school.

(ii) Since the height of the bar representing the bicycles is the highest, bicycles are used by maximum number of students. 16 students use bicycles for going to school.

(iii) 14 students use bus for going to school.

(iv) Number of students who do not use bus = Total number of students − Number of students who use bus
                                                                      = 51 − 14
                                                                      = 37 students

Page No 213:

Exercise 19A

Question 1:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A cuboid has
(a) length
(b) length and breadth only
(c) length, breadth and height
(d) thickness only

ANSWER:

A cuboid is a 3-dimensional figure. So, a cuboid has length, breadth and height. 

Hence, the correct answer is option (c).

Page No 213:

Question 2:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A dice is an example of a
(a) cuboid
(b) cube
(c) cone
(d) cylinder

ANSWER:

(b) cube

Page No 213:

Question 3:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A gas pipe is an example of a
(a) cuboid
(b) cube
(c) cone
(d) cylinder

ANSWER:

(d) cylinder

Page No 213:

Question 4:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A football is an example of a
(a) cylinder
(b) cone
(c) sphere
(d) none of these

ANSWER:

(c) sphere

Page No 213:

Question 5:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A brick is an example of a
(a) cube
(b) cuboid
(c) prism
(d) cylinder

ANSWER:

(b) cuboid

Page No 213:

Question 6:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
An ice-cream cone is an example of a
(a) cuboid
(b) cube
(c) pyramid
(d) none of these

ANSWER:

(d) None of these
Its a cone.

Page No 213:

Question 7:

Fill in the blanks:
(i) An object that occupies space is called a …… .
(ii) A cuboid has …… faces, …… edges and …… vertices.
(iii) The …… faces of a cuboid are identical.
(iv) A …… has no vertex and no edge.
(v) All the faces of a …… are identical.
(vi) A square pyramid has …… lateral triangular faces and …… edges.
(vii) A triangular pyramid has …… triangular lateral faces and …… edges.
(viii) A triangular prism has …… vertices, …… rectangular lateral faces, …… triangular bases and …… edges.

ANSWER:

(i) An object that occupies space is called a solid.
(ii) A cuboid has 6 rectangular faces, 12 edges and 8 vertices.
(iii) The opposite faces of a cuboid are identical.
(iv) A sphere has no vertex and no edge.
(v) All the faces of a cube are identical.
(vi) A square pyramid has 4 lateral triangular faces and 8 edges.
(vii) A triangular pyramid has 3 triangular lateral faces and 6 edges.
(viii) A triangular prism has 6 vertices, 3 rectangular lateral faces, 2 triangular bases and 9 edges.

Page No 213:

Question 8:

Give examples of four objects which are in the shape of:
(a) a cone
(b) a cuboid
(c) a cylinder

ANSWER:

(a) An ice cream cone, a conical tent, a clown’s cap and a conical vessel are in the shape of a cone.

(b) A wooden box, a match box, a brick and an almirah are in the shape of a cuboid.

(c) A measuring jar, a gas cylinder, a test tube and a circular pillar are in the shape of a cylinder.

Page No 218:

Exercise 20A

Question 1:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A square has
(a) one line of symmetry
(b) two lines of symmetry
(c) three lines of symmetry
(d) four lines of symmetry

ANSWER:

A square has four lines of symmetry. Hence, the correct answer is option (d).

Page No 218:

Question 2:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A rectangle is symmetrical about
(a) each one of its sides
(b) each one of its diagonals
(c) a line joining the midpoints of its opposite sides
(d) none of these

ANSWER:

(c) a line joining the midpoints of its opposite sides

Page No 218:

Question 3:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A rhombus is symmetrical about
(a) the line joining the midpoints of its opposite sides
(b) each of its diagonals
(c) perpendicular bisector of each of its sides
(d) none of these

ANSWER:

(b) each of its diagonals

Page No 218:

Question 4:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A circle has
(a) no line of symmetry
(b) one line of symmetry
(c) two lines of symmetry
(d) an unlimited number of lines of symmetry

ANSWER:

(d) an unlimited number of lines of symmetry

This is because a circle has infinite number of diameters. Also, a circle is symmetrical about each of its diameter.

Page No 218:

Question 5:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A scalene triangle has
(a) no line of symmetry
(b) one line of symmetry
(c) two lines of symmetry
(d) three lines of symmetry

ANSWER:

(a) no line of symmetry

Page No 219:

Question 6:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
ABCD is a kite in which AB = AD and BC = DC.
The kite is symmetrical about
(a) the diagonal AC
(b) the diagonal BD
(c) none of these
Figure

ANSWER:

(a) the diagonal AC

Page No 219:

Question 7:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
The letter O of the English alphabet has
(a) no line of symmetry
(b) one line of symmetry
(c) two lines of symmetry
(d) none of these

ANSWER:

(c) two lines of symmetry

Page No 219:

Question 8:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
The letter Z of the English alphabet has
(a) no line of symmetry
(b) one line of symmetry
(c) two lines of symmetry
(d) none of these

ANSWER:

(a) no line of symmetry

Page No 219:

Question 9:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
Draw the line (or lines) of symmetry of each of the following figures.
Figure

ANSWER:

(i)


(ii)


(iii)

(iv)

Page No 219:

Question 10:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
Which of the following statements are true and which are false?
(i) A parallelogram has no line of symmetry.
(ii) An angle with equal arms has its bisector as the line of symmetry.
(iii) An equilateral triangle has three lines of symmetry.
(iv) A rhombus has four lines of symmetry.
(v) A square has four lines of symmetry.
(vi) A rectanle has two lines of symmetry.
(vii) Each one of the letters H, I, O, X of the English alphabet has two lines of symmetry.

ANSWER:

(i)   True

(ii)  True

(iii) True
An equilateral triangle is symmetrical about each one of the bisectors of its interior angle. Also, it has three bisectors.

(iv) False
A rhombus has two lines of symmetry. It is symmetrical about each one of its diagonals.

(v)  True
A square is symmetrical about each one of its diagonals and the lines joining the midpoints of the opposite sides.

(vi) True
A rectangle is symmetrical about the lines joining the midpoints of the opposite sides.

(vii) True

Page No 222:

Exercise 21A

Question 1:

Find the perimeter of a rectangle in which:
(i) length = 16.8 cm and breadth = 6.2 cm
(ii) length = 2 m 25 cm and breadth = 1 m 50 cm
(iii) length = 8 m 5 dm and breadth = 6 m 8 dm

ANSWER:

We know: Perimeter of a rectangle = 2×(Length+Breadth)2×(Length+Breadth)

(i) Length = 16.8 cm
    Breadth = 6.2 cm
    Perimeter = 2×(Length+Breadth)2×(Length+Breadth)
                   = 2×(16.8+6.2) =46 cm2×(16.8+6.2) =46 cm
(ii) Length = 2 m 25 cm
                  =(200+25) cm       (1 m = 100 cm )
                  = 225 cm  
    Breadth =1 m 50 cm
                  = (100+50) cm      (1 m = 100 cm )
                  = 150 cm    
    Perimeter = 2×(Length+Breadth)2×(Length+Breadth)
                     = 2×(225+150) =750 cm2×(225+150) =750 cm
(iii) Length = 8 m 5 dm
                   = (80+5) dm   (1 m = 10 dm )
                   = 85 dm     
       Breadth = 6 m 8 dm
                     = (60+8) dm   (1 m = 10 dm )
                     = 68 dm  
    Perimeter = 2×(Length+Breadth)2×(Length+Breadth)
                   = 2×(85+68) =306 dm2×(85+68) =306 dm

Page No 222:

Question 2:

Find the cost of fencing a rectangular field 62 m long and 33 m wide at Rs 16 per metre.

ANSWER:

Length of the field = 62 m
Breadth of the field = 33 m
Perimeter of the field = 2(l + b) units
                                = 2(62 + 33) m =190 m
Cost of fencing per metre = Rs 16
Total cost of fencing = Rs (16××190) = Rs 3040

Page No 222:

Question 3:

The length and the breadth of a rectangular field are in the ratio 5 : 3. If its perimeter is 128 m, find the dimensions of the field.

ANSWER:

Let the length of the rectangle be 5x m.
Breadth of the rectangle = 3x m
Perimeter of the rectangle = 2(l + b)
                                    = 2(5x + 3x) m
                                    = (16x) m
It is given that the perimeter of the field is 128 m.

∴16x=128⇒x=12816=8∴Length =(5×8)=40mBreadth =(3×8)=24m∴16x=128⇒x=12816=8∴Length =(5×8)=40mBreadth =(3×8)=24m

Page No 222:

Question 4:

The cost of fencing a rectangular field at Rs 18 per metre is Rs 1980. If the width of the field is 23 m, find its length.

ANSWER:

Total cost of fencing = Rs 1980
Rate of fencing = Rs 18 per metre

Perimeter of the field = Total costRate=Rs 1980Rs 18/m=(198018) m=110 mTotal costRate=Rs 1980Rs 18/m=(198018) m=110 m

Let the length of the field be x metre.
Perimeter of the field = 2(x + 23) m

∴2(x+23)=110⇒(x+23)=55x=(55−23)=32∴2(x+23)=110⇒(x+23)=55x=(55-23)=32
Hence, the length of the field is 32 m.

Page No 222:

Question 5:

The length and the breadth of a rectangular field are in the ratio 7 : 4. The cost of fencing the field at Rs 25 per metre is Rs 3300. Find the dimensions of the field.

ANSWER:

Total cost of fencing = Rs 3300
Rate of fencing = Rs 25/m
Perimeter of the field = Total cost Rate of fencing=(Rs 3300Rs 25/m)=330025 m=132 mTotal cost Rate of fencing=(Rs 3300Rs 25/m)=330025 m=132 m

Let the length and the breadth of the rectangular field be 7x and 4x, respectively.
Perimeter of the field = 2(7x + 4x) = 22x

It is given that the perimeter of the field is 132 m.

∴ 22x=132⇒x=13222=6∴Length of the field =(7 × 6) m=42 mBreadth of the field =(4 × 6) m=24 m∴ 22x=132⇒x=13222=6∴Length of the field =(7 × 6) m=42 mBreadth of the field =(4 × 6) m=24 m

Page No 222:

Question 6:

Find the perimeter of a square, each of whose sides measures:
(i) 3.8 cm
(ii) 4.6 cm
(iii) 2 m 5 dm

ANSWER:

(i) Side of the square = 3.8 cm
    Perimeter of the square = (4××side)
                                    = (4××3.8) = 15.2 cm

(ii) Side of the square = 4.6 cm
     Perimeter of the square = (4××side)
                                     = (4××4.6) = 18.4 cm

(iii) Side of the square = 2 m 5 dm
                                     = (20+5) dm   (1 m = 10 dm)
                                     = 25 dm
      Perimeter of the square = (4××side)
                                       = (4××25) = 100 dm

Page No 222:

Question 7:

The cost of putting a fence around a square field at Rs 35 per metre is Rs 4480. Find the length of each side of the field.

ANSWER:

Total cost of fencing = Rs 4480
Rate of fencing = Rs 35/m
Perimeter of the field = Total costRate=Rs 4480Rs 35/m=448035 m=128 mTotal costRate=Rs 4480Rs 35/m=448035 m=128 m

Let the length of each side of the field be x metres.
Perimeter = (4x) metres
∴4x=128⇒x=1284=32 ∴4x=128⇒x=1284=32 

Hence, the length of each side of the field is 32 m.

Page No 222:

Question 8:

Each side of a square field measures 21 m. Adjacent to this field, there is a rectangular field having its sides in the ratio 4 : 3. If the perimeters of both the fields are equal, find the dimensions of the rectangular field.

ANSWER:

Side of the square field = 21m
Perimeter of the square field = (4××21) m
                                       = 84 m  

Let the length and the breadth of the rectangular field be 4x and 3x, respectively.
 Perimeter of the rectangular field = 2(4x + 3x) = 14x

 Perimeter of the rectangular field = Perimeter of the square field

∴14x=84 ⇒x=8414=6∴14x=84 ⇒x=8414=6
∴ Length of the rectangular field =(4 × 6) m=24 mBreadth of the rectangular field = (3 × 6) m=18 m∴ Length of the rectangular field =(4 × 6) m=24 mBreadth of the rectangular field = (3 × 6) m=18 m

Page No 222:

Question 9:

Find the perimeter of
(i) a triangle of sides 7.8 cm, 6.5 cm and 5.9 cm,
(ii) an equilateral triangle of side 9.4 cm,
(iii) an isosceles triangle with equal sides 8.5 cm each and third side 7 cm.

ANSWER:

(i) Sides of the triangle are 7.8 cm, 6.5 cm and 5.9 cm. 
    Perimeter of the triangle = (First side + Second side + Third Side) cm
                                      = (7.8 + 6.5 + 5.9) cm
                                      = 20.2 cm

(ii) In an equilateral triangle, all sides are equal.
     Length of each side of the triangle = 9.4 cm
     ∴∴ Perimeter of the triangle = (3 ×× Side) cm
                                            = (3 ×× 9.4) cm
                                            = 28.2 cm

(iii) Length of two equal sides = 8.5 cm
       Length of the third side = 7 cm
    ∴∴ Perimeter of the triangle = {(2 ×× Equal sides) + Third side} cm
                                           = {(2 ×× 8.5) + 7} cm
                                           = 24 cm

Page No 222:

Question 10:

Find the perimeter of
(i) a regular pentagon of side 8 cm,
(ii) a regular octagon of side 4.5 cm,
(iii) a regular decagon of side 3.6 cm,

ANSWER:

(i) Length of each side of the given pentagon = 8 cm
   ∴∴ Perimeter of the pentagon = (5××8) cm
                                                  = 40 cm

(ii) Length of each side of the given octagon = 4.5 cm
   ∴∴ Perimeter of the octagon = (8××4.5) cm
                                                = 36 cm

(iii) Length of each side of the given decagon = 3.6 cm
   ∴∴ Perimeter of the decagon = (10××3.6) cm
                                                 = 36 cm

Page No 222:

Question 11:

Find the perimeter of each of the following figures:
Figure

ANSWER:

(i) Perimeter of the figure = Sum of all the sides
                                         =(27 + 35 + 35 + 45) cm
                                         = 142 cm
(ii) Perimeter of the figure = Sum of all the sides
                                         =(18 + 18 + 18 + 18) cm
                                         = 72 cm
(iii) Perimeter of the figure = Sum of all the sides
                                         =(8 + 16 + 4 + 12 + 12 + 16 + 4) cm
                                         = 72 cm

Page No 224:

Exercise 21B

Question 1:

Find the circumference of a circle whose radius is
(i) 28 cm
(ii) 10.5 cm
(iii) 3.5 m

ANSWER:

(i) Radius, r = 28 cm

∴ Circumference of the circle, C=2πr                                                    =(2×227×28)                                =176 cmHence, the circumference of the given circle is 176 cm.∴ Circumference of the circle, C=2πr                                                    =(2×227×28)                                =176 cmHence, the circumference of the given circle is 176 cm.

(ii) Radius, r = 10.5 cm
      ∴ Circumference of the circle, C=2πr=(2×227×10.5)=66 cmHence, the circumference of the given circle is 66 cm.∴ Circumference of the circle, C=2πr=(2×227×10.5)=66 cmHence, the circumference of the given circle is 66 cm.

(iii) Radius, r = 3.5 m
     ∴ Circumference of the circle, C=2πr=(2×227×3.5)=22 mHence, the circumference of the given circle is 22 m.∴ Circumference of the circle, C=2πr=(2×227×3.5)=22 mHence, the circumference of the given circle is 22 m.

Page No 224:

Question 2:

Find the circumference of a circle whose diameter is
(i) 14 cm
(ii) 35 cm
(iii) 10.5 m

ANSWER:

(i)

Circumference=2πr                          =π(2r)                            =π× Diameter of the circle (d)       (Diameter=2×radius)⇒Circumference=Diameter×π Diameter of the given circle is 14 cm.Circumference of the given circle=14×π⇒(14×22 7)=44 cmCircumference of the given circle is 44 cm.Circumference=2πr                          =π(2r)                            =π× Diameter of the circle (d)       (Diameter=2×radius)⇒Circumference=Diameter×π Diameter of the given circle is 14 cm.Circumference of the given circle=14×π⇒(14×22 7)=44 cmCircumference of the given circle is 44 cm.

(ii)
Circumference=2πr                          =π(2r)                        =π×Diameter of the circle(d)       (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 35 cm.⇒Circumference of the given circle=35×π ⇒(35×22 7)=110 cmCircumference of the given circle is 110 cm.Circumference=2πr                          =π(2r)                        =π×Diameter of the circle(d)       (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 35 cm.⇒Circumference of the given circle=35×π ⇒(35×22 7)=110 cmCircumference of the given circle is 110 cm.

(iii)
Circumference=2πr                         =π(2r)                         =π×Diameter of the circle(d)       (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 10.5 m.Circumference of the given circle=10.5×π⇒(10.5×22 7)=33 mCircumference of the given circle is 33 m.Circumference=2πr                         =π(2r)                         =π×Diameter of the circle(d)       (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 10.5 m.Circumference of the given circle=10.5×π⇒(10.5×22 7)=33 mCircumference of the given circle is 33 m.

Page No 224:

Question 3:

Find the radius of a circle whose circumference is 176 cm.

ANSWER:

Let the radius of the given circle be r cm.
Circumference of the circle = 176 cm
Circumference = 2πr2πr
∴ 2πr=176 ⇒r=1762π⇒r=(1762×722) ⇒r=28 The radius of the given circle is 28 cm.∴ 2πr=176 ⇒r=1762π⇒r=(1762×722) ⇒r=28 The radius of the given circle is 28 cm.

Page No 224:

Question 4:

Find the diameter of a wheel whose circumference is 264 cm.

ANSWER:

Let the radius of the circle be r cm.Diameter=2×Radius=2r cmCircumference of the wheel=264 cmCircumference of the wheel=2πr∴2πr=264⇒2r=264π⇒2r=(264×722)⇒2r=84 Diameter of the given wheel is 84 cm.Let the radius of the circle be r cm.Diameter=2×Radius=2r cmCircumference of the wheel=264 cmCircumference of the wheel=2πr∴2πr=264⇒2r=264π⇒2r=(264×722)⇒2r=84 Diameter of the given wheel is 84 cm.

Page No 224:

Question 5:

Find the distance covered by the wheel of a car in 500 revolutions if the diameter of the wheel is 77 cm.

ANSWER:

Radius of the wheel =Diameter of the wheel2Diameter of the wheel2
⇒r=772cm⇒r=772cm
Circumference of the wheel =2π r2π r
=(2×227×772)=242 cm=(2×227×772)=242 cm

In 1 revolution the wheel covers a distance equal to its circumference.

∴ Distance covered by the wheel in 1 revolution=242 cm∴ Distance covered by the wheel in 500 revolutions=(500 × 242) cm                                                                =121000 cm                                                               =1210 m   (100 cm= 1m )                                                               =1.21 km   (1000 m=1 km  )∴ Distance covered by the wheel in 1 revolution=242 cm∴ Distance covered by the wheel in 500 revolutions=(500 × 242) cm                                                                =121000 cm                                                               =1210 m   (100 cm= 1m )                                                               =1.21 km   (1000 m=1 km  )

Page No 224:

Question 6:

The diameter of the wheel of a car is 70 cm. How many revolutions will it make to travel 1.65 km?

ANSWER:

Radius of the wheel(r)=Diameter of the wheel2r=702cm=35 cmCircumference of the wheel = 2πr =(2×227×35)                                   =220 cmIn one revolution, the wheel covers the distance equal to its circumference.∴ 220 cm distance =1 revolution∴1 cm distance =1220 revolution∴1km (or 100000 cm) distance =1×100000220 revolution    (∴ 1 km=100000 cm)∴1.65 km distance = 1.65×100000 220 revolutions                                        = 750   revolutions                 Thus,the wheel will make 750 revolutions to travel 1.65 km.Radius of the wheel(r)=Diameter of the wheel2r=702cm=35 cmCircumference of the wheel = 2πr =(2×227×35)                                   =220 cmIn one revolution, the wheel covers the distance equal to its circumference.∴ 220 cm distance =1 revolution∴1 cm distance =1220 revolution∴1km (or 100000 cm) distance =1×100000220 revolution    (∴ 1 km=100000 cm)∴1.65 km distance = 1.65×100000 220 revolutions                                        = 750   revolutions                 Thus,the wheel will make 750 revolutions to travel 1.65 km.

Page No 226:

Exercise 21C

Question 1:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 12 complete squares.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of complete squares ×× Area of the square
                                         =(12×1) sq cm(12×1) sq cm
                                         =12 sq cm

Page No 226:

Question 2:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 18 complete squares.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of complete squares ×× Area of the square
                                         =(18×1) sq cm(18×1) sq cm
                                         =18 sq cm

Page No 226:

Question 3:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 14 complete squares and 1 half square.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of squares ×× Area of the square
                                         =[(14 × 1) + (1 × 12)]sq cm(14 × 1) + (1 × 12)sq cm
                                         =14121412 sq cm

Page No 226:

Question 4:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 6 complete squares and 4 half squares.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of squares ×× Area of the square
                                         =[(6×1)+(4×12)] sq cm(6×1)+(4×12) sq cm
                                         =8 sq cm

Page No 226:

Question 5:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 9 complete squares and 6 half squares.
    Area of 1 small square = 1 sq cm
     ∴ Area of the figure = Number of squares ×× Area of the square
                                         =[(9 × 1) + (6 × 12)] sq cm(9 × 1) + (6 × 12) sq cm
                                         =12 sq cm

Page No 226:

Question 6:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 16 complete squares.
    Area of 1 small square = 1 sq cm
     ∴ Area of the figure = Number of squares ×× Area of a square
                                         =(16×1) sq cm(16×1) sq cm
                                         =16 sq cm

Page No 226:

Question 7:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

In the given figure, there are 4 complete squares, 8 more than half parts of squares and 4 less than half parts of squares.
We neglect the less than half parts and consider each more than half part of the square as a complete square.

                  ∴ Area = (4 + 8) sq cm
                            = 12 sq cm

Page No 226:

Question 8:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

In the given figure, there are 9 complete squares, 5 more than half parts of squares and 7 less than half parts of squares.
We neglect the less than half parts of squares and consider the more than half squares as complete squares.
∴ Area of the figure = (9 + 5) sq cm
                                      = 14 sq cm

Page No 226:

Question 9:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 14 complete squares and 4 half squares.
    Area of 1 small square = 1 sq cm
     Area of the figure = Number of squares ×× Area of one square
                                         =[(14×1)+(4×12) ]sq cm(14×1)+(4×12) sq cm
                                         =16 sq cm

Page No 229:

Exercise 21D

Question 1:

Find the area of a rectangle whose
(i) length = 46 cm and breadth = 25 cm
(ii) length = 9 m and breadth = 6 m
(iii) length = 14.5 m and breadth = 6.8 m
(iv) length = 12 m 5 cm and breadth = 60 cm
(v) length = 3.5 km and breadth = 2 km

ANSWER:

 (i) Length = 46 cm
Breadth = 25 cm
  Area of the rectangle = (Length ××Breadth) sq units
                              = (46××25) cm² = 1150 cm²
 
(ii) Length = 9 m
Breadth = 6 m
  Area of the rectangle = (Length ××Breadth) sq units
                             = (9××6) m² = 54 m²

 (iii) Length = 14.5 m
Breadth = 6.8 m
  Area of the rectangle = (Length ××Breadth) sq units
                             = (14510×681014510×6810) m² = 98601009860100 m² =98.60 m²

 (iv) Length = 2 m 5 cm
                   = (200+5) cm   (1 m = 100 cm )
                   =205cm
       Breadth = 60 cm
       Area of the rectangle = (Length ××Breadth) sq units
                                    = (205××60) cm² = 12300 cm²

 (v) Length = 3.5 km
Breadth = 2 km
  Area of the rectangle = (Length ××Breadth) sq units
                             = (3.5××2) km² = (3510×2)(3510×2) km² =7 km² 

Page No 230:

Question 2:

Find the area of a square plot of side 14 m.

ANSWER:

Side of the square plot = 14 m
Area of the square plot = (Side)² sq units
                                     = (14)² m²
                                     = 196  m²

Page No 230:

Question 3:

The top of a table measures 2 m 25 cm by 1 m 20 cm. Find its area in square metres.

ANSWER:

Length of the table = 2 m 25 cm
                         = (2 + 0.25) m     (100 cm = 1 m)
                         = 2.25 m
Breadth of the table = 1 m 20 cm
                                 = (1 + 0.20) m    (100 cm = 1 m)
                                 =1.20 m
Area of the table = (Length × Breadth) sq units
                             = (2.25 × 1.20) m²
       
                              = (225100×120100)(225100×120100) m²
                              = 2.7 m²

Page No 230:

Question 4:

A carpet is 30 m 75 cm long and 80 cm wide. Find its cost at Rs 150 per square metre.

ANSWER:

Length of the carpet = 30 m 75 cm
                           =(30 + 0.75) cm        (100 cm = 1 m)
                           = 30.75 m
Breadth of the carpet = 80 cm
                            = 0.80 m                (100 cm = 1 m)

Area of carpet = ( Length ×× breadth ) sq units

                            =(30.75 × 0.80) m2= (3075100×80100) m2=24.6 m2=(30.75 × 0.80) m2= (3075100×80100) m2=24.6 m2

Cost of 1 m² carpet= Rs 150
Cost of 24.6 m² carpet = Rs (24.6××150)
                                      =  Rs 3690

                      

Page No 230:

Question 5:

How many envelopes can be made out of a sheet of paper 3 m 24 cm by 1 m 72 cm. If each envelope requires a piece of paper of size 18 cm by 12 cm?

ANSWER:

Length of the sheet of paper = 3 m 24 cm = 324 cm
Breadth of the sheet of paper = 1 m 72 cm = 172 cm
Area of the sheet = (Length ×× Breadth)
                           =(324×172) cm2 =55728 cm2=(324×172) cm2 =55728 cm2

Length of the piece of paper required to make 1 envelope = 18 cm
Breadth of the piece of paper required to make 1 envelope = 12 cm
Area of the piece of paper required to make 1 envelope = (18××12) cm²
                                                                                   = 216 cm²

No. of envelopes that can be made =Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =55728216=258 envelopesNo. of envelopes that can be made =Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =55728216=258 envelopes

Page No 230:

Question 6:

A room is 12.5 m long and 8 m wide. A square carpet of side 8 m is laid on its floor. Find the area of the floor which is not carpeted.

ANSWER:

Length of the room = 12.5 m
Breadth of the room = 8 m
Area of the room = (Length××Breadth)
                     =(12.5×8) m2 = 100 m2=(12.5×8) m2 = 100 m2
Side of the square carpet = 8 m
Area of the carpet = (Side)²
                              = 8² m²
                              = 64 m²

Area of the floor which is not carpeted = Area of the room − Area of the carpet
                                                               = (100 − 64) m²
                                                               = 36 m²

Page No 230:

Question 7:

A lane, 150 m long and 9 m wide, is to be paved with bricks, each measuring 22.5 cm by 7.5 cm. Find the number of bricks required.

ANSWER:

Length of the road = 150 m = 15000 cm
Breadth of the road = 9 m = 900 cm
Area of the road = (Length××Breadth)
                     =15000×900  cm2=13500000 cm2=15000×900  cm2=13500000 cm2
Length of the brick = 22.5 cm
Breadth of the brick = 7.5 cm
Area of one brick = (Length××Breadth)
                             =(22.5×7.5)  cm2=168.75 cm2=(22.5×7.5)  cm2=168.75 cm2

Number of bricks = Area of the roadArea of one brick=13500000168.75=80000 bricksNumber of bricks = Area of the roadArea of one brick=13500000168.75=80000 bricks

Page No 230:

Question 8:

A room is 13 m long and 9 m broad. Find the cost of carpeting the room with a carpet 75 cm broad at the rate of Rs 65 per metre.

ANSWER:

Length of the room = 13 m
Breadth of the room = 9 m
Area of the room = (13××9) m² = 117 m²

Let length of required carpet be x m.
Breadth of the carpet = 75 cm
                            = 0.75 m         (100 cm = 1 m)
Area of the carpet = (0.75××x) m²
                       = 0.75x m²
For carpeting the room:
Area covered by the carpet = Area of the room
     ⇒0.75x=117⇒x=1170.75⇒x=117×43⇒x=156 m⇒0.75x=117⇒x=1170.75⇒x=117×43⇒x=156 m

So, the length of the carpet is 156 m.
Cost of 1 m carpet = Rs 65
Cost 156 m carpet = Rs (156××65)
                              = Rs 10140

Page No 230:

Question 9:

The length and the breadth of a rectangular park are in the ratio 5 : 3 and its perimeter is 128 m. Find the area of the park.

ANSWER:

Let the length of the rectangular park be 5x.
∴ Breadth of the rectangular park = 3x
Perimeter of the rectangular field = 2(Length + Breadth)
                                                      =2(5x + 3x)
                                                      = 16x

It is given that the perimeter of rectangular park is 128 m.
⇒16x=128⇒x=12816⇒x=8 Length of the park=(5×8) m                                            =40 mBreadth of the park=(3×8) m                                     =24 m⇒16x=128⇒x=12816⇒x=8 Length of the park=(5×8) m                                            =40 mBreadth of the park=(3×8) m                                     =24 m

Area of the park = (Length ×× Breadth) sq units
                           
                          =(40×24) m2=960 m2=(40×24) m2=960 m2

Page No 230:

Question 10:

Two plots of land have the same perimeter. One is a square of side 64 m and the other is a rectangle of length 70 m. Find the breadth of the rectangular plot. Which plot has the greater area and by how much?

ANSWER:

Side of the square plot = 64 m
Perimeter of the square plot = (4×Side) m =(4×64) m=256 m(4×Side) m =(4×64) m=256 m
Area of the square plot = (Side)²
= 64² m²
= 4096 m²

Let the breadth of the rectangular plot be x m.
Perimeter of the rectangular plot = 2(l+b)  m
= 2(70+x) m

Perimeter of the rectangular plot = Perimeter of the square plot   (Given)
⇒2(70+x)=256⇒140+2x=256⇒2x=256−140⇒2x=116⇒x=1162=58⇒2(70+x)=256⇒140+2x=256⇒2x=256-140⇒2x=116⇒x=1162=58
So, the breadth of the rectangular plot is 58 m.
Area of the rectangular plot = (Length × Breadth)=(70 × 58) m2=4060 m2(Length × Breadth)=(70 × 58) m2=4060 m2
Area of the square plot − Area of the rectangular plot
= (4096 − 4060)
= 36 m²
Area of the square plot is 36 m² greater than the rectangular plot.

Page No 230:

Question 11:

The cost of cultivating a rectangular field at Rs 35 per square metre is Rs 71400. If the width of the field is 40 m, find its length. Also, find the cost of fencing the field at Rs 50 per metre.

ANSWER:

Total cost of cultivating the field = Rs 71400
Rate of cultivating the field = Rs 35/m²


Area of the field=Total cost of cultivating the fieldRate of cultivating=Rs 71400Rs 35/m2=2040 m2Area of the field=Total cost of cultivating the fieldRate of cultivating=Rs 71400Rs 35/m2=2040 m2

Let the length of the field be x m.

Area of the field = (Length × Width) m2=(x × 40) m2 =40x m2It is given that the area of the field is 2040 m2.⇒40x=2040⇒x=204040=51∴Length of the field = 51 m Area of the field = (Length × Width) m2=(x × 40) m2 =40x m2It is given that the area of the field is 2040 m2.⇒40x=2040⇒x=204040=51∴Length of the field = 51 m 

Perimeter of the field = 2(l+b)
= 2(51+40) m
= 182 m
Cost of fencing 1 m of the field = Rs 50
Cost of fencing 182 m of the field = Rs (182××50)
= Rs 9100

Page No 230:

Question 12:

The area of a rectangle is 540 cm² and its length is 36 cm. Find its width and perimeter.

ANSWER:

Let the width of the rectangle be x cm.
Length of the rectangle = 36 cm
Area of the rectangle = (Length × WidthLength × Width) = (36 × x36 × x) cm²
It is given that the area of the rectangle is 540 cm².

⇒36 × x= 540⇒x=54036⇒x=15∴ Width of the rectangle =15 cm⇒36 × x= 540⇒x=54036⇒x=15∴ Width of the rectangle =15 cm

Perimeter of the rectangle = 2(Length + Width) cm
= 2(36 + 15) cm
= 102 cm

Page No 230:

Question 13:

A marble tile measures 12 cm × 10 cm. How many tiles will be required to cover a wall of size 4 m by 3 m? Also, find the total cost of the tiles at Rs 22.50 per tile.

ANSWER:

Length of the wall = 4 m = 400 cm
Breadth of the wall = 3 m = 300 cm
Area of the wall = (400×300) cm² = 120000 cm²

Length of the tile = 12 cm
Breadth of the tile = 10 cm
Area of one tile = (12×10) cm² = (120) cm²

Number of tiles required to cover the wall=Area of the wallArea of one tile=120000120=1000 tilesNumber of tiles required to cover the wall=Area of the wallArea of one tile=120000120=1000 tiles
Cost of 1 tile = Rs 22.50
Cost of 1000 tiles = (1000 × 22.50) = Rs 22500

Thus, the total cost of the tiles is Rs 22500.

Page No 230:

Question 14:

Find the perimeter of a rectangle whose area is 600 cm² and breadth is 25 cm.

ANSWER:

Let the length of the rectangle be x cm.
Breadth of the rectangle is 25 cm.
Area of the rectangle = (Length × Breadth) cm²
                                   = (x×25) cm²
                                   =25x cm²

It is given that the area of the rectangle is 600 cm².
⇒25x=600⇒x=60025=24⇒25x=600⇒x=60025=24
So, the length of the rectangle is 24 cm.
Perimeter of the rectangle = 2(Length + Breadth) units
                                           = 2(25 + 24) cm
                                           = 98 cm

Page No 230:

Question 15:

Find the area of a square whose diagonal is 52–√52 cm.

ANSWER:

Area of the square ={12× (Diagonal)2} sq units 12× (Diagonal)2 sq units 
 = {12×(52–√)2} cm2={12×(5)2×(2–√)2} cm2={12×25×2} cm2=(12×50) cm2= 25 cm2 = 12×(52)2 cm2=12×(5)2×(2)2 cm2=12×25×2 cm2=(12×50) cm2= 25 cm2

Page No 230:

Question 16:

Calculate the area of each one of the shaded regions given below:
Figure

ANSWER:

(i) Area of rectangle ABDC = Length ×× Breadth
                                         = AB××AC                    (AC = AE − CE)
                                         = (1×8)m21×8m2
                                         = 8 m²
   Area of rectangle CEFG = Length ×× Breadth
                                         = CG××GF               (CG = GD + CD)           
                                         = (9×2)m29×2m2
                                         = 18 m²
   Area of the complete figure = Area of rectangle ABDC + Area of rectangle CEFG
                                                 = (8 + 18) m²
                                                 =  26 m²


(ii) Area of rectangle AEDC = Length ×× Breadth
                                         = ED ×× CD              
                                         = (12×2)m212×2m2
                                         = 24 cm²
   Area of rectangle FJIH = Length ×× Breadth
                                         = HI ×× IJ                   
                                         = (1×9)m21×9m2
                                         = 9 m²
Area of rectangle ABGF = Length ×× Breadth
                                         = AB ×× AF                                  {(AB = FJ − GJ) and AF = EH − (EA + FH)}                 
                                         = (7×1.5)m27×1.5m2
                                         = 10.5 m²

   Area of the complete figure = Area of rectangle AEDC + Area of rectangle FJIH + Area of rectangle ABGF
                                         = (24 + 9 + 10.5) m²
                                         = 43.5 m²



(iii) Area of the shaded portion = Area of the complete figure − Area of the unshaded figure
                                           = Area of rectangle ABCD − Area of rectangle GBFE
                                           =(CD××AD) − (GB××BF)
                                           ={(12×9)−(7.5×10)}m2(12×9)-(7.5×10)m2                                   (BF = BC − FC)
                                           =(108 − 75) m²

                                          =33 m²

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Question 17:

Calculate the area of each one of the shaded regions given below (all measures are given in cm):
Figure

ANSWER:

(i) Area of square BCDE= (Side)²
                                        = (CD)²
                                        = (3)² cm²        
                                        = 9 cm²
      Area of rectangle ABFK = Length × BreadthLength × Breadth
                                             = AK××AB             [(AB = AC − BC) and (AK = AL + LK)
                                             = (5××1) cm²  
                                             = 5 cm²

     Area of rectangle MLKG = Length × BreadthLength × Breadth
                                             = ML ×× MG
                                             = (2 ×× 3) cm²
                                             = 6 cm²
     Area of rectangle JHGF= Length × BreadthLength × Breadth
                                             = JH××HG
                                             = (2××4) cm²
                                             = 8 cm²
       Area of the figure = Area of rectangle ABFK + Area of rectangle MLKG + Area of rectangle JHGF + Area of square BCDE
                               = (9 + 5 + 6 + 8) cm²
                               = 28 cm²
                   
(ii) Area of rectangle CEFG= Length × BreadthLength × Breadth
                                             = EF××CE
                                             = (1××5) cm²          (CE = EA − AC)
                                             = 5 cm²
      Area of rectangle ABDC = Length × BreadthLength × Breadth
                                             = AB××BD
                                             = (1××2) cm²  
                                             = 2 cm²

     Area of rectangle HIJG = Length × BreadthLength × Breadth
                                             = HI ×× IJ
                                             = (1××2) cm²
                                             = 2 cm²
       Area of the figure = Area of rectangle CEFG + Area of rectangle HIJG + Area of rectangle ABDC
                               = (5+2+2) cm²
                               = 9 cm²       
                      
(iii) In the figure, there are 5 squares, each of whose sides are 6 cm in length.
     Area of the figure = 5 ×× Area of square
                                   = 5××(side)²
                                   = 5××(6)² cm²
                                   = 180 cm²

Page No 231:

Exercise 21E

Question 1:

The sides of a rectangle are in the ratio 7 : 5 and its perimeter is 96 cm. The length of the rectangle is
(a) 21 cm
(b) 28 cm
(c) 35 cm
(d) 14 cm

ANSWER:

(b) 28 cm

Let the length and the breadth of the rectangle be 7x cm and 5x cm, respectively.
It is given that the perimeter of the rectangle is 96 cm.
Perimeter of the rectangle = 2(7x+5x) cm

⇒2(7x+5x)=96=2(12x)=96=24x=96⇒x=9624=4∴ Length =(7×4)cm=28 cm⇒2(7x+5x)=96=2(12x)=96=24x=96⇒x=9624=4∴ Length =(7×4)cm=28 cm

Page No 231:

Question 2:

The area of a rectangle is 650 cm² and its breadth is 13 cm. The perimeter of the rectangle is
(a) 63 cm
(b) 130 cm
(c) 100 cm
(d) 126 cm

ANSWER:

(d) 126 cm
Let length of the rectangle be L cm.
Area of the rectangle = 650 cm²
Area of the rectangle = (L×13L×13) cm²
⇒(L×13)=650⇒L=65013=50 Length of the rectangle is 50 cm⇒(L×13)=650⇒L=65013=50 Length of the rectangle is 50 cm

Perimeter of the rectangle = 2(Length + Breadth) cm = 2(50+13) cm = 126 cm

Page No 231:

Question 3:

The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 22.50 per metre is
(a) Rs 2430
(b) Rs 2340
(c) Rs 2400
(d) Rs 3340

ANSWER:

(b) Rs 2340
Perimeter of the rectangular field = 2(Length + Breadth)
                                                  = 2(34 + 18) m = 104 m
Cost of fencing 1 metre = Rs 22.50
Cost of fencing 104 m = Rs (22.50××104) = Rs 2340

Page No 231:

Question 4:

The cost of fencing a rectangular field at Rs 30 per metre is Rs 2400. If the length of the field is 24 m, then its breadth is
(a) 8 m
(b) 16 m
(c) 18 m
(d) 24 m

ANSWER:

(b) 16 m
Total cost of fencing = Rs 2400
Rate of fencing = Rs 30/m
Perimeter of the rectangular field = Total costRate=Rs 2400Rs 30/m=80 mTotal costRate=Rs 2400Rs 30/m=80 m
Let the breadth of the rectangular field be x m.
Perimeter of the rectangular field = 2(24 + x) m
⇒2(24+x)=80⇒48+2x=80⇒2x=(80−48)⇒2x=32⇒x=322=16So, the breadth of the rectangular field is 16 m.⇒2(24+x)=80⇒48+2x=80⇒2x=(80-48)⇒2x=32⇒x=322=16So, the breadth of the rectangular field is 16 m.

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Question 5:

The area of a rectangular carpet is 120 m² and its perimeter is 46 m. The length of its diagonal is
(a) 15 m
(b) 16 m
(c) 17 m
(d) 20 m

ANSWER:

(c) 17 m
Let the length and the breadth of the rectangle be L m and B m, respectively.

Area of the rectangular carpet = (L×BL×B) m²
⇒LB=120           … (i)⇒LB=120           … (i)
Perimeter of the rectangular carpet = 2(L+B)2(L+B)
⇒2(L+B)=46⇒(L+B)=462⇒(L+B)=23      …(ii)⇒2(L+B)=46⇒(L+B)=462⇒(L+B)=23      …(ii)

Diagonal of the rectangle = L2+B2−−−−−−−√ m                                   =(L+B)2−2LB−−−−−−−−−−−−−√ m                                                =(23)2−240−−−−−−−−−√ m                            (from equations (i) and (ii))                                  =529−240−−−−−−−−√ m                                  =289−−−√ m=17 mDiagonal of the rectangle = L2+B2 m                                   =(L+B)2-2LB m                                                =(23)2-240 m                            (from equations (i) and (ii))                                  =529-240 m                                  =289 m=17 m

Page No 231:

Question 6:

The length of a rectangle is three times its width and the length of its diagonal is 610−−√610 cm. The perimeter of the rectangle is
(a) 48 cm
(b) 36 cm
(c) 24 cm
(d) 2410 −−−√2410  cm

ANSWER:

(a) 48 cm
Let the width  and the length of the rectangle be x cm and 3x cm, respectively.

Applying Pythagoras theorem:

(Diagonal)2=(Length)2+(Width)2⇒(610−−√)2=(3x)2+(x)2⇒360=9×2+x2⇒360=10×2⇒x2=36010⇒x2=36⇒x=±6Since the width cannot be negative, we will neglect −6.(Diagonal)2=(Length)2+(Width)2⇒(610)2=(3x)2+(x)2⇒360=9×2+x2⇒360=10×2⇒x2=36010⇒x2=36⇒x=±6Since the width cannot be negative, we will neglect -6.

So, width of the rectangle is 6 cm.
Length of the rectangle = (3×6)=18 cm3×6=18 cm
Perimeter of the rectangle = 2(Length + Breadth) = 2(18 + 6) = 48 cm

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Question 7:

If the ratio between the length and perimeter of a rectangular plot is 1 : 3, then the ratio between the length and breadth of the plot is
(a) 1 : 2
(b) 2 : 1
(c) 3 : 2
(d) 2 : 3

ANSWER:

(b) 2 : 1
Let the breadth of the plot be b cm.

Let the length of the plot be x cm.
Perimeter of the plot = 3x cm

Perimeter of the plot =2(Length + Breadth)= 2(x + b) cm
⇒2(x+b)=3x2x+2b=3x⇒2b=3x−2x⇒2b=x⇒b=x2∴Ratio of the length and the breadth of the plot =x(x2)=xx × 2 = 21∴Ratio of the length and the breadth of the plot =2:1⇒2(x+b)=3x2x+2b=3x⇒2b=3x-2x⇒2b=x⇒b=x2∴Ratio of the length and the breadth of the plot =x(x2)=xx × 2 = 21∴Ratio of the length and the breadth of the plot =2:1

Page No 232:

Question 8:

The length of the diagonal of a square is 20 cm. It area is
(a) 400 cm²
(b) 200 cm²
(c) 300 cm²
(d) 1002–√1002 cm²

ANSWER:

(b) 200 cm²
Area of the square = {12×(Diagonal)2 } sq units12×(Diagonal)2  sq units
={12×(20)2 } cm2={12×(20)×(20)} cm2=(20×10) cm2=200 cm2=12×(20)2  cm2=12×(20)×(20) cm2=(20×10) cm2=200 cm2

Page No 232:

Question 9:

The cost of putting a fence around a square field at Rs 25 per metre is Rs 2000. The length of each side of the field is
(a) 80 m
(b) 40 m
(c) 20 m
(d) none of these

ANSWER:

(c) 20 m
Let one side of the square field be x m.
Total cost of fencing a square field = Rs 2000
Rate of fencing the field = Rs 25/m

Perimeter of the square field=Total cost of fencing the fieldRate of fencing the field=Rs 2000Rs 25/m=200025 m=80 mPerimeter of the square field=Total cost of fencing the fieldRate of fencing the field=Rs 2000Rs 25/m=200025 m=80 m

Perimeter of the square field = (4×side4×side) = 4x m
⇒4x=80⇒x=804⇒x=20Each side of the field is 20 m.⇒4x=80⇒x=804⇒x=20Each side of the field is 20 m.

Page No 232:

Question 10:

The diameter of a circle is 7 cm. Its circumference is
(a) 44 cm
(b) 22 cm
(c) 28 cm
(d) 14 cm

ANSWER:

(b) 22 cm
Radius=Diameter 2=72 cmCircumference of the circle=2πr=(2×227×72) cm=22 cmRadius=Diameter 2=72 cmCircumference of the circle=2πr=(2×227×72) cm=22 cm

Page No 232:

Question 11:

The circumference of a circle is 88 cm. Its diameter is
(a) 28 cm
(b) 42 cm
(c) 56 cm
(d) none of these

ANSWER:

(a) 28 cm
Circumference of the circle is 88 cm.Let the radius be r cm.It is given that the circumference of the circle is (2πr) cm.⇒2πr=88⇒2×227×r=88⇒r=12×722×88⇒r=14∴ Radius=14 cmDiameter=(2× Radius)=(2 × 14) cm =28 cmCircumference of the circle is 88 cm.Let the radius be r cm.It is given that the circumference of the circle is (2πr) cm.⇒2πr=88⇒2×227×r=88⇒r=12×722×88⇒r=14∴ Radius=14 cmDiameter=(2× Radius)=(2 × 14) cm =28 cm

Page No 232:

Question 12:

The diameter of a wheel of a car is 70 cm. How much distance will it cover in making 50 revolutions?
(a) 350 m
(b) 110 m
(c) 165 m
(d) 220 m

ANSWER:

(b) 110 m
Radius of the wheel=Diameter2=702=35 cmCircumference of the wheel =2πr=(2×227×35) cm =220 cmThe distance covered by the wheel in one revolution is equal to its circumference.Distance covered by the wheel in 1 revolution = 220 cm∴Distance covered by the wheel in 50 revolution = (50 × 220)cm=11000 cm=110 mRadius of the wheel=Diameter2=702=35 cmCircumference of the wheel =2πr=(2×227×35) cm =220 cmThe distance covered by the wheel in one revolution is equal to its circumference.Distance covered by the wheel in 1 revolution = 220 cm∴Distance covered by the wheel in 50 revolution = (50 × 220)cm=11000 cm=110 m

Page No 232:

Question 13:

A lane 150 m long and 9 m wide is to be paved with bricks, each measuring 22.5 cm by 7.5 cm. How many bricks are required?
(a) 65000
(b) 70000
(c) 75000
(d) 80000

ANSWER:

(d) 80000
Length of the road = 150 m = 15000 cm
Breadth of the road = 9 m = 900 cm
Area of the road = (Length × Breadth)
= (15000 × 900) cm²
= 13500000 cm²

Length of the brick = 22.5 cm
Breadth of the brick = 7.5 cm
Area of one brick = (Length × Breadth)
= ( 22.5 × 7.5 ) cm²
= 168.75 cm²

Number of bricks = Area of the roadArea of one brick =13500000 cm2168.75 cm2=80000 bricksNumber of bricks = Area of the roadArea of one brick =13500000 cm2168.75 cm2=80000 bricks

Page No 232:

Question 14:

A room is 5 m 40 cm long and 4 m 50 cm broad. Its area is
(a) 23.4 m²
(b) 24.3 m²
(c) 25 m²
(d) 98.01 m²

ANSWER:

(b) 24.3 m²

Length of the room = 5 m 40 cm = 5.40 m
Breadth of the room = 4 m 50 cm = 4.50 m

Area of the room = (Length × Breadth)=(5.40 × 4.50) m2=(540100×450100)m2=(275×92)m2=24310m2=24.3 m2Area of the room = (Length × Breadth)=(5.40 × 4.50) m2=540100×450100m2=275×92m2=24310m2=24.3 m2

Page No 232:

Question 15:

How many envelopes can be made out of a sheet of paper 72 cm by 48 cm, if each envelope requires a paper of size 18 cm by 12 cm?
(a) 4
(b) 8
(c) 12
(d) 16

ANSWER:

(d) 16

Length of the sheet of paper = 72 cm
Breadth of the sheet of paper = 48 cm
Area of the sheet = (Length × Breadth)
⇒ ( 72 × 48 ) cm²  = 3456 cm²

Length of the piece of paper required to make 1 envelope = 18 cm
Breadth of the piece of paper required to make 1 envelope = 12 cm
Area of the piece of paper required to make 1 envelope = (18 × 12) cm²
= 216 cm²

No. of envelopes that can be made = Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =3456216=16 envelopesNo. of envelopes that can be made = Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =3456216=16 envelopes

Page No 233:

Exercise 21F

Question 1:

Find the perimeter of the following shapes:
(i) a triangle whose sides are 5.4 cm, 4.6 cm and 6.8 cm
(ii) a regular hexagon of side 8 cm
(iii) an isosceles triangle with equal sides 6 cm each and third side 4.5 cm.

ANSWER:

(i) Sides of the triangle are 5.4 cm, 4.6 cm and 6.8 cm. 
    Perimeter of the triangle = (First side + Second side + Third Side)
                                   = (5.4 + 4.6 + 6.8) cm = 16.8 cm

(ii) Length of each side of the given hexagon = 8 cm
   ∴ Perimeter of the hexagon = (6 × 8) cm = 48 cm

(iii) Length of the two equal sides = 6 cm
     Length of the third side = 4.5 cm
    ∴ Perimeter of the triangle = {(2 × equal sides) + third side} cm = (2 × 6) + 4.5 = 16.5 cm

Page No 233:

Question 2:

The perimeter of a rectangular field is 360 m and its breadth is 75 m. Find its length.

ANSWER:

Let the length of the rectangle be x m.
Breadth of the rectangle = 75 m
Perimeter of the rectangle = 2(Length + Breadth)
                                  = 2(x + 75) m = (2x + 150) m
It is given that the perimeter of the field is 360 m.
⇒2x+150=360⇒2x=360−150⇒2x=210⇒x=2102=105 ⇒2x+150=360⇒2x=360-150⇒2x=210⇒x=2102=105 
So, the length of the rectangle is 105 m.

Page No 233:

Question 3:

The length and breadth of a rectangular field are in the ratio 5 : 4. If its perimeter is 108 m, find the dimensions of the field.

ANSWER:

Let the length of the rectangle be 5x m.
Breadth of the rectangle = 4x m
Perimeter of the rectangle = 2(Length + Breadth)
                                  = 2(5x + 4x) m = 18x m
It is given that the perimeter of the field is 108 m.
∴ 18 x = 108
⇒ x = 10818=610818=6
∴ Length of the field = ( 5 × 6 )m = 30 m
Breadth of the field = ( 4 × 6 )m = 24 m

Page No 233:

Question 4:

Find the area of a square whose perimeter is 84 cm.

ANSWER:

Let one side of the square be x cm.
Perimeter of the square = (4×side)=(4×x) cm =4x cm(4×side)=(4×x) cm =4x cm
It is given that the perimeter of the square is 84 cm.
⇒4x=84⇒x=844=21Thus, one side of the square is  21 cm.Area of the square = (Side)2=(21)2 cm2= 441cm2⇒4x=84⇒x=844=21Thus, one side of the square is  21 cm.Area of the square = (Side)2=(21)2 cm2= 441cm2

Page No 233:

Question 5:

The area of a room is 216 m² and its breadth is 12 m. Find the length of the room.

ANSWER:

Let the length of the room be x m.
Breadth of the room = 12 m
Area of the room = (Length × Breadth) = (x × 12) m²
It is given that the area of the room is 216 m².
⇒ x × 12 = 216
⇒ x = 21612=1821612=18
∴ Length of the rectangle = 18 m

Page No 233:

Question 6:

Find the circumference of a circle of radius 7 cm. [Take π = 227π = 227]

ANSWER:

Radius(r) of the given circle = 7 cm
Circumference of the circle, C = 2 πr
                                                 = (2×227×7) cm= 44 cm= 2×227×7 cm= 44 cm
Hence, the circumference of the given circle is 44 cm.

Page No 233:

Question 7:

The diameter of a wheel of a car is 77 cm. Find the distance covered by the wheel in 500 revolutions.

ANSWER:

Radius of the wheel =Diameter of the wheel2Diameter of the wheel2
⇒ r = 772772cm
Circumference of the wheel = 2 πr
= (2×227×772)2×227×772 cm
= 242 cm

In 1 revolution, the wheel covers a distance equal to its circumference.
∴ Distance covered by the wheel in 1 revolution = 242 cm
∴ Distance covered by the wheel in 500 revolutions = ( 500 × 242 ) cm
                                                                               = 121000 cm       (100 cm =1 m)
                                                                               = 1210 m

Page No 233:

Question 8:

Find the diameter of a wheel whose circumference is 176 cm.

ANSWER:

Let the radius be r cm.
Diameter = 2 × Radius(r) = 2r cm
Circumference of the wheel = 2πr
∴ 2πr = 176
⇒ 2r=176π⇒ 2r=176×722=56⇒ 2r=176π⇒ 2r=176×722=56
⇒ 2r = 56
Thus, the diameter of the given wheel is 56 cm.

Page No 233:

Question 9:

Find the area of a rectangle whose length is 36 cm and breadth 15 cm.

ANSWER:

Length of the rectangle = 36 cm 
Breadth of the rectangle = 15 cm
Area of the rectangle = (Length × Breadth) sq units
                             = (36 × 15) cm² = 540 cm²

Page No 233:

Question 10:

Perimeter of a square of side 16 cm is
(a) 256 cm
(b) 64 cm
(c) 32 cm
(d) 48 cm

ANSWER:

(b) 64 cm
Side of the square = 16 cm
 Perimeter of the square = (4 × side)
                               = (4 × 16) cm 
                               = 64 cm

Page No 233:

Question 11:

The area of a rectangle is 240 m² and its length is 16 m. Then, its breadth is
(a) 15 m
(b) 16 m
(c) 30 m
(d) 40 m

ANSWER:

(a) 15 m

Let the breadth of the rectangle be x m.
Length of the rectangle = 16 m
Area of rectangle = (Length × Breadth) = (16 × x) m²
It is given that the area of the rectangle is 240 m².
⇒ 16 × x = 240
⇒ x = 24016=1524016=15
So, the breadth of the rectangle is 15 m.

Page No 233:

Question 12:

The area of a square lawn of side 15 m is
(a) 60 m²
(b) 225 m²
(c) 45 m²
(d) 120 m²

ANSWER:

(b) 225 m²

Side of the square lawn = 15 mArea of the square lawn = (Side)2 sq units=(15)2 m2=225 m2Side of the square lawn = 15 mArea of the square lawn = (Side)2 sq units=(15)2 m2=225 m2

Page No 233:

Question 13:

The area of a square is 256 cm². The perimeter of the square is
(a) 16 cm
(b) 32 cm
(c) 48 cm
(d) 64 cm

ANSWER:

(a) 16 cm
Let one side of the square be x cm.
Area of the square = (Side )² cm² = x² cm²
It is given that the area of the square is 256 cm².
⇒ x² = 256
⇒ x = 256−−−√=±16256=±16
We know that the side of a square cannot be negative.
So, we will neglect −16.
Therefore, the side of the square is 16 cm.

Perimeter of the square = (4×side)=(4×16)cm=64 cm4×side=4×16cm=64 cm

Page No 233:

Question 14:

The area of a rectangle is 126 m² and its length is 12 m. The breadth of the rectangle is
(a) 10 m
(b) 10.5 m
(c) 11 m
(d) 11.5 m

ANSWER:

(b) 10.5 m
Let the breadth of the rectangle be x m.
Length of the rectangle = 12 m
Area of the rectangle = 126 m²
Area of the rectangle = (length×breadth)sq units=(12×x)m2=12x m2length×breadthsq units=(12×x)m2=12x m2
It is given that the area of the rectangle is 126 m².
⇒12x=126⇒x=12612=10.5So, the breadth of the rectangle is 10.5 m.⇒12x=126⇒x=12612=10.5So, the breadth of the rectangle is 10.5 m.

Page No 233:

Question 15:

Fill in the blanks.
(i) A polygon having all sides equal and all angles equal is called a …… polygon.
(ii) Perimeter of a square = …… × side.
(iii) Area of a rectangle = (……) × (……).
(iv) Area of a square = …… .
(v) If the length of a rectangle is 5 m and its breadth is 4 m, then its area is …… .

ANSWER:

(i) A polygon having all sides equal and all angles equal is called a regular polygon
(ii) Perimeter of a square = 4 × side
(iii) Area of a rectangle = (length) × (breadth)
(iv) Area of a square = (side)²
(v) If the length of a rectangle is 5 m and its breadth is 4 m, then its area is 20 m²
Area of a rectangle = (length) × (breadth) = (5×4) m² = 20 m²

Page No 233:

Question 16:

Match the following:

(a) Area of a rectangle	(i) πr2
(b) Area of a square	(ii) 4 × side
(c) Perimeter of a rectangle	(iii) l × b
(d) Perimeter of a square	(iv) (side)2
(e) Area of a circle	(v) 2(l + b)

ANSWER:

(a) Area of a rectangle	(iii) l × b
(b) Area of a square	(iv) (side)2
(c) Perimeter of a rectangle	(v) 2(l + b)
(d) Perimeter of a square	(ii) 4 × side
(e) Area of a circle	(i) πr2

Exercise 22A

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Question 1:

Define the terms:
(i) Data
(ii) Raw data
(iii) Array
(iv) Tabulation of data
(v) Observations
(vi) Frequency of an observation
(vii) Statistics

ANSWER:

(i) Data: It refers to the information in the form of numerical figures.

The marks obtained by 5 students of a class in a unit test are 34, 45, 65, 67, 87.

We call it the data related to the marks obtained by 5 students of a class in a unit test.
             
(ii) Raw Data: Data obtained in the original form is called raw data.

(iii) Array: Arranging the numerical figures in an ascending or a descending order is called an array.

(iv) Tabulation of data: Arranging the data in a systematic form in the form of a table is called tabulation or presentation of the data.

(v) Observations: Each numerical figure in a data is called an observation.

(vi) Frequency of an observation: The number of times a particular observation occurs is called its frequency.

(viii) Statistics: It is the science that deals with the collection, presentation, analysis and interpretation of numerical data.

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Question 2:

The number of children in 25 families of a colony are give below:
2, 0, 2, 4, 2, 1, 3, 3, 1, 0, 2, 3, 4, 3, 1, 1, 1, 2, 2, 3, 2, 4, 1, 2, 2.
Represent the above data in the form of a frequency distribution table.

ANSWER:

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Question 3:

The sale of shoes of various sizes at a shop on a particular day is given below:
6, 9, 8, 5, 5, 4, 9, 8, 5, 6, 9, 9, 7, 8, 9, 7, 6, 9, 8, 6, 7, 5, 8, 9, 4, 5, 8, 7.
Represent the above data in the form of a frequency distribution table.

ANSWER:

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Question 4:

Construct a frequency table for the following:
3, 2, 5, 4, 1, 3, 2, 2, 5, 3, 1, 2, 1, 1, 2, 2, 3, 4, 5, 3, 1, 2, 3.

ANSWER:

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Question 5:

Construct a frequency table for the following:
7, 8, 6, 5, 6, 7, 7, 9, 8, 10, 7, 6, 7, 8, 8, 9, 10, 5, 7, 8, 7, 6.

ANSWER:

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Question 6:

Fill in the blanks:
(i) Data means information in the form of …… .
(ii) Data obtained in the …… form is called raw data.
(iii) Arranging the numerical figures in an ascending or a descending order is called an ……. .
(iv) The number of times a particular observation occurs is called its …… .
(v) Arranging the data in the form of a table is called …… .

ANSWER:

(i) Data means information in the form of numerical figures.
(ii) Data obtained in the original form is called raw data.
(iii) Arranging the numerical figures in an ascending or a descending order is called an  array.
(iv) The number of times a particular observation occurs is called its frequency.
(v) Arranging the data in the form of a table is called tabulation of data.

Page No 192:

Exercise 15A

Question 1:

Which of the following are simple closed figures?
Figure

ANSWER:

(a) It is a simple closed figure because it does not intersect itself
(b) It is a simple closed figure because it does not intersect itself.
(c) It is not a simple closed figure because it intersects itself.
(d) It is a simple closed figure because it does not intersect itself.
(e) It is not a simple closed figure because it intersects itself.
(f) It is a simple closed figure because it does not intersect itself.

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Question 2:

Which of the following are polygons?
Figure

ANSWER:

(a) It is formed by four line segments and we know that a simple closed figure formed by three or more line segments is called a polygon. So, it is a polygon named quadrilateral.

(b) It is formed by three line segments and we know that a simple closed figure formed by three or more line segments is called a polygon. So, it is polygon named triangle.

(c) It is formed by twelve line segments and we know that a simple closed figure formed by three or more line segments is called a polygon. So, it is polygon.

(d) It is not a polygon because it contains curves.

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Question 3:

Fill in the blanks:
(i) A polygon is a simple closed figure formed by more than …… line segments.
(ii) A polygon formed by three line segments is called a ……. .
(iii) A polygon formed by four line segments is called a …… .
(iv) A triangle has …… sides and …… angles.
(v) A quadrilateral has …… sides and …… angles.
(vi) A figure which ends at the starting point is called a ………. .

ANSWER:

(i) A polygon is a simple closed figure formed by more than two line segments.
(ii) A polygon formed by three line segments is called a triangle.
(iii) A polygon formed by four line segments is called a quadrilateral .
(iv) A triangle has three sides and three angles.
(v) A quadrilateral has four sides and four angles.
(vi) A figure which ends at the starting point is called a closed figure.