NCERT Solutions Class 8 Maths Chapter 6 Cubes and Cube Roots Exercise 6.2 Read More »
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]]>1). Find the cube root of each of the following numbers by prime factorisation method.
(i) 64 (ii) 512
(iii) 10648 (iv) 27000
(v) 15625 (vi) 13824
(vii) 110592 (viii) 46656
(ix) 175616 (x) 91125
Solution:
(i) 64
64 = 2 X 2 X 2 X 2 X 2 X 2
64 = 2 X 2 X 2 X 2 X 2 X 2
64 = 2^{3} X 2^{3}
^{3}√64 = ^{3}√2^{3} X 2^{3}
= 2 X 2
= 4
Ans: ^{3}√64 = 4
(ii) 512
512 = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2
512 = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2
512 = 2^{3} X 2^{3 }X 2^{3}
^{3}√512 = ^{3}√2^{3} X 2^{3 }X 2^{3}
= 2 X 2 X 2
= 8
Ans: ^{3}√512 = 8
(iii) 10648
10648 = 2 X 2 X 2 X 11 X 11 X 11
10648 = 2 X 2 X 2 X 11 X 11 X 11
10648 = 2^{3} X 11^{3}
^{3}√10648 = ^{3}√2^{3} X 11^{3}
= 2 X 11
= 22
Ans: ^{3}√10648= 22
(iv) 27000
27000= 2 X 2 X 2 X 3 X 3 X 3 X 5 X 5 X 5
27000= 2 X 2 X 2 X 3 X 3 X 3 X 5 X 5 X 5
27000= 2^{3} X 3^{3 }X 5^{3}
^{3}√27000 = ^{3}√2^{3} X 3^{3 }X 5^{3}
= 2 X 3 X 5
= 30
Ans: ^{3}√27000 = 30
(v) 15625
15625= 5 X 5 X 5 X 5 X 5 X 5
15625= 5 X 5 X 5 X 5 X 5 X 5
15625= 5^{3} X 5^{3}
^{3}√15625 = ^{3}√5^{3} X 5^{3}
= 5 X 5
= 25
Ans: ^{3}√15625 = 25
(vi) 13824
13824= 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3 X 3
13824= 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3 X 3
13824= 2^{3} X 2^{3 }X 2^{3} X 3^{3}
^{3}√13824 = ^{3}√2^{3} X 2^{3 }X 2^{3} X 3^{3}
= 2 X 2 X 2 X 3
= 24
Ans: ^{3}√13824 = 24
(vii) 110592
110592= 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2X 3 X 3 X 3
110592= 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2X 3 X 3 X 3
110592= 2^{3} X 2^{3} X 2^{3 }X 2^{3} X 3^{3}
^{3}√110592 = ^{3}√2^{3} X 2^{3} X 2^{3 }X 2^{3} X 3^{3}
= 2 X 2 X 2 X 2 X 3
= 48
Ans: ^{3}√110592 = 48
(viii) 46656
46656= 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3 X 3 X 3 X 3 X 3
46656= 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3 X 3 X 3 X 3 X 3
46656= 2^{3} X 2^{3} X 3^{3 }X 3^{3}
^{3}√46656 = ^{3}√2^{3} X 2^{3} X 3^{3 }X 3^{3}
= 2 X 2 X 3 X 3
= 36
Ans: ^{3}√46656 = 36
(ix) 175616
175616= 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 7 X 7 X 7
175616= 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 7 X 7 X 7
175616= 2^{3} X 2^{3 }X 2^{3 }X 7^{3}
^{3}√175616 = ^{3}√2^{3} X 2^{3 }X 2^{3 }X 7^{3}
= 2 X 2 X 2 X 7
= 56
Ans: ^{3}√175616 = 56
(x) 91125
91125= 3 X 3 X 3 X 3 X 3 X 3 X 5 X 5 X 5
91125= 3 X 3 X 3 X 3 X 3 X 3 X 5 X 5 X 5
91125= 3^{3} X 3^{3 }X 5^{3}
^{3}√91125 = ^{3}√3^{3} X 3^{3 }X 5^{3}
= 3 X 3 X 5
= 45
Ans: ^{3}√91125 = 45
2). State true or false.
i). Cube of any odd number is even.
Ans: False
ii). A perfect cube does not end with two zeroes.
Ans: True
iii). If square of a number ends with 5, then its cube ends with 25.
Ans: True
iv). There is no perfect cube which ends with 8.
Ans: False
v). The cube of a two digit number may be a three digit number.
Ans: False
vi). The cube of a two digit number may have seven or more digits.
Ans: False
vii). The cube of a single digit number may be a single digit number.
Ans: False
Click here for the solutions of Std 8 Maths
1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
5). Squares and Square Roots
6). Cubes and Cube Roots
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]]>1). Which of the following are not perfect cubes?
(i) 216 (ii) 128
(iii) 1000 (iv) 100
(v) 46656
Solution:
(i) 216
216 = 2 X 2 X 2 X 3 X 3 X 3
216 = 2 X 2 X 2 X 3 X 3 X 3
Group of factors 2 and 3 is formed
Therefore, 216 is a perfect cube.
(ii) 128
128 = 2 X 2 X 2 X 2 X 2 X 2 X 2
128 = 2 X 2 X 2 X 2 X 2 X 2 X 2
in this factorisation, factor 2 cannot be formed in groups of 3 Therefore, 128 is not a perfect cube.
(iii) 1000
1000 = 2 X 2 X 2 X 5 X 5 X 5
1000 = 2 X 2 X 2 X 5 X 5 X 5
in this factorisation, group of three of factors 2 and 5 is formed
Therefore, 1000 is a perfect cube.
(iv) 100
100 = 2 X 2 X 5 X 5
in this factorisation, factors 2 and 5cannot be formed in groups of 3
Therefore, 100 is not a perfect cube.
(v) 46656
46656 = 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3 X 3 X 3 X 3 X 3
46656 = 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3 X 3 X 3 X 3 X 3
in this factorisation, group of three of factors 2 and 3 formed therefore, 46656 is a perfect cube.
2). Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243 (ii) 256
(iii) 72 (iv) 675
(v) 100
Solution:
(i) 243
243 = 3 X 3 X 3 X 3 X 3
243 = 3 X 3 X 3 X 3 X 3
Here one 3 is required to make the group of three
The least number multiplied to obtain a perfect cube = 3
(ii) 256
256 = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2
256 = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2
Here one 2 is required to make the group of three
The least number multiplied to obtain a perfect cube = 2
(iii) 72
72 = 2 X 2 X 2 X 3 X 3
72 = 2 X 2 X 2 X 3 X 3
Here one 3 is required to make the group of three
The least number multiplied to obtain a perfect cube = 3
(iv) 675
675 = 3 X 3 X 3 X 5 X 5
675 = 3 X 3 X 3 X 5 X 5
Here one 5 is required to make the group of three
The least number multiplied to obtain a perfect cube = 5
(v) 100
100 = 2 X 2 X 5 X 5
Here one 2 and one 5 is required to make the group of three
The least number multiplied to obtain a perfect cube = 2 X 5
= 10
3). Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81 (ii) 128
(iii) 135 (iv) 192
(v) 704
Solution:
(i) 81
81 = 3 X 3 X 3 X 3
81 = 3 X 3 X 3 X 3
Here 3 is the number by which 81 is divided to get a perfect cube
The least number to be divided to obtain a perfect cube = 3
(ii) 128
128 = 2 X 2 X 2 X 2 X 2 X 2 X 2
128 = 2 X 2 X 2 X 2 X 2 X 2 X 2
Here 2 is the number by which 128 is divided to get a perfect cube
The least number to be divided to obtain a perfect cube = 2
(iii) 135
135 = 3 X 3 X 3 X 5
135 = 3 X 3 X 3 X 5
Here 5 is the number by which 135 is divided to get a perfect cube
The least number to be divided to obtain a perfect cube = 5
(iv) 192
192 = 2 X 2 X 2 X 2 X 2 X 2 X 3
192 = 2 X 2 X 2 X 2 X 2 X 2 X 3
Here 3 is the number by which 192 is divided to get a perfect cube
The least number to be divided to obtain a perfect cube = 3
(v) 704
704 = 2 X 2 X 2 X 2 X 2 X 2 X 11
704 = 2 X 2 X 2 X 2 X 2 X 2 X 11
Here 11 is the number by which 704 is divided to get a perfect cube
The least number to be divided to obtain a perfect cube = 11
4). Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. how much such cuboids will he need to form a cube?
Sides of cuboid of plasticine are 5 cm, 2 cm and 5 cm
Volume of cuboid = 5 X 5 X 2
To make it perfect cube we have to make a group of three of factors 5 and 2.
Cuboids need to form a cube = 5 X 2 X 2
= 20
Click here for the solutions of Std 8 Maths
1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
5). Squares and Square Roots
6). Cubes and Cube Roots
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]]>NCERT Solutions Class 8 Maths Chapter 5 Squares and Square Roots Exercise 5.4 Read More »
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]]>1). Find the square root of each of the following numbers by Division method.
(i) 2304 (ii) 4489
(iii) 3481 (iv) 529
(v) 3249 (vi) 1369
(vii) 5776 (viii) 7921
(ix) 576 (x) 1024
(xi) 3136 (xii) 900
Solution:
(i) 2304
√2304 = 48
(ii) 4489
√4489 = 67
(iii) 3481
√3481 = 59
(iv) 529
√529 = 23
(v) 3249
√3249 = 57
(vi) 1369
√1369 = 37
(vii) 5776
√5776 = 76
(viii) 7921
√7921 = 89
(ix) 576
√576 = 24
(x) 1024
√1024 = 32
(xi) 3136
√3136 = 56
(xii) 900
√300 = 30
2). Find the number of digits in the square root of each of the following numbers (without any calculation).
(i) 64 (ii) 144
(iii) 4489 (iv) 27225
(v) 390625
Solution:
(i) 64
Number of digits in 64 is 2
Here, n = 2 (even)
Number of digits in square root = ^{n}/_{2} = ^{2}/_{2} = 1
(ii) 144
Number of digits in 144 is 3
Here, n = 3 (even)
Number of digits in square root = ^{n}^{+1}/_{2} = ^{3+1}/_{2} = ^{4}/_{2} = 2
(iii) 4489
Number of digits in 4489 is 4
Here, n = 4 (even)
Number of digits in square root = ^{n}/_{2} = ^{4}/_{2} = 2
(iv) 27225
Number of digits in 27225 is 5
Here, n = 5 (odd)
Number of digits in square root = ^{n}^{+1}/_{2} = ^{5+1}/_{2} = ^{6}/_{2} = 3
(v) 390625
Number of digits in 390625 is 6
Here, n = 6 (even)
Number of digits in square root = ^{n}/_{2} = ^{6}/_{2} = 3
3). Find the square root of the following decimal numbers.
(i) 2.56 (ii) 7.29
(iii) 51.84 (iv) 42.25
(v) 31.36
Solution:
(i) 2.56
√2.56 = 1.6
(ii) 7.29
√7.29 = 2.7
(iii) 51.84
√51.84 = 7.2
(iv) 42.25
√42.25 = 6.5
(v) 31.36
√31.36 = 5.6
4). Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402 (ii) 1989
(iii) 3250 (iv) 825
(v) 4000
Solution:
(i) 402
Here remainder is 2
Therefore, 2 is the least number to be subtracted
402 – 2 = 400
√400 = 20
(ii) 1989
Here remainder is 53
Therefore, 53 is the least number to be subtracted
1989 – 53 = 1936
√1936 = 44
(iii) 3250
Here remainder is 1
Therefore, 1 is the least number to be subtracted
3250 – 1 = 3249
√3249 = 57
(iv) 825
Here remainder is 41
Therefore, 41 is the least number to be subtracted
825 – 41 = 784
√784 = 28
(v) 4000
Here remainder is 31
Therefore, 31 is the least number to be subtracted
4000 – 31 = 3969
√3969 = 63
5). Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 525 (ii) 1750
(iii) 252 (iv) 1825
(v) 6412
Solution:
(i) 525
Here remainder is 41 and quotient is 22
Square of 22 is less than 525
Next number is 23
Square of 23 = 529
The least number to be added = 529 – 525 = 4
525 + 4 = 529
√529 = 23
(ii) 1750
Here remainder is 69 and quotient is 41
Square of 41 is less than 1750
Next number is 42
Square of 42 = 1764
The least number to be added = 1764 – 1750 = 14
√1764 = 42
(iii) 252
Here remainder is 27 and quotient is 15
Square of 15 is less than 252
Next number is 16
Square of 16 = 256
The least number to be added = 256 – 252 = 4
√256 = 16
(iv) 1825
Here remainder is 61 and quotient is 42
Square of 42 is less than 1825
Next number is 43
Square of 43 = 1849
The least number to be added = 1849 – 1825 = 24
√1849 = 43
(v) 6412
Here remainder is 12 and quotient is 80
Square of 80 is less than 6412
Next number is 81
Square of 81 =6561
The least number to be added = 6561 – 6412 = 149
√6561 = 81
6). Find the length of the side of a square whose area is 441 m^{2}.
Given: Area of square = 441 m^{2}
Area of square = side^{2}
Side^{2} = 441
taking square root on both the sides
√side^{2} = √441
side = 21
Ans: Side of the square = 21m
7). In a right triangle ABC, ÐB = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC
(b) If AC = 13 cm, BC = 5 cm, find AB
Solution:
(a) If AB = 6 cm, BC = 8 cm, find AC
In ∆ABC by Pythagoras theorem
AC^{2} = AB^{2} + BC^{2}
AC^{2} = 6^{2} + 8^{2}
= 36 + 64
= 100
taking square root on both the sides
√AC^{2} = √100
AC = 10
(b) If AC = 13 cm, BC = 5 cm, find AB
In ∆ABC by Pythagoras theorem
AB^{2} + BC^{2 }= AC^{2}
AB^{2} + 5^{2 }= 13^{2}
AB^{2} + 25 = 169
AB^{2} = 169 – 25
AB^{2} ^{ }= 144
taking square root on both the sides
√AB^{2} = √144
AB = 12
8). A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Total plants = 1000
Number of columns = number of rows
1000 is not a perfect square.
Here remainder is 39 and quotient is 31
Square of 31 is less than 1000
Next number is 32
Square of 32 = 1024
The least number to be added = 1024 – 1000 = 24
Ans: Gardener requires 24 more plants
9). There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.
Total number of students = 500
Number of rows = number of columns
500 is not a perfect square
Here remainder is 16
500 – 16= 484
√484 = 22
Ans: students left out in this arrangement are 16
Click here for the solutions of Std 8 Maths
1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
5). Squares and Square Roots
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]]>
1). What could be the possible ‘one’s’ digits of the square root of each of the following
numbers?
(i) 9801 (ii) 99856
(iii) 998001 (iv) 657666025
Solution:
(i) 9801
The possible one’s digit of the square root of 9801 may be 1 or 9. (ii) 99856
The possible one’s digit of the square root of 99856 may be 4 or 6.
(iii) 998001
The possible one’s digit of the square root of 998001 may be 1 or 9.
(iv) 657666025
The possible one’s digit of the square root of 657666025 is 5.
2). Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153 (ii) 257
(iii) 408 (iv) 441
Solution:
(i) 153
We know that numbers ending with 2, 3, 7 and 8 are not perfect squares.
Therefore 153 is not a perfect square.
(ii) 257
We know that numbers ending with 2, 3, 7 and 8 are not perfect squares.
Therefore 257 is not a perfect square.
(iii) 408
We know that numbers ending with 2, 3, 7 and 8 are not perfect squares.
Therefore 408 is not a perfect square.
(iv) 441
441 is a perfect square.
3). Find the square roots of 100 and 169 by the method of repeated subtraction.
(i) 100
100 – 1 = 99
99 – 3 = 96
96 – 5 = 91
91 – 7 = 84
84 – 9 = 75
75 – 11 = 64
64 – 13 = 51
51 – 15 = 36
36 – 17 = 19
19 – 19 = 0
In the tenth step we got 0 therefore square root of 100 is 10.
(ii) 169
169 – 1 = 168
168 – 3 = 165
165 – 5 = 160
160 – 7 = 153
153 – 9 = 144
144 – 11 = 133
133 – 13 = 120
120 – 15 = 105
105 – 17 = 98
98 – 19 = 79
79 – 21 = 58
58 – 23 = 25
25 – 25 = 0
In the thirteenth step we got 0 therefore square root of 169 is 13.
4). Find the square roots of the following numbers by the Prime Factorisation Method.
(i) 729 (ii) 400
(iii) 1764 (iv) 4096
(v) 7744 (vi) 9604
(vii) 5929 (viii) 9216
(ix) 529 (x) 8100
Solution:
(i) 729
729 = 3 X 3 X 3 X 3 X 3 X 3
= 3 X 3 X 3 X 3 X 3 X 3
= 3^{2} X 3^{2} X 3^{2}
√729 = 3 X 3 X 3 = 27
(ii) 400
400 = 2 X 2 X 2 X 2 X 5 X 5
= 2 X 2 X 2 X 2 X 5 X 5
= 2^{2} X 2^{2} X 5^{2}
√400 = 2 X 2 X 5 = 20
(iii) 1764
1764 = 2 X 2 X 3 X 3 X 7 X 7
= 2 X 2 X 3 X 3 X 7 X 7
= 2^{2} X 3^{2} X 7^{2}
√1764 = 2 X 3 X 7 = 42
(iv) 4096
4096 = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2
= 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2
= 2^{2} X 2^{2} X 2^{2 }X 2^{2} X 2^{2} X 2^{2}
√4096 = 2 X 2 X 2 X 2 X 2 X 2 = 64
(v) 7744
7744 = 2 X 2 X 2 X 2 X 2 X 2 X 11 X 11
= 2 X 2 X 2 X 2 X 2 X 2 X 11 X 11
= 2^{2} X 2^{2} X 2^{2 }X 11^{2}
√7744 = 2 X 2 X 2 X 11 = 88
(vi) 9604
9604 = 2 X 2 X 7 X 7 X 7 X 7
= 2 X 2 X 7 X 7 X 7 X 7
= 2^{2} X 7^{2} X 7^{2}
√9604 = 2 X 7 X 7 = 98
(vii) 5929
5929 = 7 X 7 X 11 X 11
= 7 X 7 X 11 X 11
= 7^{2} X 11^{2}
√5929 = 7 X 11 = 77
(viii) 9216
9216 = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3
= 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3
= 2^{2} X 2^{2} X 2^{2 }X 2^{2} X 2^{2} X 3^{2}
√4096 = 2 X 2 X 2 X 2 X 2 X 3 = 96
(ix) 529
529 = 23 X 23
= 23 X 23
= 23^{2}
√529 = 23
(x) 8100
8100 = 2 X 2 X 3 X 3 X 3 X 3 X 5 X 5
= 2 X 2 X 3 X 3 X 3 X 3 X 5 X 5
= 2^{2} X 3^{2} X 3^{2 }X 5^{2}
√8100 = 2 X 3 X 3 X 5 = 90
5). For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.
(i) 252 (ii) 180
(iii) 1008 (iv) 2028
(v) 1458 (vi) 768
Solution:
(i) 252
Prime factors of 252 = 2 X 2 X 3 X 3 X 7
Here 7 has no pair
7 is the smallest whole number by which 252 is multiplied to get a perfect square number.
New square number = 252 X 7 = 1764
√1764 = 2 X 3 X 7 = 42
(ii) 180
Prime factors of 180 = 2 X 2 X 3 X 3 X 5
Here 5 has no pair
5 is the smallest whole number by which 180 is multiplied to get a perfect square number.
New square number = 180 X 5 = 900
√900 = 2 X 3 X 5 = 30
(iii) 1008
Prime factors of 1008 = 2 X 2 X 2 X 2 X 3 X 3 X 7
Here 7 has no pair
7 is the smallest whole number by which 1008 is multiplied to get a perfect square number.
New square number = 1008 X 7 = 7056
√7056 = 2 X 2 X 3 X 7 = 84
(iv) 2028
Prime factors of 2028 = 2 X 2 X 3 X 13 X 13
Here 3 has no pair
3 is the smallest whole number by which 2028 is multiplied to get a perfect square number.
New square number = 2028 X 3 = 6084
√6084 = 2 X 3 X 13 = 78
(v) 1458
Prime factors of 1458 = 2 X 3 X 3 X 3 X 3 X 3 X 3
Here 2 has no pair
2 is the smallest whole number by which 1458 is multiplied to get a perfect square number.
New square number = 1458 X 2 = 2916
√2916 = 2 X 3 X 3 X 3 = 54
(vi) 768
Prime factors of 768 = 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 3
Here 3 has no pair
3 is the smallest whole number by which 768 is multiplied to get a perfect square number.
New square number = 768 X 3 = 2304
√2304 = 2 X 2 X 2 X 2 X 3 = 48
6). For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.
(i) 252 (ii) 2925
(iii) 396 (iv) 2645
(v) 2800 (vi) 1620
Solution:
(i) 252
Prime factors of 252 = 2 X 2 X 3 X 3 X 7
Here 7 has no pair
7 is the smallest whole number by which 252 is divided to get a perfect square number.
New square number = 252 ÷ 7 = 36
√36 = 6
(ii) 2925
Prime factors of 2925 = 3 X 3 X 5 X 5 X 13
Here 13 has no pair
13 is the smallest whole number by which 2925 is divided to get a perfect square number.
New square number = 2925 ÷ 13 = 225
√225 = 15
(iii) 396
Prime factors of 396 = 2 X 2 X 3 X 3 X 11
Here 11 has no pair
11 is the smallest whole number by which 396 is divided to get a perfect square number.
New square number = 396 ÷ 11 = 36
√36 = 6
(iv) 2645
Prime factors of 2645 = 5 X 23 X 23
Here 5 has no pair
5 is the smallest whole number by which 2645 is divided to get a perfect square number.
New square number = 2645 ÷ 5 = 529
√529 = 23
(v) 2800
Prime factors of 2800 = 2 X 2 X 2 X 2 X 5 X 5 X 7
Here 7 has no pair
7 is the smallest whole number by which 2800 is divided to get a perfect square number.
New square number = 2800 ÷ 7 = 400
√400 = 20
(vi) 1620
Prime factors of 1620 = 2 X 2 X 3 X 3 X 3 X 3 X 5
Here 5 has no pair
5 is the smallest whole number by which 1620 is divided to get a perfect square number.
New square number = 1620 ÷ 5 = 324
√324 = 18
7). The students of Class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Amount donated = 2401
Number of students = amount donated by each student
Number of students = √2401
= √(7 X 7 X 7 X 7)
= √(7^{2} X 7^{2})
= 7 X 7
= 49
Ans: number of students = 49
8). 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Number of plants = 2025
Number of plants in a row = number of rows
Number of plants in each row = √2025
= √(3 X 3 X 3 X 3 X 5 X 5)
= √(3^{2} X 3^{2 }X 5^{2})
= 3 X 3 X 5
= 45
Ans: number of plants in each row = 45
9). Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Smallest number divisible by each of the numbers 4, 9 and 10 is
LCM of 4 , 9 , 10 = 180
Prime factors of 180 = 2 X 2 X 3 X 3 X 5
Here, 5 has no pair.
The required smallest square number = 180 X 5 = 900
10). Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Smallest number divisible by each of the numbers 8, 15 and 20 is
LCM of 8, 15 and 20 = 120
Prime factors of 120 = 2 X 2 X 2 X 3 X 5
Here, 2 X 3 X 5 has no pair.
The required smallest square number = 120 X 30 = 3600
Click here for the solutions of Std 8 Maths
1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
5). Squares and Square Roots
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]]>1). Find the square of the following numbers.
(i) 32 (ii) 35
(iii) 86 (iv) 93
(v) 71 (vi) 46
Solution:
(i) 32
32 = 30 + 2
(a+b)^{2} = a^{2} + 2ab + b^{2}
32^{2} = (30 + 2)^{2}
= (30)^{2} + 2 X 30 X 2 + (2)^{2}
= 900 + 120 + 4
= 1024
(ii) 35
35 = 30 + 5
(a+b)^{2} = a^{2} + 2ab + b^{2}
35^{2} = (30 + 5)^{2}
= (30)^{2} + 2 X 30 X 5 + (5)^{2}
= 900 + 300 + 25
= 1225
(iii) 86
86 = 80 + 6
(a+b)^{2} = a^{2} + 2ab + b^{2}
86^{2} = (80 + 6)^{2}
= (80)^{2} + 2 X 80 X 6 + (6)^{2}
= 6400 + 960 + 36
= 7396
(iv) 93
93 = 90 + 3
(a+b)^{2} = a^{2} + 2ab + b^{2}
93^{2} = (90 + 3)^{2}
= (90)^{2} + 2 X 90 X 3 + (3)^{2}
= 8100 + 540 + 9
= 8649
(v) 71
71 = 70 + 1
(a+b)^{2} = a^{2} + 2ab + b^{2}
71^{2} = (70 + 1)^{2}
= (70)^{2} + 2 X 70 X 1 + (1)^{2}
= 4900 + 140 + 1
= 5041
(vi) 46
46 = 40 + 6
(a+b)^{2} = a^{2} + 2ab + b^{2}
46^{2} = (40 + 6)^{2}
= (40)^{2} + 2 X 40 X 6 + (6)^{2}
= 1600 + 480 + 36
= 2116
2). Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14
(iii) 16 (iv) 18
Solution:
(i) 6
We can get Pythagorean triplets by using general form 2m, m^{2} – 1,
m^{2} + 1.
Let m^{2} – 1 = 6
m^{2} = 6 + 1
m^{2} = 7
m cannot be an integer.
Let m^{2} + 1 = 6
m^{2 }= 6 – 1
m^{2} = 5
m cannot be an integer
therefore, let 2m = 6
m = 3
m^{2} – 1 = 3^{2} – 1 = 9 – 1 = 8
m^{2} + 1 = 3^{2} + 1 = 9 + 1 = 10
Ans: The Pythagorean triplets are 6, 8, 10.
(ii) 14
We can get Pythagorean triplets by using general form 2m, m^{2} – 1,
m^{2} + 1.
Let m^{2} – 1 = 14
m^{2} = 14 + 1
m^{2} = 15
m cannot be an integer.
Let m^{2} + 1 = 14
m^{2 }= 14 – 1
m^{2} = 13
m cannot be an integer
therefore, let 2m = 14
m = 7
m^{2} – 1 = 7^{2} – 1 = 49 – 1 = 48
m^{2} + 1 = 7^{2} + 1 = 49 + 1 = 50
Ans: the Pythagorean triplets are 14, 48, 50.
(iii) 16
We can get Pythagorean triplets by using general form 2m, m^{2} – 1,
m^{2} + 1.
let 2m = 16
m = 8
m^{2} – 1 = 8^{2} – 1 = 64 – 1 = 63
m^{2} + 1 = 8^{2} + 1 = 64 + 1 = 65
Ans: the Pythagorean triplets are 16, 63, 65.
(iv) 18
We can get Pythagorean triplets by using general form 2m, m^{2} – 1,
m^{2} + 1.
let 2m = 18
m = 9
m^{2} – 1 = 9^{2} – 1 = 81 – 1 = 80
m^{2} + 1 = 9^{2} + 1 = 81 + 1 = 82
Ans: the Pythagorean triplets are 18, 80, 82
Click here for the solutions of Std 8 Maths
1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
5). Squares and Square Roots
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]]>1). What will be the unit digit of the squares of the following numbers?
(i) 81 (ii) 272
(iii) 799 (iv) 3853
(v) 1234 (vi) 26387
(vii) 52698 (viii) 99880
(ix) 12796 (x) 55555
Solution:
(i) 81
The unit digit in the square of 81 is 1.
(ii) 272
The unit digit in the square of 272 is 4.
(iii) 799
The unit digit in the square of 799 is 1.
(iv) 3853
The unit digit in the square of 3853 is 9.
(v) 1234
The unit digit in the square of 1234 is 6. (vi) 26387
The unit digit in the square of 26387 is 9.
(vii) 52698
The unit digit in the square of 52698 is 4.
(viii) 99880
The unit digit in the square of 99880 is 0.
(ix) 12796
The unit digit in the square of 12796 is 6.
(x) 55555
The unit digit in the square of 55555 is 5.
2). The following numbers are obviously not perfect squares. Give reason.
(i) 1057 (ii) 23453
(iii) 7928 (iv) 222222
(v) 64000 (vi) 89722
(vii) 222000 (viii) 505050
Solution:
(i) 1057
The digit in the unit’s place is 7.
Therefore, it is not a perfect square.
(ii) 23453
The digit in the unit’s place is 3.
Therefore, it is not a perfect square.
(iii) 7928
The digit in the unit’s place is 8.
Therefore, it is not a perfect square.
(iv) 222222
The digit in the unit’s place is 2.
Therefore, it is not a perfect square.
(v) 64000
The number contains three 0’s at the end.
Therefore, it is not a perfect square.
(vi) 89722
The digit in the unit’s place is 2.
Therefore, it is not a perfect square.
(vii) 222000
The number contains three 0’s at the end.
Therefore, it is not a perfect square.
(viii) 505050
The number contains three 0’s at the end.
Therefore, it is not a perfect square.
3). The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826
(iii) 7779 (iv) 82004
Solution:
(i) 431
Square of 431 is odd.
(ii) 2826
Square of 2826 is even.
(iii) 7779
Square of 7779 is odd.
(iv) 82004
Square of 82004 is even.
4). Observe the following pattern and find the missing digits.
11^{2} = 121
101^{2} = 10201
1001^{2} = 1002001
100001^{2} = 1 ……… 2 ……… 1
10000001^{2} = ………………………
Solution:
11^{2} = 121
101^{2} = 10201
1001^{2} = 1002001
100001^{2} = 10000200001
10000001^{2} =100000020000001
5). Observe the following pattern and supply the missing numbers.
11^{2} = 1 2 1
101^{2} = 1 0 2 0 1
10101^{2} = 102030201
1010101^{2} = ………………………
…………^{2} = 10203040504030201
Solution:
11^{2} = 1 2 1
101^{2} = 1 0 2 0 1
10101^{2} = 102030201
1010101^{2} = 1020304030201
101010101^{2} = 10203040504030201
6). Using the given pattern, find the missing numbers.
1^{2} + 2^{2} + 2^{2} = 3^{2}
2^{2} + 3^{2} + 6^{2} = 7^{2}
3^{2} + 4^{2} + 12^{2} = 13^{2}
4^{2} + 5^{2} + _^{2} = 21^{2}
5^{2} + _^{2} + 30^{2} = 31^{2}
6^{2} + 7^{2} + _^{2} = __^{2}
Solution:
1^{2} + 2^{2} + 2^{2} = 3^{2}
2^{2} + 3^{2} + 6^{2} = 7^{2}
3^{2} + 4^{2} + 12^{2} = 13^{2}
4^{2} + 5^{2} +20^{2} = 21^{2}
5^{2} + 6^{2} + 30^{2} = 31^{2}
6^{2} + 7^{2} + 42^{2} = 43^{2}
7). Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
(i) 1 + 3 + 5 + 7 + 9
We know that sum of n odd numbers = n^{2}
1 + 3 + 5 + 7 + 9 = 5^{2} = 25 (n = 5)
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
We know that sum of n odd numbers = n^{2}
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 = 10^{2} (n = 10)
= 100
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
We know that sum of n odd numbers = n^{2}
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 + 21 + 23 = 12^{2}
= 144 (n = 12)
8). (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Solution:
(i) Express 49 as the sum of 7 odd numbers.
n = 11
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) Express 121 as the sum of 11 odd numbers.
n = 11
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9). How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26
(iii) 99 and 100
Solution:
(i) 12 and 13
We know that between the squares of n and (n+1) there are 2n nonperfect square numbers.
2n = 2 X 12 = 24
24 numbers lie between squares of 12 and 13.
(ii) 25 and 26
We know that between the squares of n and (n+1) there are 2n nonperfect square numbers.
2n = 2 X 25 = 50
50 numbers lie between squares of 25 and 26.
(iii) 99 and 100
We know that between the squares of n and (n+1) there are 2n nonperfect square numbers.
2n = 2 X 99 = 198
198 numbers lie between squares of 99 and 100.
Click here for the solutions of Std 8 Maths
1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
5). Squares and Square Roots
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]]>1). List the outcomes you can see in these experiments.
(a) Spinning a wheel (b) Tossing two coins together
Solution:
(a) Spinning a wheel
Possible outcomes = { A, B, C, D, E }
(b) Tossing two coins together
Possible outcomes = { HH, HT, TH, TT }
2). When a die is thrown, list the outcomes of an event of getting
(i) (a) a prime number (b) not a prime number.
Solution:
(a) a prime number
Possible outcomes = { 2, 3, 5 }
(b) not a prime number.
Possible outcomes = { 1, 4, 6 }
(ii) (a) a number greater than 5
Possible outcomes = { 6 }
(b) a number not greater than 5.
Possible outcomes = { 1, 2, 3, 4, 5 }
3). Find the.
(a) Probability of the pointer stopping on D in (Question 1(a))?
(b) Probability of getting an ace from a well shuffled deck of 52 playing cards?
(c) Probability of getting a red apple. (See figure below)
Solution:
(a) Probability of the pointer stopping on D in (Question 1(a))?
Possible outcomes = { A, B, C, D, E }
n(S)= 5
Let A be the event of the pointer stopping on D.
Expected outcome = D
n(A)= 1
p(A)= ^{n(A)}/_{n(S)}
= ^{1}/_{5}
(b) Probability of getting an ace from a wellshuffled deck of 52 playing cards?
Possible outcomes = 52
n(S)= 52
Let A be the event the card is an ace
Expected outcome = Ace card
n(A)= 4
p(A)= ^{n(A)}/_{n(S)}
= ^{4}/_{52}
= ^{1}/_{13}
(c) Probability of getting a red apple. (See figure below)
Possible outcomes = { G, R, R, R, G, R, G }
n(S)= 7
Let A be the event of getting a red apple.
Expected outcome = Red apple = { R, R, R, R }
n(A)= 4
p(A)= ^{n(A)}/_{n(S)}
= ^{4}/_{7}
4). Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of.
(i) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv) getting a 1digit number?
Solution:
Possible outcomes = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }
n(S) = 10
(i) getting a number 6?
Let A be the event of getting number 6
Expected outcome = { 6 }
n(A) = 1
p(A)= ^{n(A)}/_{n(S)}
= ^{1}/_{10}
(ii) getting a number less than 6?
Let B be the event of getting number less than 6
Expected outcome = { 1, 2, 3, 4, 5 }
n(B) = 5
p(B)= ^{n(B)}/_{n(S)}
= ^{5}/_{10}
= ^{1}/_{2}
(iii) getting a number greater than 6?
Let C be the event of getting number greater than 6
Expected outcome = { 7, 8, 9, 10 }
n(C) = 4
p(C)= ^{n(C)}/_{n(S)}
= ^{4}/_{10}
_{ }= ^{2}/_{5}
(iv) getting a 1digit number?
Let D be the event of getting a 1digit number
Expected outcome = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }
n(D) = 9
p(D)= ^{n(D)}/_{n(S)}
= ^{9}/_{10}
5). If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non blue sector?
Possible outcomes = { G, G, G, B, R }
n(S) = 5
let A be the event of getting a green sector
Expected outcomes = { G, G, G }
n(A) = 3
p(A)= ^{n(A)}/_{n(S)}
= ^{3}/_{5}
Let B be the event of getting a non blue sector
Expected outcome = { G, G, G, R }
n(B) = 4
p(B)= ^{n(B)}/_{n(S)}
= ^{4}/_{5}
6). Find the probabilities of the events given in Question 2.
Possible outcomes = { 1, 2, 3, 4, 5, 6 }
n(S) = 6
(a) a prime number
Let A be the event of getting a prime number
Possible outcomes = { 2, 3, 5 }
n(A) = 3
p(A)= ^{n(A)}/_{n(S)}
= ^{3}/_{6}
_{ }= ^{1}/_{2}
(b) not a prime number.
Let B be the event of getting not a prime number
Possible outcomes = { 1, 4, 6 }
n(A) = 3
p(A)= ^{n(A)}/_{n(S)}
= ^{3}/_{6}
_{ }= ^{1}/_{2}
(ii) (a) a number greater than 5
Let C be the event of getting number greater than 5
Expected outcome = { 6 }
n(C) = 1
p(C)= ^{n(C)}/_{n(S)}
= ^{1}/_{6}_{ }
(b) a number not greater than 5.
Let D be the event of getting a number not greater than 5
Expected outcome = { 1, 2, 3, 4, 5 }
n(D) = 5
p(D)= ^{n(D)}/_{n(S)}
= ^{5}/_{6}
Click here for the solutions of Std 8 Maths
1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
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]]>1). A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey. From this pie chart answer the following:
(i) If 20 people liked classical music, how many young people were surveyed?
20 people represent 10%
Let the total number of people surveyed be x
10% X x = 20
x = (20 X 100)/10
x = 2000/10
x = 200
Ans: Total young people surveyed = 200
(ii) Which type of music is liked by the maximum number of people?
Ans: Light music is liked by the maximum number of people.
(iii) If a cassette company were to make 1000 CD’s, how many of each type would they make?
Total CD’s = 1000
Light music = 40% X 1000
= (40 X 1000)/100
= 40000/100
= 400
Folk music = 30% X 1000
= (30 X 1000)/100
= 30000/100
= 300
Classical music = 10% X 1000
= (10 X 1000)/100
= 10000/100
= 100
Semi cClassical music = 20% X 1000
= (20 X 1000)/100
= 20000/100
= 200
2). A group of 360 people were asked to vote for their favourite season from the three seasons rainy, winter and summer.
Season  No of Votes 
Summer  90 
Rainy  120 
Winter  150 
(i) Which season got the most votes?
(ii) Find the central angle of each sector.
(iii) Draw a pie chart to show this information.
Solution:
(i) Which season got the most votes?
Ans: Winter season got the most votes.
(ii) Find the central angle of each sector.
Season  No of Votes  Central angle 
Summer  90  90 
Rainy  120  120 
Winter  150  150 
Total  360 
Central angle for summer = (90/360) X 360
= 90
Central angle for Rainy = (120/360) X 360
= 120
Central angle for Winter = (150/360) X 360
= 150
(iii) Draw a pie chart to show this information.
3). Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.
Colours  No. of People 
Blue  18 
Green  9 
Red  6 
Yellow  3 
Total  36 
Solution:
Colours  No. of People  fraction  Central Angle 
Blue  18  ^{18}/_{36} = ½

½ X 360 = 180^{0} 
Green  9  ^{9}/_{36} = ¼

¼ X 360 = 90^{0} 
Red  6  ^{6}/_{36} = ^{1}/_{6}

^{1}/_{6} X 360 = 60^{0} 
Yellow  3  ^{3}/_{36} = ^{1}/_{12}

^{1}/_{12} X 360 = 30^{0} 
Total  36 
Central angle for Blue = 180^{0}
Central angle for Green = 90^{0}
Central angle for Red = 60^{0}
Central angle for Yellow = 30^{0}
4). The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.
(i) In which subject did the student score 105 marks?
(Hint: for 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle?)
(ii) How many more marks were obtained by the student in Mathematics than in Hindi?
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
(Hint: Just study the central angles).
Solution:
(i) In which subject did the student score 105 marks?
(Hint: for 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle?)
For marks 540 the central angle is 360
Therefore, for 105 marks central angle = (^{360}/_{540} )X 105
= (^{2}/_{3} )X 105
= 2 X 35
= 70^{0}
Corresponding subjet to 70^{0} is Hindi.
(ii) How many more marks were obtained by the student in Mathematics than in Hindi?
Marks obtained in Mathematics = (^{90}/_{360} )X 540
= (^{1}/_{40} )X 540
= 1 X 135
= 135
Marks obtained in Hindi = 105
More marks obtained in Mathematics than Hindi = 135 – 105
= 30
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
(Hint: Just study the central angles).
Central angle for Social Science and Mathematics = 65 + 90
= 155
Central angle for Science and Hindi = 80 + 70
= 150
5). The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.
Languages  Hindi  English  Marathi  Tamil  Bengal  Total 
Number of Students  40  12  9  7  4  72 
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1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
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]]>1). State whether True or False.
(a) All rectangles are squares
(b) All rhombuses are parallelograms
(c) All squares are rhombuses and also rectangles
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Answer:
(a) All rectangles are squares
Ans: False
(b) All rhombuses are parallelograms
Ans: True
(c) All squares are rhombuses and also rectangles
Ans: True
(d) All squares are not parallelograms.
Ans: False
(e) All kites are rhombuses.
Ans: False
(f) All rhombuses are kites.
Ans: True
(g) All parallelograms are trapeziums.
Ans: True
(h) All squares are trapeziums.
Ans: True
2). Identify all the quadrilaterals that have.
(a) four sides of equal length (b) four right angles
Solution:
(a) four sides of equal length
Ans: Rhombus, square.
(b) four right angles
Ans: Rectangle, square.
3). Explain how a square is.
(i) a quadrilateral (ii) a parallelogram (iii) a rhombus (iv) a rectangle
Solution:
(i) a quadrilateral
Ans: A square has four sides. Therefore, it is a quadrilateral.
(ii) a parallelogram
Ans: Opposite sides of a square are parallel to each other. Therefore, it is a parallelogram.
(iii) a rhombus
Ans: A square is a parallelogram with all four sides equal. Therefore, it is a rhombus.
(iv) a rectangle
Ans: A square is a parallelogram with each angle being a right angle. Therefore, it is a rectangle.
4). Name the quadrilaterals whose diagonals.
(i) bisect each other (ii) are perpendicular bisectors of each other (iii) are equal
Solution:
(i) bisect each other
Ans: Parallelogram. rectangle, rhombus, square.
(ii) are perpendicular bisectors of each other
Ans: Rhombus, square.
(iii) are equal
Ans: Square, rectangle.
5). Explain why a rectangle is a convex quadrilateral.
Ans: Both the diagonals of a rectangle lie in it’s interior. Therefore, it is a convex quadrilateral.
6). ABC is a rightangled triangle and O is the midpoint of the side
opposite to the right angle. Explain why O is equidistant from A,
B and C. (The dotted lines are drawn additionally to help you).
In the □ABCD opposite sides are parallel.
Therefore, □ABCD is a parallelogram.
Diagonals of parallelogram bisect each other.
Hence, O is equidistant from A, B and C.
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1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
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]]>1). Given a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD = …… (ii) Ð DCB = ……
(iii) OC = …… (iv) m Ð DAB + m Ð CDA = ……
Answer:
(i) AD = BC
The opposite sides of a parallelogram are congruent.
(ii) Ð DCB = Ð DAB
The opposite angles of a parallelogram are congruent.
(iii) OC = OA
The diagonals of a parallelogram bisect each other.
(iv) m Ð DAB + m Ð CDA = 180^{0}
The adjacent angles of a parallelogram are supplementary.
2). Consider the following parallelograms. Find the values of the unknowns x, y, z.
(i) The adjacent angles of a parallelogram are supplementary.
Ð ABC + Ð BCD = 180^{0}
100^{0} + x = 180^{0}
x = 180 – 100
x = 80^{0}
the opposite angles of a parallelogram are congruent
Ð ADC = Ð ABC
but Ð ABC = 100^{0}
Ð ADC = y = 100^{0}
Similarly ÐBCD = ÐBAD
but Ð BCD = 80^{0}
Ð BAD = z = 80^{0}
(ii)
The adjacent angles of a parallelogram are supplementary.
50^{0} + x = 180^{0}
x = 180^{0 }– 50^{0}
x = 130^{0}
the opposite angles of a parallelogram are congruent
x = y
but x = 130^{0}
y = 130^{0}
z and x are corresponding angles
z = x
but x = 130^{0}
z = 130^{0}
(iii)
diagonals are intersecting at right angles
The given quadrilateral is a rhombus
x = 90^{0 } vertically opposite angles
x + y + 30^{0} = 180^{0} angle sum property of triangle
90^{0 }+ y + 30^{0} = 180^{0}
y + 120^{0} = 180^{0}
y = 180^{0} – 120^{0}
y = 60^{0}
z = y alternate angles
but y = 60^{0}
z = 60^{0}
(iv)
The adjacent angles of a parallelogram are supplementary.
80^{0} + x = 180^{0}
x = 180^{0 }– 80^{0}
x = 100^{0}
the opposite angles of a parallelogram are congruent
y = 80^{0}
z = y alternate angles
but y = 80^{0}
z = 80^{0}
(v)
the opposite angles of a parallelogram are congruent
y = 112^{0}^{ }
x + y + 40^{0} = 180^{0} angle sum property of triangle
x + 112^{0 }+ 40^{0} = 180^{0}
x + 152^{0} = 180^{0}
x = 180^{0} – 152^{0}
x = 28^{0}
z = x alternate angles
but x = 28^{0}
z = 28^{0}
3). Can a quadrilateral ABCD be a parallelogram if
(i) Ð D + Ð B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) Ð A = 70° and Ð C = 65°?
Solution:
(i) Ð D + Ð B = 180°?
A quadrilateral can be a parallelogram if
i). The opposite angles are congruent and
ii). The adjacent angles are supplementary
as the information is incomplete
ABCD may or may not be a parallelogram
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
A quadrilateral can be a parallelogram if the opposite sides are congruent
Here AB = DC but AD ≠ BC
ABCD is not a parallelogram
(iii) Ð A = 70° and Ð C = 65°?
A quadrilateral can be a parallelogram if the opposite angles are congruent
ÐA ≠ ÐC
ABCD is not a parallelogram
4). Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
5). The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.
Let the common multiple be x
The measures of the two adjacent angles are 3x and 2x
The adjacent angles of a parallelogram are supplementary
3x + 2x = 180^{0}
5x = 180^{0}
x = 180/5
x = 36^{0}
3x = 3 X 36 = 108^{0}
2x = 2 X 36 = 72^{0}
6). Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Let the two equal adjacent angles be x
The adjacent angles of a parallelogram are supplementary
x + x = 180^{0}
2x = 180^{0}
x = 180/2
x = 90^{0}
opposite angles of parallelogram are equal
therefore, each of the angle of parallelogram is 90^{0}
7). The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
y = 40^{0} alternate angles
y + z = 70^{0} Exterior angle property
40^{0} + z = 70^{0}
z = 70^{0 }– 40^{0}
z = 30^{0}
^{ Ð}POH + 70^{0} = 180^{0} linear pair
ÐPOH = 180^{0 }– 70^{0}
ÐPOH = 110^{0}
ÐPOH = x opposite angles
x = 110^{0}
8). The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
i).
opposite sides of a parallelogram are congruent
GS = UN and GU = SN
GS = UN
3x = 18
x = 18/3
x = 6 cm
GU = SN
3y – 1 = 26
3 y = 26 + 1
3 y = 27
y = 27/3
y = 9 cm
ii).
The diagonals of a parallelogram bisect each other
x + y = 16 and y + 7 = 20
y + 7 = 20
y = 20 – 7
y = 13 cm
x + y = 16
x + 13 = 16
x = 16 – 13
x = 3 cm
9). In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Adjacent angles of a parallelogram are supplementary
ÐRKS + ÐKSI = 180^{0}
120^{0} + ÐKSI = 180^{0}
ÐKSI = 180^{0 }– 120^{0}
ÐKSI = 60^{0}
Similarly, ÐCLU = ÐUEC opposite angles of parallel
ÐUEC = 70^{0}
70^{0} + 60^{0} + x = 180^{0}
130^{0} + x = 180^{0}
x = 180^{0 }– 130^{0}
x = 50^{0}
^{ }10). Explain how this figure is a trapezium. Which of its two sides are parallel? (Fig 3.26)
ÐM + ÐN = 100^{0} + 80^{0} = 180^{0}
One pair of adjacent angles are supplementary
□KLMN is trapezium
KL II MN
11). Find mÐC in Fig 3.27 if AB II DC.
AB II DC and BC is transversal
ÐB + ÐC = 180^{0 }interior angles
120^{0} + ÐC = 180^{0}
ÐC = 180^{0 }– 120^{0}
ÐC = 60^{0}
^{ }12). Find the measure of ÐP and ÐS if SP II RQ in Fig 3.28. (If you find mÐR, is there more than one method to find mÐP?)
SP II RQ and SR is transversal
ÐR + ÐS = 180^{0 }interior angles
90^{0} + ÐS = 180^{0}
ÐS = 180^{0 }– 90^{0}
ÐS = 90^{0}^{ }
ÐP + ÐQ + ÐR + ÐS = 360^{0 }
the sum of the angles of a quadrilateral is 360^{0}
ÐP + 130^{0}+ 90^{0 }+ 90^{0} = 360^{0}
ÐP + 310^{0} = 360^{0}
ÐP = 360^{0 }– 310^{0}
ÐS = 50^{0}
Click here for the solutions of Std 8 Maths
1). Rational Numbers
2). Linear Equations in One Variable
3). Understanding Quadrilaterals
4). Data Handling
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