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These NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots InText Questions

Try These (Page No. 111)

Question 1.
Find the one’s digit of the cube of each of the following numbers.
(i) 3331
(ii) 8888
(iii) 149
(iv) 1005
(v) 1024
(vi) 77
(vii) 5022
(viii) 53
Solution:
(i) 3331
unit digit of the number = 1
unit digit of the cube of the number = 1

(ii) 8888
unit digit of the number = 8
unit digit of the cube of the number = 2

(iii) 149
unit digit of the number = 9
unit digit of the cube of the number = 9

(iv) 1005
unit digit of the number = 5
unit digit of the cube of the number = 5

(v) 1024
unit digit of the number = 4
unit digit of the cube of the number = 4

(vi) 77
unit digit of the number = 7
unit digit of the cube of the number = 3

(vii) 5022
unit digit of the number = 2
unit digit of the cube of the number = 8

(viii) 53
unit digit of the number = 3
unit digit of the cube of the number = 7

Question 2.
Express the following numbers as the sum of odd numbers using the above pattern?
(a) 6³
(b) 8³
(c) 7³
Solution:
(a) n = 6 and (n – 1) = 5
We start with (6 × 5) + 1 = 31
We have
6³ = 31 + 33 + 35 + 37 + 39 + 41 = 216

(b) n = 8 and (n – 1) = 7
We start with (8 × 7) + 1 = 57
We have
8³ = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512

(c) n = 7 and (n – 1) = 6
We start with (7 × 6) + 1 = 43
We have
7³ = 43 + 45 + 47 + 49 + 51 + 53 + 53 + 55 = 343

Question 3.
Consider the following pattern.
2³ – 1³ = 1 + 2 × 1 × 3
3³ – 2³ = 1 + 3 × 2 × 3
4³ – 3³ = 1 + 4 × 3 × 3
Using the above pattern, find the value of the following.
(i) 7³ – 6³
(ii) 12³ – 11³
(iii) 20³ – 19³
(iv) 51³ – 50³
Solution:
(i) 7³ – 6³ = 1 + 7 × 6 × 3 = 1 + 126 = 127
(ii) 12³ – 11³ = 1 + 12 × 11 × 3 = 1 + 396 = 397
(iii) 20³ – 19³ = 1 + 20 × 19 × 3 = 1 + 1140 = 1141
(iv) 51³ – 50³ = 1 + 51 × 50 × 3 = 1 + 7650 = 7651
Note: In the prime factorisation of any number, if each other appears three times, then the number is a perfect cube.

Try These (Page No. 112)

Question 4.
Which of the following are perfect cubes?
1. 400
2. 3375
3. 8000
4. 15625
5. 9000
6. 6859
7. 2025
8. 10648
Solution:
1. We have
400 = 2 × 2 × 2 × 2 × 5 × 5
2 × 5 × 5 remain after grouping in triples.
400 is not a perfect cube.

2. We have
3375 = 3 × 3 × 3 × 5 × 5 × 5
The prime factors appear in triples.
3375 is a perfect cube.

3. We have
8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5
The prime factor of 8000 can be grouped into triples and no factor is left over.
8000 is a perfect cube.

4. We have
15625 = 5 × 5 × 5 × 5 × 5 × 5
The Prime factors of 15625 can be grouped into triples and no factor is left over.
15625 is a perfect cube.

5. We have
9000 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5
The prime factors of 9000 cannot be grouped into tripes (because 3 × 3 are leftover).
9000 is not a perfect cube.

6. We have
6859 = 19 × 19 × 19
The prime factors of 6859 can be grouped into triples and no factor is left over.
6859 is a perfect cube.

7. We have
2025 = 3 × 3 × 3 × 3 × 5 × 5
We do not get triples of prime factors of 2025 and 3 × 5 × 5.
2025 is not a perfect cube.

8. We have
10648 = 2 × 2 × 2 × 11 × 11 × 11
The prime factor of 10648 can be grouped into triples and no factor is left over.
10648 is a perfect cube.

These NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Exercise 7.2

Question 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
On grouping the factors in triplets, we get
64 = 2³ × 2³
64 = (2 × 2)³ = 4³
\(\sqrt[3]{64}\) = 4

(ii) 512
On grouping the factors in triplets, we get
512 = 2³ × 2³ × 2³
= (2 × 2 × 2)³
= 8³
\(\sqrt[3]{512}\) = 8

(iii) 10648
On grouping the factors in triplets, we get
10648 = 2³ × 11³
= (2 × 11)³
= 22³
\(\sqrt[3]{10648}\) = 22

(iv) 27000
On grouping the factors in triplets, we get
27000 = 2³ × 3³ × 5³
= (2 × 3 × 5)³
= 30³
\(\sqrt[3]{27000}\) = 30

(v) 15625
On grouping the factors in triplets, we get
15625 = 5³ × 5³
= (5 × 5)³
= 25³
\(\sqrt[3]{15625}\) = 25

(vi) 13824
On grouping the factors in triplets, we get
13824 = 2³ × 2³ × 2³ × 3³
= (2 × 2 × 2 × 3)³
= 24³
\(\sqrt[3]{13824}\) = 24

(vii) 110592
On grouping the factors in triplets, we get
110592 = 2³ × 2³ × 2³ × 2³ × 3³
= (2 × 2 × 2 × 2 × 3)³
= 48³
\(\sqrt[3]{110592}\) = 48

(viii) 46656
On grouping the factors in triplets, we get
46656 = 2³ × 2³ × 3³ × 3³
= (2 × 2 × 3 × 3)³
= 36³
\(\sqrt[3]{46656}\) = 36

(ix) 175616
On grouping the factors in triplets, we get
175616 = 2³ × 2³ × 2³ × 7³
= (2 × 2 × 2 × 7)³
= 56³
\(\sqrt[3]{175616}\) = 56

(x) 91125
On grouping the factors in triplets, we get
91125 = 3³ × 3³ × 5³
= (3 × 3 × 5)³
= 45³
\(\sqrt[3]{91125}\) = 45

Question 2.
State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If the square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two-digit number may be a three-digit number.
(vi) The cube of a two-digit number may have seven or more digits.
(vii) The cube of a single-digit number may be a single-digit number.
Solution:
(i) False
(ii) True
(iii) False [15² = 225; 15³ = 3375]
(iv) False [12³ = 1728]
(v) False [10³ = 1000]
(vi) False [99³ = 970299]
(vii) True [2³ = 8]

Question 3.
You are told that 1,331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
(i) Separating the given number 1331 into two groups.
1 and 331
331 ends in 1
unit digit of the cube root = 1
Tens digit of the cube root = 1
\(\sqrt[3]{1331}\) = 11

(ii) Cube root of 4913
Separating the given number 4913 in two groups i.e. 4 and 913
In this case, 913 has three-digit and 4 has only one digit
The digit 3 is at its own place. We take the one’s place of the required cube root as 7.
Take the other group i.e. 4, a cube of 1 is 1, and a cube of 2 is 8. 4 lies between 1 and 8.
The smaller number among 1 and 2 is 1
The one place of 1 is 1 itself.
Take 1 as ten’s place of the cube root of 4913.
\(\sqrt[3]{4913}\) = 17

(iii) Cube root of 12167
Separating 12167 in two groups i.e. 12 and 167
The digit 7 is at its one’s place. We take the one’s place of required cube root as 3.
The unit digit of the cube root = 3
Take the other group i.e. 12 Cube of 2 is 8 and cube of 3 is 27. 12 lies between 8 and 27
The smaller among 2 and 3 is 2
The one place is 2 itself.
Take 2 as ten’s place of the cube root of 12167.
\(\sqrt[3]{12167}\) = 23

(iv) Cube root of 32768
Separating 32768 in two groups i.e. 32 and 768
Take 768
The digit 8 is at its one’s place so, the one’s place of the required cube root is 2.
Take the other group i.e. 32
The cube of 3 is 27 and the cube of 4 is 64.
32 lies between 27 and 64.
The smaller number between 3 and 4 is 3
Take 3 as ten’s place of the cube root of 32768
\(\sqrt[3]{32768}\) = 32

These NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Exercise 7.1

Question 1.
Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Solution:
(i) 216
216 = 2³ × 3³
Each prime factors appear in triplets, so 216 is a perfect cube.

(ii) 128
128 = 2³ × 2³ × 2
2 remains after grouping in triplets, so 128 is not a perfect cube.

(iii) 1000
1000 = 2³ × 5³
Each prime factor appears in triplets so 1000 is a perfect cube.

(iv) 100
100 = 2² × 5²
As we do not get any triplets, So 100 is not a perfect cube.

(v) 46656
46656 = 2³ × 2³ × 3³ × 3³
Each prime factor appears in triplets, so 46656 is a perfect cube.

Question 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 = 3³ × 3²
On grouping the factors in triplets, the prime factor 3 does not appear in a group of three.
To make it a perfect cube, we need one more 3.
The number 243 should be multiplied by 3 to get a perfect cube.

(ii) 256
256 = 2³ × 2³ × 2²
On grouping the factors in triplets, the prime factor 2 does not appears in a group of three.
To make it a perfect cube, we need one more 2.
The number 256 should be multiplied by 2 to get a perfect cube.

(iii) 72
72 = 2³ × 3²
On grouping, the factors in triplets the prime factor 3 does not appear in a group of 3.
To make it a perfect cube, we multiply it by 3.
The number 72 should be multiplied by 3 to get a perfect cube.

(iv) 675
675 = 3³ × 5²
On grouping the factors in triplets, the prime factor 5 does not appear in a group of three.
To make it a perfect cube, we multiply it by 5.
The number 675 should be multiplied by 5 to get a perfect cube.

(v) 100
100 = 2² × 5²
On grouping the factors in triplets, the prime factors 2 and 5 do not appear in a group of three.
To make it a perfect cube we need one 2 and one 5 (2 × 5 = 10)
The number 100 should be multiplied by 10 to get a perfect cube.

Question 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 = 3³ × 3
On grouping the factors in triplets, we find that three is no triplet of 3.
To make it a perfect cube, we divide it by 3.

(ii) 128
128 = 2³ × 2³ × 2
On grouping the factors in triplets, we find that there is no triplet of 2.
To make it a perfect cube, we divide it by 2.

(iii) 135
135 = 3³ × 5
On grouping the factors in triplets, the prime factor 5 does not appear in a group of three.
To make it a perfect cube, 135 should be divided by 5.

(iv) 192
192 = 2³ × 2³ × 3
On grouping the factors in triplets, the prime factor 3 does not appear in a group of three.
To make it a perfect cube, 192 should be divided by 3.

(v) 704
704 = 2³ × 2³ × 11
On grouping the factors in triplets, the prime factor 11 does not appear in a group of three.
To make it a perfect cube, 704 should be divided by 11.

Question 4.
Parikshit makes a cuboid of plasticine on sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Solution:
Sides of the cuboid are 5 cm, 2 cm, and 5 cm
Volume of the cuboid = 5 cm × 2 cm × 5 cm
To form it as a cube, its dimension should be in the group of triplets.
Since there is only one 2 and only two 5 in the prime factorization.
We need 2 × 2 × 5 = 20 such cuboids to make a cube.

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.3

Question 1.
Find the value of the unknown x in the following diagrams:


Answer:
(i) x + 50° + 60°= 180°
(angle sum property of a triangle)
x + 110° = 180°
x = 180°- 110°
= 70°
∠x = 70°

(ii) x + 90° + 30°= 180°
(angle sum property of a triangle)
x + 120° = 180°
x = 180° – 120°
= 60°
∠x = 60°

(iii) 30° + 110° + x = 180°
(angle sum property of a triangle)
140°+ x = 180°
x = 180° – 140°
= 40°
∠x =40°

(iv) x + x + 50°= 180°
(angle sum property of a triangle)
2x + 50° = 180°
2x = 180° – 50°
2x = 130°
x = \(\frac{130^{\circ}}{2}\) = 65°
∠x = 65°

(v) x + x + x = 180°
(angle sum property of a triangle)
3x = 180°
x = \(\frac{180^{\circ}}{3}\)
x = 60°
∠x = 60°

(vi) x + 2x + 90° = 180°
(angle sum property of a triangle)
3x + 90° = 180°
3x = 180° – 90°
= 90°
x = \(\frac{90^{\circ}}{3}\)
= 30°
∴ ∠x = 30°

Question 2.
Find the values of the unknown x and y in the following diagrams:

Answer:
(i) ∠y + 120°= 180°
(linear pair of angles)
∠y = 180° – 120°
= 60°
∠x + ∠y + 50° = 180°
(using the angle sum property of a triangle)
∠x + 60° + 50° = 180°
∠x + 110° = 180°
∠x = 180°- 110°
= 70°
Thus, ∠x = 70° and ∠y = 60°

(ii) ∠y = 80°
(vertically opposite angles are equal)
∠x + ∠y + ∠50° = 180°
(using the angle sum property of a triangle)
∠x + 80° + 50°= 180°
∠x + 130° = 180°
∠x = 180° – 130°
= 50°
Thus, ∠x = 50° and ∠y = 80°

(iii) 50° + 60° + ∠y = 180°
(using the angle sum property of a triangle)
110° + ∠y – 180°
∠y = 180°- 110°
= 70°
∠x and ∠y form a linear pair.
∠x + ∠y = 180°
∠x + 70° = 180°
∠x = 180° – 70°
= 110°
Thus, ∠x =110° and ∠y = 70°

(iv) ∠x = 60°
(vertically opposite angles)
∠x+ ∠y + 30° = 180°
(angle sum property of a triangle)
60° + ∠y + 30° = 180°
∠y + 90° = 180°
∠y = 180° – 90°
∠y = 90°
Thus, ∠x = 60° and ∠y = 90°

(v) ∠y = 90°
(vertically opposite angles are equal)
∠x + ∠x + ∠y = 180°
(angle sum property of triangle)
2∠x + 90° = 180°
2x = 180° – 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\) = 45°
Thus, ∠y = 90°,∠x = 45°

(vi) One angle of the triangle = y
Each of the other two angles is equal to their vertically opposite angle ‘x’
∠x + ∠x + ∠y = 180°
(angle sum property of a triangle)
2x + y = 180°
2x + x = 180°
3x = 180°
x = \(\frac{180^{\circ}}{3}\)
x = 60°
y = x
(Vertically opposite angles) y = 60°
Thus, x = 60° and y = 60°

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots InText Questions

Try These (Page No. 90)

Question 1.
Find the perfect square numbers between
(i) 30 and 40 and
(ii) 50 and 60.
Solution:
(i) Since,
1 × 1 = 1
2 × 2 = 4
3 × 3 = 9
4 × 4 = 16
5 × 5 = 25
6 × 6 = 4
7 × 7 = 49
Thus 36 is a perfect square number between 30 and 40.

(ii) Since, 7 × 7 = 49 and 8 × 8 = 64.
It means there is no perfect number between 49 and 64 and thus there is no perfect number between 50 and 60.

Question 2.
Can we say whether the following numbers are perfect squares? How do we know?
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 1069
(vi) 2061
Write five numbers that you can decide by looking at their ones digit that they are not square numbers.
Solution:
(i) 1057
∵ The ending digit is 7 (which is not one of 0, 1, 4, 5, 6, or 9).
∴ 1057 cannot be a square number.

(ii) 23453
∵ The ending digit is 3 (which is not one of 0, 1, 4, 5, 6, or 9).
∴ 23453 cannot be a square number.

(iii) 7928
∵ The ending digit is 8 (which is not one of 0, 1, 4, 5, 6, or 9).
∴ 7928 cannot be a square number.

(iv) 222222
∵ The ending digit is 2 (which is not one of 0, 1, 4, 5, 6, or 9).
∴ 222222 cannot be a square number.

(v) 1069
∵ The ending digit is 9
∴ It may or may not be a square number.
Also, 30 × 30 = 900
31 × 31 = 961
32 × 32 = 1024
33 × 33 = 1089
i.e., No natural number between 1024 and 1089 which is a square number.
1069 cannot be a square number.

(iv) 2061
∵ The ending digit is 1
∴ It may or may not be a square number.
∵ 45 × 45 = 2025 and 46 × 46 = 2116
i.e., No natural number between 2025 and 2116 which is a square number.
∴ 2061 is not a square number.
We can write many numbers which do not end with 0, 1, 4, 5, 6, or 9. (i.e., which are not square numbers)
Five such numbers can be: 1234, 4312, 5678, 87543, 1002007

Question 3.
Write five numbers that you cannot decide just by looking at their units digit (or one’s place) whether they are square numbers or not.
Solution:
Property 1.
Any natural number ending in 1, 4, 5, 6, or 9 can be or cannot be a square number.
Five such numbers are: 56790, 3671, 2454, 76555, 69209
Property 2.
If a number has 1 or 9 in the unit’s place, then its square ends in 1.
For example: (1)² = 1, (9)² = 81, (11)² = 121, (9)² = 361, (21)² = 441

Try These (Page No. 91)

Question 4.
Which of 123² , 77², 82², 161², 109² would end with the digit 1?
Solution:
The squares of those numbers end in 1 which end in either 1 or 9.
∴ The squares of 161 and 109 would end in 1.

Question 5.
Which of the following numbers would have digit 6 at the unit place.
(i) 19²
(ii) 24²
(iii) 26²
(iv) 36²
(v) 34²
Solution:
(i) 19²: Unit place digit = 9.
19² would not have unit digit as 6.

(ii) 24²: Unit place digit = 4.
24² would not have unit digit as 6.

(iii) 26²: Unit place digit = 6.
26² would have 6 as unit place.

(iv) 36²: Unit place digit = 6.
36² would end in 6

(v) 34²: Since the unit place digit is 4
34² would have unit place digit as 6.

Try These (Page No. 92)

Question 6.
What will be the “one’s digit” in the square of the following numbers?
(i) 1234
(ii) 26387
(iii) 52698
(iv) 99880
(v) 21222
(vi) 9106
Solution:
(i) 19²: Ending digit = 4 and 4² = 16.
(1234)² will have 6 as the ones digit.

(ii) 24²: Ending digit is 7 and 7² = 49.
(26387)² will have 9 as the ones digit.

(iii) 26²: Ending digit is 8, and 8² = 64.
(52698)² will end in 4.

(iv) 36²: Ending digit is 0.
(99880)² will end in 0.

(v) 2² = 4
Ending digit of (21222)² is 4.

(v) 6² = 36
Ending digit of (9106)² is 6.

Question 7.
The square of which of the following numbers would be an odd number/an even number? Why?
(i) 727
(ii) 158
(iii) 269
(iv) 1980
Solution:
(i) 727
Since 727 is an odd number.
Its square is also an odd number.

(ii) 158
Since 158 is an even number.
Its square is also an even number.

(iii) 269
Since 269 is an even number.
Its square is also an odd number.

(iv) 1980
Since 1980 is an even number.
Its square is also an even number.

Question 8.
What will be the number of zeros in the square of the following numbers?
(i) 60
(ii) 400
Solution:
(i) In 60, a number of zero is 1. Its square will have 2 zeros.
(ii) There are 2 zeros in 400. Its square will have 4 zeros.

Try These (Page No. 94)

Question 9.
How many natural numbers lie between 92 and 102 Between 112 and 122?
Solution:
(a) Between 92 and 102
Here, n = 9 and n + 1 = 10
Natural numbers between 92 and 102 are (2 × n) or 2 × 9, i.e. 18.

(b) Between 112 and 122
Here, n = 11 and n + 1 = 12
Natural numbers between 11² and 12² are (2 × n) or 2 × 11, i.e. 22.

Question 10.
How many non-square numbers lie between the following pairs of numbers:
(i) 100² and 101²
(ii) 90² and 91²
(iii) 1000² and 1001²
Solution:
(i) Between 100² and 101²
Here, n = 100
n × 2 = 100 × 2 = 200
200 non-square numbers lie between 100² and 101²

(ii) Between 90² and 91²
Here, n = 90
2 × n = 2 × 90 = 180
180 non-square numbers lie between 90² and 91².

(iii) Between 1000² and 1001²
Here, n = 1000
2 × n = 2 × 1000 = 2000
2000 non-square numbers lie between 1000² and 1001².

Question 11.
Find whether each of the following numbers is a perfect square or not?
(i) 121
(ii) 55
(iii) 81
(iv) 49
(v) 69
Solution:
If a natural number cannit be expressed as a sum of successive off natural numbers starting from 1, then it is not a perfect square.
(i) 121
Since, 121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
i.e., 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.
Thus, 121 is a square.

(ii) 55
Since, 55 – 1 = 54
54 – 3 = 51
51 – 5 = 46
46 – 7 = 39
39 – 9 = 30
30 – 11 = 19
19 – 13 = 6
6 – 15 = -9
Since 55 cannot be expressed as the sum of successive odd numbers starting from 1.
55 is not a perfect square.

(iii) 81
Since, 81 – 1 = 80
80 – 3 = 77
77 – 5 = 72
72 – 7 = 65
65 – 9 = 56
56 – 11 = 45
45 – 13 = 32
32 – 15 = 17
17 – 17 = 0
81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
Thus, 81 is a perfect square.

(iv) 49
Since, 49 – 1 = 48
48 – 3 = 45
45 – 5 = 40
40 – 7 = 33
33 – 9 = 24
24 – 11 = 13
13 – 13 = 0
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
Thus, 49 is a perfect square.

(iv) 69
Since, 69 – 1 = 68
68 – 3 = 65
65 – 5 = 60
60 – 7 = 53
53 – 9 = 44
44 – 11 = 33
33 – 13 = 20
20 – 15 = 5
5 – 17 = -12
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
69 cannot be expressed as the sum of consecutive odd numbers starting from 1.
Thus, 69 is not a perfect square.

Try These (Page No. 95)

Question 12.
Express the following as the sum of twp consecutive integers.
(i) 21²
(ii) 13²
(iii) 11²
(iv) 19²
Solution:
(i) n = 21
(n² – 1)/2 = (441 – 1)/2 = 440/2 = 220
(n² + 1)/2 = (441 + 1)/2 = 442/2 = 221
n² = (n2 – 1)/2 + (n2 + 1)/2
21² = 220 + 221 = 441

(ii) n = 13
(n² + 1)/2 = (132 + 1)/2 = (169 + 1)/2 = 85
(n² – 1)/2 = (132 – 1)/2 = (169 – 1)/2 = 84
13² = 85 + 84 = 169

(iii) n = 11
(n² + 1)/2 = (112 + 1)/2 = 61
(n² – 1)/2 = (112 – 1)/2 = 60
11² = 60 + 61 = 121

(iv) n = 19
(n² – 1)/2 = (192 – 1)/2 = 180
(n² + 1)/2 = (192 + 1)/2 = 181
19² = 180 + 181 = 361

Question 13.
Do you think the reverse is also true, i.e., is the sum of any two consecutive positive integers is a perfect square of a number? Give example to support your answer.
Solution:
No, it is not always true.
For example:
(i) 5 + 6 = 11, 11 is not a perfect square.
(ii) 21 + 22 = 43, 43 is not a perfect square.
Examples:
9² – 8² = 81 – 64 = 17 = 9 + 8
10² – 9² = 100 – 81 = 19 = 10 + 9
15² – 14² = 225 – 196 = 29 = 15 + 14
101² – 100² = 10201 – 10000 = 201 = 101 + 100
For Example,
10 × 12 = (11 – 1) × (11 + 1) = 11² – 1
11 × 12 = (12 – 1) × (12 + 1) = 11² – 1
25 × 27 = (26 – 1) × (26 + 1) = 26² – 1

Question 14.
Write the square, making use of the above pattern.
(i) 111111²
(ii) 1111111²
Solution:
using the above pattern we can write
(i) (111111)² = 12345654321
(ii) (1111111)² = 1234567654321

Question 15.
Can you find the square of the following numbers using the above pattern?
(i) 6666667²
(ii) 66666667²
Solution:
using the above pattern we can write
(i) (6666667)² = 44444448888889
(ii) (66666667)² = 4444444488888889

Try These (Page No. 97)

Question 16.
Find the square of the following numbers containing 5 in unit place.
(i) 15
(ii) 95
(iii) 105
(iv) 205
Solution:
(i) (15)² = 1 × (1 + 1) × 100 + 25
= 1 × 2 × 100 + 25
= 200 + 25
= 225

(ii) (95)² = 9 × (1 + 1) × 100 + 25
= 9 × 10 × 100 + 25
= 9000 + 25
= 9025

(iii) (105)² = 10 × (10 + 1) × 100 + 25
= 10 × 11 × 100 + 25
= 1100 + 25
= 11025

(iv) (205)² = 20 × (20 + 1) × 100 + 25
= 20 × 21 × 100 + 25
= 4200 + 25
= 42025

Try These (Page No. 99)

Question 17.
(i) 11² = 121. What are the square roots of 121?
(ii) 14² = 196. What are the square roots of 196?
Solution:
(i) The square root of 121 is 11.
(ii) The square root of 196 is 14.

Try These (Page No. 100)

Question 18.
By repeated subtraction of numbers starting from 1, find whether the following numbers are perfect squares the number is a perfect square, then square root.
(i) 121
(ii) 55
(iii) 36
(iv) 49
(v) 90
Solution:
(i) Subtracting the successive odd number from 121,
We have 121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
√121 = 11 [we had to subtract the first 11 odd numbers.]

(ii) 55 – 1 = 54
54 – 3 = 51
51 – 5 = 46
46 – 7 = 39
39 – 9 = 30
30 – 11 = 19
19 – 13 = 6
6 – 15 = -9
and we do not reach to 0.
55 is not a perfect square.

(iii) 36 – 1 = 35
35 – 3 = 32
32 – 5 = 27
27 – 7 = 20
20 – 9 = 11
11 – 11 = 0
and we do have obtained 0 after subtracting 6 successive odd numbers.
36 is a perfect square.
Thus, √36 = 6.

(iv) We have
49 – 1 = 48
48 – 3 = 45
45 – 5 = 40
40 – 7 = 33
33 – 9 = 24
24 – 11 = 13
13 – 13 = 0
We have obtained 0 after successive subtraction of 7 odd numbers.
49 is a perfect square Thus, √49 = 7.

(v) We have
90 – 1 = 89
89 – 3 = 86
86 – 5 = 81
81 – 7 = 74
74 – 9 = 65
65 – 11 = 54
54 – 13 = 41
41 – 15 = 26
26 – 17 = 9
9 – 19 = -10
Since we can not reach 0 after subtracting successive odd numbers.
90 is not a perfect square.

Try These (Page No. 105)

Question 19.
Without calculating square roots, find the number of digits in the square root of the following numbers.
(i) 25600
(ii) 100000000
(iii) 36864
Solution:
(i) 25600
n = 5 [an odd number]
Its square root will have \(\frac{n+1}{2}\)
i.e. \(\frac{5+1}{2}=\frac{6}{2}\) = 3 digits

(ii) 100000000
n = 9 (odd number)
Number of digits of its square root
\(\frac{n+1}{2}=\frac{9+1}{2}=\frac{10}{2}=5\) digits

(iii) 36864
n = 5 (odd number)
Number of digits in its square root
\(\frac{n+1}{2}=\frac{5+1}{2}=\frac{6}{2}=3\) digits

Try These (Page No. 107)

Question 20.
Estimate the value of the following the nearest whole number.
(i) √180
(ii) √100
(iii) √350
(iv) √500
Solution:
(i) √80
10² = 100, 9² = 81, 8² = 64
and 80 is between 64 and 81.
i.e., 64 < 80 < 81
or 8² < 80 < 9²
or 8 < √80 < 9
Thus, √80 lies between 8 and 9.

(ii) √1000
We know that
30² = 900, 31² = 961, 32² = 1024
1000 lies between 961 and 1024.
i.e., 916 < 1000 < 1024
or 31² < 1000 < 32²
or 31² < √1000 < 32.
Thus, √1000 lies between 31 and 32.

(iii) √350
We have 18² = 324, 19² = 361
Since 350 lies between 324 and 316.
i.e., 324 < 350 < 361
or 18² < 350 < 19²
or 18 < √350 < 19.
Thus, √350 lies between 18 and 19.

(iv) √500
22² = 484 and 23² = 529
Since, 500 lies between 484 and 529.
or 484 < 500 < 529
or 22² < 500 < 23²
or 22 < √350 < 23.
Thus, √500 lies between 22 and 23.

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.2

Question 1.
Find the value of the unknown exterior angle x in the following diagrams:


Answer:
(i) x = 50° + 70°
= 120°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(ii) x = 45° + 65°
= 110°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(iii) x = 30° + 40°
= 70°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(iv) x = 60° + 60°
= 120°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(v) x = 50° + 50°
= 100°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(vi) x = 30° + 60°
= 90°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

Question 2.
Find the value of the unknown interior angle x in the following figures:

Answer:
(i) x + 50°= 115°
x = 115° – 50°
= 65°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(ii) x + 70° = 100°
x = 100° – 70°
= 30°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(iii) x + 90° = 125°
x = 125° – 90°
= 35°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(iv) x + 60° = 120°
x = 120° – 60°
= 60°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(v) x + 30° = 80°
x = 80° – 30°
= 50°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(vi) x + 35° = 75°
x = 75° – 35°
= 40°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)