Explore topic-wise InterviewSolutions in .

Three non-collinear points form a triangle.

i. Co-ordinate of the point E is 9.

 Co-ordinate of the point D is -7. 

 Since, 9 > -7 

∴ d(D, E) = 9 – (-7) = 9 + 7 = 16 

∴ l(DE) = 16 …(i) 

Co-ordinate of the point A is -3.

Co-ordinate of the point B is 5.

 Since, 5 > -3 

∴ d(A, B) = 5 – (-3) = 5 + 3 = 8 

∴ l(AB) = 8 …(ii)

∴ l(DE) ≠ l(AB) …[From (i) and (ii)] 

∴ seg DE and seg AB are not congruent

ii. Co-ordinate of the point B is 5.

 Co-ordinate of the point C is 2. 

 Since, 5 > 2 

∴ d(B, C) = 5 – 2 = 3 

∴ l(BC) = 3 …(i) 

Co-ordinate of the point A is -3. 

Co-ordinate of the point D is -7.

 Since, -3 > -7 

∴ d(A, D) = -3 – (-7) = -3 + 7 = 4 

∴ l(AD) = 4 . ..(ii) 

∴ l(BC) ≠ l(AD) … [From (i) and (ii)] 

∴ seg BC and seg AD are not congruent.

 iii. Co-ordinate of the point E is 9.

 Co-ordinate of the point B is 5.

Since, 9 > 5 

∴ d(B, E) = 9 – 5 = 4 

∴ l(BE) = 4 …(i) 

Co-ordinate of the point A is -3. 

Co-ordinate of the point D is -7.

 Since, -3 > -7

 ∴ d(A, D) = -3 – (-7) = 4 

∴ l(AD) = 4 …(ii) 

∴ l(BE) =l(AD) …[From (i) and (ii)] 

∴ seg BE and seg AD are congruent. 

i.e, seg BE ≅ seg AD

Here, given number is \(\overline{67\times19}\) . 

We know that a number is divisible by 11 if and only if the difference between the sum of odd and even place digits is a multiple of 11. 

∴ Sum of even placed digits – Sum of odd placed digits = 0, 11, 22,… 

∴ (6 + x + 9) – ( 7+1) = 0, 11, 22,… 

∴ x + 7 is a multiple of 11 

∴ x + 7 = 11 ∴ x = 4

i. Co-ordinate of the point E is 9. 

Co-ordinate of the point D is -7. 

Since, 9 > -7 

∴ d(D, E) = 9 – (-7) = 9 + 7 = 16 

∴ l(DE) = 16 …(i)

Co-ordinate of the point A is -3. 

Co-ordinate of the point B is 5. 

Since, 5 > -3 

∴ d(A, B) = 5 – (-3) = 5 + 3 = 8 

∴ l(AB) = 8 …(ii) 

∴ l(DE) ≠ l(AB) …[From (i) and (ii)] 

∴ seg DE and seg AB are not congruent.

ii. Co-ordinate of the point B is 5. 

Co-ordinate of the point C is 2. 

Since, 5 > 2 

∴ d(B, C) = 5 – 2 = 3 

∴ l(BC) = 3 …(i) 

Co-ordinate of the point A is -3. 

Co-ordinate of the point D is -7. 

Since, -3 > -7 

∴ d(A, D) = -3 – (-7) = -3 + 7 = 4 

∴ l(AD) = 4 . ..(ii) 

∴ l(BC) ≠ l(AD) … [From (i) and (ii)] 

∴ seg BC and seg AD are not congruent.

iii. Co-ordinate of the point E is 9. 

Co-ordinate of the point B is 5. 

Since, 9 > 5 

∴ d(B, E) = 9 – 5 = 4 

∴ l(BE) = 4 …(i) 

Co-ordinate of the point A is -3. 

Co-ordinate of the point D is -7. 

Since, -3 > -7 

∴ d(A, D) = -3 – (-7) = 4 

∴ l(AD) = 4 …(ii) 

∴ l(BE) =l(AD) …[From (i) and (ii)] 

∴ seg BE and seg AD are congruent. 

i.e, seg BE ≅ seg AD

i. Co-ordinate of first point is 3. 

Co-ordinate of second point is 6. 

Since, 6 > 3 

∴ Distance between the points = 6 – 3 = 3

ii. Co-ordinate of first point is -9. 

Co-ordinate of second point is -1. 

Since, -1 > -9 

∴ Distance between the points = -1 – (-9) = -1 + 9 = 8

iii. Co-ordinate of first point is -4. 

Co-ordinate of second point is 5. 

Since, 5 > -4 

∴ Distance between the points = 5 – (-4) = 5 + 4 = 9

iv. Co-ordinate of first point is 0. 

Co-ordinate of second point is -2. 

Since, 0 > – 2 

∴ Distance between the points = 0 – (-2) = 0 + 2 = 2

v. Co-ordinate of first point is x + 3. 

Co-ordinate of second point is x – 3. 

Since, x + 3 > x – 3

∴ Distance between the points = x + 3 – (x – 3) 

= x + 3 – x + 3 = 3 + 3 = 6 

vi. Co-ordinate of first point is -25. 

Co-ordinate of second point is -47. 

Since, -25 > -47 

∴ Distance between the points = -25 – (-47) = -25 + 47 = 22

vii. Co-ordinate of first point is 80. 

Co-ordinate of second point is -85. 

Since, 80 > -85 

∴ Distance between the points = 80 – (-85) = 80 + 85 = 165

We know that if sum of digits of a number is divisible by 9, then the number is divisible by 9. 

Here, 

6 + 4 + y + 3 = multiple of 9 

y + 13= 0, 9, 18, … 

Hence, 

y + 13 = 18 

∴ y = 5 

The number will be 6453.

(C) 7

Since, 5 > -2

∴ d(A, B) = 5 – (-2) = 5+2 = 7

For making 521 a perfect square, we have to add 8 on it as: 

521 + 8 = 529 

And we know that,

\(\sqrt{529}=23\)

Hence, option (D) is correct

For making 178 a perfect square we have to subtract 9 from it as: 

178 – 9 = 169 

And we know that,

\(\sqrt{169}=13\)

Therefore, option (C) is correct

Correct option is (C) one or three

Correct option is (C) one

Here, given number is \(\overline{3\times2}\) . 

We know that a number is divisible by 11 if and only if the difference between the sum of odd and even place digits is a multiple of 11. 

∴ Sum of even placed digits – Sum of odd placed digits = 0, 11, 22,… 

∴ x – ( 3+2) = 0, 11, 22,… 

∴ x – 5 is a multiple of 11 

∴ x – 5 = 0 ∴ x = 5

Here it is given that \(\overline{66784\times}\) is divisible by 9. 

We know that if a number is divisible by 9, then sum of digits must be a multiple of 9. 

∴ 6 + 6 + 7 + 8 + 4 + x = multiple of 9 

∴ x + 31 = 0,9,18,27,….. But ‘x’ is a digit, 

hence, 

‘x’ can have value between 0 to 9. 

∴ x + 31 = 36. 

∴ x = 5

Here it is given that \(\overline{35a64}\) is divisible by 3.

We know that if a number is divisible by 3, then sum of digits must be a multiple of 3. 

∴ 3 + 5+a+6+4 = multiple of 3 

∴ a + 18 = 0,3,6,9,12,15,….. 

But ‘a’ is a digit, hence, ‘a’ can have value between 0 to 9. 

∴ a + 18 = 18 which gives a = 0. 

∴ a + 18 = 21 which gives a = 3. 

∴ a + 18 = 24 which gives a = 6. 

∴ a + 18 = 27 which gives a = 9. 

∴ a = 0,3,6,9

(B) 8

d(P, R) = d(P, Q) + d(Q, R) 

∴ 10 = 2 + d(Q, R) 

∴ d(Q, R) = 8

For unit’s place, 

We have two conditions, when 7 + B ≤ 9 and 7 + B > 9 

For 7 + B ≤ 9 

7 + B = A 

∴ A – B = 7 …(1) 

In ten’ place, 

B + A = 8 …(2) 

Solving 1 and 2 simultaneously, 

2A = 15 which means A = 7.5 which is not possible 

Hence, our condition 7 + B ≤ 9 is wrong. 

∴ 7 + B > 9 is correct condition 

Hence, carrying one in ten’s place and subtracting 10 from unit’s place, 

7 + B – 10 = A 

∴ B – A = 3 … (3) 

For ten’s place, 

B + A + 1 = 8 

∴ B + A =7 …(4) 

Solving 3 and 4 simultaneously, 

2B = 10 

∴ B = 5 and A = 2

Students can also reach out to Sarthaks to get Important Questions for Class 8 Maths for all other chapters. Sarthaks eConnect is a platform that provides study materials for students. Multiple Choice Questions for Class 8 are given here. Each problem carries four options, in which one is the correct answer. You can solve the Class 8 Maths MCQ Questions of square and square roots with Answers for your extensive revision for exams.

Students have to find a solution for each problem and choose the correct answer. These important questions with solutions will definitely benefit all students with revision and mastering the topic. you will learn the concepts of square and square roots. There are many important questions given for the chapter that will give you an idea of what kind of question you can expect in exams.

Practice MCQ Question for Class 8 Maths chapter-wise

1. A perfect square number between 30 and 40 is

(a) 36
(b) 32
(c) 33
(d) 39

2. Find perfect square numbers between 50 and 60

(a) 50, 54
(b) 52,58
(c) 56,59
(d) None of the above

3. Which of the following is a perfect square number?

(a) 1067
(b) 7828
(c) 4333
(d) 625

4. Which of 132², 87², 72², and 209² would end with digit 1?

(a) 132²
(b) 87²
(c) 72²
(d) 209²

5. Which of 105², 216², 333², and 111² would end with digit 1?

(a) 105²
(b) 216²
(c) 333²
(d) 111²

6. What will be the number of zeros in the square of the number 60?

(a) 1
(b) 2
(c) 3
(d) 4

7. The square of which of the following numbers will be even? 11, 111, 1111, 112

(a) 11
(b) 111
(c) 1111
(d) 112

8. The square of which of the following numbers will be odd ? 10, 100, 1000, 99

(a) 10
(b) 100
(c) 1000
(d) 99

9. Write the correct answer from the given four options. How many natural numbers lie between 5² and 6²?

(a) 9
(b) 10
(c) 11
(d) 12

10. How many numbers lie between the squares of 20 and 21

(a) 10
(b) 40
(c) 5
(d) 60

11. How many non-square numbers lie between the pairs of numbers 100² and 101² :

(a) 200
(b) 400
(c) 500
(d) 600

12. How many non-square numbers lie between the pairs of numbers 90² and 91² :

(a) 200
(b) 400
(c) 180
(d) 240

13. What will be the unit digit of the squares of the number 81?

(a) 1
(b) 2
(c) 4
(d) 6

14. The unit digit in the square of the number 1333 is

(a) 3
(b) 6
(c) 9
(d) 1

15. The unit digit in the square of the number 166 is

(a) 22
(b) 34
(c) 36
(d) 28

16. Which of the following is not a Pythagorean triplet?

(a) 3, 4, 5
(b) 6, 8, 10
(c) 5, 12, 13
(d) 2, 3, 4

17. The square root of 441/961 is:

(a) 21/39
(b) 37/21
(c) 21/31
(d) 11/13

18. If M is a square number, then the next immediate square number is

(a) \(M+2\sqrt M+1\)
(b) M² + 2M
(c) M + 5
(d) none of these

19. If a square number ends in 6, the preceding figure is

(a) an odd number
(b) a composite number
(c) an even number
(d) a prime number

20. The smallest number by which 396 must be multiplied so that the product becomes a perfect square is:

(a) 5
(b) 11
(c) 3
(d) 2

21. A number added to its square gives 56. The number is:

(a) 12
(b) 9
(c) 8
(d) 7

22 .\(\sqrt{\frac{1}{16}+\frac{1}{9}}\) is 

(a) 5/12
(b) 7/12
(c) 3/12
(d) none of these

23. Which of the following will have 6 at the unit place?

(a) 19²
(b) 11²
(c) 24²
(d) 13²

24. The value of 9² – 1 is equal to:

(a) 81
(b) 80
(c) 79
(d) None of the above

25. The Pythagorean triples whose smallest number is 8:

(a) 8, 16 17
(b) 8, 17, 18
(c) 8, 15, 17
(d) 8, 15, 16

Answer:

 1. Answer: (a) 36

Explanation:  Since, 

1 x 1 = 1

2 x 2 = 4

3 x 3 = 9

4 x 4 = 16

5 x 5 = 25

6 x 6 = 36

7 x 7 = 49

Thus, 36 is a perfect square number between 30 and 40.

2. Answer: (d) None of the above

Explanation: By squaring, we get as 

1 → 1

2 → 4

3 → 9

4 → 16

5 → 25

6 → 36

7 → 49

8 → 64

From, this we can say that there is no perfect square between 50 and 60

3. Answer: (d) 625

Explanation: Perfect square numbers end with, 0, 1, 4, 5, 6, or 9 at the unit’s place.

4. Answer: (d) 209²

Explanation:  If a number has 1 or 9 in the unit’s place, then its square ends in 1.

5. Answer:(d) 111²

Explanation: If a number has 1 or 9 in the unit’s place, then its square ends in 1.

6. Answer: (b) 2

Explanation: In 60, the number of zero is 1

∴ Its square will have 2 zeros.

7. Answer: (d) 112

Explanation: ∵ 112 is even

∴ Its square will be even.

8. Answer: (d) 99

Explanation: ∵ 99 is odd

∴ Its square will be odd.

9. Answer: (b) 10

Explanation: Square of 5 = 25 and 

Square of 6 = 36 

Natural numbers lie between 25 and 36 = 26, 27, 28, 29, 30, 31, 32, 33, 34, 35.

10. Answer: (b) 40

Explanation:  20² = 400

21² = 441

∴number of numbers

=(441−400)−1

=40

11. Answer: (a) 200

Explanation:  Between 100² and 101². Here, n=100 

∴  n x 2 = 100 x 2 

= 200

∴ 200 non-square numbers lie between 100² and 101² .

12. Answer: (c) 180

Explanation:  Between 90² and 91²

Here, n=90

∴  2 x n = 2 x 90 or 180

∴  180 non-square numbers lie between 90 and 91.

13. Answer: (a) 1

Explanation:  1² = 1. So, the unit digit will be 1.

14. Answer: (c) 9

Explanation:  3² = 9. So, the unit digit will be 9.

15. Answer: (c) 36

Explanation:  6² = 36. So, the unit digit will be 36.

16. Answer: (d) 2, 3, 4

Explanation: 3² + 4² = 5² 

6² + 8² = 10² 

5² + 12² = 13² 

2² + 3² ≠ 4² is not a Pythagorean triplet.

17. Answer: (c) 21/31

Explanation: The square root of 441/961 is

\(\frac{\sqrt{144}}{\sqrt{961}}=\frac{21}{31}\)

18. Answer:  (a) \(M+2\sqrt M+1\)

Explanation:  \(M+2\sqrt M+1\)

Take M = 4, a Square number.

\(M+2\sqrt M+1\) = \(4+2\sqrt4+1\)

= 4 + 2 × 2 +1

= 9

19. Answer: (a) an odd number

Explanation: Let us take some square numbers ending with 6. 

Eg: 136,196,256,676,....

By observation, we can say that the preceding figure are odd numbers.

20. Answer: (b) 11

Explanation: 396 = 2 × 2 × 3 × 3 × 11

So 396 should be multiplied by 11 to make the product a perfect square.

21. Answer: (d) 7

Explanation: A number added to its square gives 56. Let the number be x.

Hence, 

x² +x = 56

⇒ x² + x − 56 = 0

⇒ x²  + 8x − 7x − 56 = 0

⇒ x(x+8)−7(x+8)=0

⇒ (x+8)(x−7) = 0

Hence, x=−8 or x=7

22. Answer: (a) 5/12

Explanation: \(\sqrt{\frac{1}{16}+\frac{1}{9}}\)

\(\sqrt{\frac{25}{144}}\)

= 5/12

23. Answer: (c) 24²

Explanation: 24² = 24 x 24 

= 576

24. Answer: (b) 80

Explanation: 9² - 1

= 81 – 1 

= 80

25. Answer: (c) 8, 15, 17

Explanation: The general form of Pythagorean triplets is 2m, m²-1, m²+1

Given, 2m = 8

So, m = 4

m² – 1

= 4² - 1

= 16-1 

= 15

m²+1

= 4² + 1

= 16+1 

= 17

Click here Practice MCQ Question for Squares and Square Roots Class 8

(i) (√59.29 – √5.29)/ (√59.29 + √5.29)

= √59.29 

= √5929/ √100

= 77/10

= 7.7

Now, √5.29 

= √5.29/ √100

= 23/10

= 2.3

So, (7.7 – 2.3)/ (7.7 + 2.3)

= 54/10

= 0.54

(ii) (√0.2304 + √0.1764)/ (√0.2304 – √0.1764)

  = √0.2304 

= √2304/ √10000

= 48/100

= 0.48

Now, √0.1764 

= √1764/ √10000

= 42/100

= 0.42

So, (0.48 + 0.42)/ (0.48 – 0.42)

= 0.9/0.06

= 15

The area of the remaining portion = 1764 – 784 = 980 m² 

It is divided into 5, equal square parts 

∴ Each square part = \(\frac{980}{5}\) = 196 m² 

Area of each square part = 196 m² 

(side)² = 196 

side = √196 

side = 14. 

The length of each side of the square = 14m 

Perimeter = 4 side = 4 × 14 = 56m

If B is multiplied by 4 then only 0 satisfies the above condition. 

Hence, for unit place to satisfy the above condition, we have, B = 0. 

Similarly for ten’s place, only 0 satisfies. 

But, AB cannot be 00 as 00 is not a two digit number. 

So, A and B cannot be equal to 0 

Hence, there is no solution satisfying the condition \(4\times\overline{AB}A = \overline{CAB}\)

(i) 3048625 = 3375 x 729

Taking cube root of the whole, we get,

= \(\sqrt[3]{3048625}\) = \(\sqrt[3]{3375\times729}\)

We know that,

= \(\sqrt[3]{ab}\) = \(\sqrt[3]{a\times}\) \(\sqrt[3]{b}\)

= \(\sqrt[3]{3048625}\) = \(\sqrt[3]{3375}\) x \(\sqrt[3]{729}\)

Now by prime factorization,

= \(\sqrt[3]{3\times3\times3\times5\times5\times5}\) x \(\sqrt[3]{9\times9\times9}\)

= \(\sqrt[3]{3^3\times5^3}\times\sqrt[3]{9^3}\) 

= \(\sqrt[3]{3^3}\times\) \(\sqrt[3]{5^3}\times\sqrt[3]{9^3}\)

= \(3\times5\times9 = 135.\)

(ii) 20346417 = 9261 x 2197

Taking cube root of the whole,

= \(\sqrt[3]{20346417}\) = \(\sqrt[3]{9261\times2197}\)

We know that,

=  \(\sqrt[3]{ab}\) = \(\sqrt[3]{a\times}\) \(\sqrt[3]{b}\)

= \(\sqrt[3]{9261\times2197}\) =  \(\sqrt[3]{9261}\times\) \(\sqrt[3]{2197}\)

Now by prime factorization,

= \(\sqrt[3]{3\times3\times3\times7\times7\times7}\) \(\times\sqrt[3]{13\times13\times13}\)

= \(\sqrt[3]{3^3\times7^3}\times\) \(\sqrt[3]{13^3}\)

= \(\sqrt[3]{3^3}\times\) \(\sqrt[3]{7^3}\times\) \(\sqrt[3]{13^3}\)

= 3 × 7 × 13 = 273.

(iii) 210644875 = 42875 x 4913

Taking cube root of the whole,

= \(\sqrt[3]{210644875}\) = \(\sqrt[3]{42875\times4913}\)

We know that,

=  \(\sqrt[3]{ab}\) = \(\sqrt[3]{a\times}\) \(\sqrt[3]{b}\)

= \(\sqrt[3]{42875\times4913}\) = \(\sqrt[3]{42875\times}\) \(\sqrt[3]{4913}\)

Now by prime factorization,

= \(\sqrt[3]{5\times5\times5\times7\times7\times7}\) x \(\sqrt[3]{17\times17\times17}\)

= \(\sqrt[3]{5^3\times7^3}\times\) \(\sqrt[3]{13^3}\)

= \(\sqrt[3]{5^3}\times\) \(\sqrt[3]{7^3}\times\) \(\sqrt[3]{17^3}\)

= \(5\times7\times17 = 595.\)

(iv) 57066625 = 166375 x 343

Taking cube root of the whole, we get,

= \(\sqrt[3]{57066625}\) = \(\sqrt[3]{166375\times343}\)

We know that,

=  \(\sqrt[3]{ab}\) = \(\sqrt[3]{a\times}\) \(\sqrt[3]{b}\)

Now by prime factorization method,

= \(\sqrt[3]{5\times5\times5\times11\times11\times11}\)\(\times\sqrt[3]{7\times7\times7}\)

= \(\sqrt[3]{5^3\times11^3\times}\) \(\sqrt[3]{7^3}\)

= \(\sqrt[3]{5^3}\times\sqrt[3]{11^3}\times\) \(\sqrt[3]{7^3}\)

= \(5\times7\times11 = 385.\)

Given, perimeter of the square field = 96 m

We know that, perimeter of square = 4 × side

Then,

96 = 4 × side

Side = 96/4

Side = 24 m

So, the length of the side of square = 24 m

Now,

Area of the square field = (side)²

= 24²

= 576 m²

∴The area of the square field is 576m².

Given, 

Volume of a cube = 474.552 cubic metres 

V = 8³,

S = side of the cube 

So, 

8³ = 474.552 cubic metres

= 8 = \(\sqrt[3]{474.552}\) = \(\sqrt[3]{\frac{474552}{1000}}\) = \(\frac{\sqrt[3]{474552}}{\sqrt[3]{1000}}\)

On factorising 474552 into prime factors, we get:

474552 = 2×2×2×3×3×3×13×13×13

On grouping the factors in triples of equal factors, we get: 

474552 = {2×2×2}×{3×3×3}×{13×13×13} 

Now taking 1 factor from each group we get:

\(\sqrt[3]{474.552}\) = \(\sqrt[3]{{\{2\times2\times2\times\}}\{\times3\times3\times3\}\{13\times13\times13\}}\)

= \(2\times3\times13\) = 78

Also,

\(\sqrt[3]{1000} = 10\)

∴ \(8 = \frac{\sqrt[3]{474552}}{\sqrt[3]{1000}}\) = \(\frac{78}{10} = 7.8\)

So, length of the side is 7.8m.

From the question it is given that,

Volume of the cube = 512 cm³

We know that,

Volume of cube = side³

512 = side³

By taking cube root on both the side,

³√512 = side

Side = ³√(8 × 8 × 8)

Side = ³√ (8)³

Side = 8 cm

∴The length of each side of a cube is 8 cm.

From the question it is given that,

Length of one side of the a cube = 15 m

We know that, volume of cube = (side)³

= 15³

= 15 × 15 × 15

= 3375 m³

∴The volume of cube is 3375 m³.

False.

Let a² be 7²

So, 7² = 49

Then, 7³ = 343

True.

Square root of a number x is denoted by √x.

Correct option is : (a) 88

63, 64, 67, 72, 79, ……….. 

63, (63 + 1), (64 + 3), (67 + 5), (72 + 7),…. 

So, next number is (79 + 9) = 88

Let  the ratio 2:3:4 be 2a, 3a, and 4a.

So according to the question:

(2a)³+ (3a)³+(4a)³ = 0.334125

8a³+27a³+64a³ = 0.334125

99a³ = 0.334125

a³ = 334125/1000000×99

= 3375/1000000

a = ∛(3375/1000000)

= ∛((15×15×15)/ 100×100×100)

= 15/100

= 0.15

Hence ,the numbers are:

2×0.15 = 0.30

3×0.15 = 0.45

= 4×0.15 

= 0.6

False.

Let us take -3

Now, square of the -3 i.e. -3² = 9

Then, cube of the same number i.e. -3³ = -27

9

Resolving 72 into prime factors, we get
72=2 x 2 x 2 x 3 x 3

Grouping the factors in triplets of equal factors, we get

72 = (2 x 2 x 2) x 3 x 3

Clearly, if we divide 72 by 3 x 3, the quotient would be 2 x 2 x 2, which is a perfect cube. 

Hence, the least number by which 72 be divided to make it, a perfect cube, is 9.

From the question it is given that,

Difference of two perfect cubes = 189

The cube root of the smaller of the two numbers = 3

So, cube of smaller number = 3³

= 3 × 3 × 3

= 27

Let us assume the cube root of larger number be a³

Then,

As per the condition given in the question,

a³ – 27 = 189

a³ = 189 + 27

a³ =216

By taking cube root on both side,

a = ³√216

a = ³√(6 × 6 × 6)

a = ³√(6³)

a = 6

∴The cube root of the larger number is 6.

False.

Let us consider LHS = ³√ (8 + 27)

= ³√ (35)

Then, RHS = ³√8 + ³√27

= 2 + 3

= 5

By comparing LHS and RHS

LHS ≠ RHS

³√ (35) ≠ 5

True.

Let us assume the perfect square has 9 digits, then its square root has 5 digits

i.e. (9 + 1)/2 = 10/2 = 5

Given, {(5² + (12²)^1/2)}³

= {(25 + (144)^1/2)}³

= {(25 + (12)}³

= {37}³

= 37 × 37 × 37

= 50,653

(i) Unit place digit in the square root of 9604 = 2,8

(ii) Unit place digit in the square root of 65536 = 4,6

(iii) Unit place digit in die square root of 998001 = 1,9

(iv) Unit place digit in the square root of 60481729 = 3,7

(i) 10 + 11 =21 numbers

(ii) 17 + 18 = 35 numbers

(iii) 30 +31 =61 numbers

(i) 100² – 99² 

= 100 + 99 

= 199 

(ii) 111² - 109² 

= 111² – 110² + 110² - 109² 

= (111 + 110) + (110 + 109) 

= 440 

(iii) 99² - 96² = 99² – 98² + 98² – 97² + 97² - 96² 

= (99 + 98) + (98 + 92) + (97 + 96) 

= 585

False.

Let us take -27

Cube root of -27 i.e. ³√-27 

= ³√((-3) × (-3) × (-3))

= -3

(a) m³

Given in the question,

³√n = m

n = m³

Let us assume the three number be 2a, 3a, 4a

Then,

Given, sum of cube of three numbers is 0.334125

i.e. (2a)³ + (3a)³ + (4a)³ = 0.334125

8a³ + 27a³ + 64a³ = 0.334125

99a³ = 0.334125

a³ = 0.334125/99

a³ = 0.003375

a³ = 3375/1000000

a = ³√(3375/1000000)

a = ³√((15 × 15 × 15)/(10 × 10 × 10 × 10 × 10 × 10))

a = ³√(15³/10⁶)

a = 15/1000

a = 0.015

∴The numbers are, 2a = 2 × 0.015 = 0.03

3a = 3 × 0.015 = 0.045

4a = 4 × 0.015 = 0.06

By using distributive law,

We have,

72 = 70 + 2

So,

72² = (70 + 2)²

= (70 + 2) (70 + 2)

= 70 (70 + 2) + 2 (70 + 2)

= ((70 × 70) + (70 × 2)) + ((2 × 70) + (2 × 2))

= 4900 + 140 + 140 + 4

= 5184

∴The square of the given number i.e. 72² = 5184

Given, {(6² + (8²)^1/2)}³

= {(36 + (64)^1/2)}³

= {(36 + (8)}³

= {44}³

= 44 × 44 × 44

= 85,184

(i) (8, 15, 17) 

L.H.S = 8² + 15² = 289 

R.H.S = 17² = 289 

L.H.S = R.H.S 

So, it is Pythagoras 

(ii) (18, 80, 82) 

L.H.S = 18² + 80² = 6724 

R.H.S = 82² = 6724 

L.H.S = R.H.S 

So, it is Pythagoras 

(iii) (14, 48, 51) 

L.H.S = 14² + 48² = 2500 

R.H.S = 51² = 2601 

L.H.S ≠ R.H.S 

So, it is not Pythagoras 

(iv) (10, 24, 26) 

L.H.S = 10² + 24² = 676 

R.H.S = 26² = 676 

L.H.S = R.H.S 

So, it is Pythagoras 

(v) (16, 63, 65) 

L.H.S = 16² + 63² = 4225 

R.H.S = 65² = 4225 

L.H.S = R.H.S 

So, it is Pythagoras 

(vi) (12, 35, 38) 

L.H.S = 12² + 35² = 1369 

R.H.S = 38² = 1444 

L.H.S ≠ R.H.S 

So, it is Pythagoras

(i) (9)² + (12)² = 81 + 144 = 225 = (15)²
So (9)² + (12)²= (15)²
or 9, 12, 15 Pythagorean triplets.

(ii) (7)² + (11)² = 49 + 121 = 170
∴ (7)² + (11)² ≠ (13)²
or 7, 11, 13 Pythagorean triplets.

(iii) (10)² + (24)² = 100 + 576 = 676 = (26)²
∴ (10)² + (24)² = (26)²
or 10, 24, 26 Pythagorean triplets.

False.

There are only 3 perfect cubes between 1 and 100.

2 × 2 × 2 = 8

3 × 3 × 3 = 27

4 × 4 × 4 = 64

False

³√8000 = 20

By using distributive law,

We have,

101 = 100 + 1

So,

101² = (100 + 1)²

= (100 + 1) (100 + 1)

= 100 (100 + 1) + 1 (100 + 1)

= ((100 × 100) + (100 × 1)) + ((1 × 100) + (1 × 1))

= 10000 + 100 + 100 + 1

= 10201

∴The square of the given number i.e. 101² = 10201

Let us assume PQRS is perfect square,

Where, P = even, Q = even, R = odd, S = even

So, 8836 is the perfect square.

There are 2n natural numbers between n² and (n + 1)².