In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions pdf, free KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions**

## Karnataka State Syllabus Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

Question 2.

Choose the correct option:

(a) The number of perfect squares from 1 to 500 is

A. 1

B. 16

C. 22

D. 25

Solution:

C. 22

(b) The last digit of a perfect square can never be

A. 1

B. 3

C. 5

D. 9

Solution:

B. 3

(c) If a number ends in 5 zeros, its square ends in

A. 5 zeros

B. 8 zeros

C. lo zeros

D. 12 zeros

Solution:

C. 10 zeros

(d) Which could be the remainder among the following when a perfect square is devided by 8?

A. 1

B. 3

C. 5

D. 7

Solution:

A. 1

(e) The 6th triangular number is

A. 6

B. 10

C. 21

D. 28

Solution:

C. 21

Question 3.

Consider all integers froni – 10 to 5 and square each of them. How many distinct numbers you get?

Solution:

(-10)^{2} = 100, (-9)^{2} = 81, (-8)^{2}

= 64,(-7)^{2} = 49

(-6)^{2} = 36, (5)^{2} = 25, (-4)^{2} 16, (-3)^{2} =9,

(2)^{2} =4, (-1)^{2} = 1, 0^{2} = 0, 1^{2} = 1, 2^{2} = 4, 3^{2} = 9, 4^{2} = 16, 5^{2} = 25

There are 11 distinct numbers.

Question 5.

Write all the numbers from 400 to 425 which end in 2, 3, 7 or 8. Check if any of these in a perfect square.

Solution:

402, 403, 407, 408, 412, 413, 417, 418, 422, 423

None of these is a perfect square.

Question 7.

Suppose x^{2} + y^{2} = Z^{2}

(i) If x = 4 and y = 3 find z

(ii) If x = 5 and z = 13 find y

(iii) If y = 15 and z = 17 find x.

Solution:

(i) x = 4, y = 3

x^{2} + y^{2} = z2

4^{2} + 3^{2} = z^{2}

16 + 9 = z^{2}

25 = z^{2}

∴ z = √25 = 5

(ii) x = 5, z = 13

x^{2} + y^{2} = z^{2}

5^{2} + y^{2} = 13^{2}

25+ y^{2} =169

y^{2} =169 – 25

y^{2} = 144

y = √144 = 12

(iii) y = 15, z = 17

x^{2}+ y^{2} = z^{2}

x^{2} + 15^{2} = 17^{2}

x^{2} + 225 = 289

x^{2} = 289 – 225

x^{2} = 64

x = √ 64 = 8

Question 9.

Define a new addition e on (be set of all natural numbers by m * n = m^{2} + n^{2}

(i) Is N closed under e?

(ii) Is * commutative on N?

(iii) Is * associative on N?

(iv) Is there an identity element in N with respect * ?

Solution:

(i) Let m 2 and n 5

m * n = m^{2} + n^{2}

2 * 5 = 2^{2} + 5^{2} = 4 + 25 = 29 ∈ N

∀ m, n ∈ N, m^{2} + n^{2} ∈ N

∴ * is closed under N.

(ii) Let m = 4 and n = 7

m * n = m^{2} + n^{2} = 4^{2} + 7^{2} = 16 + 49 = 65

n * m = n^{2} + m^{2} = 7^{2} + 4^{2} = 49 + 16 = 65

∴ m * n = n * m ∀ m, n ∈ N

Hence * is commutative on N.

(iii) Lei m = 2, n = 3 and p = 4.

m * (n * P) = 2 * (3*4)

= 2*(3^{2} + 4^{2}) = 2*(9 + 16)

= 2*25 = 2^{2} + 25^{2} = 4 + 625

= 629 ….(i)

(m * n) * p = (2 * 3)*4

=(2^{2} + 3^{2}) * 4 = (4 * 9) * 4

= 13 * 4 = 13^{2} + 4^{2}.

= 169 + 16 = 185 …(ii)

From (i) and (ii) m*(n*p) ≠ (m*n)*p

∴ * is not associative on N.

(iv) Let K be the identity element then m * k = m, m^{2} + k^{2} = m^{2} which means

k^{2} = m^{2} – m^{2} = O

K = 0, O does not belong to N.

There is no identity element in N with respect to *.

Question 10.

(Exploration) Find all the perfect squares from 1 to 500 each of which is a sum of two perfect squares.

Solution:

Perfect squares from 1 to 500 are 1, 4, 9, 1, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289. 324, 361, 400, 441, 484.

25 = 9 + 16

100 =36 + 64

169 = 25 + 144

289 = 225 + 64

Question 11.

Supposes the area of a square field is 7396 find its perimeter.

Solution:

Let each side of the square be

l^{2} = 7396 = 2 × 2 × 43 × 43 = 2 × 43 × 2 × 43

l^{2} =7396 = 86 × 86.

∴ √7396 = 86 = l

Perimeter = 4l = 4 × 86 = 344 m.

Question 12.

Can 1010 be written as a difference of two perfect squares?

Solution:

If a^{2} – b^{2} = 1010 for any two integers a and b then either both ‘a’ and ‘b’ are odd or both even. Hence a^{2} – b^{2} is divisible by 4. But 1010 is not divisible by 4. Hence 1010 is not the difference of two perfect squares.

Question 13.

What are the remainders when a perfect cube is divided by 7?

Solution:

When the perfect cubes 8, 27, 64, 125, 216, 343, 512, 729, 1000 are divided by 7 the remainders respectively are 1, 6, 1, 6, 6, 0, 1, 1, 6. Hence the remainders are 0. 1 or 6.

Question 14.

What is the least perfect square which leaves the remainder 1 when divided by 7 as welt as by 11?

Solution:

The least number divisible by both 7 and 11 is 7 × 11 = 77. When 1 is added to 77 we get 78. But 78 is not a perfect square. (77 × 2) + 1 = 154 + 1 = 155 is not a perfect square. In the same way, continue we find that

(77 × 15) + 1 = 1155 + 1 = 1156 in a perfect square.

∴ The number required is 1156 = 342

Question 15.

Find two smallest perfect squares whose product is a perfect cube.

Solution:

4 and 16 are perfect squares. 4 × 16 = 64 is a perfect cube.

Question 16.

Find a proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs? By considering 16 as a factor of 48 and 48 as a factor of 240, we can write 16 + 240 = 16^{2}.

Solution:

Multiple of 48 is 48 and considering it as 48l where l = m (3m + 2), m – 1, 2, 3 …. we can get infinite numbers.

**All Chapter KSEEB Solutions For Class 8 maths**

—————————————————————————–**All Subject KSEEB Solutions For Class 9**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share kseebsolutionsfor.com to your friends.

**Best of Luck!!**