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

Question 1.

Match the following numbers ¡n the column A with their squares ¡n the column B.

Solution:

(1) (a) 25

(2) (e) 64

(3) (f) 4

(4) (c) 36

(5) (d) 484

(6) (b) 144

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 4.

Write the digit in units place when the following numbers are squared.

Solution:

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 6.

Find the sum of the digits of (1111111111)^{2}.

Solution:

Observe the pattern

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 8.

A sum of Rs. 2304 is equally distributed among several people. Each gets as many rupees as the number of persons. How much does each one get?

Solution:

Let the number of persons be x. Each person get rupees x.

∴ x × x = 2304

x^{2} = 2304

x = √2304

2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 16 × 3 × 16 × 3

2304 = 48 × 48

√2304 = 48 Each person gets Rs. 48

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.