Class 8th mathematics Squares and Square Roots chapter 6 exercise 6.4 solution

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Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.1 solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.2 solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.3 solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.4 solution 

SQUARE AND SQUARE ROOTS (Chapter-6) Exercise 6.4 SOLUTION

EXERCISE 6.4 

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- 

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- A perfect square is of n-digits, then its square root will have n2 digits if n is even or n+12  if n is odd. 

(i) In 64,  n = 2 (even)

Number of digits in √64 = 22 = 1

(ii) In 144,  n = 3 (odd)

Number of digits in √144 = 42 = 2

(iii) In 4489,  n = 4 (even)

Number of digits in √4489 = 42 = 2

(iv) In 27225,  n = 5 (odd)

Number of digits in √27225 = 62 = 3

(v) In 390625,  n = 6 (even)

Number of digits in √390625 = 62 = 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-

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, the remainder is 02

New number = 402 – 02 = 400
Therefore, 

√400 = 20

Hence, the least required number which should be subtracted from 402 to get a perfect square is 02.

(ii) 1989 

Here, the remainder is 53

New number = 1989 – 53 = 1936
Therefore, 

√1936 = 44

Hence, the least required number which should be subtracted from 1989 to get a perfect square is 53.

(iii) 3250 

Here, the remainder is 01

New number = 3250– 01 = 3249
Therefore, 

√3249 = 57

Hence, the least required number which should be subtracted from 3250 to get a perfect square is 01.

(iv) 825 

Here, the remainder is 41 

New number = 825 – 41 = 784
Therefore, 

√784 = 28

Hence, the least required number which should be subtracted from 825 to get a perfect square is 41.

(v) 4000 

Here, the remainder is 31 

New number = 4000 – 31 = 3969
Therefore, 

√3969= 63

Hence, the least required number which should be subtracted from 4000 to get a perfect square is 31.

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, the falling short is 04

New number = 525 + 04 = 529
Therefore, 

√529 = 23

Hence, the least required number which should be added to 525 to get a perfect square is 04.

(ii) 1750 

Here, the falling short is 14

New number = 1750 + 14 = 1764
Therefore, 

√1764 = 42

Hence, the least required number which should be added to 1750 to get a perfect square is 14.

(iii) 252 

Here, the falling short is 04

New number = 252 + 04 = 256
Therefore, 

√256 = 16

Hence, the least required number which should be added to 252 to get a perfect square is 04.

(iv) 1825 

Here, the falling short is 24

New number = 1825 + 24 = 1849
Therefore, 

√1849 = 43

Hence, the least required number which should be added to 1825 to get a perfect square is 24.

(v) 6412

Here, the falling short is 149

New number = 6412 + 149 = 6561
Therefore, 

√6561 = 81

Hence, the least required number which should be added to 6412 to get a perfect square is 149.

6. Find the length of the side of a square whose area is 441 m². 

Solution-

As we know that the length of all side of all square is equal.

Let the side of the square is x

The area of the square = side ✕ side = x ✕ x =x²

Then,

x² = 441

x = √441

x = 21

Hence, the length of the side of a square is 21. 

7. In a right triangle ABC, ∠B = 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

First we draw a figure of right triangle ABC

 According to the Pythagoras theorem

we know that AC² = AB² + BC²

AC² = 6² + 8² = 36 + 64 = 100

AC = √100 = 10

Hence, AC is 10 cm.

(b) If AC = 13 cm, BC = 5 cm

According to the Pythagoras theorem

we know that AB² = AC² ー BC²

AB² =13²  ー 5² = 169 ー  25 = 144

AB = √144 = 12

Hence, AB is 12 cm.

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. 

Solution -

Let the number of rows = x.
And the number of columns = x.
Total number of plants = x × x = x2

We know that 1000 is not a perfect square so we need to add sum more plant to make a perfect square.

Then,

1000 + y = x2

And we also know that the perfect square number just greater than 1000 is 1024.

So, 

1000 + y = 1024

x2 = 1024

x = √1024

x = 32

And 

1000 + y = 1024

y = 1024 - 1000 = 24

Hence, the minimum number of plants he needs more for this is 24.

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.  

Solution- 

Let the number of children in a row = x. 

And also that of in a column = x.

Total number of students = x × x = x2 

We know that 500 is not a perfect square so we need to add sum more plant to make a perfect square.

Then,

1000 - y = x2

And we also know that the perfect square number just less than 500 is 484.

So, 

500 - y = 484

x2 = 484

x = √484

x = 22

And 

500 - y = 484

y = 500 - 484 = 16

Hence, to stand in such a manner that the number of rows is equal to number of columns. 16 children would be left out in this arrangement.

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Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.1 solution

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