The DAV Class 8 Maths Book Solutions and **DAV Class 8 Maths Chapter 1 Brain Teasers** Solutions of Squares and Square Roots offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 1 Brain Teasers Solutions

Question 1A.

Tick (✓) the correct option.

(i) The difference between the squares of two consecutive numbers is equal to their

(a) difference

(b) sum

(c) product

(d) quotient

Solution:

(b) sum

(ii) What will be the digit in the thousands place of (1111)^{2} ?

(a) 3

(b) 4

(c) 2

(d) 1

Solution:

(b) 4

[(1111)^{2} = 1111 × 1111 = 1234321]

(iii) Perfect squares cannot have 2, 3, ___ and ___ in its ones place.

(a) 1, 7

(b) 5, 6

(c) 7, 8

(d) 7, 9

Solution:

(c) 7, 8

(iv) The smallest number by which 72 must be divided to make it a perfect square is

(a) 4

(b) 5

(c) 3

(d) 2

Solution:

(d) 2

[72 = 2 × 2 × 2 × 3 × 3]

(v) The square root of 3.052009 has ___ decimal places.

(a) 3

(b)4

(c) 5

(d) 1

Solution:

(a) 3

[\(\sqrt{3.052009}\)= 1.747]

Question 1B.

Answer the following questions.

(i) How many non-square numbers are there between 13^{2} and 14^{2} ?

(ii) Write the first four triangular numbers.

(iii) Is 5, 7, 9 a Pythagorean triplets? Why? Justify.

(iv) Find √9 by repeated subtraction method.

(v) Find the measure of the side of a square handkerchief of area 324 cm^{2}.

Solution:

(i) There are twenty-six non-square numbers between 13^{2} and 14^{2} (i.e. 2 × 13 = 26).

(ii) First four triangular numbers are 1, 3, 6 and 10.

(iii) 5, 7, 9 is not a Pythagorian triplet as 5^{2} + 7^{2} ≠ 9^{2}.

(iv) Using repeated subtraction method, we have

Step 1: 9 – 1 = 8

Step 2: 8 – 3 = 5

Step 3: 5 – 5 = 0

We have subtracted 1, 3 and 5, the successive odd numbers from 9 and obtained 0.

Question 2.

Find the square root of 10, correct to four places of decimal.

Solution:

\(\sqrt{10}\) = 3.16227 = 3.163.

Question 3.

Find the values of \(\sqrt{3.1428}\) and \(\sqrt{0.31428}\) correct to three decimal places.

Solution:

\(\sqrt{3.1428}\) = 1.7727 = 1.773

\(\sqrt{0.31428}\) = 0.5606 = 0.561

Question 4.

Simplify :

(i) \(\frac{\sqrt{0.0441}}{\sqrt{0.000441}}\)

(ii) \(\sqrt{49}+\sqrt{0.49}+\sqrt{0.0049}\)

Solution:

(i) \(\frac{\sqrt{0.0441}}{\sqrt{0.000441}}\)

\(\frac{\sqrt{0.0441}}{\sqrt{0.000441}}=\frac{0.21}{0.021}\) = 10

(ii) \(\sqrt{49}+\sqrt{0.49}+\sqrt{0.0049}\)

= 7 + 0.7 + 0.07 = 7.77

Question 5.

The area of a square field is 101 \(\frac{1}{400}\) m^{2}. Find the length of one side of the field.

Solution:

101 \(\frac{1}{400}\) = \(\frac{40401}{400}\)

Side of the square field = \(\sqrt{\frac{40401}{400}}\)

= \(\frac{201}{20}=10 \frac{1}{20}\) m.

Question 6.

What is that number which when multiplied by itself gives 227.798649?

Solution:

Let the required decimal be x

x × x = 227.798649

x^{2} = 227.798649

∴ x = \(\sqrt{227.798649}\) = 15.093

Hence the required decimal = 15.093.

Question 7.

In a lecture hall, 8649 students are sitting in such a manner that there are as many students in a row as there are rows in the lecture hail. How many students are there in each row of the lecture hall?

Solution:

Let the number of students in lecture hail be x

∴ Number of rows = x

Total number of students = x × x = x^{2}

:. x^{2} = 8649

x = \(\sqrt{8649}\) = 93.

Hence the required number of students = 93.

Question 8.

A General wishing to draw up his 64019 men in the form of a square found that he had 10 men extra. Find the numebr of men in the front row.

Solution:

Number of men = 64019

Number of extra men = 10

∴ Number of remaining men = 64019 – 10 = 64009

Number of men in front row = \(\sqrt{64009}\) = 253

Hence the required number of men = 253.

### DAV Class 8 Maths Chapter 1 HOTS

Question 1.

The cost of levelling a square lawn at 15 per square metre is 19,935. Find the cost of fencing the lawn at 22 per metre.

Solution:

Total cost of levelling = 19,935

Cost of levelling per square metre = 15

∴ Area of square lawn = \(\frac{19,935}{15}\) = 1329

Side of square lawn = \(\sqrt{1329}\) = 36.455

Perimeter of square lawn = 36.455 × 4 = 145.82

Thus, the cost of fencing the lawn at 22 per metre = 145.82 × 22 = 3208.04.

Question 2.

If √2 = 1.414, √5 = 2.236 and √3 = 1.732, find the value of:

(i) \(\sqrt{72}+\sqrt{48}\)

(ii) \(\sqrt{\frac{125}{64}}\)

Solution:

(i) \(\sqrt{72}+\sqrt{48}=\sqrt{2 \times 2 \times 2 \times 3 \times 3}+\sqrt{2 \times 2 \times 2 \times 2 \times 3}\)

= 2 × 3 × √2 + 2 × 2 × √3

= 6√2 + 4√3

= 6 × 1.414 + 4 × 1.732 (Given √2 = 1.414 and √3 = 1.732)

= 8.484 + 6.928

= 15.412.

(ii) \(\sqrt{\frac{125}{64}}=\sqrt{\frac{5 \times 5 \times 5}{2 \times 2 \times 2 \times 2 \times 2 \times 2}}\)

= \(\frac{5 \times \sqrt{5}}{2 \times 2 \times 2}=\frac{5 \times 2.236}{8}\)

= \(\frac{11.18}{8}\) = 1.3975.

### DAV Class 8 Maths Chapter 1 Enrichment Questions

Question 1.

The product of two numbers is 1296. If one number is 16 times the other, find the number.

Solution:

Let the other number be x.

Then, the first number = 16x

Now, according to question,

16x × x = 1296

⇒ x^{2} = \(\frac{1296}{16}\)

⇒ x^{2} = 81

∴ x = \(\sqrt{81}\) = 9

⇒ other number = 9

and first number = 16 × 9 = 144

Thus, the required numbers are 144 and 9.

Question 2.

Find the value of \(\sqrt{50625}\) and hence the value of \(\sqrt{506.25}\) + \(\sqrt{5.0625}\).

Solution:

\(\sqrt{50625}\) = \(\sqrt{3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 5}\)

= 3 × 3 × 5 × 5 = 225

Now,

\(\sqrt{506.25}\) + \(\sqrt{5.0625}\) = \(\sqrt{\frac{50625}{100}}+\sqrt{\frac{50625}{10000}}\)

= \(\sqrt{\left(\frac{225}{10}\right)^2}+\sqrt{\left(\frac{225}{100}\right)^2}\)

= \(\frac{225}{10}+\frac{225}{100}\)

= 22.5 + 2.5 = 24.75

∴ \(\sqrt{10000}\) = 2 × 2 × 5 × 5 = 100

∴ \(\sqrt{100}\) = 2 × 5 = 10

Question 3.

Write a Pythagorean triplet if one number is 14.

Solution:

Let the smallest number be 14, so that

2m = 14

⇒ m = \(\frac{14}{2}\) = 7.

Now, m^{2} – 1 = 7^{2} – 1

= 49 – 1 = 48 and,

m^{2} + 1 = 7^{2} + 1

= 49 + 1 = 50.

Thus, 14, 48 and 50 are the Pythagorian triplet.

Additional Questions:

Question 1.

Find \(\sqrt{361 \times 225}\).

Solution:

\(\sqrt{361 \times 225}=\sqrt{361} \times \sqrt{225}\)

[∵ \(\sqrt{}\) = √a × √b]

= 19 × 15 = 285

Question 2.

Find \(\sqrt{1 \frac{56}{169}}\).

Solution:

\(\sqrt{1 \frac{56}{169}}=\sqrt{\frac{225}{169}}=\sqrt{\frac{225}{169}}=\frac{15}{13}\)

Question 3.

Find \(\sqrt{4096}\).

Solution:

\(\sqrt{4096}\) = 64

Question 4.

Find \(\sqrt{17.64}\)

Solution:

\(\sqrt{17.64}\) = 4.2

Question 5.

Find \(\sqrt{0.119025}\)

Solution:

\(\sqrt{0.119025}\) = 0.

Question 6.

Find the square root of 7 upto three places of decimals.

Solution:

√7 = 2.645

Question 7.

Find square root of 0.1 upto three places of decimals.

Solution:

\(\sqrt{0.1}\) = 0.316.

Question 8.

Find the number of digits in the square root of the following numbers (without any calculation).

(i) 144

(ii) 390625

Solution:

(i) 144 has 3 digits (odd).

∴ Number of digits in its square root = \(\frac{n+1}{2}=\frac{3+1}{2}=\frac{4}{2}\) = 2

(ii) 390625 has 6 digits (even)

∴ Number of digits in its square root = \(\frac{n}{2}=\frac{6}{2}\) = 3.

Question 9.

Find the square root of 250 by estimation.

Solution:

The two nearest perfect square number to 250 are 225 and 256.

∴ 225 < 250 < 256 ⇒ 15^{2} < 250 < 16

Let us try for 15.5.

15.5 × 15.5 = 240.25

∴ 15.5^{2} < 250 < 16

Now 15.7 × 15.7 = 246.49

∴ 15.7^{2} < 250 < 162

and 15.8 × 15.8 = 249.64

and 15.9 × 15.9 = 252.81

∴ 15.8^{2} < 250 < 15.9^{2}

Hence, \(\sqrt{250}\) = 15.85 (approximate)

Question 10.

The area of a square plot is 1000 m^{2}. Find the estimated length of the side of the plot.

Solution:

The nearest perfect square number to 1000 are 961 and 1024.

∴ 961 < 1000 < 1024

⇒ 31^{2} < 1000 < 32^{2}

Let us try for 31.4.

31.4 × 31.4 = 985.96 and 31.8 × 31.8 = 1011.24

∴ 31.4^{2} < 1000 < 31.8^{2}

31.6^{2} < 1000 < 31.7^{2}

∴ Estimated value of \(\sqrt{1000}\) = 31.65 m.

Question 11.

How many natural numbers lie between 25^{2} and 26^{2}?

Solution:

Here n = 25 and n + 1 = 26

∴ Natural numbers between 25^{2} and 26^{2}

= 25 + 26 = 51

Question 12.

Find square root of 11025 by prime factorisation.

Solution:

11025 = 3 × 3 × 5 × 5 × 7 × 7

\(\sqrt{11025}\) = 3 × 5 × 7 = 105

Question 13.

Find the value of \(\sqrt{\frac{243}{867}}\) using prime factorisation method.

Solution:

Question 14.

Find the least number of six-digits which is a perfect square. Also, find the square root of this number.

Solution:

The least six-digit number is 100000.

Hence, the least six-digit perfect square number is

100000 + 489 = 100489 and \(\sqrt{100489}\) = 317.

Question 15.

Find the square root of 147.1369.

Solution:

Hence, \(\sqrt{147.1369}\) = 12.13

Question 16.

Find the square root of 2\(\frac{1}{5}\) correct to 3 places of decimal.

Solution:

\(\sqrt{2 \frac{1}{5}}=\sqrt{\frac{11}{5}}=\sqrt{2.2}\)

∴ \(\sqrt{2.2}\) = 1.483.

Multiple Choice Questions:

Question 1.

How many natural numbers lie between 35^{2} and 36^{2}?

(a) 70

(b) 71

(c) 37

(d) 34

Solution:

(b) 71

Question 2.

How many digits are there in the square root of 1234321?

(a) 3

(b) 5

(c) 4

(d) 6

Solution:

(c) 4

Question 3.

The number of non-perfect squares between 25 and 36 is

(a) 8

(b) 9

(c) 10

(d) 11

Solution:

(c) 10

Question 4.

The value of 1 + 3 + 5 + 7 + 9 + …………… is

(a) 12^{2}

(b) 11^{2}

(c) 10^{2}

(d) 13^{2}

Solution:

(b) 11^{2}

Question 5.

The smallest number by which 9408 must be divided to get a perfect square is

(a) 4

(b) 5

(c) 3

(d) 1

Solution:

(c) 3

Question 6.

The square root of 1521 is

(a) 31

(b) 41

(c) 39

(d) 49

Solution:

(c) 39

Question 7.

The least number that must be added to 5607 to make the sum a perfect square is

(a) 18

(b) 17

(c) 19

(d) none of these

Solution:

(a) 18

Question 8.

The square root of 0.1 is

(a) 0.123

(b) 0.361

(c) 0.136

(d) none of these

Solution:

(d) none of these

Question 9.

The value of \(\sqrt{49}+\sqrt{0.49}\) is

(a) 7.07

(b) 7.007

(c) 0.7

(d) 7.7

Solution:

(d) 7.7

Question 10.

A perfect square number leaves 0 or 1 as remainder on division by

(a) 7

(b) 2

(c) 3

(d) 1

Solution:

(c) 3