DAV Class 8 Maths Chapter 1 Brain Teasers Solutions

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 13² and 14² ?
(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 cm².
Solution:
(i) There are twenty-six non-square numbers between 13² and 14² (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 5² + 7² ≠ 9².
(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 \(\frac{1}{400}\) m². 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
x² = 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 = x²
:. x² = 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
⇒ x² = \(\frac{1296}{16}\)
⇒ x² = 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, m² – 1 = 7² – 1
= 49 – 1 = 48 and,
m² + 1 = 7² + 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 ⇒ 15² < 250 < 16
Let us try for 15.5.
15.5 × 15.5 = 240.25
∴ 15.5² < 250 < 16
Now 15.7 × 15.7 = 246.49
∴ 15.7² < 250 < 162
and 15.8 × 15.8 = 249.64
and 15.9 × 15.9 = 252.81
∴ 15.8² < 250 < 15.9²
Hence, \(\sqrt{250}\) = 15.85 (approximate)

Question 10.
The area of a square plot is 1000 m². 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
⇒ 31² < 1000 < 32²
Let us try for 31.4.
31.4 × 31.4 = 985.96 and 31.8 × 31.8 = 1011.24
∴ 31.4² < 1000 < 31.8²
31.6² < 1000 < 31.7²
∴ Estimated value of \(\sqrt{1000}\) = 31.65 m.

Question 11.
How many natural numbers lie between 25² and 26²?
Solution:
Here n = 25 and n + 1 = 26
∴ Natural numbers between 25² and 26²
= 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 35² and 36²?
(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) 12²
(b) 11²
(c) 10²
(d) 13²
Solution:
(b) 11²

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