Unit Test ( Solutions): Square and Square Roots | Mathematics (Maths) Class 8 PDF Download

Time: 1 hour

M.M.: 30

Attempt all questions.

Question numbers 1 to 5 carry 1 mark each.

Question numbers 6 to 8 carry 2 marks each.

Question numbers 9 to 11 carry 3 marks each.

Question numbers 12 & 13 carry 5 marks each.

Q1: A rectangular field has a length of 24 meters and a width of 7 meters. Calculate the diagonal of the field using the Pythagorean theorem and verify if the result is correct. (1 Mark)
Ans:
Diagonal = √(24² + 7²) = √(576 + 49) = √625 = 25 meters
Verification: 25² = 625, which confirms the calculation is correct.

Q2: Use the identity and find the square of 189. (1 Mark)

(a – b)² = a² – 2ab + b²
Ans:
189 = (200 – 11)²

= 40000 – 2 x 200 x 11 + 11²

= 40000 – 4400 + 121

= 35721

Q3: The sum of the first 5 odd natural numbers is equal to the square of which number? (1 Mark)
(i) 2
(ii) 3
(iii) 4
(iv) 5

Ans: (iv)
The sum of the first 5 odd natural numbers is 1 + 3 + 5 + 7 + 9 = 25, which is equal to 5².

Q4: Find the number of non-square numbers between the squares of 10 and 11. (1 Mark)
(i) 21
(ii) 19
(iii) 20
(iv) 18

Ans: (iii)
There are 20 non-square numbers between 10² = 100 and 11² = 121.

Q5: The sum of two consecutive triangular numbers is a square number. Find the sum of 10 and 11 triangular numbers and verify this statement. (1 Mark)
Ans:
10th triangular number = 10(10+1)/2 = 55
11th triangular number = 11(11+1)/2 = 66
Sum = 55 + 66 = 121 = 11², which is a perfect square.

Q6: Find the smallest square number that is divisible by each of the numbers 8, 15 and 20. (2 Marks)
Ans:

L.C.M of 8, 15 and 20 is (2×2×5×2×3) = 120.

120 = 2×2×3×5×2 = (2×2)×3×5×2

Here, 3, 5 and 2 cannot be paired.

Therefore, we need to multiply 120 by (3×5×2) i.e. 30 to get a perfect square.

Hence, the smallest squared number which is divisible by numbers 8, 15 and 20 = 120×30 = 3600

Q7: Show that the sum of two consecutive natural numbers is 13². (2 Marks)

Ans:

Let 2n + 1 = 13

So, n = 6

So, ( 2n + 1)² = 4n² + 4n + 1

= (2n² + 2n) + (2n² + 2n + 1)

Substitute n = 6,

(13)² = ( 2 x 6² + 2 x 6) + (2 x 6² + 2 x 6 + 1)

= (72 + 12) + (72 + 12 + 1)

= 84 + 85

Q8: How many numbers lie between squares of the following numbers? (2 Marks)
(i) 12 and 13
(ii) 25 and 26

Ans: As we know, between n² and (n+1)², the number of non–perfect square numbers are 2n.

(i) Between 12² and 13² there are 2×12 = 24 natural numbers.

(ii) Between 25² and 26² there are 2×25 = 50 natural numbers.

Q9: Find the smallest whole number by which 1008 should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained. (3 Marks)

Ans:

Let us factorise the number 1008.

1008 = 2×2×2×2×3×3×7

= (2×2)×(2×2)×(3×3)×7

Here, 7 cannot be paired.

Therefore, we will multiply 1008 by 7 to get a perfect square.

New number so obtained = 1008×7 = 7056

Now, let us find the square root of 7056

7056 = 2×2×2×2×3×3×7×7

⇒ 7056 = (2×2)×(2×2)×(3×3)×(7×7)

⇒ 7056 = 2²×2²×3²×7²

⇒ 7056 = (2×2×3×7)²

Therefore;

⇒ √7056 = 2×2×3×7 = 84

Q10: Write a Pythagorean triplet whose one member is: (3 Marks)
(i) 6
(ii) 14
(iii) 16
(iv) 18
Ans:

We know, for any natural number m, 2m, m²–1, m²+1 is a Pythagorean triplet.

(i) 2m = 6

⇒ m = 6/2 = 3

m²–1= 3² – 1 = 9–1 = 8

m²+1= 3²+1 = 9+1 = 10

Therefore, (6, 8, 10) is a Pythagorean triplet.

(ii) 2m = 14

⇒ m = 14/2 = 7

m²–1= 7²–1 = 49–1 = 48

m²+1 = 7²+1 = 49+1 = 50

Therefore, (14, 48, 50) is not a Pythagorean triplet.

(iii) 2m = 16

⇒ m = 16/2 = 8

m²–1 = 8²–1 = 64–1 = 63

m²+ 1 = 8²+1 = 64+1 = 65

Therefore, (16, 63, 65) is a Pythagorean triplet.

(iv) 2m = 18

⇒ m = 18/2 = 9

m²–1 = 9²–1 = 81–1 = 80

m²+1 = 9²+1 = 81+1 = 82

Therefore, (18, 80, 82) is a Pythagorean triplet.

Q11. Find the square roots of 100 and 169 by the method of repeated subtraction. (3 Marks)
Ans:

Let us find the square root of 100 first.

1. 100 – 1 = 99
2. 99 – 3 = 96
3. 96 – 5 = 91
4. 91 – 7 = 84
5. 84 – 9 = 75
6. 75 – 11 = 64
7. 64 – 13 = 51
8. 51 – 15 = 36
9. 36 – 17 = 19
10. 19 – 19 = 0

Here, we have performed a subtraction ten times.

Therefore, √100 = 10

Now, the square root of 169:

1. 169 – 1 = 168
2. 168 – 3 = 165
3. 165 – 5 = 160
4. 160 – 7 = 153
5. 153 – 9 = 144
6. 144 – 11 = 133
7. 133 – 13 = 120
8. 120 – 15 = 105
9. 105 – 17 = 88
10. 88 – 19 = 69
11. 69 – 21 = 48
12. 48 – 23 = 25
13. 25 – 25 = 0

Here, we have performed subtraction thirteen times.

Therefore, √169 = 13

Q12: 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row. (5 Marks)
Ans:

Let the number of rows be, x.

Therefore, the number of plants in each row = x.

Total many contributed by all the students = x×x = x²

Given, x² = Rs.2025

x² = 3×3×3×3×5×5

⇒ x² = (3×3)×(3×3)×(5×5)

⇒ x² = (3×3×5)×(3×3×5)

⇒ x² = 45×45

⇒ x = √(45×45)

⇒ x = 45

Therefore,

Number of rows = 45

Number of plants in each rows = 45

Q13: Find the least number which must be added to 1750 so as to get a perfect square. Also, find the square root of the obtained number. (5 Marks)
Ans:

Using long division method:

Here, (41)² < 1750 > (42)²

We can say 1750 is ( 164 – 150 ) = 14 less than (42)².

Therefore, if we add 14 to 1750, it will be a perfect square.

New number = 1750 + 14 = 1764

Therefore, the square root of 1764 is as follows:

∴√1764 = 42