Chapter 6 of Class 8 Mathematics is about ‘Squares and Square Roots’. In this chapter Squares and square roots, both concepts are exactly opposite to each other. Squares are the numbers produced after multiplying a value by itself. At the same time, the square root of a number is a value that, on getting multiplied by itself, gives its original value. Hence, both are converse and inverse methods. For example, the square of 2 is 4, and at the same time, the square root of 4 is 2.
If n is a particular number, then its square is represented by n raised to the power 2, i.e., n2 and its square root is expressed as ‘√n’, where ‘√’ is called radical. The value under the root symbol is said to be radicand.
The square numbers are widely explained in terms of the area of a square shape. The shape of a square has all its sides equal. Therefore, the area of the square is equal to (side x side) or side2. Hence, if the square’s side length is 3cm, its area is 32= 9 sq. cm.
Properties of Square Numbers
The square numbers are the values produced when we multiply a number by itself. Some of the properties are:
- A square of 1 is equal to 1
- Square of positive numbers are positive
- A square of negative numbers is also positive. For example, (-3)2 = 9
- The square of zero is zero
- The square of a number’s root equals the value under the root. For example, (√3)2 = 3
- The unit place of the square of any even number will have an even number only.
- If a number has 1 or 9 in the unit’s place, its square ends in 1.
- If a number has 4 or 6 in the unit’s place, its square ends in 6.
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Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions
Important Questions Class 8 Mathematics Chapter 6 – With Solutions
Given below are a few of the questions and their answers from our Chapter 6 Class 8 Mathematics important questions.
Question 1:Show that the sum of two consecutive natural numbers is 13².
Answer 1: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
Question 2: What would be the square root of the number 625 using the identity
(a +b)² = a² + b² + 2ab?
Answer 2: (625)²
= (600 + 25)²
= 600² + 2 x 600 x 25 +25²
= 360000 + 30000 + 625
= 390625
Question 4:Use the following identity and find the square of 189.
(a – b)² = a² – 2ab + b²
Answer 4: 189 = (200 – 11) 2
= 40000 – 2 x 200 x 11 + 112
= 40000 – 4400 + 121
= 35721
Question 5: Find the smallest whole number from which 1008 should be multiplied in order to obtain a perfect square number. Also, find out the square root of the square number so obtained.
Answer 5:
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
Question 6: The sides of a right triangle whose hypotenuse is 17cm are _________ and _________.
Answer 6:
For each natural number, m > 1, 2m, m2 –1 and m2 + 1 form a Pythagorean triplet.
Now,
m² + 1 = (2m)² + (m2 – 1)²
Where,
m² + 1 = 17
m² = 17 – 1
m² = 16
m = √16
m = 4
Then,
2m = 2 × 4
= 8
And,
m² – 1 = 4² – 1
= 16 -1
= 15
Question 7: √(1.96) = _________.
Answer 7: We have,
= √(1.96)
= √(196/100)
= √((14 × 14)/(10 × 10))
= √(142 / 102)
= 14/10
= 1.4
Question 8:If m is the required square of a natural number given by n, then n is
(a) the square of m
(b) greater than m
(c) equal to m
(d) √m
Answer 8: √m
n² = m
Then,
= n = √m
Question 9:There are _________ perfect squares between 1 and 100.
Answer 9: There are 8 perfect squares between 1 and 100.
2 × 2 = 4
3 × 3 = 9
4 × 4 = 16
5 × 5 = 25
6 × 6 = 36
7 × 7 = 49
8 × 8 = 64
9 × 9 = 81
Question 10: By what least number should the given number be divided to get a perfect square number? In each of the following cases, find the number whose square is the new number 1575.
Answer 10: A method for determining the prime factors of a given number, such as a composite number, is known as prime factorisation.
Given 1575,
Resolve 1575 into prime factors, we get
1575 = 3 X 3 X 5 X 5 X 7 = (3² X 5² X 7)
To obtain a perfect square, we have to divide the above equation by 7
Then we get, 3380 = 3 X 3 X 5 X 5
New number = (9 X 25) = (3² X 5² )
Taking squares on both sides of the above equation, we get
∴ New number = (3 X 5)² = (15)²
Therefore, the new number is a square of 15
Question 11:Show that each of the numbers is a perfect square. In each case, find the number whose square is the given number:
7056
Answer 11: 7056,
A perfect square is always expressed as a product of pairs of prime factors.
Resolving 7056 into prime factors, we obtain
7056 = 11 X 539
= 12 X 588
= 12 X 7 X 84
= 84 X 84
= (84)²
Thus, 84 is the number whose square is 5929
Therefore,7056 is a perfect square.
Question 12: Without adding, find the sum of the following:
(1+3+5+7+9+11+13+15+17+19+21+23)
Answer 12: (1+3+5+7+9+11+13+15+17+19+21+23)
As per the given property of perfect square, for any natural number n, we
have some of the first n odd natural numbers=n²
But here n=12
By applying the above the law, we get
thus, (1+3+5+7+9+11+13+15+17+19+21+23) = 12² = 144
Question 13: Using the formula (a – b)²=(a² – 2ab + b²), evaluate:
(196)²
Answer 13: (196)²
We have (a – b)² = (a² – 2ab + b²)
(196)² can be written as 200-4
So here, a=200 and b=4
Using the formula,
(200 – 4)² = (200² – 2 X 200 X 4 + 4² )
On simplifying, we get
(200 – 4)² = (40000 – 1600 +16)
(196)² =38416
Question 14: By what least number should the number be divided to obtain a number with a perfect square? In this, in each case, find the number whose square is the new number 4851.
Answer 14: The number is a perfect square if and only if the prime factorization creates pairs; it is not exactly a perfect square if it is not paired up.
Given 4851,
Resolving 4851 into prime factors, we obtain
4851 = 3 X 3 X 7 X 7 X 11
= (32 X 72 X 11)
To obtain a perfect square, we need to divide the above equation by
11
we obtain, 9075 = 3 X 3 X 7 X 7
The new number = (9 X 49)
= (3² X 7² )
Taking squares on both sides from the above equation, we obtain
∴ The new number = (3 X 7)²
= (21)²
Therefore, the new number is a square of 21
Question 15: By what least number should the number be divided to obtain a perfect square number? In such a case, find the number whose square is the new number 4500.
Answer 15: The number is exactly a perfect square if and only if the prime factorization creates pairs; or else, it is not a perfect square number.
As per the given 4500,
Resolving 4500 into prime factors, we obtain
3380 = 2 X 2 X 3 X 3 X 5 X 5 X 5
= (2² X 3² X 5² X 5)
To obtain a perfect square, we need to divide the above equation by 5
Then we obtain, 4500 = 2 X 2 X 3 X 3 X 5 X 5
The new number = (4 X 9 X 25)
= (2² X 3² X 5² )
Taking squares on both sides from the above equation, we obtain
∴ The new number = (2 X 3 X 5)²
= (30)²
Therefore, the new number is a square of 30
Question 16: Write a Pythagorean triplet whose one member is:
(i) 6
(ii) 14
(iii) 16
(iv) 18
Answer 16: Any natural number m, 2m, m2–1, m2+1 is a Pythagorean triplet.
(i) 2m = 6
m = 6/2
m = 3
m²–1= 3² – 1 = 9–1 = 8
m²+1= 3²+1 = 9+1 = 10
Thus, (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
Thus, (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
Thus, (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
Thus, (18, 80, 82) is a Pythagorean triplet.
Question 17: How many numbers lie between the
squares of the following numbers?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100
Answer 17: 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.
(iii) Between 992 and 1002, there are 2×99 =198 natural numbers.
Question 18: 2025 plants are to be planted in a garden in a way that each of the rows contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Answer 18:
Let the number of rows be x.
Thus, 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 row = 45
Question 19: The digit at the one’s place of the number 572 is _________.
Answer 19: The digit at the one’s place of the number 572 is 9.
We see that 3 or 7 at the unit’s place ends in 9.
= 572
= 57 × 57
Question 20: Give a reason to show that the number given below is a perfect square:
5963
Answer 20:The unit digit of the square numbers will be 0, 1, 4, 5, 6, or 9 if we examine the squares of numbers from 1 to 10. Thus, the unit digit for all perfect squares will be 0, 1, 4, 5, 6, or 9, and none of the square numbers will end in 2, 3, 7, or 8.
Given 5963
We have the property of a perfect square, i.e. a number ending in 3 is never
a perfect square.
Therefore the given number 5963 is not a perfect square.
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