CBSE Class 8 Mathematics Squares and Square Roots Assignment Set 14

Mathematics Assignment No. 01 - Squares and Sq. Root

Q1. Is 256 a perfect square?
Answer: Yes, 256 is a perfect square.
Prime factorization of 256:
\(256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8 = (2^4)^2 = 16^2\)
Since 256 can be expressed as the square of an integer (16), it is a perfect square.
In simple words: A number is a perfect square if it can be written as the product of an integer multiplied by itself.

Exam Tip: Memorize squares of numbers up to 25 to quickly identify perfect squares.

Q2. Show that 3600 is a perfect square.
Answer: Prime factorization of 3600:
\(3600 = 36 \times 100 = (6^2) \times (10^2) = (6 \times 10)^2 = 60^2\)
Since 3600 is the square of 60, it is a perfect square.
In simple words: Breaking down numbers into powers of 10 and single-digit squares makes it easy to prove they are perfect squares.

Exam Tip: For numbers ending in even zeros, pair the zeros and check if the remaining number is a perfect square.

Q3. Is 2100 a perfect square, explain?
Answer: No, 2100 is not a perfect square.
Prime factorization of 2100:
\(2100 = 21 \times 100 = 3 \times 7 \times 10^2\)
Since the prime factors 3 and 7 do not occur in pairs (they have an odd exponent of 1), 2100 cannot be expressed as the square of an integer.
In simple words: A number with an odd number of trailing zeros or whose prime factors cannot be grouped into pairs is never a perfect square.

Exam Tip: Any number ending with an odd number of zeros (like one zero, three zeros) is never a perfect square.

Q4. Using prime factorization method, find the square root of the following numbers if exist
(i) 2048
(ii) 4096
(iii) 3549
(iv) 11025
Answer:
(i) For 2048:
\(2048 = 2^{11} = 2 \times (2^5)^2\)
Since all prime factors do not form complete pairs, square root does not exist as a rational number.
(ii) For 4096:
\(4096 = 2^{12} = (2^6)^2 = 64^2\)
\(\sqrt{4096} = 64\)
(iii) For 3549:
Prime factorization of 3549 = \(3 \times 7 \times 169 = 3 \times 7 \times 13^2\)
Since 3 and 7 do not form pairs, square root does not exist as a rational number.
(iv) For 11025:
\(11025 = 3^2 \times 5^2 \times 7^2 = (3 \times 5 \times 7)^2 = 105^2\)
\(\sqrt{11025} = 105\)
In simple words: Group prime factors in pairs of two. If any factor is left unpaired, the number is not a perfect square and its square root is not a whole number.

Exam Tip: Always verify divisibility by 3, 5, and 11 systematically when doing prime factorization.

Q5. Find Pythagorean triplet, one of whose members is
(i) 6
(ii) 4
(iii) 8
Answer:
A Pythagorean triplet is given in the form \(2m, m^2 - 1, m^2 + 1\).
(i) If one member is 6 (let \(2m = 6\), so \(m = 3\)):
\(m^2 - 1 = 3^2 - 1 = 9 - 1 = 8\)
\(m^2 + 1 = 3^2 + 1 = 9 + 1 = 10\)
Triplet: (6, 8, 10)
(ii) If one member is 4 (let \(2m = 4\), so \(m = 2\)):
\(m^2 - 1 = 2^2 - 1 = 4 - 1 = 3\)
\(m^2 + 1 = 2^2 + 1 = 4 + 1 = 5\)
Triplet: (4, 3, 5)
(iii) If one member is 8 (let \(2m = 8\), so \(m = 4\)):
\(m^2 - 1 = 4^2 - 1 = 16 - 1 = 15\)
\(m^2 + 1 = 4^2 + 1 = 16 + 1 = 17\)
Triplet: (8, 15, 17)
In simple words: Take the given even number as \(2m\), find \(m\), and then calculate \(m^2 - 1\) and \(m^2 + 1\).

Exam Tip: Remember that the smallest member is not always \(2m\), but for standard textbook problems of this type, equating the given even number to \(2m\) works directly.

Q6. Express the perfect square number as sum of odd numbers
(i) 49
(ii) 64
Answer:
The sum of the first \(n\) odd natural numbers is \(n^2\).
(i) For 49 (\(7^2\)):
\(49 = 1 + 3 + 5 + 7 + 9 + 11 + 13\)
(ii) For 64 (\(8^2\)):
\(64 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15\)
In simple words: Write down consecutive odd numbers starting from 1 until the count of numbers equals the square root of the given number.

Exam Tip: The number of terms in the sum always equals the square root of the target number.

Q7. Without actual addition, find the sum
(i) \(1 + 3 + 5 + 7 + 9\)
(ii) \(1 + 3 + 5 + 7 + 9 + 11 + 13 + \dots + 21\)
Answer:
(i) Counting the terms, there are 5 consecutive odd numbers starting from 1.
Sum = \(n^2 = 5^2 = 25\)
(ii) Counting the terms from 1 to 21, the number of terms \(n = \frac{21 + 1}{2} = 11\).
Sum = \(n^2 = 11^2 = 121\)
In simple words: Count how many odd numbers are added together, and square that count.

Exam Tip: For an arithmetic progression of consecutive odd numbers starting from 1, the number of terms is \(n = \frac{\text{Last Term} + 1}{2}\).

Q8. How many numbers lie between squares of the following numbers?
(i) 10 and 11
(ii) 20 and 21
Answer:
Non-square numbers between \(n\) and \(n+1\) is given by \(2n\).
(i) For 10 and 11 (\(n = 10\)):
Number of non-square numbers = \(2 \times 10 = 20\)
(ii) For 20 and 21 (\(n = 20\)):
Number of non-square numbers = \(2 \times 20 = 40\)
In simple words: Take the smaller number and multiply it by 2 to find how many non-square numbers lie between their squares.

Exam Tip: Always use the formula \(2n\) where \(n\) is the smaller of the two consecutive integers.

Q9. Using the pattern, find the missing number.
(i) \(3^2 + 4^2 + 12^2 = 13^2\)
(ii) \(4^2 + 5^2 + \dots = 21^2\)
(iii) \(5^2 + \dots + 30^2 = 31^2\)
(iv) \(6^2 + 7^2 + 42^2 = \dots\)
Answer:
Based on the given pattern \(a^2 + b^2 + (a \times b)^2 = (a \times b + 1)^2\):
(ii) Missing terms for \(4^2 + 5^2 + \dots = 21^2\):
Here \(a = 4, b = 5\), so \(a \times b = 20\).
Missing term = \(20^2\)
(iii) Missing terms for \(5^2 + \dots + 30^2 = 31^2\):
Here \(a = 5\), and product is 30, so \(b = 6\).
Missing term = \(6^2\)
(iv) Missing terms for \(6^2 + 7^2 + 42^2 = \dots\):
Here \(a = 6, b = 7\), and \(a \times b = 42\).
Resulting square = \((42 + 1)^2 = 43^2\)
In simple words: The third number is the product of the first two numbers, and the answer on the right side is one more than that product squared.

Exam Tip: Look at the relationship between the first two bases and the last base in completed rows to identify the multiplicative pattern.

Q10. Without actual finding the squares of the numbers, find the value of
(i) \(100^2 - 99^2\)
(ii) \(26^2 - 25^2\)
Answer:
Using the algebraic identity \(a^2 - b^2 = (a - b)(a + b)\):
(i) \(100^2 - 99^2 = (100 - 99)(100 + 99) = 1 \times 199 = 199\)
(ii) \(26^2 - 25^2 = (26 - 25)(26 + 25) = 1 \times 51 = 51\)
In simple words: For consecutive squares, the difference between them is simply the sum of the two numbers.

Exam Tip: Whenever subtracting squares of consecutive numbers, just add the two numbers together to get the direct answer.