CBSE Class 8 Mathematics Squares and Square Roots Assignment Set 06

REVISION ON (SQUARE AND SQUARE ROOT)

Q1 Is 2100 a perfect square?
Answer: Prime factorization of 2100:
\(2100 = 2^2 \times 3 \times 5^2 \times 7\)
Here, the prime factors 3 and 7 do not have pairs.
Therefore, 2100 is not a perfect square.
In simple words: A number is a perfect square only if all its prime factors can be grouped in pairs. Since 3 and 7 are left unpaired, 2100 is not a perfect square.

Exam Tip: Always find the prime factorization and check if every exponent is an even number.

Q2 Find a Pythagorean triplet, whose one of the member is 16.
Answer: For any natural number \(m > 1\), a Pythagorean triplet is given by \(2m, m^2 - 1, m^2 + 1\).
Let us take the first member \(2m = 16\).
Then, \(m = \frac{16}{2} = 8\).
The second member is \(m^2 - 1 = 8^2 - 1 = 64 - 1 = 63\).
The third member is \(m^2 + 1 = 8^2 + 1 = 64 + 1 = 65\).
Thus, the Pythagorean triplet is 16, 63, 65.
In simple words: Use the standard formula \(2m\), \(m^2 - 1\), and \(m^2 + 1\) to find the three numbers of the Pythagorean triplet.

Exam Tip: Check your answer by verifying that the sum of the squares of the two smaller numbers equals the square of the largest number: \(16^2 + 63^2 = 256 + 3969 = 4225 = 65^2\).

Q3 How many numbers lie between the square of the numbers 20 and 21.
Answer: For any natural number \(n\), the non-square numbers lying between the squares of numbers \(n\) and \(n + 1\) is given by \(2n\).
Here, \(n = 20\) and \(n + 1 = 21\).
Number of numbers lying between \(20^2\) and \(21^2\) = \(2 \times 20 = 40\).
In simple words: The count of non-square numbers between two consecutive squares \(n^2\) and \((n+1)^2\) is always twice the smaller number (\(2n\)).

Exam Tip: Do not actually calculate \(20^2\) (400) and \(21^2\) (441); directly apply the formula \(2n\) to save time.

Q4 Find the value of \(100^2 - 99^2\) without actual solving the square?
Answer: Using the algebraic identity \(a^2 - b^2 = (a - b)(a + b)\):
Here, \(a = 100\) and \(b = 99\).
\(100^2 - 99^2 = (100 - 99)(100 + 99)\)
\(= (1)(199) = 199\)
In simple words: Instead of multiplying large numbers, use the difference of squares formula to make the subtraction very simple.

Exam Tip: Remember that the difference between the squares of two consecutive numbers is always equal to their sum (\((n+1)^2 - n^2 = 2n + 1\)). Here, \(100 + 99 = 199\).

Q5 Find the square root of 56644 by prime factorization method.
Answer: Prime factorization of 56644:
\(56644 = 2 \times 2 \times 137 \times 137\)
\(56644 = 2^2 \times 137^2\)
\(\sqrt{56644} = \sqrt{2^2 \times 137^2}\)
\(= 2 \times 137 = 274\)
In simple words: Break the number into prime factors, make pairs of identical factors, and take one factor from each pair to find the square root.

Exam Tip: Group identical prime factors into pairs and multiply them together to obtain the final square root.

Q6 Find the square root of 290521 by long division method.
Answer: Steps for long division method for 290521:
1. Group digits in pairs from right to left: \(\overline{29}\overline{05}\overline{21}\).
2. Find the largest number whose square is less than or equal to 29, which is 5 (\(5 \times 5 = 25\)).
3. Subtract 25 from 29 to get remainder 4, and bring down the next pair 05 (New dividend is 405).
4. Double the divisor 5 to get 10. Find a digit \(\) \) such that \(10\) \times \ \) is close to 405. For \(\) = 3 \), \(103 \times 3 = 309\).
5. Subtract 309 from 405 to get 96, and bring down 21 (New dividend is 9621).
6. Double the new divisor 53 to get 106. Find a digit \(\) \) such that \(106\) \times \ \) equals 9621. For \(\) = 9 \), \(1069 \times 9 = 9621\).
7. Remainder becomes 0.
Therefore, \(\sqrt{290521} = 539\).
In simple words: Pair the digits from the right, find matching divisors step-by-step, and bring down pairs until the remainder is zero.

Exam Tip: Always place bars over pairs of digits starting from the decimal point towards the left for whole numbers.

Q7 Find the square root of 147.1369 by long division method.
Answer: Steps for long division method for 147.1369:
1. Group digits in pairs: for whole part \(\overline{1}\overline{47}\), for decimal part \(\overline{13}\overline{69}\).
2. For 1, \(1 \times 1 = 1\). Subtract and bring down 47. Remainder is 47.
3. Double 1 to get 2. Taking 22 \(\times\) 2 = 44. Subtract 44 from 47 to get 3. Bring down decimal part 13 (Dividend is 313, place decimal point in quotient).
4. Double 12 to get 24. Taking 241 \(\times\) 1 = 241. Subtract 241 from 313 to get 72. Bring down 69 (Dividend is 7269).
5. Double 121 to get 242. Taking 2423 \(\times\) 3 = 7269. Remainder is 0.
Therefore, \(\sqrt{147.1369} = 12.13\).
In simple words: Apply long division on decimals by pairing digits leftwards from the decimal point and rightwards after the decimal point.

Exam Tip: Be very careful with the placement of the decimal point in the quotient as soon as you cross the decimal in the dividend.

Q8 Simplify: \(\sqrt{144 \times 169}\)
Answer: \(\sqrt{144 \times 169} = \sqrt{144} \times \sqrt{169}\)
\(= 12 \times 13\)
\(= 156\)
In simple words: Split the square root of the product into the product of individual square roots to solve easily.

Exam Tip: Property used here is \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\).

Q9 Find the least number which should be subtracted from 5634 to make it a perfect square?
Answer: Using long division method to find the square root of 5634:
1. Group digits: \(\overline{56}\overline{34}\)
2. \(7 \times 7 = 49\). Remainder = \(56 - 49 = 7\). Bring down 34 (New dividend = 734).
3. Double 7 to get 14. For digit 5, \(145 \times 5 = 725\).
4. Remainder = \(734 - 725 = 9\).
Since the remainder is 9, it means \(75^2\) is less than 5634 by 9.
Therefore, the least number to be subtracted is 9.
In simple words: Find the square root using long division. The leftover remainder is the exact amount you need to subtract to make the number a perfect square.

Exam Tip: The number to be subtracted is always equal to the final remainder obtained in the long division process.

Q10 An army general arranges his soldiers in such a way that the number of rows is same as the number of columns. In such an arrangement 55 soldiers are left behind. If the total number of students are 6455, find the number of soldiers in each row.
Answer: Total number of soldiers given = 6455
Number of soldiers left behind = 55
Number of soldiers arranged in the square form = \(6455 - 55 = 6400\)
Since the number of rows equals the number of columns, let the number of rows be \(x\).
Total soldiers in square arrangement = \(x^2 = 6400\)
\(x = \sqrt{6400} = 80\)
Therefore, the number of soldiers in each row is 80.
In simple words: Subtract the leftover soldiers from the total count, then find the square root of the remaining number to get the count per row.

Exam Tip: Whenever rows equal columns, the total arrangement forms a square number.

Q11 Find the least number which should be added to 9999 to make it a perfect square?
Answer: Using long division method for 9999:
1. Group digits: \(\overline{99}\overline{99}\)
2. \(9 \times 9 = 81\). Remainder = \(99 - 81 = 18\). Bring down 99 (Dividend = 1899).
3. Double 9 to get 18. For digit 9, \(189 \times 9 = 1701\).
4. Remainder = \(1899 - 1701 = 198\).
This shows that \(99^2 < 9999 < 100^2\).
Specifically, \(99^2 = 9999 - 198 = 9801\), and the next square is \(100^2 = 10000\).
Therefore, the least number to be added = \(10000 - 9999 = 1\).
In simple words: Find the next consecutive square after the current number and subtract the given number from it.

Exam Tip: For addition problems, find the square of (\(\text{quotient} + 1\)) and subtract the original number from it.

Q12 Find the smallest number by which 1458 should be multiplied to make it a perfect square and hence find the square root of the number.
Answer: Prime factorization of 1458:
\(1458 = 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3\)
\(1458 = 2 \times 3^6 = 2 \times (3^3)^2\)
Grouping prime factors in pairs: \(1458 = (2) \times (3 \times 3) \times (3 \times 3) \times (3 \times 3)\)
Here, the prime factor 2 is left without a pair.
Therefore, 1458 must be multiplied by 2 to make it a perfect square.
New number = \(1458 \times 2 = 2916\)
Square root of the new number = \(\sqrt{2916} = \sqrt{2^2 \times 3^6} = 2 \times 3^3 = 2 \times 27 = 54\).
In simple words: Find the prime factors, find the factor that lacks a pair, and multiply by it to complete the pairs.

Exam Tip: The multiplier is always the unpaired prime factor from the prime factorization tree.

Q13 Find the smallest number by which 2925 should be divided to make it a perfect square and hence find the square root of the number.
Answer: Prime factorization of 2925:
\(2925 = 3 \times 3 \times 5 \times 5 \times 13\)
\(2925 = 3^2 \times 5^2 \times 13\)
Here, the prime factor 13 does not have a pair.
Therefore, 2925 must be divided by 13 to make it a perfect square.
New number = \(\frac{2925}{13} = 225\)
Square root of the new number = \(\sqrt{225} = 15\).
In simple words: Factorize the number and divide by the leftover unpaired prime factor to obtain a perfect square.

Exam Tip: The divisor is always the unpaired prime factor that remains after pairing up all other identical factors.

Q14 Without adding, find the sum of
a) \(1 + 3 + 5 + 7 =\)
Answer: The sum of the first \(n\) odd natural numbers is equal to \(n^2\).
Here, number of terms \(n = 4\).
Sum = \(4^2 = 16\).
In simple words: Count how many odd numbers are given and square that count.

Exam Tip: This property applies only when adding consecutive odd numbers starting from 1.

b) \(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 =\)
Answer: Counting the number of terms \(n\), there are 15 consecutive odd numbers starting from 1.
Sum = \(n^2 = 15^2 = 225\).
In simple words: Total terms are 15, so the sum is simply \(15 \times 15 = 225\).

Exam Tip: Verify the count of terms carefully before squaring.

c) express 49 as the sum of 7 odd numbers.
Answer: Since 49 is \(7^2\), it can be expressed as the sum of the first 7 consecutive odd numbers.
\(49 = 1 + 3 + 5 + 7 + 9 + 11 + 13\)
In simple words: Write down the first 7 odd numbers starting from 1. Their sum will equal 49.

Exam Title: Reverse of the property: Any square number \(n^2\) equals the sum of the first \(n\) odd natural numbers.