RS Aggarwal Class 8 Mathematics Solutions Chapter 3 Squares and Square Roots

Exercise 3A

Question 1. Is 441 a perfect square? Give reasons.
Answer: By breaking 441 into prime factors, we find that \( 441 = 49 \times 9 = 7 \times 7 \times 3 \times 3 = 7 \times 3 \times 7 \times 3 = 21 \times 21 = (21)^2 \). Since 441 can be expressed as a product of two equal numbers, it is a perfect square.
In simple words: When you multiply the same number (21) by itself, you get 441. This makes it a perfect square.

Exam Tip: Always resolve numbers into prime factors to check if all prime factors appear in pairs - this confirms a perfect square.

Question 2. Is 576 a perfect square? Give reasons.
Answer: Breaking 576 into prime factors gives \( 576 = 64 \times 9 = 8 \times 8 \times 3 \times 3 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 24 \times 24 = (24)^2 \). Since 576 equals the product of two equal numbers, it is a perfect square.
In simple words: Since 24 multiplied by 24 gives 576, it is a perfect square.

Exam Tip: When prime factors pair up completely with no remainder, the number is always a perfect square.

Question 3. Is 11025 a perfect square? Give reasons.
Answer: Breaking 11025 into prime factors: \( 11025 = 441 \times 25 = 40 \times 9 \times 5 \times 5 = 7 \times 7 \times 3 \times 3 \times 5 \times 5 = 7 \times 5 \times 3 \times 7 \times 5 \times 3 = 105 \times 105 = (105)^2 \). Since 11025 is the product of two equal numbers, it is a perfect square.
In simple words: Since 105 times 105 equals 11025, it is a perfect square.

Exam Tip: Break down large numbers systematically into prime factors to reveal the square root easily.

Question 4. Is 1176 a perfect square? Give reasons.
Answer: Breaking 1176 into prime factors: \( 1176 = 7 \times 168 = 7 \times 21 \times 8 = 7 \times 7 \times 3 \times 2 \times 2 \times 2 \). Since 1176 cannot be written as a product of two equal numbers, it is not a perfect square.
In simple words: The prime factors of 1176 do not pair up evenly, so it cannot be a perfect square.

Exam Tip: If any prime factor appears an odd number of times, the number is not a perfect square.

Question 5. Is 5625 a perfect square? Give reasons.
Answer: Breaking 5625 into prime factors: \( 5625 = 225 \times 25 = 9 \times 25 \times 25 = 3 \times 3 \times 5 \times 5 \times 5 \times 5 \times 5 = 3 \times 5 \times 5 \times 3 \times 5 \times 5 = 75 \times 75 = (75)^2 \). Since 5625 is the product of two equal numbers, it is a perfect square.
In simple words: Since 75 times 75 gives 5625, it is a perfect square.

Exam Tip: Look for repeated prime factors that can be grouped into pairs to find the square root.

Question 6. Is 9075 a perfect square? Give reasons.
Answer: Breaking 9075 into prime factors: \( 9075 = 25 \times 363 = 5 \times 5 \times 3 \times 11 \times 11 = 55 \times 55 \times 3 \). Since 9075 cannot be written as a product of two equal numbers, it is not a perfect square.
In simple words: The prime factors do not pair up completely, so 9075 is not a perfect square.

Exam Tip: If even one prime factor has an odd exponent, the entire number fails to be a perfect square.

Question 7. Is 4225 a perfect square? Give reasons.
Answer: Breaking 4225 into prime factors: \( 4225 = 25 \times 169 = 5 \times 5 \times 13 \times 13 = 5 \times 13 \times 5 \times 13 = 65 \times 65 = (65)^2 \). Since 4225 equals the product of two equal numbers, it is a perfect square.
In simple words: Since 65 multiplied by 65 gives 4225, it is a perfect square.

Exam Tip: Always verify your answer by squaring the derived root to confirm it matches the original number.

Question 8. Is 1089 a perfect square? Give reasons.
Answer: Breaking 1089 into prime factors: \( 1089 = 9 \times 121 = 3 \times 3 \times 11 \times 11 = 3 \times 11 \times 3 \times 11 = 33 \times 33 = (33)^2 \). Since 1089 is the product of two equal numbers, it is a perfect square.
In simple words: Since 33 times 33 equals 1089, it is a perfect square.

Exam Tip: For smaller numbers, you can also check by estimating the square root and verifying by multiplication.

Question 9. Find the largest 2-digit perfect square.
Answer: The smallest three-digit number is 100, which is itself a perfect square. Its square root is 10. The number just before 10 is 9. The square of 9 equals \( (9)^2 = 81 \). Therefore, the greatest two-digit number that is a perfect square is 81.
In simple words: When you square the number 9, you get 81, which is the largest two-digit perfect square.

Exam Tip: To find the largest n-digit perfect square, find the square root of the boundary and work backwards.

Question 10. Find the greatest 3-digit perfect square.
Answer: The greatest three-digit number is 999. The square root of 999 is approximately 31.61. The largest whole number less than or equal to 31.61 is 31. When we square 31, we get \( 31^2 = 31 \times 31 = 961 \). Therefore, the greatest three-digit perfect square is 961.
In simple words: Since 31 times 31 gives 961, and any larger whole number squared exceeds 999, 961 is the greatest three-digit perfect square.

Exam Tip: Always verify the result by squaring the number to ensure it falls within the required range.

Exercise 3A, Question 1. Which of the following numbers are perfect squares?
(i) 441
(ii) 576
(iii) 11025
(iv) 1176
(v) 5625
(vi) 9075
(vii) 4225
(viii) 1089

Exercise 3B

Question 1. Without actually performing the division, state whether the following numbers are perfect squares or not.
(i) 5372
(ii) 5963
(iii) 8457
(iv) 9468
(v) 360
(vi) 64000
(vii) 2500000
Answer: Using properties of perfect squares, we can determine whether a given number is a perfect square or not.
(i) 5372 - A number ending with 2 cannot be a perfect square. Thus, 5372 is not a perfect square.
(ii) 5963 - A number ending with 3 cannot be a perfect square. Thus, 5963 is not a perfect square.
(iii) 8457 - A number ending with 7 cannot be a perfect square. Thus, 8457 is not a perfect square.
(iv) 9468 - A number ending with 8 cannot be a perfect square. Thus, 9468 is not a perfect square.
(v) 360 - Any number ending with an odd count of zeros cannot be a perfect square. Hence, 360 is not a perfect square.
(vi) 64000 - Any number ending with an odd count of zeros cannot be a perfect square. Hence, 64000 is not a perfect square.
(vii) 2500000 - Any number ending with an odd count of zeros cannot be a perfect square. Hence, 2500000 is not a perfect square.
In simple words: Perfect squares can only end in 0, 1, 4, 5, 6, or 9. If a number ends in 2, 3, 7, or 8, or has an odd number of trailing zeros, it cannot be a perfect square.

Exam Tip: Use properties of last digits and trailing zeros to quickly eliminate non-perfect squares without division.

Question 2. Which of the following numbers are perfect squares of even numbers?
(i) 196
(ii) 441
(iii) 900
(iv) 625
(v) 324
Answer: The square of an even number is always even. Therefore, among the given squares, only even numbers will be perfect squares of even numbers.
(i) 196 - This is an even number. Thus, it must be a perfect square of an even number.
(ii) 441 - This is an odd number. Thus, it is not a perfect square of an even number.
(iii) 900 - This is an even number. Thus, it must be a perfect square of an even number.
(iv) 625 - This is an odd number. Thus, it is not a perfect square of an even number.
(v) 324 - This is an even number. Thus, it is a perfect square of an even number.
In simple words: When you multiply an even number by itself, you always get an even result. So look for even numbers in the list.

Exam Tip: Remember that the square of an odd number is always odd, and the square of an even number is always even.

Question 3. Which of the following numbers are perfect squares of odd numbers?
(i) 484
(ii) 961
(iii) 7396
(iv) 8649
(v) 4225
Answer: Following the property that the square of an odd number is also an odd number, we will identify which numbers in the given list are perfect squares of odd numbers.
(i) 484 - This is an even number. Thus, it is not a perfect square of an odd number.
(ii) 961 - This is an odd number. Thus, it is a perfect square of an odd number.
(iii) 7396 - This is an even number. Thus, it is not a perfect square of an odd number.
(iv) 8649 - This is an odd number. Thus, it is a perfect square of an odd number.
(v) 4225 - This is an odd number. Thus, it is a perfect square of an odd number.
In simple words: When you multiply an odd number by itself, the result is always odd. So select the odd numbers from the list.

Exam Tip: Odd squares are always odd, and even squares are always even - use this parity rule to filter quickly.

Question 4. Find the sum of the first 7 odd natural numbers.
Answer: Using the formula that the sum of the first n odd natural numbers equals \( n^2 \):
(i) \( (1 + 3 + 5 + 7 + 9 + 11 + 13) = 7^2 = 49 \)
(ii) \( (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19) = 10^2 = 100 \)
(iii) \( (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23) = 12^2 = 144 \)
In simple words: The sum of the first n odd numbers always equals n squared. This is a useful pattern to remember.

Exam Tip: This rule eliminates the need for actual addition - simply square the count of odd numbers.

Question 5. Express 81 as the sum of consecutive odd numbers starting from 1.
Answer: Using the result that the sum of the first n odd natural numbers equals \( n^2 \):
(i) Expressing 81 as a sum of 9 odd numbers:
\( 81 = (9)^2 \)
\( n = 9 \)
\( 81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 \)
(ii) Expressing 100 as a sum of 10 odd numbers:
\( 100 = (10)^2 \)
\( n = 10 \)
\( 100 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 \)
In simple words: To express a perfect square as a sum of odd numbers, take consecutive odd numbers starting from 1 until you reach the required count.

Exam Tip: Always verify by confirming that the count of odd numbers equals the square root of the target number.

Question 6. For every number m > 1, the Pythagorean triplet is \( (2m, m^2 - 1, m^2 + 1) \). Using this formula, find four Pythagorean triplets.
Answer: Using the formula for Pythagorean triplets:
(i) When \( 2m = 6 \):
\( m = 3, m^2 = 9 \)
\( m^2 - 1 = 9 - 1 = 8 \)
\( m^2 + 1 = 9 + 1 = 10 \)
Thus, the Pythagorean triplet is [6, 8, 10].
(ii) When \( 2m = 14 \):
\( m = 7, m^2 = 49 \)
\( m^2 - 1 = 49 - 1 = 48 \)
\( m^2 + 1 = 49 + 1 = 50 \)
Thus, the Pythagorean triplet is [14, 48, 50].
(iii) When \( 2m = 16 \):
\( m = 8, m^2 = 64 \)
\( m^2 - 1 = 64 - 1 = 63 \)
\( m^2 + 1 = 64 + 1 = 65 \)
Thus, the Pythagorean triplet is [16, 63, 65].
(iv) When \( 2m = 20 \):
\( m = 10, m^2 = 100 \)
\( m^2 - 1 = 100 - 1 = 99 \)
\( m^2 + 1 = 100 + 1 = 101 \)
Thus, the Pythagorean triplet is [20, 99, 101].
In simple words: Substitute different values of m into the formula to generate sets of three numbers that satisfy the Pythagorean theorem.

Exam Tip: Always verify that \( a^2 + b^2 = c^2 \) holds for the triplet you find.

Question 7. Using the identity \( [(n+1)^2 - n^2] = (n+1) + n \), find the value of each expression.
(i) \( (38)^2 - (37)^2 \)
(ii) \( (75)^2 - (74)^2 \)
(iii) \( (92)^2 - (91)^2 \)
(iv) \( (105)^2 - (104)^2 \)
(v) \( (141)^2 - (140)^2 \)
(vi) \( (218)^2 - (217)^2 \)
Answer: Using the given identity:
(i) \( (38)^2 - (37)^2 = 38 + 37 = 75 \)
(ii) \( (75)^2 - (74)^2 = 75 + 74 = 149 \)
(iii) \( (92)^2 - (91)^2 = 92 + 91 = 183 \)
(iv) \( (105)^2 - (104)^2 = 105 + 104 = 209 \)
(v) \( (141)^2 - (140)^2 = 141 + 140 = 281 \)
(vi) \( (218)^2 - (217)^2 = 218 + 217 = 435 \)
In simple words: The difference between the squares of consecutive numbers always equals their sum.

Exam Tip: This algebraic identity saves time - never expand the squares; simply add the two numbers.

Question 8. Find the values using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \).
(i) \( 310^2 \)
(ii) \( 508^2 \)
(iii) \( 630^2 \)
Answer:
(i) \( 310^2 = (300 + 10)^2 = (300)^2 + 2(300)(10) + (10)^2 = 90000 + 6000 + 100 = 96100 \)
(ii) \( 508^2 = (500 + 8)^2 = (500)^2 + 2(500)(8) + (8)^2 = 250000 + 8000 + 64 = 258064 \)
(iii) \( 630^2 = (600 + 30)^2 = (600)^2 + 2(600)(30) + (30)^2 = 360000 + 36000 + 900 = 396900 \)
In simple words: Break the number into two parts, apply the formula, and add the three terms together.

Exam Tip: Choose the breakdown strategically to make calculations easier - typically separate the last digit(s).

Question 9. Find the values using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \).
(i) \( (196)^2 \)
(ii) \( (689)^2 \)
(iii) \( (891)^2 \)
Answer:
(i) \( (196)^2 = (200 - 4)^2 = (200)^2 - 2(200)(4) + (4)^2 = 40000 - 1600 + 16 = 38416 \)
(ii) \( (689)^2 = (700 - 11)^2 = (700)^2 - 2(700)(11) + (11)^2 = 490000 - 15400 + 121 = 474721 \)
(iii) \( (891)^2 = (900 - 9)^2 = (900)^2 - 2(900)(9) + (9)^2 = 810000 - 16200 + 81 = 793881 \)
In simple words: Express the number as a difference, apply the formula, and compute the three components.

Exam Tip: Use subtraction when the number is close to a round number from above.

Question 10. Find the values using the identity \( (a + b)(a - b) = a^2 - b^2 \).
(i) \( 69 \times 71 \)
(ii) \( 94 \times 106 \)
Answer:
(i) \( 69 \times 71 = (70 - 1) \times (70 + 1) = (70)^2 - (1)^2 = 4900 - 1 = 4899 \)
(ii) \( 94 \times 106 = (100 - 6) \times (100 + 6) = (100)^2 - (6)^2 = 10000 - 36 = 9964 \)
In simple words: When two numbers are equidistant from a central value, use the difference of squares formula for quick multiplication.

Exam Tip: Look for numbers that are symmetric around a round number - this method is much faster than direct multiplication.

Question 11. Find the values using the identity \( (a - b)(a + b) = a^2 - b^2 \).
(i) \( 88 \times 92 \)
(ii) \( 78 \times 82 \)
Answer:
(i) \( 88 \times 92 = (90 - 2) \times (90 + 2) = (90)^2 - (2)^2 = 8100 - 4 = 8096 \)
(ii) \( 78 \times 82 = (80 - 2) \times (80 + 2) = (80)^2 - (2)^2 = 6400 - 4 = 6396 \)
In simple words: When multiplying two numbers symmetric around a middle value, find the difference of their squares from that middle value.

Exam Tip: Identify the central value first, then apply the difference of squares formula efficiently.

Question 12. State whether each of the following statements is true or false.
(i) The square of an even number is even.
(ii) The square of an odd number is odd.
(iii) The square of a proper fraction is smaller than the given fraction.
(iv) \( n^2 = \) the sum of first n odd natural numbers.
Answer:
(i) True - The square of an even number is always even.
(ii) True - The square of an odd number is always odd.
(iii) True - The square of a proper fraction (less than 1) is always smaller than the original fraction.
(iv) True - The sum of the first n odd natural numbers always equals \( n^2 \).
In simple words: All four statements reflect fundamental properties of perfect squares that hold universally.

Exam Tip: Memorize these four core properties as they appear frequently in square root problems.

Question 13. State whether each of the following is true or false with reasons.
(i) The number of digits in a square can be odd.
(ii) A prime number cannot be a perfect square.
(iii) The sum of two perfect squares is a perfect square.
(iv) The difference between two perfect squares is a perfect square.
(v) All numbers ending with an even number of zeros are perfect squares.
Answer:
(i) True - The count of digits in a square can be odd. For instance, 121 (which is \( 11^2 \)) has three digits.
(ii) True - A prime number, defined as one indivisible except by 1 and itself, cannot be a perfect square because a perfect square must have an even count of each prime factor.
(iii) False - Consider the example: \( 4 + 9 = 13 \). Both 4 and 9 are perfect squares (of 2 and 3 respectively), but their sum (13) is not a perfect square.
(iv) False - Consider the example: \( 36 - 25 = 11 \). Both 36 and 25 are perfect squares, yet their difference (11) is not a perfect square.
(v) True - Numbers that end with an even count of zeros are perfect squares.
In simple words: Statements (i), (ii), and (v) are correct based on square number properties. Statements (iii) and (iv) fail because sums and differences of perfect squares don't always yield perfect squares.

Exam Tip: Always provide counterexamples when a statement is false - this demonstrates deep understanding.

Exercise 3C

Question 1. Find \( 23^2 \) using the column method.
Answer: Using the column method with \( a = 2 \) and \( b = 3 \):

\( 23^2 = 529 \)
In simple words: Calculate \( a^2 \), \( 2ab \), and \( b^2 \) separately, then combine the results by handling carries.

Exam Tip: The column method is systematic and reduces arithmetic errors - particularly useful for two-digit numbers.

Question 2. Find \( 35^2 \) using the column method.
Answer: Using the column method with \( a = 3 \) and \( b = 5 \):

\( 35^2 = 1225 \)
In simple words: Column method involves writing each component value, then adding columns with carries.

Exam Tip: When carries occur between columns, adjust the values accordingly to get the correct final result.

Question 3. Determine \( 52^2 \) using the column method.
Answer: Using the column method with \( a = 5 \) and \( b = 2 \):

\( 52^2 = 2704 \)
In simple words: Work through each column separately, handle carries, and combine to form the final answer.

Exam Tip: Practice with various two-digit numbers to master the column technique quickly.

Question 4. Calculate \( 96^2 \) using the column method.
Answer: Using the column method with \( a = 9 \) and \( b = 6 \):

\( 96^2 = 9216 \)
In simple words: The column technique requires careful handling of carries between adjacent columns.

Exam Tip: Double-check your carries as they are the most common source of errors in this method.

Question 5. Find \( 67^2 \) using the column method with visual representation.
Answer: The diagram shows the column method breakdown for calculating \( 67^2 \).

\( 67^2 = 4489 \)
In simple words: Break 67 into 6 and 7, calculate the three component values, manage carries, and combine the results.

Exam Tip: Visual representation helps track the calculation flow - draw the layout if needed for clarity.

Question 6. Find \( 86^2 \) using the column method with visual representation.
Answer: The diagram illustrates the column method for \( 86^2 \).

\( 86^2 = 7396 \)
In simple words: Separate 86 into 8 and 6, compute the three values, handle carries correctly, and determine the final answer.

Exam Tip: The visual method makes it easier to see where carries propagate and reduces careless mistakes.

Question 7. Determine \( 137^2 \) using the column method with visual representation.
Answer: The diagram shows the column method computation for \( 137^2 \).

\( 137^2 = 18769 \)
In simple words: For three-digit numbers, expand and compute using the same column principle - break into parts, square and multiply, then combine.

Exam Tip: The column method extends to any number of digits; the principle remains consistent throughout.

Question 8. Calculate \( 256^2 \) using the column method with visual representation.
Answer: The diagram represents the column method calculation for \( 256^2 \).

\( 256^2 = 65536 \)
In simple words: The column method works uniformly - apply it systematically regardless of how many digits the number has.

Exam Tip: Organize your work in columns to prevent arithmetic errors - this structured approach is reliable and fast.

Exercise 3D

Question 1. Find the square root of 225 using the prime factorisation method.
Answer: By breaking down into prime factors:
\( 225 = 3 \times 3 \times 5 \times 5 \)
\( \sqrt{225} = 3 \times 5 = 15 \)

Exam Tip: Group prime factors in pairs; each pair contributes one factor to the square root. Verify by multiplying: \( 15 \times 15 = 225 \).

Question 2. Find the square root of 441 using the prime factorisation method.
Answer: By breaking down into prime factors:
\( 441 = 3 \times 3 \times 7 \times 7 \)
\( \therefore \sqrt{441} = 3 \times 7 = 21 \)

Exam Tip: Always pair identical prime factors; the product of one from each pair gives the square root.

Question 3. Find the square root of 729 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)
\( \therefore \sqrt{729} = 3 \times 3 \times 3 = 27 \)

Exam Tip: For numbers with repeated prime factors (like six 3s), pair them systematically: three pairs give three factors in the square root.

Question 4. Find the square root of 1296 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \)
\( \therefore \sqrt{1296} = 2 \times 2 \times 3 \times 3 = 36 \)

Exam Tip: Write out all prime factors before pairing; this avoids counting mistakes.

Question 5. Find the square root of 2025 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5 \)
\( \therefore \sqrt{2025} = 3 \times 3 \times 5 = 45 \)

Exam Tip: Check your answer: \( 45 \times 45 = 2025 \) confirms the result.

Question 6. Find the square root of 4096 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \)
\( \therefore \sqrt{4096} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64 \)

Exam Tip: For large numbers with many identical factors, count them carefully and divide the count by 2 to find how many appear in the square root.

Question 7. Find the square root of 7056 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 7056 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7 \)
\( \therefore \sqrt{7056} = 2 \times 2 \times 3 \times 7 = 84 \)

Exam Tip: Ensure every prime appears an even number of times; if any prime has an odd count, the number is not a perfect square.

Question 8. Find the square root of 8100 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 8100 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \)
\( \therefore \sqrt{8100} = 2 \times 3 \times 3 \times 5 = 90 \)

Exam Tip: Multiply factors from each pair in the correct order to avoid arithmetic errors.

Question 9. Find the square root of 9216 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 9216 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \)
\( \therefore \sqrt{9216} = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 96 \)

Exam Tip: Double-check by multiplying your result back: \( 96 \times 96 \) should equal 9216.

Question 10. Find the square root of 11025 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 11025 = 3 \times 3 \times 5 \times 5 \times 7 \times 7 \)
\( \therefore \sqrt{11025} = 3 \times 5 \times 7 = 105 \)

Exam Tip: For numbers with three different prime factors, pair each type separately and multiply all three.

Question 11. Find the square root of 15876 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 15876 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7 \)
\( \therefore \sqrt{15876} = 2 \times 3 \times 3 \times 7 = 126 \)

Exam Tip: Organize prime factors from smallest to largest to avoid missing any or counting them twice.

Question 12. Find the square root of 17424 using the prime factorisation method.
Answer: Breaking down into prime factors:
\( 17424 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 11 \times 11 \)
\( \therefore \sqrt{17424} = 2 \times 2 \times 3 \times 11 = 132 \)

Exam Tip: When dividing during factorization, verify each division leaves no remainder.

Question 13. Find the smallest number by which 252 must be multiplied to make it a perfect square.
Answer: Breaking down into prime factors:
\( 252 = 2 \times 2 \times 3 \times 3 \times 7 \)

The factor 7 appears an odd number of times (once). To make all prime factors appear an even number of times, we must multiply by 7.

New number \( = 252 \times 7 = 1764 \)
\( \therefore \sqrt{1764} = 2 \times 3 \times 7 = 42 \)

Exam Tip: Identify any prime with an odd exponent; multiply by that prime to pair it up.

Question 14. Find the smallest number by which 2925 must be divided to make it a perfect square.
Answer: Breaking down into prime factors:
\( 2925 = 3 \times 3 \times 5 \times 5 \times 13 \)

The factor 13 appears an odd number of times (once). The smallest number by which the given number must be divided to form a perfect square is 13.

New number \( = 2925 \div 13 = 225 \)
\( \sqrt{225} = 3 \times 5 = 15 \)

Exam Tip: Find the prime with an odd exponent; divide by it to remove the unpaired factor.

Question 15. In a school, there are 1225 plants. The plants are to be arranged in rows with the same number of plants in each row. Find the number of rows and the number of plants in each row.
Answer: Let the number of rows be \( x \). Then the number of plants in each row is also \( x \).

Total number of plants \( = (x \times x) = x^2 = 1225 \)
\( x^2 = 1225 = 5 \times 5 \times 7 \times 7 \)
\( x = \sqrt{1225} = 5 \times 7 = 35 \)

Therefore, the number of rows is 35 and the number of plants in each row is also 35.

Exam Tip: Set up an equation from the problem statement; finding the square root gives the required arrangement.

Question 16. Students of a class collected Rs 1156 to buy books for the school library. Each student gave as many rupees as the number of students in the class. How many students are there in the class?
Answer: Let the number of students be \( x \). Each student gave Rs \( x \).

Total amount contributed \( = x \times x = x^2 = 1156 \)

\( 1156 = 2 \times 2 \times 17 \times 17 \)
\( x = \sqrt{1156} = 2 \times 17 = 34 \)

Therefore, the strength of the class is 34.

Exam Tip: When each person contributes an amount equal to the number of people, the total is the square of that number.

Question 17. Find the smallest number that must be multiplied to 6, 9, 15, and 20 to get a perfect square.
Answer: The smallest number divisible by each of these numbers is their least common multiple (L.C.M).

L.C.M of 6, 9, 15, 20 = 180

Breaking down into prime factors:
\( 180 = 2 \times 2 \times 3 \times 3 \times 5 \)

The factor 5 appears an odd number of times. To make it a perfect square, we must multiply it by 5.

Required number \( = 180 \times 5 = 900 \)

Exam Tip: First find the L.C.M, then check which prime factors have odd exponents; multiply by those primes.

Question 18. Find the smallest number that must be multiplied to 8, 12, 15, and 20 to get a perfect square.
Answer: The smallest number divisible by each of these numbers is their least common multiple (L.C.M).

L.C.M of 8, 12, 15, 20 = 120

Breaking down into prime factors:
\( 120 = 2 \times 2 \times 2 \times 3 \times 5 \)

To form a perfect square, we need to multiply the number by \( 2 \times 3 \times 5 = 30 \).

Required number \( = 120 \times 30 = 3600 \)

Exam Tip: Multiply the L.C.M by the product of all primes with odd exponents to get the perfect square.

Exercise 3E

Question 1. Find the square root of 576 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 24 \\ \hline 2 & 5 \, 76 \\ & 4 \\ \hline 44 & 1 \, 76 \\ & 1 \, 76 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{576} = 24 \)

Exam Tip: In the long division method, pair digits from the right; bring down pairs and find the quotient systematically.

Question 2. Find the square root of 1444 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 38 \\ \hline 3 & 14 \, 44 \\ & 9 \\ \hline 68 & 5 \, 44 \\ & 5 \, 44 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{1444} = 38 \)

Exam Tip: After finding the first digit, always double it to form the next divisor; this helps locate the next quotient digit.

Question 3. Find the square root of 4489 using the long division method.
Answer: Using the long division method, the square root of 4489 is found by systematically pairing digits, finding quotient digits, and doubling the divisor at each step. The final answer is \( \sqrt{4489} = 67 \).

Exam Tip: Verify by squaring: \( 67 \times 67 = 4489 \).

Question 4. Find the square root of 4489 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 67 \\ \hline 6 & 44 \, 89 \\ & 36 \\ \hline 127 & 8 \, 89 \\ & 8 \, 89 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{4489} = 67 \)

Exam Tip: The remainder becomes zero when the division is exact, confirming a perfect square.

Question 5. Find the square root of 7056 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 84 \\ \hline 8 & 70 \, 56 \\ & 64 \\ \hline 164 & 6 \, 56 \\ & 6 \, 56 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{7056} = 84 \)

Exam Tip: Group digits in pairs starting from the right; this prevents errors in the long division process.

Question 6. Find the square root of 9025 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 95 \\ \hline 9 & 90 \, 25 \\ & 81 \\ \hline 185 & 9 \, 25 \\ & 9 \, 25 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{9025} = 95 \)

Exam Tip: Once you bring down all digit pairs and reach a remainder of zero, you have found the complete square root.

Question 7. Find the square root of 11440 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 107 \\ \hline 1 & 1 \, 14 \, 40 \\ & 1 \\ \hline 207 & 14 \, 40 \\ & 14 \, 49 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{11440} = 107 \)

Exam Tip: For larger numbers, maintain focus on pairing, finding each quotient digit, and bringing down the next pair.

Question 8. Find the square root of 14161 using the long division method.
Answer: Using the long division method, the square root of 14161 is 119.

Exam Tip: Check: \( 119 \times 119 = 14161 \) confirms the answer.

Question 9. Find the square root of 10404 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 102 \\ \hline 1 & 1 \, 04 \, 04 \\ & 1 \\ \hline 202 & 04 \, 04 \\ & 04 \, 04 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{10404} = 102 \)

Exam Tip: When a pair of digits is brought down, work with the combined remainder and new pair together.

Question 10. Find the square root of 17956 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 134 \\ \hline 1 & 1 \, 79 \, 56 \\ & 1 \\ \hline 23 & 79 \\ & 69 \\ \hline 264 & 10 \, 56 \\ & 10 \, 56 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{17956} = 134 \)

Exam Tip: Always verify your result by squaring it back to check if you get the original number.

Question 11. Find the square root of 19600 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 140 \\ \hline 1 & 1 \, 96 \, 00 \\ & 1 \\ \hline 24 & 96 \\ & 96 \\ \hline 280 & 00 \\ & 00 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{19600} = 140 \)

Exam Tip: When the final pairs are zeros, the quotient will end in zero as well.

Question 12. Find the square root of 92416 using the long division method.
Answer: Using the long division method, the square root of 92416 is 304.

Exam Tip: Verify: \( 304 \times 304 = 92416 \) to ensure accuracy.

Question 13. Find the smallest number that should be subtracted from 2509 to make it a perfect square.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 50 \\ \hline 5 & 25 \, 09 \\ & 25 \\ \hline 100 & 09 \\ & 0 \\ \hline & 9 \end{array} \]

The obtained quotient is 50 with a remainder of 9. Therefore, the number that should be subtracted from the given number to make it a perfect square is 9.

Exam Tip: The remainder in the long division method is the number to subtract from the original number to achieve the nearest perfect square.

Question 14. Find the smallest number that should be added to 7581 to make it a perfect square.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 87 \\ \hline 8 & 75 \, 81 \\ & 64 \\ \hline 167 & 11 \, 81 \\ & 11 \, 69 \\ \hline & 12 \end{array} \]

The obtained quotient is 87 with a remainder of 12. Therefore, the number that should be subtracted from the given number to make it a perfect square is 12.

Perfect square \( = 7581 - 12 = 7569 \)

Its square root is 87.

Exam Tip: To find the number to subtract, look at the remainder; to find the number to add, compute the next perfect square and find the difference.

Question 15. Find the smallest number that must be added to 5203 to make it a perfect square.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 78 \\ \hline 7 & 62 \, 03 \\ & 49 \\ \hline 148 & 13 \, 03 \\ & 11 \, 84 \\ \hline & 1 \, 19 \end{array} \]

To obtain a perfect square greater than the given number, we take the square of the next natural number of the obtained quotient, i.e., 78.
\( 79^2 = 6241 \)

Number that should be added to the given number to make it a perfect square \( = 6241 - 5203 = 38 \)

The perfect square thus obtained is 6241 and its square root is 79.

Exam Tip: When adding, square the next integer quotient and subtract the original number to find the amount to add.

Question 16. Find the smallest four-digit number that is a perfect square.
Answer: Using the long division method, the smallest four-digit number is 1000.

\[ \begin{array}{c|c} & 31 \\ \hline 3 & 10 \, 00 \\ & 9 \\ \hline 61 & 1 \, 00 \\ & 61 \\ \hline & 39 \end{array} \]

1000 is not a perfect square. By the long division method, the obtained square root is between 31 and 32. Squaring the next integer (32) will give the next perfect square.

\( 32^2 = 1024 \)

Therefore, 1024 is the smallest four-digit perfect square.

Also, \( \sqrt{1024} = 32 \)

Exam Tip: For the smallest n-digit perfect square, find the square root of the smallest n-digit number, round up to the next integer, then square it.

Question 17. Find the largest five-digit number that is a perfect square.
Answer: The largest five-digit number is 99999.

Using the long division method, the obtained square root is between 316 and 317. Squaring the smaller number, i.e., 316, will give the perfect square that would be less than 99999.

\( 316^2 = 99856 \)

99856 is the required number. Its square root is 316.

Exam Tip: To find the largest n-digit perfect square, find the square root of the largest n-digit number, round down, and square it.

Question 18. Find the largest five-digit number that is a perfect square.
Answer: The largest five-digit number is 99999.

Using the long division method, the obtained square root is between 316 and 317. Squaring the smaller number, i.e., 316, will give the perfect square that would be less than 99999.

\( 316^2 = 99856 \)

99856 is the required number. Its square root is 316.

Exam Tip: Always work with the floor of the square root to find the largest perfect square below a given number.

Question 19. A rectangular field has an area of 60025 square metres. The dimensions are 245 metres by 245 metres. A man cycles around the field at a speed of 18 km/h. How long does it take to cycle around the entire field?
Answer: Area of the square field \( = 60025 \) m²

Length of each side of the square field \( = \sqrt{60025} = 245 \) m

Perimeter of the field \( = 4 \times 245 = 980 \) m \( = \frac{980}{1000} \) km

The man is cycling at a speed of 18 km/h.

Time \( = \frac{\text{Distance travelled}}{\text{Speed}} = \frac{\frac{980}{1000}}{18} = \frac{980}{1000 \times 18} \) hr \( = \frac{980}{18000} \) hr \( = \frac{980 \times 60 \times 60}{18000} \) sec \( = 98 \times 2 \) sec \( = 196 \) sec \( = 3 \) min \( 16 \) sec

Exam Tip: Break the problem into steps: find the side from area, calculate perimeter, then use distance = speed × time formula with consistent units.

Exercise 3F

Question 1. Find the square root of 1.69 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 1.3 \\ \hline 1 & 1 \, 69 \\ & 1 \\ \hline 23 & 69 \\ & 69 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{1.69} = 1.3 \)

Exam Tip: For decimal numbers, pair digits on both sides of the decimal point carefully; the decimal point in the answer falls after pairing.

Question 2. Find the square root of 33.64 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 5.8 \\ \hline 5 & 33 \, 64 \\ & 25 \\ \hline 108 & 8 \, 64 \\ & 8 \, 64 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{33.64} = 5.8 \)

Exam Tip: Group digits in pairs starting from the decimal point; this ensures the decimal point falls in the correct position in the answer.

Question 3. Find the square root of 156.25 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 12.5 \\ \hline 1 & 1 \, 56 \, 25 \\ & 1 \\ \hline 22 & 56 \\ & 44 \\ \hline 245 & 12 \, 25 \\ & 12 \, 25 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{156.25} = 12.5 \)

Exam Tip: After locating the decimal point in the quotient, continue the division process exactly as for whole numbers.

Question 4. Find the square root of 75.69 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 8.7 \\ \hline 8 & 75 \, 69 \\ & 64 \\ \hline 167 & 11 \, 69 \\ & 11 \, 69 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{75.69} = 8.7 \)

Exam Tip: Verify: \( 8.7 \times 8.7 = 75.69 \) confirms the answer.

Question 5. Find the square root of 9.8596 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 3.14 \\ \hline 3 & 9 \, 85 \, 96 \\ & 9 \\ \hline 61 & 85 \\ & 61 \\ \hline 624 & 24 \, 96 \\ & 24 \, 96 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{9.8596} = 3.14 \)

Exam Tip: For numbers with more decimal places, ensure you pair all digits before and after the decimal point consistently.

Question 6. Find the square root of 10.0489 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 3.17 \\ \hline 3 & 10 \, 04 \, 89 \\ & 9 \\ \hline 61 & 1 \, 04 \\ & 61 \\ \hline 627 & 43 \, 89 \\ & 43 \, 89 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{10.0489} = 3.17 \)

Exam Tip: Double-check by multiplying: \( 3.17 \times 3.17 = 10.0489 \).

Question 7. Find the square root of 1.0816 using the long division method.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 1.04 \\ \hline 1 & 1 \, 08 \, 16 \\ & 1 \\ \hline 204 & 08 \, 16 \\ & 08 \, 16 \\ \hline & 0 \end{array} \]

\( \therefore \sqrt{1.0816} = 1.04 \)

Exam Tip: Numbers slightly above 1 will have square roots close to 1 as well; this helps verify reasonableness.

Question 8. Find the square root of 0.2916 using the long division method.
Answer: Using the long division method, the square root of 0.2916 is 0.54.

Exam Tip: Check: \( 0.54 \times 0.54 = 0.2916 \).

Question 9. Find the square root of 3 correct to two decimal places.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 1.732 \\ \hline 1 & 3 \, 00 \, 00 \, 00 \\ & 1 \\ \hline 27 & 2 \, 00 \\ & 1 \, 89 \\ \hline 343 & 11 \, 00 \\ & 10 \, 29 \\ \hline 3462 & 71 \, 00 \\ & 69 \, 24 \\ \hline & 1 \, 76 \end{array} \]

\( \sqrt{3} = 1.732 \)
\( \Rightarrow \sqrt{3} = 1.73 \) (correct up to two decimal places)

Exam Tip: Calculate one extra decimal place beyond what is required, then round appropriately.

Question 10. Find the square root of 2.8 correct to two decimal places.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 1.673 \\ \hline 1 & 2 \, 80 \, 00 \, 00 \\ & 1 \\ \hline 26 & 1 \, 80 \\ & 1 \, 56 \\ \hline 327 & 2 \, 40 \, 0 \\ & 2 \, 289 \\ \hline 3343 & 11 \, 100 \\ & 10 \, 029 \\ \hline & 1 \, 071 \end{array} \]

\( \therefore \sqrt{2.8} = 1.673 \)
\( \Rightarrow \sqrt{2.8} = 1.67 \) (correct up to two decimal places)

Exam Tip: For decimals less than 3, the square root will fall between 1 and 2; use this to estimate reasonableness.

Question 11. Find the square root of 0.9 correct to two decimal places.
Answer: Using the long division method:

\[ \begin{array}{c|c} & 0.948 \\ \hline 0 & 0 \, 90 \, 00 \, 00 \, 00 \\ & 81 \\ \hline 184 & 9 \, 00 \\ & 7 \, 36 \\ \hline 1888 & 1 \, 64 \, 00 \\ & 15 \, 104 \\ \hline & 1 \, 296 \end{array} \]

\( \therefore \sqrt{0.9} = 0.948 \)
\( \Rightarrow \sqrt{0.9} = 0.95 \) (correct up to two decimal places)

Exam Tip: Numbers between 0 and 1 have square roots larger than the original number; verify this property as a sanity check.

Question 12. Find the length of a side of a square whose area is 46.24 square metres.
Answer: Area of the rectangle \( = (13.6 \times 3.4) = 46.24 \) sq m

Therefore, area of the square is 46.24 sq m.

Length of each side of the square \( = \sqrt{46.24} \) m

Using long division method:

\[ \begin{array}{c|c} & 6.8 \\ \hline 6 & 46 \, 24 \\ & 36 \\ \hline 128 & 10 \, 24 \\ & 10 \, 24 \\ \hline & 0 \end{array} \]

\( \sqrt{46.24} = 6.8 \)

Therefore, the length of a side of the square is 6.8 metres.

Exam Tip: Always verify that the given area is correct before computing the side length; squaring the result back should yield the original area.

Exercise 3G

Question 1. Simplify: \( \sqrt{\frac{16}{81}} \)
Answer: We find that \( \sqrt{16} = 4 \) and \( \sqrt{81} = 9 \).

\[ \sqrt{\frac{16}{81}} = \frac{\sqrt{16}}{\sqrt{81}} = \frac{4}{9} \]

Exam Tip: To simplify the square root of a fraction, take the square root of the numerator and the square root of the denominator separately.

Question 2. Simplify: \( \sqrt{\frac{64}{225}} \)
Answer: Using the long division method, we determine that \( \sqrt{64} = 8 \) and \( \sqrt{225} = 15 \).

\[ \sqrt{\frac{64}{225}} = \frac{\sqrt{64}}{\sqrt{225}} = \frac{8}{15} \]

Exam Tip: Always verify your square root values by checking that they multiply back to give the original number.

Question 3. Simplify: \( \sqrt{\frac{121}{256}} \)
Answer: Using the division method, we get \( \sqrt{121} = 11 \) and \( \sqrt{256} = 16 \).

\[ \sqrt{\frac{121}{256}} = \frac{\sqrt{121}}{\sqrt{256}} = \frac{11}{16} \]

Exam Tip: Factorizing larger numbers helps you find their square roots quickly and accurately.

Question 4. Simplify: \( \sqrt{\frac{625}{729}} \)
Answer: Using the long division method, we obtain \( \sqrt{625} = 25 \) and \( \sqrt{729} = 27 \).

\[ \sqrt{\frac{625}{729}} = \frac{\sqrt{625}}{\sqrt{729}} = \frac{25}{27} \]

Exam Tip: When the numerator and denominator are both perfect squares, the result will always be a simple fraction.

Question 5. Simplify: \( \sqrt{3\frac{13}{36}} \)
Answer: First, we transform the mixed number into an improper fraction:
\[ \sqrt{3\frac{13}{36}} = \sqrt{\frac{121}{36}} \]
Taking the square root of both parts:
\[ = \frac{\sqrt{121}}{\sqrt{36}} \]
\[ = \frac{11 \times 11}{\sqrt{36}} \]
\[ = \frac{11}{6} = 1\frac{5}{11} \]

Exam Tip: Always convert mixed numbers to improper fractions before taking square roots to avoid calculation errors.

Question 6. Simplify: \( \sqrt{4\frac{73}{324}} \)
Answer: We convert the mixed number to an improper fraction:
\[ \sqrt{4\frac{73}{324}} = \sqrt{\frac{1369}{324}} \]
Using the long division method, \( \sqrt{1369} = 37 \) and by prime factorization, \( \sqrt{324} = \sqrt{2 \times 2 \times 9 \times 9} = 2 \times 9 = 18 \).
\[ \therefore \sqrt{4\frac{73}{324}} = \frac{37}{18} = 2\frac{1}{18} \]

Exam Tip: When dealing with improper fractions under a square root, factorize the denominator to identify perfect square pairs.

Question 7. Simplify: \( \sqrt{\frac{625}{729}} \)
Answer: Applying the method of extracting square roots from the numerator and denominator separately:
\[ \sqrt{\frac{625}{729}} = \frac{\sqrt{625}}{\sqrt{729}} = \frac{25}{27} \]

Exam Tip: Remember that the square root of a fraction can be split into the square root of the numerator divided by the square root of the denominator.

Question 8. Simplify: \( \frac{\sqrt{80}}{\sqrt{405}} \)
Answer: We combine the square roots:
\[ \frac{\sqrt{80}}{\sqrt{405}} = \sqrt{\frac{80}{405}} = \sqrt{\frac{16}{81}} = \frac{\sqrt{16}}{\sqrt{81}} = \frac{4}{9} \]

Exam Tip: When dividing square roots, combine them first into a single fraction, then simplify.

Question 9. Simplify: \( \frac{\sqrt{1183}}{\sqrt{2023}} \)
Answer: We join the square roots into a single fraction:
\[ \frac{\sqrt{1183}}{\sqrt{2023}} = \sqrt{\frac{1183}{2023}} = \sqrt{\frac{169}{289}} = \frac{\sqrt{169}}{\sqrt{289}} = \frac{\sqrt{13 \times 13}}{\sqrt{17 \times 17}} = \frac{13}{17} \]

Exam Tip: Always reduce the fraction to its lowest terms before finding the square root of the reduced form.

Question 10. Simplify: \( \sqrt{98} \times \sqrt{162} \)
Answer: We merge the square roots and then simplify:
\[ \sqrt{98} \times \sqrt{162} = \sqrt{98 \times 162} = \sqrt{2 \times 7 \times 7 \times 2 \times 9 \times 9} = 2 \times 7 \times 9 = 126 \]

Exam Tip: When multiplying square roots, combine them first and then factorize to identify pairs that simplify easily.

Exercise 3H

Question 1. Which of the following is not a perfect square?
(a) Some value
(b) Some value
(c) 5478
(d) Some value
Answer: (c) 5478
In simple words: According to the property that numbers ending in 2, 3, 7, or 8 cannot be perfect squares, the number 5478 - which ends in 8 - is definitely not a perfect square.

Exam Tip: Remember the basic property: perfect squares can only end in 0, 1, 4, 5, 6, or 9 - never 2, 3, 7, or 8.

Question 2. Which of the following is not a perfect square?
(a) Some value
(b) Some value
(c) Some value
(d) 2222
Answer: (d) 2222
In simple words: Based on the property that numbers ending in 2, 3, 7, or 8 cannot be perfect squares, 2222 - which ends in 2 - is not a perfect square.

Exam Tip: Use the unit digit test as your first quick check to eliminate impossible perfect squares.

Question 3. Which of the following is not a perfect square?
(a) 1843
(b) Some value
(c) Some value
(d) Some value
Answer: (a) 1843
In simple words: Since 1843 ends in 3, and numbers ending in 2, 3, 7, or 8 are never perfect squares, this cannot be a perfect square.

Exam Tip: The unit digit property is a powerful shortcut for MCQs - check it before doing any calculations.

Question 4. Which of the following is not a perfect square?
(a) Some value
(b) 4787
(c) Some value
(d) Some value
Answer: (b) 4787
In simple words: The number 4787 has a unit digit of 7. Numbers with unit digits 2, 3, 7, or 8 can never be perfect squares.

Exam Tip: When you see a number ending in 2, 3, 7, or 8, immediately eliminate it as a perfect square candidate.

Question 5. Which of the following is not a perfect square?
(a) Some value
(b) Some value
(c) 81000
(d) Some value
Answer: (c) 81000
In simple words: According to the property that a number ending with an odd count of zeroes is never a perfect square, 81000 - which has three zeroes (an odd number) - is not a perfect square.

Exam Tip: A perfect square must have an even number of trailing zeroes (0, 2, 4, 6...), never an odd count.

Question 7. If the square of a positive number is smaller than the number itself, which of the following statements is true?
Answer: (b) The number is smaller than the fraction 1
In simple words: When you square a number between 0 and 1, the result becomes even smaller. For example, 0.5 squared is 0.25, which is less than 0.5.

Exam Tip: Test with decimal or fractional values between 0 and 1 to understand this property clearly.

Question 8. What is the value of \( n^2 \) if n is a natural number?
Answer: (c) \( n^2 \)
In simple words: When you square any natural number n, you get n raised to the power 2, which is written as \( n^2 \).

Exam Tip: This is asking you to recognize the notation - the square of n is always expressed as \( n^2 \).

Question 9. Which set represents a Pythagorean triplet?
Answer: (d) (8, 15, 17)
In simple words: A Pythagorean triplet is a set of three numbers where the sum of the squares of the two smaller numbers equals the square of the largest number. For m = 4: we have 2m = 8, m² - 1 = 15, and m² + 1 = 17. Since \( 8^2 + 15^2 = 64 + 225 = 289 = 17^2 \), this is a valid Pythagorean triplet.

Exam Tip: Use the formula \( (2m, m^2 - 1, m^2 + 1) \) for natural numbers m > 1 to generate Pythagorean triplets quickly.

Question 10. Find the value of x if \( (176 - 7) = 169 \) and \( \sqrt{169} = 13 \).
Answer: (c) 7
In simple words: When you subtract 7 from 176, you get 169, and the square root of 169 is 13. This shows that x has a value of 7.

Exam Tip: Always verify that your subtraction is correct before finding the square root of the result.

Question 11. Simplify: 526 + 3 = 529 and \( 529 = 23^2 \). Find the next perfect square after 529.
Answer: (a) 3
In simple words: The working shows that 526 + 3 = 529, and 529 equals 23 squared. This indicates the answer is 3.

Exam Tip: When finding consecutive perfect squares, remember that if n² is a perfect square, the next one is (n+1)².

Question 12. Find the square root using the long division method: 15370 + 6 = 15376 and \( \sqrt{15376} = 124 \).
Answer: (b) 6
In simple words: By adding 6 to 15370, we get 15376, whose square root is 124. This confirms the value is 6.

Exam Tip: Use the long division method systematically to find square roots of large numbers step by step.

Question 13. Find the square root of 0.9409 using the long division method.
Answer: (d) 0.94
In simple words: Using the long division method on 0.9409, we obtain 0.94 as the square root. The working shows each step of the division process until the remainder becomes zero.

Exam Tip: For decimal numbers, apply long division with careful attention to decimal places to ensure your answer is accurate.

Question 14. Find the square root of 0.1 using the long division method.
Answer: (c) 0.316
In simple words: Applying the long division method to find the square root of 0.1, we arrive at approximately 0.316.

Exam Tip: When working with decimal square roots, be prepared for non-terminating decimals and round appropriately as instructed.

Question 15. Simplify: \( \sqrt{0.9} \times \sqrt{1.6} \)
Answer: (b) 1.2
In simple words: When you multiply the square roots together, \( \sqrt{0.9} \times \sqrt{1.6} = \sqrt{1.44} = 1.2 \), since 1.2 multiplied by itself gives 1.44.

Exam Tip: Combine square roots before evaluating - this often leads to easier calculations with perfect decimals.

Question 16. Simplify: \( \sqrt{\frac{288}{128}} \)
Answer: (c) \( \frac{3}{2} \)
In simple words: Reducing the fraction inside the square root, we get \( \sqrt{\frac{288}{128}} = \sqrt{\frac{9}{4}} = \frac{\sqrt{9}}{\sqrt{4}} = \frac{3}{2} \) after canceling common factors from numerator and denominator.

Exam Tip: Always simplify the fraction before taking the square root to reduce calculation complexity.

Question 17. Simplify: \( \sqrt{2\frac{1}{4}} \)
Answer: (b) \( 1\frac{1}{2} \)
In simple words: Converting the mixed number to an improper fraction gives \( \sqrt{\frac{9}{4}} = \frac{\sqrt{9}}{\sqrt{4}} = \frac{3}{2} = 1\frac{1}{2} \).

Exam Tip: Always convert mixed numbers to improper fractions first before applying square root operations.

Question 18. What is the square of 14?
Answer: (a) 196
In simple words: When an even number is squared, the outcome will always be an even number. Since 14 is even, its square, 196, is also even.

Exam Tip: Remember that the square of any even number is always even, and the square of any odd number is always odd.

Question 19. What is the square of 37?
Answer: (c) 1369
In simple words: When an odd number is squared, the result is always an odd number. Since 37 is odd, squaring it gives 1369, which is also odd.

Exam Tip: Use this property as a quick check - if you square an odd number, your answer must also be odd.

Question 20. Find the square root of 0.2809 using the long division method.
Answer: Using the long division method on 0.2809, we arrive at \( \sqrt{0.2809} = 0.53 \).

Exam Tip: When applying long division to decimal square roots, carefully track the decimal point position throughout each step.