NCERT Solutions Class 8 Maths Square Square Roots Ex 6.1

Understanding Square Numbers & Their Key Properties

A square number, or a perfect square, is a number that can be expressed as the product of an integer with itself. For instance, 3 × 3 = 9, so 9 is a perfect square. In Exercise 6.1, we study several fascinating properties of these numbers. First, consider the unit digits. If you list the squares of numbers from 1 to 10 (1, 4, 9, 16, 25, 36, 49, 64, 81, 100), you will notice that all perfect squares end with the digits 0, 1, 4, 5, 6, or 9 at their unit's place. A number ending in 2, 3, 7, or 8 is never a perfect square! Second, look at the number of zeros at the end. A perfect square can only end with an even number of zeros (like two, four, or six zeros). If a number ends with an odd number of zeros (like 10, 4000, or 900000), it cannot be a perfect square. Finally, there are exactly 2n non-perfect square numbers lying between the squares of consecutive natural numbers n and (n+1).

4 Steps to Solve Exercise 6.1 Problems Quickly

1. Step 1: Predict the Unit Digit of a Square — Look only at the last digit of the given number. Square that single digit. The unit digit of this squared result is the unit digit of the overall large square (e.g., for 81, the unit digit is 1. Since 1² = 1, the square of 81 ends in 1).
2. Step 2: Check Perfect Square Eligibility — Examine the ending digit of any number. If it ends in 2, 3, 7, 8, or contains an odd count of trailing zeros, state immediately that it is not a perfect square.
3. Step 3: Count Non-Square Numbers Between n² and (n+1)² — Identify the smaller number as 'n'. Use the formula: Total non-square integers = 2n. For example, between 12² and 13², there are 2 × 12 = 24 non-square numbers.
4. Step 4: Express Squares as Sums of Odd Numbers — Remember that the sum of the first 'n' consecutive odd natural numbers is always equal to n². Thus, you can represent 25 (which is 5²) as 1 + 3 + 5 + 7 + 9 without doing any addition.

Exam Secrets for Exercise 6.1

Save precious exam time with these shortcuts:

• Don't calculate full products: If a CBSE question asks for the unit's digit of the square of a giant number like 26387, DO NOT multiply 26387 by itself. Just square the last digit 7 to get 49, and identify 9 as the answer.
• Odd/Even Rule: The square of an even number is always even (e.g., 6² = 36), and the square of an odd number is always odd (e.g., 9² = 81). CBSE often tests this rule in multiple-choice questions!
• Verify Zeros: Keep in mind that 400 is a perfect square because it has an even number of zeros (two), but 4000 is not because it contains three zeros (odd).

Practice Questions with Solutions

• Q: What will be the unit's digit of the square of 55555? A: Step 1: Identify the unit digit of the given number, which is 5. Step 2: Square this unit digit: 5 × 5 = 25. Step 3: Find the unit digit of the product 25, which is 5. Final answer: The unit's digit of the square of 55555 is 5.
• Q: State whether the number 7928 can be a perfect square. Give a clear mathematical reason. A: Step 1: Look at the unit's digit of the number 7928. The unit digit is 8. Step 2: Recall that perfect squares must end with the digits 0, 1, 4, 5, 6, or 9. Step 3: Since the number ends in 8, it does not satisfy this criteria. Final answer: No, 7928 cannot be a perfect square because it ends with the digit 8.
• Q: How many non-square numbers lie between 12² and 13²? A: Step 1: Find the value of 'n' by taking the smaller of the two consecutive numbers. Here, n = 12. Step 2: Apply the formula for the number of non-square integers between n² and (n+1)², which is 2n. Step 3: Calculate the value: 2 × 12 = 24. Final answer: There are 24 non-square numbers lying between 12² and 13².
• Q: Without adding, find the sum of: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15. A: Step 1: Count the total number of terms in the series. There are 8 consecutive odd numbers starting from 1. Step 2: Use the property that the sum of the first n consecutive odd natural numbers is n². Step 3: Here, n = 8. Calculate 8² = 64. Final answer: The sum of the given consecutive odd numbers is 64.

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