NCERT Solutions Class 8 Maths Chapter 1 — Squares, Cubes & Their Roots

Solution:

Step 1: A perfect square cannot end with the digits 2, 3, 7, or 8.

Step 2: Let's check the unit digit of each option:

Step 3: 3136 ends in 6 (can be a perfect square, e.g., 56²).

Step 4: 4096 ends in 6 (can be a perfect square, e.g., 64²).

Step 5: 5249 ends in 9 (can be a perfect square, e.g., 73²).

Step 6: 6723 ends in 3. Since a perfect square cannot end in 3, 6723 cannot be a perfect square.

Solution:

Step 1: Let's take an example: For n=2, n² = 4 and (n+1)² = 9.

Step 2: The numbers between 4 and 9 are 5, 6, 7, 8. There are 4 non-perfect square numbers.

Step 3: Using the formula 2n, we get 2 × 2 = 4.

Step 4: This property states that there are 2n non-perfect square numbers between the squares of 'n' and '(n+1)'.

Solution:

Step 1: The sum of the first 'n' odd natural numbers is equal to n².

Step 2: In this case, we need the sum of the first 6 odd natural numbers, so n = 6.

Step 3: Therefore, the sum is 6².

Step 4: 6² = 36.

Solution:

Step 1: The prime factorization of 196 is indeed 2 × 2 × 7 × 7.

Step 2: To find the square root, we group the prime factors into pairs: (2 × 2) and (7 × 7).

Step 3: Then, we take one factor from each pair and multiply them: 2 × 7 = 14.

Step 4: All of Ravi's steps and his conclusion are correct according to the prime factorization method for square roots.

Solution:

Step 1: We need to check which set satisfies the condition a² + b² = c².

Step 2: A) (3, 4, 6): 3² + 4² = 9 + 16 = 25. But 6² = 36. So, 25 ≠ 36.

Step 3: B) (8, 15, 17): 8² + 15² = 64 + 225 = 289. And 17² = 289. So, 289 = 289. This is a Pythagorean triplet.

Step 4: C) (6, 8, 9): 6² + 8² = 36 + 64 = 100. But 9² = 81. So, 100 ≠ 81.

Step 5: D) (7, 24, 26): 7² + 24² = 49 + 576 = 625. But 26² = 676. So, 625 ≠ 676.