Example 8 - Chapter 5 Class 8 Squares and Square Roots
Last updated at April 16, 2024 by Teachoo
Chapter 5 Class 8 Squares and Square Roots
Concept wise
- Square numbers
- Properties of square numbers
- Sum of consecutive odd numbers
- Numbers between square numbers
- Pattern Solving
- Finding square of large numbers
- Pythagorean triplets
- Square root
- Finding Square root through repeated subtraction
- Finding Square root through prime factorisation
- Checking if perfect square by prime factorisation
- Smallest number multiplied to get perfect square
- Smallest number divided to get perfect square
-
Smallest square number divisible by numbers
- Finding number of digits in square root (without calculation)
- Finding square root by division method - Integers
- Least number subtracted to get a perfect square
- Least number added to get a perfect square
- Finding square root by division method - Decimals
- Statement Questions
Transcript
Example 8 Find the smallest square number which is divisible by each of the numbers 6, 9 and 15. Smallest square number divisible by 6, 9, 15 = L.C.M of 6, 9, 15 or Multiple of L.C.M Finding L.C.M 6, 9, 15 L.C.M of 6, 9, 15 = 2 × 3 × 3 × 5 = 90 Now, checking if 90 is a perfect square or not Checking if 90 is a perfect square 90 = 2 × 3 × 3 × 5 Since 2 & 5 do not occur in pairs Thus, we cannot calculate square root of 90 ∴ 90 is not a perfect square. So, we multiply by 2 and 5 to make pairs Thus, our number becomes 90 × 2 × 5 = 2 × 3 × 3 × 5 × 2 × 5 900 = 2 × 2 × 3 × 3 × 5 × 5 Now, it becomes a perfect square. Thus, required number is 900
