Class 8 RD Sharma Solutions Chapter 3 Squares and Square Roots Exercise 3.1 | Set 2 (original) (raw)
Last Updated : 23 Jul, 2025
Exercise 3.1 | Set 2 from Chapter 3 of RD Sharma's Class 8 mathematics textbook delves into the fundamental concepts of squares and square roots, building upon the knowledge introduced in Set 1. This exercise set aims to deepen students' understanding of perfect squares, square roots, and their properties.
The questions are designed to enhance computational skills, logical thinking, and pattern recognition. By working through these problems, students will develop a strong foundation in squares and square roots, which is crucial for future mathematical concepts such as algebra, geometry, and higher-level arithmetic.
**Question 9. Find the greatest number of two digits which is a perfect square.
**Solution:
Greatest two-digit number is 99
99 = 81+18
= 9×9 + 18
18 is the remainder
Perfect square number is 99 - 18 = 81
Therefore, the greatest number of two digits which is perfect square is 81
**Question 10. Find the least number of three digits which is a **perfect square.
**Solution:
Least three-digit number is 100
100 = 10 × 10
100 itself is the square of 10
Therefore, the least number of three digits which is perfect square is 100
**Question 11. Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.
**Solution:
Prime factorization of 4851
4851 = 3×3×7×7×11
By grouping the prime factors
= (3×3) × (7×7) × 11
11 is left out
Therefore, the smallest number by which 4851 must be multiplied so that the product becomes a perfect square is 11.
**Question 12. Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.
**Solution:
Prime factorization of 28812
28812 = 2×2×3×7×7×7×7
By grouping the prime factors
= (2×2) × 3 × (7×7) × (7×7)
3 is left out
Therefore, the smallest number by which 28812 must be divided so that the quotient becomes a perfect square is 3.
**Question 13. Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number.
**Solution:
Prime factorization of 1152
1152 = 2×2×2×2×2×2×2×3×3
By grouping the prime factors
= (2×2) × (2×2) × (2×2) × (3×3) × 2
Therefore, the smallest number by which 1152 must be divided so that the quotient becomes a perfect square is 2.
The number after division, 1152/2 = 576
Prime factors for 576 = 2×2×2×2×2×2×3×3
By grouping the prime factors
= (2×2) × (2×2) × (2×2) × (3×3)
= (2×2×2×3) × (2×2×2×3)
= 242
Therefore, the resulting number is square of 24.
Read: Chapter 3 Squares and Square Roots - Exercise 3.1 | Set 1
Summary
Exercise 3.1 | Set 2 in Chapter 3 of RD Sharma's Class 8 mathematics textbook offers a comprehensive exploration of squares and square roots, building upon the foundational concepts introduced in Set 1. Through a diverse range of problems, students are challenged to apply various methods for finding square roots, identify perfect squares, and solve real-world applications involving squares and square roots.
The exercise set emphasizes both computational accuracy and conceptual understanding, encouraging students to recognize patterns and relationships between numbers and their squares. By tackling questions that involve estimation, decimal square roots, and the properties of perfect squares, students develop critical thinking skills and mathematical intuition.