RS Aggarwal Class 8 Maths Chapter 3 Ex 3.5 Solutions: This exercise deals with the topics related to the long division method which provides the means for detecting the square root of a perfect square by the long division method. These solutions assist the students to practice regularly while learning the fundamentals as they provide each answer to the questions from the textbook. The solutions are solved in a precise manner that is easily understandable by the students to excel in the final exams.
RS Aggarwal Class 8 Maths Chapter 3 Ex 3.5 Solutions are very helpful for the students to clear all their doubts and understand the basics of this chapter more efficiently. The students must refer to these solutions while preparing for the Maths final exam as well as while doing their homework. They can also be able to solve questions that mostly come in the Maths final exam. These solutions are formulated as per the latest syllabus & the guidelines of CBSE that cover detailed illustrations with all solutions.
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Download RS Aggarwal Class 8 Maths Chapter 3 Ex 3.5 Solutions
RS Aggarwal Class 8 Maths Chapter 3 Ex 3.5 Solutions
Important Definition for RS Aggarwal Class 8 Maths Chapter 3 Ex 3.5 Solutions
- Long division method with the help of an example. All the given problems are presented step by step with detailed explanations so that the students can easily revise all the topics & formulae during exam days. These solutions are very beneficial to grasp the concepts in a simple & structured manner & enhance their analytical skills to achieve excellent marks in the exam.
Assuming 484 as the number whose square root is to be evaluated
- Place a bar over the pair of numbers beginning from the unit place or the Right-hand side of the number. In case, if the students have the total number of digits as an odd number, the leftmost digit will also have a bar, 4¯¯¯ 84¯¯¯¯¯
- Consider the largest number as the divisor whose square is less than or equal to the number on the extreme left of the number. The digit on the left is the dividend. They need to divide & write the quotient. Here, the quotient is 2 & the remainder is 0.
- Next, we then bring down the number, which is under the bar, to the right side of the remainder. Here, in this case, we bring down 84. Now, 84 is our new dividend.
- They need to double the value of the quotient & enter on the right side with a blank space.
- They need to select the largest digit for the unit place of the divisor (4_) such that the new number. When multiplied by the new digit at the unit’s place, is equal to or less than the dividend (84).
- In this case, 42 × 2 = 84
Therefore, the new digit is 2.
The remainder is 0, & they have no number left for the division.
Thus, √484= 22
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