Contents
Introduction:
Welcome to Online Tutorial Classes! We are excited to present a valuable resource for CBSE students studying Class 8 Mathematics. In this post, we will delve into the NCERT class 8 Mathematics Chapter 5 ‘Squares and Square Roots’ Exercise 5.2 solutions from the NCERT curriculum and provide you with a comprehensive notes and worksheets to help you master this important topic.
NCERT Class 8 Mathematics Chapter 5, “Squares and Square Roots,” covers fundamental concepts such as squares, square roots, and their applications, including methods for finding squares and square roots, patterns in squares and Pythagorean triplets.
NCERT Maths book:
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NCERT class 8 Mathematics Chapter 5 ‘Squares and Square Roots’ Exercise 5.2 Overview
NCERT Class 8 Mathematics Chapter 5, titled ‘Squares and Square Roots’, exercise 5.2 provides a comprehensive introduction to the concept of squares and square roots. The chapter begins by explaining the properties of square numbers, including patterns in squares and the relationship between square numbers and triangular numbers. It then delves into the concept of square roots, illustrating how to find the square root of a number through repeated subtraction, prime factorisation, and the long division method. The chapter also discusses the characteristics of perfect squares and how to determine whether a number is a perfect square. It further explores the application of squares and square roots in real-life scenarios, such as calculating the area of a square plot or arranging students in a square formation. The chapter concludes with exercises that reinforce the concepts learned and encourage students to find patterns and relationships in squares and square roots.
NCERT class 8 Maths Chapter 5 ‘Squares and Square Roots’ Exercise 5.2 Solutions
Question 1
Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
Solution:
(i) 32²
We know that (a + b)² = a² + b ² + 2ab
= (30 + 2)²
=30² + 2² + 2 x 30 x 2
= 900 + 4 + 120 = 1024
(ii) 35²
= (30 + 5)²
= 30² + 5² + 2 x 30 x 5
= 900 + 25 + 300
= 1225
(iii) 86²
= (80 + 6)²
= 80² + 6² + 2 x 80 x 6
= 6400 + 36 + 960
= 7396
(iv) 93²
= (90 + 3)²
= 90² + 3² + 2 x 90 x 3
= 8100 + 9 + 540
= 8649
(v) 71²
= (70 + 1)²
= 70² + 1² + 2 x 70 x 1
= 4900 + 1 = 140
= 5041
(vi) 46²
= (40 + 6)²
= 40² + 6²+ 2 x 40 x 6
= 1600 + 36 + 480
= 2116
Question 2
Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv) 18
Solution:
For any natural number m > 1, we have (2m)² + (m²– 1)² = (m² + 1)², So, 2m, (m² – 1) and (m² + 1) forms a Pythagorean triplet.
∴ (2m)² + (m²– 1)² = (m² + 1)² is a Pythagorean equation
(i) Replacing the value of 2m with 6 or m with 3 we have
(2 x 3)² + (3²-1)² = (3² + 1)²
⇒ 6² + 8² = 10²
Therefore 6, 8 and 10 are Pythagorean triplet
(ii) 2m = 14 or m = 7 we have
(2 x 7)² + (7² – 1)² = (7² + 1)²
⇒ 14² + 48² = 50²
Therefore 14, 48 and 50 are Pythagorean triplet.
(iii) 2m = 16 or 8 = m
(2 x 8)² + (8² – 1)² = (8² +1)²
⇒ 16² + 63² = 65²
Therefore 16, 63 and 65 are Pythagorean triplet.
(iv) 2m = 18 or m = 9
(2 x 9)² + (9² – 1)² = (9² + 1)²
⇒ 18² + 80² = 82²
Therefore 18, 80 and 82 are Pythagorean triplet.
Accessing All Chapters of NCERT Clas 8 Mathematics:
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NCERT class 8 Mathematics Notes
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