Question 5 - Page 49 - Pythagorean triples - Chapter 2 Class 8 - The Baudhayana-Pythagoras Theorem (Ganita Part 2)

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Question 5 - Page 49 How do we generate more primitive triples?We use our equation 〖(𝒏−𝟏)〗^𝟐 + (𝟐𝒏−𝟏)=𝒏^𝟐 Here, 2n – 1 is an odd square number Our odd square numbers are 1, 9, 25, 49, 81, 121, 169, 225… Thus, we put 2n – 1 as Odd square number and find value of n Taking 25 as odd square number Thus, 2n – 1 = 25 2n = 25 + 1 2n = 26 n = 26/2 n = 13 So, our equation becomes 〖(𝑛−1)〗^2 + (2𝑛−1)=𝑛^2 〖(13−1)〗^2 + 25=13^2 12^2 +25=13^2 〖𝟏𝟐〗^𝟐 +𝟓^𝟐=〖𝟏𝟑〗^𝟐 Thus, our Pythagorean triplet is (5, 12, 13)