Question 1 - Figure it out - Page 50 - Chapter 2 Class 8 - The Baudhayana-Pythagoras Theorem (Ganita Part 2)

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Question 1 Find 5 more Baudhāyana triples using this idea.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 We already used 9 and 25. Let's use the next five odd perfect squares: 49, 81, 121, 169, 225. Taking 49 as odd square number Thus, 2n – 1 = 49 2n = 49 + 1 2n = 50 n = 50/2 n = 25 So, our equation becomes 〖(𝑛−1)〗^2 + (2𝑛−1)=𝑛^2 〖(25−1)〗^2 + 49=25^2 24^2 +49=25^2 〖𝟐𝟒〗^𝟐 +𝟕^𝟐=〖𝟐𝟓〗^𝟐 Thus, our Pythagorean triplet is (7, 24, 25) Taking 81 as odd square number Thus, 2n – 1 = 81 2n = 81 + 1 2n = 82 n = 82/2 n = 41 So, our equation becomes 〖(𝑛−1)〗^2 + (2𝑛−1)=𝑛^2 〖(41−1)〗^2 + 81=41^2 40^2 +81=41^2 〖𝟒𝟎〗^𝟐 +𝟗^𝟐=〖𝟒𝟏〗^𝟐 Thus, our Pythagorean triplet is (9, 40, 41) Taking 121 as odd square number Thus, 2n – 1 = 121 2n = 121 + 1 2n = 122 n = 122/2 n = 61 So, our equation becomes 〖(𝑛−1)〗^2 + (2𝑛−1)=𝑛^2 〖(61−1)〗^2 +121=61^2 60^2 +121=61^2 〖𝟔𝟎〗^𝟐 +〖𝟏𝟏〗^𝟐=〖𝟔𝟏〗^𝟐 Thus, our Pythagorean triplet is (11, 60, 61) Taking 169 as odd square number Thus, 2n – 1 = 169 2n = 169 + 1 2n = 170 n = 170/2 n = 85 So, our equation becomes 〖(𝑛−1)〗^2 + (2𝑛−1)=𝑛^2 〖(85−1)〗^2 + 169=85^2 84^2 +169=85^2 〖𝟖𝟒〗^𝟐 +〖𝟏𝟑〗^𝟐=〖𝟖𝟓〗^𝟐 Thus, our Pythagorean triplet is (13, 84, 85) Taking 225 as odd square number Thus, 2n – 1 = 225 2n = 225 + 1 2n = 226 n = 226/2 n = 113 So, our equation becomes 〖(𝑛−1)〗^2 + (2𝑛−1)=𝑛^2 〖(113−1)〗^2 + 225=113^2 112^2 +225=113^2 〖𝟏𝟏𝟐〗^𝟐 +〖𝟏𝟓〗^𝟐=〖𝟏𝟏𝟑〗^𝟐 Thus, our Pythagorean triplet is (15, 112, 113)