Question

Check if (8, 15, 17) and (6n, 8n, 10n) are Pythagorean triplets

Answer 1

$\underline{ \pink{ Pythagorean \: Triplet : }}$

$Three \: natural \: numbers \:a,b \:and \: c\\are \:said \:to \: form \: a \: Pythagorean \: Triplet \\(m,n,p), if \: a^{2} + b^{2} = c^{2}$

$i ) Here, Given \: (8,15,17 )$

$Let \: a = 8 , \: b = 15 \: and \: c = 17$

$\blue{a^{2} + b^{2}}\\ = 8^{2} + 15^{2} \\= 64 + 225 \\= 289 \\= 17^{2} \\\green {= c^{2}}$

$\green { \therefore (8,15,17) \: is \:: Pythagorean \: Triplet }$

$ii ) Here, Given \: (6n,8n,10n )$

$Let \: a = 6n , \: b = 8n \: and \: c = 10n$

$\blue{ a^{2} + b^{2}}\\ = (6n)^{2} + (8n)^{2} \\= 36n^{2} + 64n^{2} \\= 100n^{2} \\= (10n)^{2} \\\green{= c^{2}}$

$\green { \therefore (6n , 8n , 10n) \: is \:: Pythagorean \: Triplet }$

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