When any three numbers are given and we have to check that these numbers are pythagorean triplet or not then we have to put the value of all that three numbers into the pythagoras theorem formula. Pythagoras theorem states that, the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. Hypotenuse is always the longest side in the right angled triangle.
Step-by-step explanation:
Option B) 5,12,13 is the answer
When any three numbers are given and we have to check that these numbers are pythagorean triplet or not then we have to put the value of all that three numbers into the pythagoras theorem formula. Pythagoras theorem states that, the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. Hypotenuse is always the longest side in the right angled triangle.
[tex] \bf \underline{ Pythagoras \: theorem} :[/tex]
[tex] \underline{ \boxed{\sf (Hypotenuse)^{2} = (Sum \: of \: square \: of \: the \: two \: numbers) }}[/tex]
[tex]\bf \underline{\underline{\maltese\: Solution }}[/tex]
[tex] \bf (A) \: \: 3, \: 4, \: 5[/tex]
[tex] \sf Length \: of \: longest \: side = 5 [/tex]
[tex] \sf \implies So, \: hypotenuse = 5[/tex]
[tex] \sf By \: pythagoras \: theorem, [/tex]
[tex] \sf \implies (5)^{2} = (3)^{2} + (4)^{2}[/tex]
[tex] \sf \implies 25 = 9 + 16[/tex]
[tex] \sf \implies 25 = 25[/tex]
[tex] \sf \underline{ Hence, \: it \: forms \: a \: pythagorean \: triplet.}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
[tex] \bf (B) \: \: 5, \: 12, \: 13[/tex]
[tex] \sf Length \: of \: longest \: side = 13[/tex]
[tex] \sf \implies So, \: hypotenuse = 13[/tex]
[tex] \sf By \: pythagoras \: theorem, [/tex]
[tex] \sf \implies (13)^{2} = (5)^{2} + (12)^{2}[/tex]
[tex] \sf \implies 169 = 25 + 144[/tex]
[tex] \sf \implies 169 = 169[/tex]
[tex] \sf \underline{ Hence, \: it \: forms \: a \: pythagorean \: triplet.}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
[tex] \bf (C) \: \: 15, \: 20, \: 25[/tex]
[tex] \sf Length \: of \: longest \: side = 25[/tex]
[tex] \sf \implies So, \: hypotenuse = 25[/tex]
[tex] \sf By \: pythagoras \: theorem, [/tex]
[tex] \sf \implies (25)^{2} = (15)^{2} + (20)^{2}[/tex]
[tex] \sf \implies 625 = 225 + 400[/tex]
[tex] \sf \implies 625 = 625[/tex]
[tex] \sf \underline{ Hence, \: it \: forms \: a \: pythagorean \: triplet.}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
[tex] \bf(D) \: \: 7, \: 24, \: 26[/tex]
[tex] \sf Length \: of \: longest \: side = 26[/tex]
[tex] \sf \implies So, \: hypotenuse = 26[/tex]
[tex] \sf By \: pythagoras \: theorem, [/tex]
[tex] \sf LHS \rightarrow (26)^{2} = 676 [/tex]
[tex] \sf RHS \rightarrow (7)^{2} + (24)^{2} [/tex]
[tex] \sf \implies 49 + 576 = 625[/tex]
[tex] \sf LHS \not= RHS[/tex]
[tex] \sf \implies 676 \not=625[/tex]
[tex] \sf \underline{ Hence, \: it \:doesn’t \: forms \: a \: pythagorean \: triplet.}[/tex]