Respuesta :
Answer:
[tex]\huge\boxed{9,40,46}[/tex]
Step-by-step explanation:
Let's check it using Pythagorean Theorem:
[tex]c^2 = a^2 + b^2[/tex]
Where c is the longest sides, a and b are rest of the 2 sides
1) 9 , 40 , 46
=> [tex]c^2 = a^2 + b^2[/tex]
=> [tex]46^2 = 9^2 + 40^2[/tex]
=> 2116 = 81 + 1600
=> 2116 ≠ 1681
So, this is not a Pythagorean Triplet
2) 16, 30 and 34
=> [tex]c^2 = a^2 + b^2[/tex]
=> [tex]34^2 = 16^2 + 30^2[/tex]
=> 1156 = 256 + 900
=> 1156 = 1156
No need to check more as we've found the one which is not a Pythagorean Triplet.
Answer:
[tex] \boxed{ \huge{ \boxed{ \sf{ \blue{9 , \: 40 \:, 46 \: }}}}}[/tex]
Option A is the correct option.
Step-by-step explanation:
1. Let h , p and b are the hypotenuse , perpendicular and base of a right - angled triangle respectively.
From Pythagoras theorem,
[tex] \sf{ {h}^{2} = {p}^{2} + {b}^{2} }[/tex]
Here, we know that the hypotenuse is always greater than perpendicular and base,
h = 46 , p = 40 , b = 9
⇒[tex] \sf{ {46}^{2} = {40}^{2} + {9}^{2} }[/tex]
⇒[tex]2116 = 1600 + 81[/tex]
⇒[tex] \sf{2116 ≠ 1681}[/tex]
Thus , the relation [tex] \sf{ {h}^{2} = {p}^{2} + {b}^{2} }[/tex] is not satisfied by h = 46 , p = 40 , b = 9
So, The set of numbers 9 , 40 , 46 is not Pythagorean triple.
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2. 16 , 30 , 34
h = 34 , p = 30 , b = 16
[tex] \sf{ {h}^{2} = {p}^{2} + {b}^{2} }[/tex]
⇒[tex] \sf{ {34}^{2} = {30}^{2} + {16}^{2} }[/tex]
⇒[tex] \sf{1156 = 900 + 256}[/tex]
⇒[tex] \sf{1156 = 1156}[/tex]
The relation [tex] \sf{ {h}^{2} = {p}^{2} + {b}^{2} }[/tex] is satisfied by the particular values of h , p and b i.e h = 34 , p = 30 , b = 16
So, the set of numbers 16 , 30 , 34 is a Pythagorean triple.
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3. 10, 24 , 26
h = 26 , p = 24 , b = 10
[tex] \sf{ {h}^{2} = {p}^{2} + {b}^{2} }[/tex]
⇒[tex] \sf{ {26}^{2} = {24}^{2} + {10}^{2} }[/tex]
⇒[tex] \sf{676 = 576 + 100}[/tex]
⇒[tex] \sf{676 = 676}[/tex]
The relation [tex] \sf{ {h}^{2} = {p}^{2} + {b}^{2} }[/tex] is satisfied by the particular values of h , p and h i.e h = 26 , p = 24 , b = 10
So, the set of numbers 10, 24 , 26 is the Pythagorean triple.
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4. 50 , 120 , 130
h = 130 , p = 120 , b = 50
[tex] \sf{ {h}^{2} = {p}^{2} + {b}^{2} }[/tex]
⇒[tex] \sf{ {130}^{2} = {120}^{2} + {50}^{2} }[/tex]
⇒[tex] \sf{16900 = 14400 + 2500}[/tex]
⇒[tex] \sf{16900 = 16900}[/tex]
The relation [tex] \sf{ {h}^{2} = {p}^{2} + {b}^{2} }[/tex] is satisfied by the particular values of h , p and b i.e h = 130 , p = 120 , b = 50
So, the set of numbers 50, 120 , 130 is the Pythagorean triple.
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In this way, to satisfy the Pythagoras Theorem , the hypotenuse ( h ) , perpendicular ( p ) and the base ( b ) of a right - angles triangle should have the particular values in order. These values of h , p and b are called Pythagorean triple.
Hope I helped!
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