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Deffense [45]
2 years ago
7

Pick out the set of numbers that is not Pythagorean triple

Mathematics
2 answers:
Ann [662]2 years ago
7 0

Answer:

<h3>\boxed{ \huge{ \boxed{ \sf{ \blue{9  , \: 40 \:, 46 \: }}}}}</h3>

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,

\sf{ {h}^{2}  =  {p}^{2}  +  {b}^{2} }

Here, we know that the hypotenuse is always greater than perpendicular and base,

h = 46 , p = 40 , b = 9

⇒\sf{ {46}^{2}  =  {40}^{2}  +  {9}^{2} }

⇒2116 = 1600 + 81

⇒\sf{2116  ≠ 1681}

Thus , the relation \sf{ {h}^{2}  =  {p}^{2}  +  {b}^{2} } is not satisfied by h = 46 , p = 40 , b = 9

So, The set of numbers 9 , 40 , 46 is not Pythagorean triple.

------------------------------------------------------

2. 16 , 30 , 34

h = 34 , p = 30 , b = 16

\sf{ {h}^{2}  =  {p}^{2}  +  {b}^{2} }

⇒\sf{ {34}^{2} =  {30}^{2}  +  {16}^{2} }

⇒\sf{1156 = 900 + 256}

⇒\sf{1156 = 1156}

The relation \sf{ {h}^{2}  =  {p}^{2}  +  {b}^{2} } 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.

------------------------------------------------------

3. 10, 24 , 26

h = 26 , p = 24 , b = 10

\sf{ {h}^{2}  =  {p}^{2}  +  {b}^{2} }

⇒\sf{ {26}^{2}  =  {24}^{2}  +  {10}^{2} }

⇒\sf{676 = 576 + 100}

⇒\sf{676 = 676}

The relation \sf{ {h}^{2}  =  {p}^{2}  +  {b}^{2} } 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.

-----------------------------------------------------

4. 50 , 120 , 130

h = 130 , p = 120 , b = 50

\sf{ {h}^{2}  =  {p}^{2}  +  {b}^{2} }

⇒\sf{ {130}^{2}  =  {120}^{2}  +  {50}^{2} }

⇒\sf{16900 = 14400 + 2500}

⇒\sf{16900 = 16900}

The relation \sf{ {h}^{2}  =  {p}^{2}  +  {b}^{2} } 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.

-----------------------------------------------------

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!

Best regards!!

Zielflug [23.3K]2 years ago
6 0

Answer:

\huge\boxed{9,40,46}

Step-by-step explanation:

<u><em>Let's check it using Pythagorean Theorem:</em></u>

c^2 = a^2 + b^2

Where c is the longest sides, a and b are rest of the 2 sides

1) 9 , 40 , 46

=> c^2 = a^2 + b^2

=> 46^2 = 9^2 + 40^2

=> 2116 = 81 + 1600

=> 2116 ≠ 1681

So, this is not a Pythagorean Triplet

2) 16, 30 and 34

=> c^2 = a^2 + b^2

=> 34^2 = 16^2 + 30^2

=> 1156 = 256 + 900

=> 1156 = 1156

No need to check more as we've found the one which is not a Pythagorean Triplet.

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