Question

Pick out the set of numbers that is not Pythagorean triple

Answer 1

Answer:

$\boxed{ \huge{ \boxed{ \sf{ \blue{9 , \: 40 \:, 46 \: }}}}}$

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!!

Answer 2

Answer:

$\huge\boxed{9,40,46}$

Step-by-step explanation:

Let's check it using Pythagorean Theorem:

$c^2 = a^2 + b^2$

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.