Which of the given numbers is not a Pythagorean triplets. *

Mathematics

1 answer:

Ludmilka [50]3 years ago

8 0

Step-by-step explanation:

By Pythagoras' Theorem,

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

where c is always the largest number.

a and b can be interchangeable between the 2nd largest and the 3rd largest numbers.

Given a = 8, b = 15 and c = 17,

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

Since c^2 = a^2 + b^2 , 8 , 15 and 17 are pythagorean triplets.

Now let's move on to 9, 40 and 41.

${a}^{2} + {b}^{2} = {9}^{2} + {40}^{2} \\ = 81 + 1600 \\ = 1681 \\ \\ {c}^{2} = {41}^{2} \\ = 1681$

Since c^2 = a^2 + b^2 , 9 , 40 and 41 are pythagorean triplets.

Last let's move on to 4,7 and 8.

${a}^{2} + {b}^{2} = {4}^{2} + {7}^{2} \\ = 16 + 49 \\ = 65 \\ \\ {c}^{2} = {8}^{2} \\ = 64$

Since a^2+b^2 IS NOT EQUAL to c^2, 4,7 and 8 ARE NOT pythagorean triplets.

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(I have attached an example for reference)

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