Write a Pythagorean triplet with the following numbers as one of the number a,4 b,8 c,12 d,22 with the formula pls

class 8

For any natural number m, we know that

2m, \mathrm{m}^{2}-1, \mathrm{m}^{2}+1m

2

−1,m

2

+1 is a Pythagorean triplet.

i. 2m = 6

\begin{array}{l} \Rightarrow \mathrm{m}=\frac{6}{2}=3 \\ \mathrm{m}^{2}-1=3^{2}-1=9-1=8 \\ \mathrm{m}^{2}+1=3^{2}+1=9+1=10 \end{array}

⇒m=

2

6

=3

m

2

−1=3

2

−1=9−1=8

m

2

+1=3

2

+1=9+1=10

∴ (6, 8, 10) is a Pythagorean triplet.

ii. 2m = 14

\begin{array}{l} \Rightarrow \mathrm{m}=\frac{14}{2}=7 \\ \mathrm{m}^{2}-1=7^{2}-1=49-1=48 \\ \mathrm{m}^{2}+1=7^{2}+1=49+1=50 \end{array}

⇒m=

2

14

=7

m

2

−1=7

2

−1=49−1=48

m

2

+1=7

2

+1=49+1=50

∴ (14, 48, 50) is not a Pythagorean triplet.

iii. 2m = 16

\begin{array}{l} \Rightarrow \mathrm{m}=\frac{16}{2}=8 \\ \mathrm{m}^{2}-1=8^{2}-1=64-1=63 \\ \mathrm{m}^{2}+1=8^{2}+1=64+1=65 \end{array}

⇒m=

2

16

=8

m

2

−1=8

2

−1=64−1=63

m

2

+1=8

2

+1=64+1=65

∴ (16, 63, 65) is a Pythagorean triplet.

iv. 2m = 18

\begin{array}{l} \Rightarrow \mathrm{m}=\frac{18}{2}=9 \\ \mathrm{m}^{2-1}=9^{2}-1=81-1=80 \\ \mathrm{m}^{2}+1=9^{2}+1=81+1=82 \end{array}

⇒m=

2

18

=9

m

2−1

=9

2

−1=81−1=80

m

2

+1=9

2

+1=81+1=82

∴ (18, 80, 82) is a Pythagorean triplet

Step-by-step explanation:

For any natural number greater than 1,(2m,m

2

−1,m

2

+1) is Pythagorean triplets.

So, if one number is 2m, then another two numbers will be m

2

−1 and m

2

+1.

Given, one number = 4

Then Pythagorean triplets:

2m=4 or m = 2

So,

m

2

−1=(2)

2

−1=4−1=3

m

2

+1=(2)

2

+1=4+1=5

Now, (3)

2

+(4)

2

=(5)

2

Or 9 + 16 = 25.

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