Question

$\huge\mathtt\purple{QUESTION}$
WRITE A PYTHAGOREAN TRIPLET WHOSE ONE NUMBER IS .
$(i) \: 14 \: (ii) \: 22$
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Answer 1

Answer:

(i) 14,48,50 (ii) 22,120,122

Step-by-step explanation:

(i) Let's assume 2 m = 14 therfore m = 7

Now m^2+1=7^2+1=49+1=50

And m^2-1=7^2-1=49-1=48

14^2+48^2 = 196 + 1304 = 2500 = 50^2

Hence the triplet is 14,48 and 50.

(ii) For any natural numbers m>1,2m,m^2-1,m^2+1 form a Pythagorean triplet.

If we take m^2+1=22, then m^2=21.

The value of m will not be an Integer

If we take m^2-1=22,then m^2=23

Again the value of m will not be an Integer.

Let 2m = 22

m = 22/2

m = 11

2m = 2×11 = 22

m^2-1 = 11^2-1 = 121-1 = 120

m^2+1 = 11^2+1 = 121+1 = 122.

Therfore,the Pythagorean triplets are 22,120 and 122.

Hope this helps you.

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Answer 2

Answer:

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