Pythagorean Triplet whose one no is 8 with method
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2
If one number is 8, then we can take other 2 numbers as 6 and 10
So by Pythagoras Theorem
H^2=B^2+P^2
10^2=6^2+8^2
100=36+64
100=100
As LHS= RHS
So the triplet (6,8,10) is forming Pythagoras
So by Pythagoras Theorem
H^2=B^2+P^2
10^2=6^2+8^2
100=36+64
100=100
As LHS= RHS
So the triplet (6,8,10) is forming Pythagoras
chetna2005:
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7
Pythagorean Triplet are in form of:
2 n , n² + 1 and n²- 1
Here,
2n = 8
n = 4
So
n²+1 = 4²+1 = 17
n²-1 = 4²-1 = 15
- Pythagorean Triplet = 8 , 15 and 17.
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