write the Pythagorean triplet whose one member is 16
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Answer:
63,16,65 is an example of a pythagorean triplet with 16 as a member.
Step-by-step explanation:
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If a² + b² = c², then a = m² - n², b = 2mn, and c = m² + n²
where a, b, c are Pythagorean triplets (counting numbers) and m, n are numbers that m > n, m ≠ n, m ≥ 2 and n ≥ 1.
Here we can consider 16 as b = 2mn because both 16 and 2mn are even numbers.
2mn = 16
mn = 16 / 2 = 8
∴ m, n are factors of 8.
8 = 8 x 1
∴ We can take m as 8 and n as 1. (Not m = 1 and n = 8 as m > n.)
a = m² - n² = 8² - 1² = 64 - 1 = 63
b = 2mn = 2 x 8 x 1 = 16 (We know)
c = m² + n² = 8² + 1² = 64 + 1 = 65
∴ 63, 16 and 65 are one Pythagorean triplet.
8 = 4 x 2
∴ Consider m = 4 and n = 2.
a = 4² - 2² = 16 - 4 = 12
b = 2 x 4 x 2 = 16
c = 4² + 2² = 16 + 4 = 20
∴ 12, 16 and 20 are another Pythagorean triplet.
We can also consider 16 as a = m² - n².
16 is a multiple of 4.
Every multiples of 4 can be written as the difference of squares of two consecutive odd/even numbers. And that odd/even numbers are of next and previous numbers of 1/4 of this multiple, i. e., the average of these two consecutive odd/even numbers is 1/4 of the multiple of 4.
E. g., consider 36.
36 / 4 = 9
9 + 1 = 10
9 - 1 = 8
∴ 36 = 10² - 8²
Here,
16 / 4 = 4
4 + 1 = 5
4 - 1 = 3
∴ 16 = 5² - 3²
So we can take m as 5 and n as 3.
a = 5² - 3² = 25 - 9 = 16
b = 2 x 5 x 3 = 30
c = 5² + 3² = 25 + 9 = 34
∴ 16, 30 and 34 are another Pythagorean triplet.
But we can't consider 16 as c = m² + n² because 16 is not a sum of squares of two counting numbers.
16 = 1 + 15 ( 15 is not a perfect square.)
16 = 4 + 12 (12 is not a perfect square.)
16 = 9 + 7 (7 is not a perfect square.)
16 = 16 + 0 (0 = 0² but 0 can't be taken.)
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