Question

m and n are real numbers and m is greater than n if m^2 + n^2 , m^2 - n^2 and 2mn are the sides of triangle then prove that the triangle is right angled triangle .Fing out two pythagorian triplets using convient values of m and n...............

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Answer 1

Answer:

Let A, B and C are the vertices of the triangle such that,

AB = m² - n²

BC = 2mn

CA = m² + n²

if we prove AB² +BC² = AC² , it means this is a right angled ∆ .

(m²-n²)² + (2mn)² = (m²+n²)²

m⁴+n⁴-2m²n² + 4m²n² = m⁴+n⁴+2m²n²

m⁴+n⁴+2m²n² = m⁴+n⁴+2m²n² [PROVED]

Now,

Let m = 4, n = 3

AB = 4² - 3² = 7

BC = 2*4*3 = 24

AC = 4²+3² = 25

7²+24² = 49+576 = 625 = = 25²

$\color{red}{Mark\: as\: Brainlist}$