Question

suppose M and n are any two numbers. If m^2 - n^2 , 2mn and m^2 + n^2 are the three sides of a triangle , then show that its a right angled and hence write any two pairs of pythagorean triplet.

Answer 1

please mark as direllent..............

Answer 2

Answer:

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

AB = m² - n²,

BC = 2mn,

CA = m² + n²,

$AB^2=(m^2-n^2)^2=m^4-2m^2n^2+n^4$

( Because, (a-b)² = a² - ab + b² ),

$BC^2=(2mn)^2=4m^2n^2$

$CA^2=(m^2+n^2)^2=m^4+2m^2n^2+n^4$

Since,

$AB^2+BC^2=m^4-2m^2n^2+n^4+4m^2n^2=m^4+n^4+2m^2n^2$

$AB^2+BC^2=CA^2$

We know that in right triangle sum of squares of two sides is equal to the square of third side.

⇒ Triangle ABC is right triangle.

Hence, proved.