Chapter 8.3, Problem 18WE

To determine

To show: a triangle with sides of lengths 2xy, $x^2-y^2$ and $x^2+y^2$ are the sides of a right triangle.

Explanation of Solution

Given Information: A triangle with sides 2xy, $x^2-y^2$ and $x^2+y^2$ with x>y

Any triangle is said to be right triangle, if one of the three angles is 90⁰. This implies that, it satisfies the Pythagoras theorem. This can be expressed for the given triangle with sides a, b and c as

  $c^2=a^2+b^2$........ (1)

  

If the triangle is acute, then $c^2<a^2+b^2$

If the triangle is obtuse, then $c^2>a^2+b^2$

In the given case, three sides are given as 2xy, $x^2-y^2$ and $x^2+y^2$ . Here, since x>y, the term $x^2+y^2$ is the greatest among the three sides. So, the relation between the sides has to be checked.

Considering the right hand side of the expression in (1) and expanding it based on the algebraic expansion ${\left(a+b\right)}^2=a^2+b^2+2ab$ , it becomes

  $\begin{array}{c}{\left(x^2-y^2\right)}^2+{\left(2xy\right)}^2=x^4+y^4-2x^2{y}^2+4x^2{y}^2\\ =x^4+y^4+2x^2{y}^2\end{array}$

Now, expanding the left-hand side of the expression in (1) for the given sides,

  ${\left(x^2+y^2\right)}^2=x^4+y^4+2x^2{y}^2$

It can be seen that, the given sides satisfies the Pythagoras theorem in (1) and hence it is proved that they always form the sides of a right triangle with $x^2+y^2$ as the hypotenuse.