Chapter A.2, Problem 56AYU

To determine

Find the lengths of the sides of a right triangle.

Answer to Problem 56AYU

  $\boxed{\begin{array}{l}a=m^2-n^2\\ b=2mn\\ c=\left(m^2+n^2\right)\end{array}}$

Explanation of Solution

Given information:

Suppose that $m$ and $n$ are positive integers with $m>n$ . If $a=m^2-n^2,b=2mn$ and $c=m^2+n^2$ ,show that $a,b,c$ are the lengths of the sides of a right triangle.(This formula can be used to find the sides of a right triangle that are integers, such as $3,4,5;5,12,13$ andso on. Such triplets of integers are called Pythagorean triples)

Calculation:

Consider the equation,

  $a=m^2-n^2,b=2mn$

  $c=m^2+n^2$

Where $m$ and $n$ are positive integers with $m>n$

Apply Pythagorean Theorem,

  $c^2=a^2+b^2$

Where $a,b,c$ are the lengths of the sides of a right triangle.

Now substitute the given values in above expression we get,

  $\begin{array}{l}{c}^2={\left(m^2-n^2\right)}^2+{\left(2mn\right)}^2\\ c^2=\left(m^4+n^4-2m^2{n}^2\right)+\left(4m^2{n}^2\right)\\ c^2=\left(m^4+n^4+2m^2{n}^2\right)\\ c^2={\left(m^2+n^2\right)}^2\\ c=\left(m^2+n^2\right)\end{array}$

Hence, the sides of right triangle is $\boxed{\begin{array}{l}a=m^2-n^2\\ b=2mn\\ c=\left(m^2+n^2\right)\end{array}}$ .