Question

Prove that when x is greater than one a triangle with side lengths a equals x squared minus one and c equals x squared plus one is a right angle. Use the Pythagorean Theorem and the given side lengths to create an equation use the equation to show that this triangle follows the rule describing right triangles. Explain your steps

Answer 1

Using the Pythagorean Theorem,

$a^2+b^2=c^2$

(See attachment)

Let us substitute the values and solve for b in terms of $x$.

$(x^2-1)^2+b^2=(x^2+1)^2$

Grouping the  $x$ terms on one side gives,

$b^2=(x^2+1)^2-(x^2-1)^2$

We now apply difference of two squares to obtain,

$b^2=((x^2+1-(x^2-1))(x^2+1+(x^2-1))$

$b^2=((x^2+1-x^2+1))(x^2+1+x^2-1))$

$b^2=(2)(2x^2)$

$b^2=4x^2$

$b=\sqrt{4x^2}$

$b=2x$,

Testing for some few values greater than 1, we can generate the Pythagorean triples as follows;

When $x=2$

$a=2^2-1=3$

$b=2(1)=2$

$c=2^2 +1=5$

When $x=3$

$a=3^2 -1=8$

$b=2(3)=6$

$c=3^2 +1=10$