Question

Prove that when x> 1, a triangle with side lengths a = x2 - 1, b = 2x, and C = x2 + 1 is a right triangle. 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

Step-by-step explanation:

a = x^2 - 1

b = 2x

c = x^2 + 1

The Pythagorean theorem states

a^2 + b^2 = c^2

Let's find a^2 and b^2 and add them to get a^2 + b^2:

a^2 = (x^2 - 1)^2 = x^4 - 2x^2 + 1

b^2 = (2x)^2 = 4x^2

a^2 + b^2 = x^4 - 2x^2 + 1 + 4x^2 = x^4 + 2x^2 + 1

Now let's find c^2:

c = x^2 + 1

c^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1

We see that both a^2 + b^2 and c^2 equal x^4 + 2x^2 + 1, so we have shown that the triangle is a right triangle.

Answer 2

We can see that a² + b² = c² = x⁴ + 2x² + 1, indicating that the triangle is a right triangle.

What is the definition of a right-angle triangle?

It's a form of a triangle with one 90-degree angle that follows Pythagoras' theorem and can be solved using the trigonometric function.

According to Pythagoras, the sum of the squares of two sides equals the square of the longest side.

Given data;

a = x² - 1

b = 2x

c = x² + 1

According to the Pythagorean theorem;

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

Let's find a^2 and b^2 and add them to get a^2 + b^2:

$\rm a^2 = (x^2 - 1)^2 \\\\ a^2 = x^4 - 2x^2 + 1 \\\\ b^2 = (2x)^2 \\\\ b^2 = 4x^2$

Left hand side:

$\rm a^2 + b^2 = x^4 - 2x^2 + 1 + 4x^2 \\\\ a^2 + b^2 =x^4 + 2x^2 + 1$

Right hand side:

$\rm c^2 = (x^2 + 1)^2 \\\\ c^2 = x^4 + 2x^2 + 1$

L.HS.=R.H.S

Hence,the given triangle is a right angled triangle.

To learn more about right-angle triangles, refer to https://brainly.com/question/3770177.

#SPJ2