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.

see explanation

Step-by-step explanation:

If the triangle is right then the square of the longest side (hypotenuse) will equal the sum of the squares of the other 2 sides.

longest side is x² + 1, then

(x² + 1)² = + 2x² + 1

The sum of the square of the other 2 sides

(x² - 1)² + (2x)²

= - 2x² + 1 + 4x² = + 2x² + 1

Hence triangle is right

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.

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb

you need to rotate this triangle about the origin 270 degrees clockwise to create the full square.