Can Pythagoras' Theorem Prove this Right Triangle Hypotenuse Equation?

In summary, using the given values for u and v, we can show that the hypotenuse w of a right triangle with legs u and v is equal to 2 times the sum of the squares of m and n, divided by the product of m-n and n. This can be proven by substituting the values into the Pythagorean theorem and simplifying the equation.
  • #1
mathdad
1,283
1
A right triangle is given. One leg is u units and the other leg is v units. The hypotenuse is given to be w units.

If u = [2(m + n)]/n and v = 4m/(m - n), show that
w = [2(m^2 + n^2)/(m - n)n.

Must I square u and v to show that
w = [2(m^2 + n^2)/(m - n)n?
 
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  • #2
According to Pythagoras, we should have:

\(\displaystyle u^2+v^2=w^2\)

Plug in the given values:

\(\displaystyle \left(\frac{2(m+n)}{n}\right)^2+\left(\frac{4m}{m-n}\right)^2=\left(\frac{2(m^2+n^2)}{(m-n)n}\right)^2\)

I would multiply through by \(\displaystyle (n(m-n))^2\) to get:

\(\displaystyle (2(m+n)(m-n))^2+(4mn)^2=(2(m^2+n^2))^2\)

\(\displaystyle (2(m^2-n^2))^2+(4mn)^2=(2(m^2+n^2))^2\)

Continuing, you will obtain an identity. :D
 
  • #3
MarkFL said:
According to Pythagoras, we should have:

\(\displaystyle u^2+v^2=w^2\)

Plug in the given values:

\(\displaystyle \left(\frac{2(m+n)}{n}\right)^2+\left(\frac{4m}{m-n}\right)^2=\left(\frac{2(m^2+n^2)}{(m-n)n}\right)^2\)

I would multiply through by \(\displaystyle (n(m-n))^2\) to get:

\(\displaystyle (2(m+n)(m-n))^2+(4mn)^2=(2(m^2+n^2))^2\)

\(\displaystyle (2(m^2-n^2))^2+(4mn)^2=(2(m^2+n^2))^2\)

Continuing, you will obtain an identity. :D

Will do so. Thanks.
 

1. What is a right triangle?

A right triangle is a type of triangle that has one angle measuring 90 degrees. This angle is often referred to as the "right angle".

2. What are the sides of a right triangle?

A right triangle has three sides: the hypotenuse, which is the longest side and is opposite the right angle, and the two shorter sides, known as the legs, which form the right angle.

3. How do you calculate the missing sides of a right triangle?

The Pythagorean theorem can be used to calculate the missing sides of a right triangle. It states that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

4. What are the trigonometric ratios used in right triangles?

The three main trigonometric ratios used in right triangles are sine, cosine, and tangent. These ratios represent the relationship between the sides of a right triangle and its angles.

5. How is a right triangle used in real life?

Right triangles are commonly used in fields such as engineering, architecture, and physics to calculate distances, angles, and forces. They are also used in navigation, surveying, and map making.

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