MHB Can Pythagoras' Theorem Prove this Right Triangle Hypotenuse Equation?

[mathdad]

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?

 

[MarkFL]

According to Pythagoras, we should have:

$$u^2+v^2=w^2$$

Plug in the given values:

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

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

$$(2(m+n)(m-n))^2+(4mn)^2=(2(m^2+n^2))^2$$

$$(2(m^2-n^2))^2+(4mn)^2=(2(m^2+n^2))^2$$

Continuing, you will obtain an identity. :D

 

[mathdad]

According to Pythagoras, we should have:

$$u^2+v^2=w^2$$

Plug in the given values:

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

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

$$(2(m+n)(m-n))^2+(4mn)^2=(2(m^2+n^2))^2$$

$$(2(m^2-n^2))^2+(4mn)^2=(2(m^2+n^2))^2$$

Continuing, you will obtain an identity. :D

Will do so. Thanks.