Can You Prove This Right Triangle Theorem Involving Variables u, v, and w?

[mathdad]

A right triangle is given. The legs are u and v. The hypotenuse is w.

If u = 2(m + n)/n, and v = 4m/(m - n), show that

w = 2(m^2 + n^2)/(m - n)n

Does this question involve a^2 + b^2 = c^2?

 

[MarkFL]

Yes, showing that:

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

will suffice to show that the triangle is a right triangle. :D

 

[HallsofIvy]

But the problem is NOT "to show that the triangle is a right triangle". We are given that the triangle is a right triangle. So we know that $$w^2= u^2+ v^2$$ and that u = 2(m + n)/n, and v = 4m/(m - n).

So [math]w^2= 4(m+ n)^2/n^2+ 16m^2/(m-n)^2[/math]

To combine those to fractions we need to get a "common denominator", [math]n^2(m-n)^2[/math]:
[math]w^2= \frac{4(m+n)^2(m-n)^2}{n^2(m-n)^2}+ \frac{16m^2n^2}{n^2(m-n)^2}[/math]

Finish that.

 

[mathdad]

Do I square u and v to show it is equivalent to w^2?

 

[MarkFL]

From what we are given in the problem, we may state:

$$w=\sqrt{u^2+v^2}$$

Next, plug in the given values for u and v and then see if you can get to the required value for w. :D

 

[mathdad]

From what we are given in the problem, we may state:

$$w=\sqrt{u^2+v^2}$$

Next, plug in the given values for u and v and then see if you can get to the required value for w. :D

I will definitely try this when time allows.

 

[I like Serena]

Supporting what HallsofIvy already said, we can choose to use the implicit information that $w^2=u^2+v^2$ or not.
But what we really need to do is prove that the numerical value of the circumference is the same as the numerical value of the area. Ultimately we can only do that by substituting and verifying.

 

[mathdad]

Thank you everyone.