Height of a right triangle without area

How do I find the height of a triangle without the area??

$$\boxed{h_c=\dfrac{a*b}{c}}$$

This link might help: http://www.wikihow.com/Find-the-Height-of-a-Triangle 

u divide ur answer they gave you by the other amount on the triangle. then divide by 1/2

Really Heureka??  

 If I had more time I would try and work out how that could be demonstrated.

I did try a little but I had no success.  

I don't remember seeing anything like it before.   

Can you show me a proof of this?

Assuming you know the length of all three sides you could make use of Heron's formula.  If the sides are of length a, b and c, and b is the base then 

$$\frac{1}{2}\times b\times h=\sqrt{s(s-a)(s-b)(s-c)}$$

where h is the height and s is the semi-perimeter (=(a+b+c)/2).  This can be rearranged to find h.

.

See the question was: Height of a right triangle without area :

$$\\(1) \quad Area = \dfrac{h_c*c}{2} \\\\ (2) \quad Area = \dfrac{a*b}{2}\\\\ \dfrac{h_c*c}{2} = \dfrac{a*b}{2} \\\\ h_c*c = a*b \\\\ h_c= \dfrac{a*b}{c} \\\\$$

Thanks Heureka,

Sorry I did not notice that we were talking only of right triangles.  

In my defence there is no 'right' in the question it is only in the title.   

ALSO note:

The height that is being found here is the perpendicular distance from the hypotenuse to the right angle vertex.

Here's another proof of this....assuming that "c" is the hypoteneuse and the height is drawn from the right angle to the hypoteneuse ..  so we have

(1/2) ab sin(90)  = (1/2)ch

 ab   = ch

 h  = ab / c