change in area over an isosceles triangle

**posted this a minute ago but in the wrong forum section***

I am trying to determine the formula for dA/dH for an isosceles triangle with a base of 50mm and a height of 30mm. I think it has something to do with the changing area of a trapezoid as the angles from teh base to the sides must remain the same.

Using that idea the A(trap) = h[(b1+b2)/2] but I'm not sure how to relate the top surface (b2) with the change in h.

Any help would be appreciated!

Thanks,

Nick