Thread: 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

Originally Posted by nsabs

**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.

Mr F says: This area of this triangle is constant so it's rate of change is zero.

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.

Mr F says: Where does the trapezoid come from??

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.
[snip]

..

The final height will be 30mm. However the height is growing from 0-30mm.

The trapazoid is coming from when say the height = 10mm. I am goign to have a base of 50mm, 2 equal sides, and a top horizontal or some length. The top horizontal will decrease from 50mm to 0mm as the shape aproaches the final triangle shape.

Sorry if thats confusing, its hard without being able to draw pictures.