Imagine a trapezoid stacked on a rectangle, with all six of the corners rounded off. I'm trying to find it's perimeter.
Here is an unrounded trapezoid (note, this is obviously not my diagram, so the letters do not match the rectangle below)
Here is a rectangle with the corners rounded off.
The stack is such that the A of the rounded rectangle is smoothly attached to the B of the trapezoid (that is, with no overhang), like so:
Note! Please read the diagram carefully! I have relabelled the sides!
The red lines show the curved top corners, and the angles for the side points.
This is a 'zoomed in' image of the joint.
I can calculate the perimeter of the rounded rectangle using the formula on this page:
$$p = 2(a + b + pi*r)$$
And the formula for the regular/isosceles trapezoid (without rounded corners) is given on this page:
$$p = a + b + 2\times\left(\sqrt{\left(\frac{a-b}{2}\right)^2 + h^2}\right)$$
However I need to round the top two obtusely angled corners, and the side obtusely angled corners and factor that in.
The angles would be $\theta_{top} = \tan{\frac{a-b}{h}}+90$ and $\theta_{side} = \tan{\frac{h}{a-b}}+90$ in degrees, so the contributions should be $\theta\times{\pi}r$.
But plugging this into Wolfram Alpha gives me a computation time-out error, making me think it's wrong.
Knowing A, B, D, R and H what is the correct formula for the perimeter?
Best Answer
This is actually fairly simple, if you know some measurements. In total, you have a sharp cornered item, that lies in the center of a rounded item:
As is easily visible, the outer circumference is the inner circumference, but also larger by the sum of the circle segments. These circle segments sum up (from symmetry) $\alpha+\alpha+\beta+\beta+\gamma+\gamma=2\times (\alpha+\beta+\gamma)=360°$. Due to the definition you had of $\gamma=90°$ we also know that $\alpha+\beta=90°$, and from symmetry, we also know $\beta+\theta=90°$. We also can turn h' into h by substracting r (which is trivial), which we only need for area or calculating c. Which is triangonometric: $$c=\sqrt{(h'-r)^2+\left(\frac {a-d}{2}\right)^2}$$
As the result the circumference $C$ is just larger by one circle with the radius r: $$C=a+2b + 2c + d + 2 \pi r$$
Likewise the area is the inner area plus the full circle area, plus the areas of thickness r that are between the inner and outer perimeter. This boils down to $$A= a \times b +\frac 1 2 (a+d)\times h + d \times r + a \times r + 2 b \times r + 2c \times r $$
not-sharing a line?
The circumference does decidedly not change at all, if the connection is made with a gap of height g to the rounded body, so that the rounded corner has a tangential connection to the trapezoid The only thing that actually changes is the nomenclature of $h_2$ and $h_2'$: in contrast to the $h_1$ and $h_1'$ the following must be true: $$h_2'+g=h_1'=h_2+r+g=h_1+r$$
How big is $g$ then? Simple sinus solves that: $g=r\sin(beta)$, where we already know that $\beta = 90°-\theta$.
Following through with the calculations from the start, the circumference is unchanged, only when the Area is calculated, the change does matter, because it relies on the value of $h_1$ and we only have $h_2$ or $h_2'$, which we can convert into the needed value via $g$.