Find the line that gives constant area in odd shape

[MD11]

Hi,

This is my first post here. I think my question is relevant to the area of calculus hence posting here.

In short: I suck at mathematics after years of trying, and I have a problem that I hope someone can help me with.

Full disclosure: this is for a commercial project, so if that is not allowed, my apologies in advance.

Imagine I have an irregular 5-sided closed shape (Figure A). I draw a line horizontally through the shape, cutting it into two areas. Let's shade the bottom area, and call this area A.

Imagine I now rotate the shape (Figure B), but my line remains in the same position horizontally on the page.

How do I compute the new position of the line in order that area A remains constant? How would I compute this line for any angle?

Many thanks in advance.

MD11.

 

[Vorde]

Well, I have an algebra-heavy way. There might be an easier way but I can't think of it right now.Lets say that the shape in figure A is a rectangle of width $W$ and height $H$ that has a triangle of width $w$ and height $h$ cut out of it. This would mean that the total area of the shape is $(W\cdot H) - (\frac{1}{2}\cdot w\cdot h)$

Assuming that the 'cut-out triangle' will always be below the shaded line in Figure A, the area of the shaded figure when the top line is at height $L_1$ is $(W\cdot L_1) - (\frac{1}{2}\cdot w\cdot h)$, and let's call this number $A$.

Now moving over to the second figure, Assuming that the 'cut-out triangle' is always over the line in this figure. The question is what is the height $L_2$ of the line such that the area under the line is equal to $A$.
Realize the height in the first figure is now the width of the second figure, and the area of the shaded region in figure 2 is $H\cdot L_2$, and we want to set it equal to $A$.
This leads to $(W\cdot L_1)-(\frac{w\cdot h}{2}) = H\cdot L_2$
Solving for $L_2$, which is what you are trying to find, you get: $L_2=\frac{(W\cdot L_1)-(\frac{w\cdot h}{2})}{H}$ where $W$ is the width in figure 1, $L_1$ is the height of the line in figure 1, $w$ is the width of the cut-out triangle, $h$ is the height of the cut-out triangle, $H$ in the height in figure 1, and $L_2$ is what you are looking for: the height of the line in the second figure.

I'm not sure how much this would change if you both removed the restriction of the position of the triangle and allowed the triangle to be at any angle. I'm not positive it'd require calculus, but it would definitely need trig, and it will definitely be more complicated, so I'll save it until I know you need it.

 

[MD11]

Hi,

Thanks for the reply!

I would need to be able to compute the line for any angle of the box, with area A remaining constant (the line need not be perpendicular to the edges of the box).

Thanks again!

MD11.

 

[SteamKing]

In the absence of using algebra, sometimes iteration can determine the position of the line such that the area A remains constant with rotation of the figure.

What is the application?