# Area between Triangle and Rectangle

1. Oct 8, 2011

### orchidthief

1. The problem statement, all variables and given/known data

Consider a large triangle, the tip is located at the origo x=0, it is sloped at an angle θ and -θ relative to the x-axis, relative to the x-axis, its dimensions in x can be considered infinite.

A large stripe/strip/band is placed on top of the triangle but perpendicular to the x axis so to speak, it has a width b and it's center is located at x=0 and it can be considered infinite in the y direction. (So one edge of the stripe at the start is at x=b/2 and the other at x=-b/2.)

This large stripe can be shifted in the x direction to gradually cover a larger and larger trapezoidal area of the triangle.

The exercise is to find a generalised expression for the area shared by the triangle and the stripe as a function of x.

2. The attempt at a solution

What I deduced was that this was a piecewise solution, while x<b/2 the area they share is:

A(x)= tanθ(x+b/2)^2 (a basic area of a triangle, that gives it an area when x=0)

And when x>b/2 it

A(x)=1/2*b*(q+p) (the area of a trapezoid)

Where q=2tanθ(x-b/2) and p=2tanθ(x+b/2) giving me a final expression in this part of:

A(x)=2bx*tanθ

Now my question is, the actual wording of the problem makes it sound like it should be possible to find one elegant expression covering the x>0 regardless of the zone, but I'm just completely lost in how one such might be found. Any help would be nice.

Last edited: Oct 8, 2011
2. Oct 8, 2011

### LCKurtz

So one side of your triangle is a line through the origin at angle θ. What is a second side, the x axis? If it's a triangle, what is the third side? And your rectangle, what it its height or are you saying it is a vertical strip. I'm having trouble picturing the trapezoid to which you refer. Even so, once you explain that, I think almost certainly you will have a multiple piece formula for your answer.

3. Oct 8, 2011

### orchidthief

Yeah the tip of the triangle is at origin. The axis down the middle of the triangle is the x-axis, and then one side of the triangle running at an angle θ and one at -θ. Can see how that would be confusing. It can be considered infinite in the x direction.

Yes, the rectangle is a stripe that is infinite in the y direction, and has width b in the x direction, centered around the y axis at the start, but then the whole thing can be shifted in the x direction. At the start the tip and the center of the stripe are both at x=0, so the area they share is just a small triangle, but as the stripe is moved in the x direction, eventually the area they share goes from being a triangle to a trapezoid. (When the stripe has moved b/2 in the x direction and the triangle starts emerging on the other side of the stripe.)

Btw, I realise this is in the precalc forum, since it seems like a fairly straightforward geometrical problem, but if calculus can be helpful then that's fair game as well.

Last edited: Oct 8, 2011
4. Oct 8, 2011

### LCKurtz

OK. With that clarification, I get the same answers you have. Good work. The answer is inherently a two piece formula and the only way to express it as a "single" formula would be to introduce an artifice like a unit step function into the expression.

5. Oct 8, 2011

### orchidthief

Yeah, didn't really have any doubts about the correctness of the answers I got, more about consolidating it into a single expression. How would one go about trying to introduce such a unit step artifice? Heavisides step function I take it? Some hints and pointers would be useful, only having to deal with one expression would certainly simplify the rest of the homework set.

6. Oct 8, 2011

### LCKurtz

Let's call the Heaviside unit step function u(x). If you want a function F(x) to equal f(x) for x < a and g(x) for x > a you could write:

F(x) = f(x) + u(x-a)(g(x) - f(x)).

When x < a the second term is 0 and when x > a the second term subtracts out the f and adds in the g. You can add a second similar term if it changes formula again for some other value of x greater than a.

7. Oct 9, 2011

### orchidthief

That makes good sense. Thanks a bunch for the help.