And if not, why ?

Thanks.

FastChef

And if not, why ?

Thanks.

11 months ago

All urls found in this thread:

https://en.wikipedia.org/wiki/Planimeter

https://en.wikipedia.org/wiki/Shoelace_formula

https://en.wikipedia.org/wiki/Planimeter

https://en.wikipedia.org/wiki/Shoelace_formula

viagrandad

@FastChef

of course. the angles and side lengths of a polygon completely define it. Here's an exercise as a sort of proof: Can you change the area of a polygon without changing the sidelengths or the angles?

11 months ago

RavySnake

@viagrandad

Here's an exercise as a sort of proof

Nope assuming the number of side is the same.

11 months ago

VisualMaster

No.

Well, it depends. Most of the time no. If you gave me a polygon with some parallel sides maybe. It's area would have to be defined in terms of its side which comprises itself in varying magnitudes in the other sides the most methinks.

11 months ago

Harmless_Venom

@FastChef

Yes. Your question is inverse-equivalent to asking : "Are there two polygons with the same angles&edges with different areas?" which is obviously not true.

11 months ago

New_Cliche

Yes. Divide the polygon into triangles. The area of any one triangle is:

A = a x b x sin (alpha) / 2, where a and b are the length of any two sides and alpha is the angle between them.

If you have any triangles without already known parameters, you can use basic geometry and the laws of sins and cosines to find them.

11 months ago

GoogleCat

Nad if not, why ?

Nahkts.

11 months ago

Boy_vs_Girl

@FastChef

This field is called computational geometry. Textbooks exist and contain the answers to your questions.

11 months ago

Crazy_Nice

Ok thanks for replies.

11 months ago

Methshot

@FastChef

yes. you could turn the whole thing into triangles and there's a formula for that. if it was just the angles you couldn't solve.

11 months ago

Boy_vs_Girl

@SomethingNew

Polygons are uniquely defined by their points, if you know all angles and edges you know all the points.

11 months ago

MPmaster

@FastChef

Sure, it's easy by employing the chemist's integral.

Just draw the polygon on a piece of paper, cut it out and weigh it.

11 months ago

TechHater

@FastChef

if you don't have vertices then you can calculate them from the edges

than you can triangulate the polygon

than you get the sum of all triangle areas

11 months ago

Raving_Cute

@VisualMaster

if you know the lengths and angles then you can calculate all the triangles that comprise that shape.

11 months ago

Evil_kitten

@FastChef

No

you need atleast one leanght

double all sides and you have another polygon with the same angles

11 months ago

Burnblaze

@Evil_kitten

that's what OP means with edges I guess.

Also you need more than one length if you're not just discribing a triangle.

consider the a polygon with 4 90° angles and one sidelength given. There's an infinite number of rectangles that fit into that discription.

10 months ago

Ignoramus

triangulation or green's theorem

10 months ago

PurpleCharger

@5mileys

this. fix one point at (0,0), the next at (L,0) where L is the length of the first edge. Go around to figure out the (x,y) coordinates of the other points. Then use a special case of green's theorem, i.e.

10 months ago

Booteefool

@Soft_member

No, pic related is a counter example. All angles are the same, but only one edge has changed.

10 months ago