Public Overrides Function GetArea() As Single Return Math.Sqrt(ShapeWidth * ShapeHeight) End FunctionWhy do I need to apply the square root?
Using Square Root to calculate Area
Author
To calculate the area of a 2D square or rectangle, you multiply the width by height. However, unless I then get the square root of this value the result is wrong!
Why do I need to apply the square root?
Why do I need to apply the square root?
Because
ShapeWidth * ShapeHeight is NOT Sqrt(ShapeWidth * ShapeHeight)
remain :
Sqrt(x) * Sqrt(x) = x
ShapeWidth * ShapeHeight is NOT Sqrt(ShapeWidth * ShapeHeight)
remain :
Sqrt(x) * Sqrt(x) = x
Why do you think that the value is wrong? That is the correct calculation for the area of a rectangle.
Author
Hmm ... does it have something to do with my coordinates? Perhaps related, why would I also need to square root the result of a distance calculation?
Public Function Distance(ByVal Point1 As PointF, ByVal Point2 As PointF) As Single Dim dx As Single = Point1.X - Point2.X Dim dy As Single = Point1.Y - Point2.Y Return Math.Sqrt(dx * dx + dy * dy) End Function
Well this is just the application of the Pythagorean theorem to calculate the distance between two points (in euclidian co-ordinates to be precise). And no, it's not (directly) related to your rectangle-area-calculation in your original post.
Work it out with some graph paper. For a 2x2 square the area is 4, two times two is four, your code is wrong.
OK the Area of a 4 sided figure, Square or Rectangle is Width * height.
If you are using coordinates you need to calculate the distance of the width and the height.
IE:
Width will be W = Sqrt((X2-X1)^2 + (Y2-Y1)^2)
Height will be H = Sqrt((X3-X1)^2 + (Y3-Y1)^2)
So area is W * H
OR
Sqrt((X2-X1)^2 + (Y2-Y1)^2) * Sqrt((X3-X1)^2 + (Y3-Y1)^2)
A more robust method that will work for any closed figure is to calculate by sum of latitudes and departures : http://www.wikihow.com/Calculate-the-Area-of-a-Polygon
If you are using coordinates you need to calculate the distance of the width and the height.
IE:
Width will be W = Sqrt((X2-X1)^2 + (Y2-Y1)^2)
Height will be H = Sqrt((X3-X1)^2 + (Y3-Y1)^2)
So area is W * H
OR
Sqrt((X2-X1)^2 + (Y2-Y1)^2) * Sqrt((X3-X1)^2 + (Y3-Y1)^2)
A more robust method that will work for any closed figure is to calculate by sum of latitudes and departures : http://www.wikihow.com/Calculate-the-Area-of-a-Polygon
Quote:Original post by Magpie
OK the Area of a 4 sided figure, Square or Rectangle is Width * height.
If you are using coordinates you need to calculate the distance of the width and the height.
IE:
Width will be W = Sqrt((X2-X1)^2 + (Y2-Y1)^2)
Height will be H = Sqrt((X3-X1)^2 + (Y3-Y1)^2)
So area is W * H
OR
Sqrt((X2-X1)^2 + (Y2-Y1)^2) * Sqrt((X3-X1)^2 + (Y3-Y1)^2)
You don't need to do it that way, but you certainly can.
Quote:A more robust method that will work for any closed figure is to calculate by sum of latitudes and departures : http://www.wikihow.com/Calculate-the-Area-of-a-Polygon
Yes, that's the method I would use, although that page doesn't explain at all why it works.
Author
Quote:Original post by Magpie
OK the Area of a 4 sided figure, Square or Rectangle is Width * height.
If you are using coordinates you need to calculate the distance of the width and the height.
IE:
Width will be W = Sqrt((X2-X1)^2 + (Y2-Y1)^2)
Height will be H = Sqrt((X3-X1)^2 + (Y3-Y1)^2)
So area is W * H
Ahaha - Yes I am using a 2D coord system. This must be why I need the square root in there, because it needs to be the distance otherwise it doesn't make sense when I convert it to real world measurements.
I can expand my actual function - this is what it is doing:
Area = Math.Sqrt((m_MaxPoint.X - m_MinPoint.X) * (m_MaxPoint.Y - m_MinPoint.Y))
Does that sit with what you have said Magpie?
This topic is closed to new replies.
Advertisement
Popular Topics
Advertisement
