Please help with formula

**oxxxis** · January 3rd 2011, 02:23 PM

I have quadrangle with coordinates x,y (3, 4) (2, 5) (8, 3) (4, 2), can someone explain how to use them in this formula
Centroid - Wikipedia, the free encyclopedia to calculate centroid.

I know how to get the area, but I have no idea how to use these coordinates in this formula. My english my not be perfect, in this case by quadrangle i mean an irregular figure where every angle may differ.

Thank you

**Also sprach Zarathustra** · January 3rd 2011, 02:36 PM

You don't need that formula...

You have to find the center quadrangle, start with drawing that shape on x-y coordinates and then try to find the center point(it includes x and y)

**oxxxis** · January 3rd 2011, 02:41 PM

I tried to divide it in two triangles, get their centroid, then (x1,y1)+(x2,y2), but it was wrong, i found this formula and my tutor told me that this is correct, but now I have to figure this out. The problem is with Xi+1, because I understand that this is sum of all these together. When i insert the last x, what should i insert for Yi+1? the first y in list?

**Archie Meade** · January 3rd 2011, 03:07 PM

Originally Posted by oxxxis

I have quadrangle with coordinates x,y (3, 4) (2, 5) (8, 3) (4, 2), can someone explain how to use them in this formula
Centroid - Wikipedia, the free encyclopedia to calculate centroid.

I know how to get the area, but I have no idea how to use these coordinates in this formula. My english my not be perfect, in this case by quadrangle i mean an irregular figure where every angle may differ.

Thank you

First, choose a point $\left(x_0,\;y_0\right)$ and label the remaining points in counterclockwise fashion.

For example:

$\left(x_0,\;y_0\right)=(8,\;3)$

$\left(x_1,\;y_1\right)=(2,\;5)$

$\left(x_2,\;y_2\right)=(3,\;4)$

$\left(x_3,\;y_3\right)=(4,2)$

$\left(x_4,\;y_4\right)=\left(x_0,\;y_0\right)$

Then

$\displaystyle\ A=\frac{1}{2}\sum_{i=0}^{n-1}\left[x_i\;y_{i+1}-x_{i+1}\;y_i\right]=$

$\frac{1}{2}\left[x_0\;y_1-x_1\;y_0\right]+\frac{1}{2}\left[x_1\;y_2-x_2\;y_1\right]+\frac{1}{2}\left[x_2\;y_3-x_3\;y_2\right]+\frac{1}{2}\left[x_3\;y_0-x_0\;y_3\right]$

Similarly for $C_x,\;\;C_y$

**oxxxis** · January 3rd 2011, 03:13 PM

Thank you very much!

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