# how to find the area of the triangles formed as a result of delaunay triangulation

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Abinaya on 3 Mar 2014
I have applied delaunay triangulation to my image. i have got 160 triangles. i need to find area of each triangle. Kindly help me with code. only after i can proceed with my prooj.i am new to matlab.
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M.S. Khan on 24 Aug 2020
How you counted the number of triangles? Please tell me the methods, regards

Roger Stafford on 3 Mar 2014
If a triangle is in two dimensional space with vertices at (x1,y1), (x2,y2), and (x3,y3), its area is given by:
area = 1/2*abs((x2-x1)*(y3-y1)-(x3-x1)*(y2-y1));
or
area = 1/2*abs(det([x1,y1,1;x2,y2,1;x3,y3,1]));
If it is in three dimensions with vertices at P1 = [x1,y1,z1], P2 = [x2,y2,z2], P3 = [x3,y3,z3], its area is:
area = 1/2*norm(cross(P2-P1,P3-P1));
These formulas are more accurate for numerical computation than those which depend on the lengths of the three sides of the triangle.
Of course you must adapt these formulas for use with your 160 triangles, either using the necessary for-loop or possibly a vectorized expression.
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Valerio Ficcadenti on 14 Aug 2022
Hi
How did you manage the polygons that are not closed? Namely, those with vertexes to infinite?
Valerio

### More Answers (1)

Christopher Rock on 2 Aug 2018
Vectorised code based off Roger Stafford's answer.
function A = triarea(t, p)
% A = TRIAREA(t, p) area of triangles in triangulation
Xt = reshape(p(t, 1), size(t)); % X coordinates of vertices in triangulation
Yt = reshape(p(t, 2), size(t)); % Y coordinates of vertices in triangulation
A = 0.5 * abs((Xt(:, 2) - Xt(:, 1)) .* (Yt(:, 3) - Yt(:, 1)) - ...
(Xt(:, 3) - Xt(:, 1)) .* (Yt(:, 2) - Yt(:, 1)));
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T is the output of delaunay()
P = [X Y]; % array with two columns that contains the coordinates of the nodes