formulas for area of triangle

area =  ah/2

there is big rectangle made up by two smaller rectangles 

the area of the first = hd  and  area of the second = he

the first rectangle is divided by two triangles. area of these triangles = hd/2

the second rectangle is divided by two triangles. area of these triangles = he/2

thanks to "ElliottMK1" for Correction

since the big triangle is made up by two triangles whose areas are hd/2 and he/2 

then area of big triangle = area of the two triangles = hd/2 + he/2 = h/2 [ d + e] = ha/2

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area = (ab/2) sin C

to proof this we know that area = ah/2 

also sin c = opposite / hypotenuse = h/b

then h = b sin C then area = (ab/2) sin C

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by use vectors

area = 

to proof this Let vectors AB and AC point respectively from A to B and from A to C

then.the area of parallelogram ABDC is  

The area of triangle ABC is half of area of parallelogram ABDC

area of  triangle ABC

now we can write tha area of triangle ABC in dot product

now we will write vector AB = (x₁,y₁) and AC = (x₂,y₂) then the area will be

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Using coordinates

let vertex A is located at the origin (0, 0) and the coordinates of the other two vertices are given by B = (xB, yB) and C = (xC, yC), then the area can be computed as ½ times the absolute value of the determinant

For three general vertices, the equation is:

In three dimensions, the area of a general triangle {A = (xA, yA, zA), B = (xB, yB, zB) and C = (xC, yC, zC)} is the Pythagorean sum of the areas of the respective projections on the three principal planes

 (i.e. x = 0, y = 0 and z = 0):

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Heron's formula.

from Pythagoras's theorem a=d+e, b²=h²+d², and c²=e²+h². then 

d = √(b²-h²) and and c² =(a-d)²+h² 

then  c²  = a²-2 X a X d + b^{2 =} a²-2 X a X √(b²-h²)+b² 

then h=√(4 X a² X b²-(a²+b²-c²)²)/(2 X a) 

then h = √((a+b+c) X (-a+b+c) X (a-b+c) X (a+b-c))/(2 X a).

area  = ah/2 = a X √((a+b+c) X (-a+b+c) X (a-b+c) X (a+b-c))/(2 X a) = 

√((a+b+c) X (-a+b+c) X (a-b+c) X (a+b-c))/4 = 

√(((a+b+c)/2) X ((-a+b+c)/2) X ((a-b+c)/2) X ((a+b-c)/2))

take s = (a+b+c)/2. then

area = √(s X (s-a) X (s-b) X (s-c))

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Using Pick's Theorem

Pick's theorem for a technique for finding the area of any arbitrary lattice polygon.
The theorem states:

where I is the number of internal lattice points and B is the number of lattice points lying inline with the border of the polygon.

there are other area formulas i will show them later 

thanks

MR HESHAM