## How to calculate the area of a triangle

The area of a triangle is the area of the portion of the plane that it encloses. There are several ways to calculate it, depending on the information you want to start from.

## Special case of the right triangle

In the case of a right-angled triangle and when we know the width and length of it, just do:

Area = (L * l) / 2

## Calculation from a height

If the triangle is right, it is immediate that its area is $S = \ dfrac {ah} {2}$

where a is the length of a different side of the hypotenuse and h is the length of the height from that side. If the triangle is not right-angled, the relation remains true, because the triangle breaks up into two right-angled triangles (as in the figure below). ## From the lengths of all three sides

For an expression of the area of a triangle whose side lengths are a, b and c and p the half-perimeter [ $p = \ dfrac {a + b + c} {2}$]

, we can use Heron's formula: $Area = \ dfrac {1} {4} \ sqrt {(a + b + c) (- a + b + c) (a-b + c) (a + bc)} = \ sqrt {p (pa) ( pb) (pc)}$

From the coordinates of the vertices

The area of the parallelogram defined by two vectors $\ overrightarrow {u}$, $\ overrightarrow {vb}$is the norm of their cross product: $S_p = \ left \ | {\ overrightarrow {u} \ wedge \ overrightarrow {v}} \ right \ |$

We can calculate the area of a triangle from this formula: $S = \ dfrac12 \ left \ | {\ overrightarrow {AB} \ wedge \ overrightarrow {AC}} \ right \ |$.

Given an orthonormal coordinate system, the area of triangle ABC can be calculated from the coordinates of the vertices.

In the plane, if the coordinates of A, B and C are given by A ( x A , y A ) , B ( x B , y B ) and C ( x C , y C ) , then the area S is the half of the absolute value of the determinant $S = \ dfrac {1} {2} \ left | \ det \ begin {pmatrix} x_B-x_A & x_C-x_A \\ y_B-y_A & y_C-y_A \ end {pmatrix} \ right | = \ dfrac {1} {2} | (x_B-x_A) (y_C-y_A) - (x_C-x_A) (y_B-y_A) |.$

The area of triangle ABC can also be calculated from the formula $S = \ dfrac {1} {2} \ left | \ det \ begin {pmatrix} x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1 \ end {pmatrix} \ right | = \ dfrac {1} {2} \ big | x_A y_C - x_A y_B + x_B y_A - x_B y_C + x_C y_B - x_C y_A \ big |.$

This method is generalized in three dimensions. The area of triangle ABC where A = ( x A , y A , z A ) , B = ( x B , y B , z B ) and C = ( x C , y C , z C ) is expressed as $S = \ frac {1} {2} \ sqrt {\ left (\ det \ begin {pmatrix} x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1 \ end {pmatrix} \ right) ^ 2 + \ left (\ det \ begin {pmatrix} y_A & y_B & y_C \\ z_A & z_B & z_C \\ 1 & 1 & 1 \ end {pmatrix} \ right) ^ 2 + \ left (\ det \ begin {pmatrix} z_A & z_B & z_C \\ x_A & x_B & x_C \\ 1 & 1 & 1 \ end {pmatrix} \ right) ^ 2}.$

Content subject to CC-BY-SA license . Source: Triangle article from Wikipedia in French ( authors )

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