You will need
- the lengths of the sides of triangles, Heron's formula, the theorem of cosines
Instruction
1
Imagine you are given a triangle ABC with sides AB = c, AC = b, BC = a. The area of this triangle can be found by Heron's formula.
The perimeter P is the sum of the lengths of its three sides: P = a+b+c. We denote its properiter over p. It will be equal to p = (a+b+c)/2.
The perimeter P is the sum of the lengths of its three sides: P = a+b+c. We denote its properiter over p. It will be equal to p = (a+b+c)/2.
2
Heron's formula for area of a triangle as follows: S = sqrt(p(p-a)(p-b)(p-c)). If you paint properiter p, we get: S = sqrt(((a+b+c)/2)((b+c-a)/2)((a+c-b)/2)((a+b-c)/2)) = (sqrt((a+b+c)(a+b-c)(a+c-b)(b+c-a)))/4.
3
It is possible to derive a formula for the area of a triangle and for other reasons, for example, applying the theorem of cosines.
Cosine theorem AC^2 = (AB^2)+(BC^2)-2*AB*BC*cos(ABC). Using the introduced notation, these expressions can also be written in the form: b^2 = (a^2)+(c^2)-2a*c*cos(ABC). Hence, cos(ABC) = ((a^2)+(c^2)-(b^2))/(2*a*c)
Cosine theorem AC^2 = (AB^2)+(BC^2)-2*AB*BC*cos(ABC). Using the introduced notation, these expressions can also be written in the form: b^2 = (a^2)+(c^2)-2a*c*cos(ABC). Hence, cos(ABC) = ((a^2)+(c^2)-(b^2))/(2*a*c)
4
The area of the triangle is also the formula S = a*c*sin(ABC)/2 two sides and the angle between them. The sine of the angle ABC can be expressed via its cosine by using the basic trigonometric identities: sin(ABC) = sqrt(1-((cos(ABC))^2). Substituting the sine in the formula for the area and describing it, you can come to the formula for the area of triangle ABC.