Search of the area of a triangle - one of the most widespread problems of school planimetry. Knowledge of three parties of a triangle is enough for determination of the area of any triangle. In special cases of isosceles and equilateral triangles it is enough to know lengths of two and one parties respectively.
It is required to you
- lengths of the parties of triangles, Heron's formula, theorem of cosines
Instruction
1. Let a triangle of ABC with the parties of AB = c, AC = b, BC = an is set. The area of such triangle can be found on Heron's formula. The perimeter of a triangle of P is the sum of lengths of its three parties: P = a+b+c. Let's designate it poluperimetr for p. It will be equal to p = (a+b+c)/2.
2. Heron's formula for the area of a triangle looks as follows: S = sqrt (p(p-a) (p-b) (p-c)). If to paint poluperimetr p, then it will turn out: 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 bring a formula for the area of a triangle and out of other reasons, for example, having applied the theorem of cosines. According to the theorem of cosines of AC^2 = (AB^2)+ (BC^2)-2*AB*BC*cos (ABC). Using the entered designations, these expressions can also be written down in a look: b^2 = (a^2)+ (c^2)-2a*c*cos (ABC). From here, cos(ABC) = ((a^2)+ (c^2) (-b^2)) / (2*a*c)
4. The area of a triangle is also on formula S = a*c*sin(ABC)/2 through two parties and a corner between them. The sine of the angle of ABC can be expressed through its cosine by means of the main trigonometrical identity: sin(ABC) = sqrt (1-(cos (ABCs)) ^2). Substituting a sine in a formula for the area and painting it, it is possible to come to a formula for the area of a triangle of ABC.