P is the half circumference; 2 * AB * √ [1-(A2+B2-C2) 2/:P = (A+B+C)/,and the diagonal lines of C are respectively A, p-c=(a+b-c)/ Helen's formula.

Suppose there is a triangle,

Cosine theorem is

cosC

=

(a 2+b 2-c 2)/, c, c, the side length is a respectively; 16]

=√[p(p-a)(p-b)(p-c)]

So; 2:

Let three sides of a triangle be a; 2; 4*√[(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)]

= 1/、b，

P-a=(-a+b+c)/, b, b, the area s of the triangle can be obtained by the following formula; 2

Prove,

p-b=(a-b+c)/4*√[(a+b)^2-c^2][c^2-(a-b)^2]

= 1/2; 2*ab*√( 1-cos^2

c)

= 1/2ab

S

= 1/2*ab*sinC

= 1/4a^2*b^2]

= 1/4*√[4a^2*b^2-(a^2+b^2-c^2)^2]

= 1/4 *√[(a+b+c)(a+b-c)(a-b+c)(-a+b+c)]

Let p=(a+b+c)/2,

The above formula = √ [(a+b+c) (a+b-c) (a-b+c) (-a+b+c)/2.

Then p=(a+b+c)/