Area A = [pic]
Proof:
Let a trilateral ABC with sides a, b and c whose area is equal to A = [pic]. Let the triangle be as follows:
Here, perimeter is the length of the sides, now as the sides are a, b & c, assume it is also the length of the sides, then the perimeter of this triangle is
P = a+b+c
And semi-perimeter i.e. half of the perimeter is
S = a+b+c/2
If we scuff a perpendicular from C to base and call it h which divides the base into two parts i.e. x and c-x, then the plat looks as follows:
The perpendicular has divided the triangle into two right-angled triangles. Now for any right-angle triangle, according to Pythagorean Theorem,
[pic] = [pic] + [pic]
If Pythagorean is employ to the right-angled triangles in the above triangle, then in the fictitious character of left over(p) right-angle triangle in the above diagram, it would give us the equality
[pic] = [pic] + [pic] where a = hypotenuse and h = height/perpendicular and x = base.
Re-writing it, the equation would become which we will call Eq.
A
[pic] = [pic] - [pic] ---------------------( Eq. A
Similarly, for the right angle triangle on the right half to triangle ABC,
[pic] = [pic] + [pic] where b = hypotenuse, h = height/perpendicular and c-x = base.
Re-writing this equation would result in
[pic] = [pic] [pic]
Expanding [pic] would give us
[pic] = [pic] ([pic] + [pic] - 2cx)
[pic] = [pic] [pic] - [pic] + 2cx --------------------------( Eq. B
As, the left hand sides of Eq.A and Eq. B are equal, we can equate them. equate Eq.A and Eq. B would give us
[pic] - [pic] = [pic] [pic] - [pic] + 2cx
Solving it further would give us
[pic] = [pic] [pic] + 2cx
Re-arranging...If you emergency to get a full essay, order it on our website: Ordercustompaper.com
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