# Area of an arbitrary triangle?

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• May 21st 2009, 09:53 AM
fardeen_gen
Area of an arbitrary triangle?
Prove that area of an arbitrary triangle is less than one fourth of the square of its perimeter.

(The question is from my sequences and series text)
• May 21st 2009, 10:46 AM
Grandad
Hello fardeen_gen
Quote:

Originally Posted by fardeen_gen
Prove that area of an arbitrary triangle is less than one fourth of the square of its perimeter.

(The question is from my sequences and series text)

Do you know the formula for the area of a triangle: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, where the semi-perimeter $s = \tfrac12(a+b+c)$?

Well, each of $(s-a), (s-b), (s-c)$ is less than $s$. So $\Delta < \sqrt{s^4} = \tfrac14(a+b+c)^2$.

Grandad