Area of an arbitrary triangle?

**fardeen_gen** · May 21st 2009, 09:53 AM

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)

**Grandad** · May 21st 2009, 10:46 AM

Hello fardeen_gen

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

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