Each side of a triangle is not greater than 2. Prove that it

[Obvious~Hat]

Each side of a triangle is not greater than 2.
Prove that its area is not greater than sqrt(3)

I'm really not sure what to do at all. I mean, I can't figure out how to work the area out when you don't have the perpendicular height. (The triangle on the diagram is not a right angled one)

 

[wjm11]

Each side of a triangle is not greater than 2.
Prove that its area is not greater than sqrt(3)

Use Heron's (also called Hero's) Formula for triangle area:

http://en.wikipedia.org/wiki/Heron%27s_formula

 

[soroban]

Hello, Obvious~Hat!

I may have a proof . . .

Each side of a triangle is not greater than 2.
Prove that its area is not greater than \(\displaystyle \sqrt{3}\)

Consider an equilateral triangle with side 2.
Call it \(\displaystyle \Delta_1\)

            C
            *
           * *
          *   *
       2 *     * 2
        *       *
       *         *
    A *  *  *  *  * B
            2

\(\displaystyle \text{Each angle is }60^o.\)

\(\displaystyle \text{The area of a triangle is: }\:A \:=\:\tfrac{1}{2}bc\sin A\)

\(\displaystyle \text{Hence, our triangle has area: }\:\Delta_1 \;=\;\tfrac{1}{2}(2)(2)\sin60^o \:=\:2\left(\tfrac{\sqrt{3}}{2}\right) \;=\;\sqrt{3}\)

If the sides are less than 2, the area will be less than \(\displaystyle \sqrt{3}.\)


\(\displaystyle \text{Note: }\angle A\text{ cannot be greater than }60^o.\)
. . . . \(\displaystyle \text{If }\angle A \,>\,60^o,\,\text{ then side }a\,>\,2.\)


\(\displaystyle \text{Suppose }\angle A \,<\,60^o\)

\(\displaystyle \text{We have a triangle with: }\:b = 2,\;c = 2,\;\text{ and }\angle A \,<\,60^o\)
Call it \(\displaystyle \Delta_2\)

                C
                *
              *  *
         2  *     *
          *        *
        * @         *
      *  *  *  *  *  * B
              2

\(\displaystyle \text{We have: }\:\theta \:<\:60^o\)

\(\displaystyle \text{Then: }\:\sin\theta \:<\:\sin60^o \quad\Rightarrow\quad \sin\theta \:<\:\tfrac{\sqrt{3}}{2}\)

\(\displaystyle \text{Multiply both sides by }\tfrac{1}{2}bc\!:\;\;\tfrac{1}{2}bc\sin\theta \:<\:\tfrac{1}{2}bc\tfrac{\sqrt{3}}{2}\)

\(\displaystyle \text{If }b=c=2\text{, then we have: }\:\underbrace{\tfrac{1}{2}(2)(2)\sin\theta}_{\text{This is }\Delta_2} \;\leq \;\underbrace{\tfrac{1}{2}(2)(2)\tfrac{\sqrt{3}}{2}}_{\text{This is }\sqrt{3}}\)

\(\displaystyle \text{Therefore: }\;\Delta_2 \:<\:\sqrt{3}\)

 

[chrisr]

[attachment=0:1y13rqsn]max area.jpg[/attachment:1y13rqsn]

The two arrows show the circle arcs against which the triangle's top apex cannot go beyond.
The radius of both circles is 2
As triangle area is 0.5(base)(perpendicular height),
the triangles maximum area corresponds to maximum base=2 and maximum height where the arcs intersect.

\(\displaystyle max\ area\ =\ h\ =\ 2Sin60^o\ =\frac{2\sqrt{3}}{2}=\sqrt{3}\)