Question

If $A$ is the area and $2s$ the sum of 3 sides of triangle then

• A. $A\le \frac{s^2}{3\left(\sqrt{3}\right)}$
• B. $A\le \frac{s^2}{2}$
• C. $A>\frac{s^2}{\left(\sqrt{3}\right)}$
• D. None of these

Answer 1

The correct option is A

$A\le \frac{s^2}{3\left(\sqrt{3}\right)}$

We have
$2s=a+b+c,A^2=s(s-a)(s-b)(s-c)\because A.M.\ge G.M.\frac{s-a+s-b+s-c}{3}\ge \sqrt[3]{3(s-a)(s-b)(s-c)}\Rightarrow \frac{3s-2s}{3}\ge {\left(\frac{A^2}{s}\right)}^{\frac{1}{3}}\Rightarrow \frac{s^3}{27}\ge \frac{A^2}{s}\Rightarrow A\le \frac{s^2}{\sqrt{3\sqrt{3}}}.$