Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, and if A is the area and $2s$ the sum of three sides of a triangle, then:

• A. $A\le \frac{s^2}{3\sqrt{3}}$
• B. $A\le \frac{s^2}{2}$
• C. $A>\frac{s^2}{\sqrt{3}}$
• D. none of these

Answer 1

The correct option is A $A\le \frac{s^2}{3\sqrt{3}}$

We know that, A.M$\ge$G.M
$\frac{(s-a)+(s-b)+(s-c)}{3}\ge {\left[(s-a)(s-b)(s-c)\right]}^{1/3}$
$\Rightarrow \frac{3s-2s}{3}\ge {\left[(s-a)(s-b)(s-c)\right]}^{1/3}$
$\Rightarrow (s-a)(s-b)(s-c)\le \frac{s^3}{3}$
$\Rightarrow \sqrt{s(s-a)(s-b)(s-c)}\le \frac{s^2}{3\sqrt{3}}$
$\Rightarrow A\le \frac{s^2}{3\sqrt{3}}$
Ans: A