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
Correct Answer: A
Solution :
We have, \[2s=a+b+c,{{A}^{2}}=s(s-a)(s-b)(s-c)\] \[\because \ \text{A}\text{.M}.\ge \text{G}\text{.M}\]. Þ \[\frac{s-a+s-b+s-c}{3}\ge \sqrt[3]{(s-a)(s-b)(s-c)}\] Þ \[\frac{3s-2s}{3}\ge \frac{{{({{A}^{2}})}^{1/3}}}{{{s}^{1/3}}}\Rightarrow \frac{{{s}^{3}}}{27}\ge \frac{{{A}^{2}}}{s}\Rightarrow A\le \frac{{{s}^{2}}}{3(\sqrt{3})}\].
You need to login to perform this action.
You will be redirected in
3 sec
