Question

lf $\Delta$ is the area and $2s$ is the perimeter of a triangle then

• A. $\Delta >\frac{s^2}{\sqrt{3}}$
• B. $\Delta \le \frac{s^2}{3\sqrt{3}}$
• C. $\Delta >\frac{s^2}{3\sqrt{3}}$
• D. $\Delta \le \frac{s^2}{2}$

Answer 1

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

${△}^2=s(s-a)(s-b)(s-c)$
$\Rightarrow 16{△}^2=2s(2s-2a)(2s-2b)(2s-2c)$
$\Rightarrow \frac{16{△}^2}{2s}=(b+c-a)(a+c-b)(a+b-c)$ ...(1)
Now taking $(b+c-a)(a+c-b)(a+b-c)$ as three numbers and using $A.M.\ge G.M.$, we get
$\frac{(b+c-a)+(a+c-b)+(a+b-c)}{3}\ge \sqrt[3]{3(b+c-a)(a+c-b)(a+b-c)}\Rightarrow {\left(\frac{a+b+c}{3}\right)}^3\ge (b+c-a)(a+c-b)(a+b-c)\Rightarrow \frac{8s^3}{27}\ge (b+c-a)(a+c-b)(a+b-c)$ ...(2)
From $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow \frac{8s^3}{27}\ge \frac{16{△}^2}{2s}\Rightarrow \frac{s^4}{27}\ge {△}^2\Rightarrow △\le \frac{s^2}{3\sqrt{3}}$