Question

If $△$ be the area of the triangle $ABC,$ and length of its two sides are $3$ and $5$.If $c$ is the third side, then

• A. $△\le \frac{(c^2+16c+64)}{12\sqrt{3}}$
• B. $△=\frac{(c^2+16c+64)}{8\sqrt{3}}$
• C. $△>\frac{(c^2+16c+64)}{4\sqrt{3}}$
• D. none of these

Answer 1

Answer: (A)

$△\le \frac{(c^2+16c+64)}{12\sqrt{3}}$

Perimeter$=2s=a+b+c$
Given:$a=3,b=5$
$\Rightarrow 2s=3+5+c$
$\Rightarrow s=\frac{8+c}{2}$
$\because$A.M$\ge$G.M
$\Rightarrow \frac{(s-a)+(s-b)+(s-c)}{3}\ge {\left[(s-a)(s-b)(s-c)\right]}^{\frac{1}{3}}$
$\Rightarrow \frac{3s-(a+b+c)}{3}\ge {\left[(s-a)(s-b)(s-c)\right]}^{\frac{1}{3}}$
$\Rightarrow \frac{3s-2s}{3}\ge {\left[(s-a)(s-b)(s-c)\right]}^{\frac{1}{3}}$
$\Rightarrow {\left(\frac{s}{3}\right)}^3\ge \frac{{△}^2}{s}$
$\Rightarrow {△}^2\le \frac{s^4}{3^3}$
$\Rightarrow △\le \frac{s^2}{3\sqrt{3}}$
$\Rightarrow △\le {\left(\frac{8+c}{2}\right)}^2\frac{1}{3\sqrt{3}}$
$\Rightarrow △\le \frac{{(8+c)}^2}{12\sqrt{3}}$
$\therefore △\le \frac{c^2+16c+64}{12\sqrt{3}}$