Looking back at each other

$By\quad Heron's\quad formula,\\ area\quad of\quad triangle=\sqrt { s(s-a)(s-b)(s-c) } \\ \\ Also,\quad by\quad AM-GM\quad inequality,\quad \\ \frac { (s-a)+(s-b)+(s-c) }{ 3 } \ge { \left[ (s-a)(s-b)(s-c) \right] }^{ \frac { 1 }{ 3 } }\\ \Rightarrow \quad \frac { s }{ 3 } \ge { \left[ (s-a)(s-b)(s-c) \right] }^{ \frac { 1 }{ 3 } }\\ \Rightarrow \quad { \left( \frac { s }{ 3 } \right) }^{ 3 }\ge (s-a)(s-b)(s-c)\\ \\ Multiplying\quad both\quad sides\quad by\quad s,\quad and\quad taking\quad square\quad root,\\ \Rightarrow \sqrt { s(s-a)(s-b)(s-c) } \le \sqrt { { s\left( \frac { s }{ 3 } \right) }^{ 3 } } \\ \Rightarrow \sqrt { s(s-a)(s-b)(s-c) } \le \frac { { s }^{ 2 } }{ 3\sqrt { 3 } } \\ \Rightarrow \sqrt { s(s-a)(s-b)(s-c) } \le \frac { 16 }{ 3\sqrt { 3 } } \quad \quad \quad \quad \quad \left\{ s=\frac { 8 }{ 2 } =4 \right\} \\ \\ \boxed { \therefore \quad max\left( Area \right) =\frac { 16 }{ 3\sqrt { 3 } } } \\$

s=(a+b+c)/2 (or s=perimeter/2) a b c are sides of triangle

A=√s(s-a)(s-b)(s-c) It will be max when (s-a)(s-b)(s-c) is max. Clearly its maximum when a=b=c=2s/3=perimeter/3 substitute and get ans

Other solution using ap >= gp>=hp A=√s(s-a)(s-b)(s-c) as AP>=GP ≤ √s((s-a+s-b+s-c)/3)³ = √s(s/3)³ = √s⁴/3³ = s²/3√3 S=8/2 or 4 Substitute and get ans

ANS=16/3√3

In $\triangle PQR$ let $b = \overline{PQ}$ , $a = \overline{PR}$ , and $\theta = \angle P$ .

The area of the triangle is $A = \tfrac12 b h = \tfrac12 b a \sin\theta.$ Now consider a small deformation of the triangle, moving points $Q$ and $R$ without changing the angle at $P$ , such that now $a \mapsto a + dx,\ \ \ b \mapsto b - dx.$ If the triangle is maximal, then the area should remain constant under such infinitesimal change. Therefore we write $0 = dA = \tfrac12 (b\:da + a\:db)\sin\theta = \tfrac12 (b\:dx - a\:dx) \sin\theta.$ This implies that $a = b$ for the maximal triangle.

It follows that the desired triangle is equilateral, with sides $8/3$ and height $8/\sqrt3$ . This gives an area of $16/3\sqrt 3$ .

Area of any shape is maximum when its symmetrical....and so in case of triangle it will be equilateral triangle. So side= a =8/3 And area=(√3a^2/4)