If A is the area and 2s is the sum of the 3 s | vedclass

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(A) For a given perimeter,an equilateral triangle has the maximum area.Let the sides be $a, b, c$. The perimeter is $2s = a + b + c$,so the semi-perim

Vedclass · Accepted Answer

$A \le \frac{s^2}{3\sqrt{3}}$

Vedclass · Answer

(A) For a given perimeter,an equilateral triangle has the maximum area.Let the sides be $a, b, c$. The perimeter is $2s = a + b + c$,so the semi-perimeter is $s$.For an equilateral triangle,$a = b = c = \frac{2s}{3}$.The maximum area $A_{max} = \frac{\sqrt{3}}{4} \left(\frac{2s}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{4s^2}{9} = \frac{s^2}{3\sqrt{3}}$.Since any triangle with perimeter $2s$ has an area $A \le A_{max}$,we have $A \le \frac{s^2}{3\sqrt{3}}$.Alternatively,using Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$.By the $AM$-$GM$ inequality,$\frac{(s-a) + (s-b) + (s-c)}{3} \ge \sqrt[3]{(s-a)(s-b)(s-c)}$.Since $(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$,we have $\frac{s}{3} \ge \sqrt[3]{(s-a)(s-b)(s-c)}$.Cubing both sides: $\frac{s^3}{27} \ge (s-a)(s-b)(s-c)$.Multiplying by $s$: $A^2 = s(s-a)(s-b)(s-c) \le \frac{s^4}{27}$.Taking the square root: $A \le \frac{s^2}{\sqrt{27}} = \frac{s^2}{3\sqrt{3}}$.

If $A$ is the area and $2s$ is the sum of the $3$ sides of a triangle,then :

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