Question

If $A$ is the area and $2s$ the sum of the sides of a triangles, then

• A. $A\le \frac{s^2}{4}$
• B. $A\le \frac{s^2}{3\sqrt{3}}$
• C. $A<\frac{s^2}{\sqrt{3}}$
• D. None of the above.

Answer 1

Answer: (A), (B)

The correct options are
A $A\le \frac{s^2}{4}$
B $A\le \frac{s^2}{3\sqrt{3}}$

Let $a,b,c$ be sides of the triangle.

Given area is $A$ sq. units

Area of a triangle is given by $\sqrt{s(s-a)(s-b)(s-c)}$

where $2s=a+b+c$

therefore, $A=\sqrt{s(s-a)(s-b)(s-c)}$

we know that $A.M.\ge G.M.$

$⟹\frac{s+(s-a)+(s-b)+(s-c)}{4}\ge \sqrt[4]{4s(s-a)(s-b)(s-c)}$

$⟹\frac{4s-(a+b+c)}{4}\ge \sqrt[4]{4s(s-a)(s-b)(s-c)}$

$⟹\frac{2s}{4}\ge \sqrt[4]{4s(s-a)(s-b)(s-c)}$

$⟹\frac{2s}{4}\ge (\sqrt{s(s-a)(s-b)(s-c)}{)}^{\frac{1}{2}}$

$⟹\frac{s}{2}\ge (A{)}^{\frac{1}{2}}$

squaring on both sides

$⟹\frac{s^2}{4}\ge A$

$⟹A\le \frac{s^2}{4}$ ----------(1)

similarly, $A.M.\ge G.M.$

$⟹\frac{x+y+z}{3}\ge \sqrt[3]{3xyz}$

let $x=s-a,y=s-b,z=s-c$

therefore, $\frac{(s-a)+(s-b)+(s-c)}{3}\ge \sqrt[3]{3(s-a)(s-b)(s-c)}$

$⟹\frac{3s-(a+b+c)}{3}\ge \sqrt[3]{3\frac{A^2}{s}}$

$⟹\frac{3s-2s}{3}\ge \sqrt[3]{3\frac{A^2}{s}}$

$⟹\frac{s}{3}\ge \sqrt[3]{3\frac{A^2}{s}}$

cubing on both sides

$⟹\frac{s^3}{27}\ge \frac{A^2}{s}$

$⟹\frac{s^4}{27}\ge A^2$

$⟹A^2\le \frac{s^4}{27}$

square root on both sides

$⟹A\le \frac{s^2}{3\sqrt{3}}$ --------(2)