Question

If $A$ is the area and $2s$ the sum of three sides of triangle, then
A. $A\le \dfrac{{{s}^{2}}}{3\sqrt{3}}$
B. $A\le \dfrac{{{s}^{2}}}{2}$
C. $A>\dfrac{{{s}^{2}}}{\sqrt{3}}$
D. None of these.

Answer 1

Hint: To solve this question, we will use the relationship between the arithmetic mean and geometric mean. We will calculate the value of arithmetic mean and geometric mean using their formulas and Heron’s formula and then substitute the values in the relation $AM\ge G.M$ and find the suitable relation from the options.

Formula Used: Heron’s formula : $A=\sqrt{s(s-a)(s-b)(s-c)}$
$A.M=\dfrac{Sum\,\,of\,\,observation}{Number\,\,of\,\,observation}$
$G.M={{(abc)}^{\dfrac{1}{n}}}$ , where $a,b,c$ are the observations and $n$ is the total observation.

Complete step by step solution: We are given a triangle with area $A$ and $2s$ the sum of three sides of the triangle. We know the relationship between arithmetic mean and geometric mean as $AM\ge G.M$.
Now we will first calculate the arithmetic mean with sides $(s-a)(s-b)(s-c)$.
$\begin{align}
  & A.M=\dfrac{Sum\,\,of\,\,observation}{Number\,\,of\,\,observation} \\
 & A.M=\dfrac{s-a+s-b+s-c}{3} \\
 & A.M=\dfrac{3s-(a+b+c)}{3}
\end{align}$
We are given $2s$ as the sum of three sides of a triangle. So,
$\begin{align}
  & A.M=\dfrac{3s-2s}{3} \\
 & =\dfrac{s}{3}
\end{align}$
We will now find the geometric mean.
$G.M={{((s-a)(s-b)(s-c))}^{\dfrac{1}{3}}}$
We will now put the values of A.M and G.M in the relation $AM\ge G.M$.
$\begin{align}
  & AM\ge G.M \\
 & \dfrac{s}{3}\ge {{\left( (s-a)(s-b)(s-c) \right)}^{\dfrac{1}{3}}}
\end{align}$
Cubing both sides of the equation.
$\dfrac{{{s}^{3}}}{27}\ge \left( (s-a)(s-b)(s-c) \right)$
From Heron’s formula $A=\sqrt{s(s-a)(s-b)(s-c)}$we will derive the value of $(s-a)(s-b)(s-c)$.
$\begin{align}
  & A=\sqrt{s(s-a)(s-b)(s-c)} \\
 & {{A}^{2}}=s(s-a)(s-b)(s-c) \\
 & \dfrac{{{A}^{2}}}{2}=(s-a)(s-b)(s-c)
\end{align}$
Now we will substitute the derived value in the relation.
$\begin{align}
  & \dfrac{{{s}^{3}}}{27}\ge \dfrac{{{A}^{2}}}{s} \\
 & \dfrac{{{s}^{4}}}{27}\ge {{A}^{2}} \\
 & \dfrac{{{s}^{2}}}{3\sqrt{3}}\ge A
\end{align}$

When the area of triangle is $A$ and sum of the sides are $2s$ then $A\le \dfrac{{{s}^{2}}}{3\sqrt{3}}$. Hence the correct option is (A).

Note: Arithmetic mean of numbers or observation can be defined as the ratio of sum of all the observations to the total number of observations while geometric mean can be defined as $n^{th}$ root of the product of the observations where $n$ is the total observations.
The geometric mean for any set of the observations will always be less than the arithmetic mean for that data set.