Question

Assertion :If $A$ is the area and $2s$ is the sum of three sides of a triangle, then $A\le \frac{s^2}{3\sqrt{3}}$ Reason: $A.M.\ge G.M.$

• A. Both the statements are TRUE and STATEMENT 2 is the correct expansion of STATEMENT 1
• B. Both the statements are TRUE but STATEMENT 2 is NOT the correct explanation of STATEMENT 1
• C. STATEMENT 1 is TRUE and STATEMENT 2 is FALSE
• D. STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

Answer 1

The correct option is A Both the statements are TRUE and STATEMENT 2 is the correct expansion of STATEMENT 1

We have, $2s=a+b+c$

Also, $A^2=s(s-a)(s-b)(s-c)$

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

$\Rightarrow \frac{s-a+s-b+s-c}{3}\ge {\left((s-a)(s-b)(s-c)\right)}^{\frac{1}{3}}$

$\Rightarrow \frac{3s-2s}{3}\ge \frac{{\left(A^2\right)}^{\frac{1}{3}}}{s^{\frac{1}{3}}}\Rightarrow \frac{s^3}{27}\ge \frac{A^2}{s}$

$\therefore A\le \frac{s^2}{3\sqrt{3}}[\because s>0]$